Basic ElectronicsICT/CU/CS/CC/01/6/B
ICT/CU/CS/CC/01/6/B · TVET CDACC Level 6

Basic Electronics Notes

Complete trainee notes for the unit of competency Apply Basic Electronic Skills, written to the TVET CDACC course outline and pitched at the depth the written assessment actually asks for. Six learning outcomes, worked examples throughout, and past-paper style exercises at the end of every topic.

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E.m.f (V) switch R (Ω) load conventional current: + → −
6learning outcomes
100 hunit duration
60+practice questions
25diagrams
5calculators
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About this unit

~6 min

This unit specifies the competencies required to apply basic electronics skills. It covers identifying electric circuits and electronic components, understanding semi-conductor theory, identifying and classifying memories, applying number systems and binary coding, and identifying emerging trends in electronics. It is a common unit of competency, so trainees across ICT Technician, Computer Science and Computer Programming Level 6 all sit it.

Unit summary. Codes vary slightly between qualifications; the content and assessment are the same.
ItemDetail
Unit of competencyApply Basic Electronic Skills
Unit codeICT/CU/CS/CC/01/6/B
LevelTVET CDACC Level 6 (Diploma)
Duration100 hours, credit factor 10
TypeCommon unit of competency
Assessment methodsPractical exercises, written, observation, oral

The six learning outcomes

#Learning outcomeSuggested assessment
1Identify electrical circuitsPractical · Written · Observation · Oral
2Identify electronic componentsPractical · Written · Observation · Oral
3Understand semi-conductor theoryPractical · Written · Observation · Oral
4Identify and classify memoryWritten · Observation · Oral
5Apply number systems and binary codingWritten · Observation · Oral
6Identify emerging trends in electronicsWritten · Observation · Oral

What the written assessment looks like

Recent CDACC papers for this unit follow a consistent shape, and the exercises in these notes are written to match it.

  • Time: 3 hours, separate answer booklet, no writing on the question paper
  • Section A — 40 marks: answer all questions. Short items of 2 to 6 marks: definitions, differences, a small sketch, a one-step calculation
  • Section B — 60 marks: answer any three of four or five scenario questions, each worth 20 marks and split into parts (a), (b), (c)
  • Scenario style: a workshop, a smart home startup, a research lab or an agricultural sensor network, with calculations and a diagram embedded in the story
How to use these notes

Read a topic, work the calculator beside it, then attempt the practice questions before opening the model answers. Tick Mark as read under each outcome as you finish it, and use Save my place in the top bar when you stop for the day. Finish with the mock paper near the end, timed at three hours.

1

Identify electrical circuits

~25 min

By the end of this outcome you should be able to

  • Define an electrical circuit and name its essential parts
  • State the basic electrical quantities, their SI units and measuring instruments
  • Apply Ohm's law, the power and energy relationships, and the resistivity formula
  • Solve series, parallel and series-parallel networks using Kirchhoff's laws
  • Distinguish a.c. from d.c. and analogue from digital circuits
  • Calculate peak, r.m.s. and average values of a sinusoidal waveform

Definition of an electrical circuit

An electrical circuit is a complete, closed conducting path through which electric current flows from a source of electromotive force, through a load that does useful work, and back to the source. Break the path anywhere and current stops — that is the whole idea behind a switch, and behind a blown fuse.

Every working circuit has four essential parts:

Source

Supplies the e.m.f. that drives charge round — cell, battery, generator, solar panel or power supply.

Load

Converts electrical energy into something useful — light, heat, motion, sound or computation.

Conductors

Wires and PCB tracks joining everything with as little resistance as possible.

Control & protection

Switches, fuses, circuit breakers and relays that open, close and safeguard the path.

Three conditions of a circuit

ConditionWhat is happeningResult
ClosedThe path is complete and unbrokenCurrent flows; the load operates normally
OpenThe path is broken — switch off, cut wire, blown fuseNo current; the load is dead
ShortAn unintended low-resistance path bypasses the loadVery large current, overheating, fire risk
cell closed switch R A ammeter V voltmeter across R Ammeter always in series · voltmeter always in parallel with the component being measured.
Figure 1.1 — A labelled schematic with cell, resistor, ammeter, voltmeter and closed switch. Papers ask you to draw exactly this, so practise it until it is automatic.
Safety

Isolate and discharge before working on a circuit. Treat mains and charged capacitors as live until you have proved otherwise with a meter. Never replace a fuse with a higher rating — the fuse protects the wiring, and upgrading it moves the failure point to the cable.

Basic electrical quantities and their units

Learn the quantity symbol, the unit, the unit symbol and the measuring instrument for each — assessors ask for all four.

The core quantities. The quantity symbol is italic (I); the unit symbol is upright (A).
QuantitySymbolSI unitUnitMeasured withPlain meaning
E.m.f. / voltageE, VvoltVVoltmeter (parallel)Electrical push per unit charge
CurrentIampereAAmmeter (series)Rate of flow of charge
ResistanceRohmΩOhmmeter / multimeterOpposition to current
PowerPwattWWattmeterRate of using energy
EnergyWjouleJEnergy meter (kWh)Total work done
ChargeQcoulombCDerived, Q = I tQuantity of electricity
CapacitanceCfaradFCapacitance meterAbility to store charge
InductanceLhenryHLCR meterOpposition to current change
FrequencyfhertzHzOscilloscope / frequency meterCycles per second
Resistivityρohm metreΩmDerived from R, A and lHow resistive the material itself is

Defining relationships

E.m.f. E = W / Q volts · 1 V = 1 joule per coulomb
Current I = Q / t amperes · 1 A = 1 coulomb per second
Resistance R = V / I ohms · 1 Ω = 1 volt per ampere
Power P = W / t = V I watts · 1 W = 1 joule per second
Energy W = P t = V I t joules · 1 kWh = 3.6 × 106 J
E.m.f. is not the same as p.d.

Electromotive force is the total energy the source gives each coulomb. Potential difference is the energy each coulomb gives up in one part of the circuit. Because the source has internal resistance r, terminal voltage is always slightly less than e.m.f. once current flows: V = E − I r.

Multiples and sub-multiples

PrefixSymbolMultiplierTypical use
teraT1012TB storage
gigaG109GHz clock speeds
megaM106MΩ insulation resistance
kilok103kΩ resistors, kW loads
millim10−3mA signal currents, mH coils
microµ10−6µF capacitors, µΩm resistivity
nanon10−9nF capacitors, ns access times
picop10−12pF at radio frequencies

Ohm's law, power and resistivity

Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant.

V = I R  →  I = V / R  →  R = V / I
P = V I = I²R = V² / R
V I R cover the one you want P V I the rest multiply or divide V = I × RI = V ÷ RR = V ÷ I P = V × IP = I²R = V²/R
Figure 1.2 — Cover the quantity you want with your thumb; what remains is the formula.

Resistivity of a conductor

Resistance depends on the material and on the shape of the conductor. This appears in papers as a straight calculation, usually about a transmission line or a cable run.

R = ρ l / A  ·  ρ = resistivity (Ωm), l = length (m), A = cross-sectional area (m²)
  • Resistance is directly proportional to length — double the run, double the resistance
  • Resistance is inversely proportional to cross-sectional area — a thicker cable has less resistance
  • Watch the units: 1 mm² = 10−6 m², and 1 µΩm = 10−6 Ωm

Worked example — past-paper style

A power distribution line uses an aluminium conductor 1.2 km long
with a cross-sectional area of 25 mm². The resistivity of aluminium
is 0.028 µΩm. Calculate the resistance of the conductor.

ρ = 0.028 µΩm = 0.028 × 10⁻⁶ Ωm = 2.8 × 10⁻⁸ Ωm
l = 1.2 km   = 1200 m
A = 25 mm²  = 25 × 10⁻⁶ m²

R = ρl / A
  = (2.8 × 10⁻⁸ × 1200) / (25 × 10⁻⁶)
  = (3.36 × 10⁻⁵) / (2.5 × 10⁻⁵)
  = 1.344 Ω
Ohm's law and power calculator

Enter any two values and leave the third blank.

Waiting for two values…
Conductor resistance calculator (R = ρl/A)

Common resistivities at 20 °C: silver 0.0159, copper 0.0168, gold 0.0244, aluminium 0.028, iron 0.097 µΩm.

Types of electrical circuits: series, parallel and network laws

Circuits are classified two ways: by how components are connected (series, parallel or series-parallel) and by the kind of supply (d.c. or a.c.).

Series — one path R1 R2 R3 Rₜ = R1+R2+R3 · same I everywhere Parallel — branches R1 R2 R3 1/Rₜ = 1/R1+1/R2+1/R3 · same V across each Quick contrast Series: current is common Parallel: voltage is common Series: one break kills everything Parallel: one break kills one branch Series: Rₜ is larger than any R Parallel: Rₜ is smaller than any R House wiring is parallel — so every socket gets full 240 V independently.
Figure 1.3 — Series and parallel connection, with the rule for total resistance in each.
FeatureSeries circuitParallel circuit
Number of pathsOne onlyTwo or more branches
CurrentSame through every componentDivides; IT = I1+I2+…
VoltageDivides; VT = V1+V2+…Same across every branch
Total resistanceR = R₁+R₂+R₃1/R = 1/R₁+1/R₂+1/R₃
If one part opensWhole circuit stopsOnly that branch stops
Typical useFuses, switches, current limitingMains outlets, lighting, distribution

Advantages of parallel circuits

Papers regularly ask for three advantages. The reliable answers are:

  1. Independent operation — each branch can be switched on or off without affecting the others
  2. Full supply voltage to every load — each appliance receives the rated 240 V rather than a share of it
  3. Fault tolerance — if one lamp burns out, the rest keep working
  4. Loads can be added without re-rating the others, up to the capacity of the supply and protective device

Kirchhoff's laws

KCL: the total current entering a junction equals the total current leaving it  ·  ΣIin = ΣIout
KVL: around any closed loop, the sum of e.m.f.s equals the sum of the p.d.s  ·  ΣE = ΣIR

Almost every circuit calculation in this unit reduces to one of these two, plus Ohm's law. Two further theorems are named in papers:

  • Superposition theorem — in a linear circuit with more than one source, the current in any branch equals the algebraic sum of the currents produced by each source acting alone, with all other sources replaced by their internal resistances.
  • Thévenin's theorem — any linear two-terminal network can be replaced by a single voltage source VTH in series with a single resistance RTH, as seen from those two terminals.

Worked example — parallel network on a 240 V supply

Three resistors of 10 Ω, 20 Ω and 30 Ω are connected in
parallel across a 240 V d.c. supply. Find Rₜ, Iₜ and the power.

1/Rₜ = 1/10 + 1/20 + 1/30
      = 6/60 + 3/60 + 2/60  =  11/60
Rₜ   = 60/11              =  5.4545 Ω   ≈ 5.45 Ω

Iₜ   = V / Rₜ = 240 / 5.4545  =  44 A
Check by branches:
  I₁ = 240/10 = 24 A
  I₂ = 240/20 = 12 A
  I₃ = 240/30 =  8 A
  ΣI = 24+12+8 = 44 A  ✓ agrees with KCL

P   = V × Iₜ = 240 × 44   =  10 560 W  =  10.56 kW

Worked example — fuse rating and energy

A lighting circuit runs from a 240 V supply protected by a 13 A fuse.
Each bulb is rated 100 W. (a) How many bulbs can run at once?
(b) Energy used in 1 hour if all are on?

(a) Current per bulb  I = P/V = 100/240 = 0.4167 A
    Number of bulbs   n = 13 / 0.4167 = 31.2
    → 31 bulbs (you cannot have a fraction of a bulb;
      a 32nd would draw 13.33 A and blow the fuse)

(b) Total power P = 31 × 100 = 3100 W
    t = 1 hour = 3600 s
    W = P × t = 3100 × 3600 = 11 160 000 J = 11.16 MJ  (3.1 kWh)

Simple a.c. and d.c. circuits

In a direct current (d.c.) circuit, charge flows in one direction only and the voltage holds a steady value. Sources are cells, batteries, solar PV panels, d.c. generators and rectified power supplies. Every electronic device runs on d.c. internally — phones, laptops, routers and microcontrollers all sit at 5 V, 3.3 V or lower.

In an alternating current (a.c.) circuit, the current periodically reverses direction and the voltage varies sinusoidally with time. Kenya's mains supply is 240 V at 50 Hz, meaning the waveform completes fifty full cycles every second. A.c. is used for generation and distribution because transformers can step it up for low-loss transmission and back down for safe use.

Direct current (d.c.) steady value one direction · battery, solar, PSU output Alternating current (a.c.) r.m.s. = 0.707 Vₘ average = 0.637 Vₘ peak Vₘ 1 cycle (T) reverses direction · mains 240 V, 50 Hz
Figure 1.4 — A steady d.c. level against one and a half cycles of a.c., with the three values you must be able to calculate.

Peak, r.m.s. and average values

Assessment questions ask you to convert between these for a sinusoidal waveform. Memorise the three factors.

r.m.s. Vrms = 0.707 × Vpeak = Vpeak / √2
average Vav = 0.637 × Vpeak = 2Vpeak / π
form factor = Vrms / Vav = 1.11  ·  peak factor = Vpeak / Vrms = 1.414

Worked example

A sinusoidal voltage has a maximum value of 120 V.
Calculate its r.m.s. and average values.

Vᵣᵐᵣ = 0.707 × 120 = 84.84 V   (or 120/√2 = 84.85 V)
Vᴡᵛ  = 0.637 × 120 = 76.44 V   (or 2×120/π  = 76.39 V)

Note: the 240 V mains figure is an r.m.s. value.
Its peak is 240 × 1.414 = 339.4 V — which is why
insulation must be rated well above 240 V.
Point of comparisond.c.a.c.
Direction of flowConstant, one wayReverses periodically
WaveformStraight lineSine wave (usually)
Frequency0 Hz50 Hz in Kenya; 60 Hz elsewhere
TransformableNo, not directlyYes, with a transformer
Long-distance transmissionLossy at low voltageEfficient — step up, transmit, step down
Opposition to flowResistance onlyImpedance: resistance plus reactance
Typical sourcesCell, battery, solar PV, rectifierAlternator, mains socket, inverter
Converted to the other byAn inverter (d.c. → a.c.)A rectifier (a.c. → d.c.)

Analogue and digital circuits

A separate classification that papers ask you to differentiate.

FeatureAnalogue circuitDigital circuit
SignalContinuously variable over a rangeTwo discrete levels only: 0 and 1
RepresentationSmooth waveformSquare pulse train
NoiseAccumulates and degrades the signalRejected as long as levels stay distinguishable
ComponentsOp-amps, transistors in active regionLogic gates, flip-flops, transistors switching
AccuracyLimited by component toleranceSet by the number of bits used
ExamplesAudio amplifier, radio receiver, thermocoupleMicroprocessor, memory, calculator, digital watch
Using a multimeter correctly

To measure resistance: switch off and isolate the circuit, select the ohms range, zero the meter by touching the probes together, then connect the probes across the component with it removed from circuit or at least one leg lifted. To measure voltage in a live circuit: select a.c. or d.c. volts on a range above the expected value, then connect the probes in parallel across the component while the circuit stays energised. To measure current: break the circuit and insert the meter in series on a current range.

Multimeter probes making contact with component leads plugged into a breadboard
Testing on a breadboard. Pressing the probe tips against wires plugged into the board gives a steadier contact than holding them in mid-air — useful when you are taking a reading with one hand and steadying the circuit with the other. Photo: Zeroping, CC BY 4.0 · Wikimedia Commons

Quick check — LO1

1. Which instrument is connected in series to measure current?

An ammeter must carry the current it measures, so it goes in the path. A voltmeter goes across (in parallel).

2. Three 6 Ω resistors are in parallel. What is the total resistance?

1/R = 3/6, so R = 2 Ω. For n equal resistors in parallel, R = R₁/n.

3. One joule per coulomb defines which unit?

E = W/Q, so 1 V = 1 J/C. Compare 1 A = 1 C/s and 1 W = 1 J/s.

4. Doubling the cross-sectional area of a cable does what to its resistance?

R = ρl/A, so resistance is inversely proportional to area.

5. The 240 V figure quoted for mains supply is which value?

Mains voltages are always quoted as r.m.s. The peak is 240 × 1.414 ≈ 339 V.

Practice questions — LO1 40 marks
  1. Define the term circuit as used in electronics and identify THREE components of an electronic circuit. (5 marks)
    Model answer
    A circuit is a complete closed conducting path through which electric current flows from a source, through a load, and back to the source.
    Three components: a source (cell or battery) that supplies e.m.f.; a load (lamp, resistor, motor) that converts the energy; conductors (wires or PCB tracks) that carry the current. A control device such as a switch is also accepted.
  2. Name THREE parameters that are measured in an electric circuit and state the instrument used for each. (6 marks)
    Model answer
    Voltage, measured in volts with a voltmeter connected in parallel. Current, measured in amperes with an ammeter connected in series. Resistance, measured in ohms with an ohmmeter or multimeter, with the circuit isolated. (Power with a wattmeter is also acceptable.)
  3. Differentiate between an analogue circuit and a digital circuit. (4 marks)
    Model answer
    An analogue circuit handles signals that vary continuously over a range of values and are represented as smooth waveforms; noise accumulates and degrades them. A digital circuit handles signals at two discrete levels only, 0 and 1, represented as square pulses; noise is rejected provided the two levels remain distinguishable. Examples: audio amplifier versus microprocessor.
  4. State Ohm's law. An electric iron has a resistance of 50 Ω and a current of 3.2 A flows through it. Calculate the voltage across it and the power dissipated. (6 marks)
    Model answer
    Ohm's law: the current through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant.
    V = I R = 3.2 × 50   = 160 V
    P = V I = 160 × 3.2 = 512 W   (or P = I²R = 3.2² × 50 = 512 W)
  5. Draw a labelled schematic diagram of an electric circuit comprising a cell, a resistor, an ammeter, a voltmeter and a closed switch. (7 marks)
    Model answer
    See Figure 1.1 above. Marks are given for: correct cell symbol with long and short plates; correct zig-zag or rectangle resistor symbol; ammeter drawn as a circled A in series in the main loop; voltmeter drawn as a circled V in parallel across the resistor; switch shown closed; a complete unbroken loop; all five items labelled.
  6. Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel across a 24 V d.c. supply. Calculate the total resistance, the total current, and the current in the 6 Ω branch. (6 marks)
    Model answer
    1/Rₜ = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12
    Rₜ   = 2 Ω
    Iₜ   = V/Rₜ = 24/2 = 12 A
    I(6Ω) = 24/6 = 4 A
  7. A copper cable of length 500 m has a cross-sectional area of 2.5 mm². Taking the resistivity of copper as 0.0168 µΩm, calculate its resistance. (4 marks)
    Model answer
    ρ = 0.0168 × 10⁻⁶ = 1.68 × 10⁻⁸ Ωm
    A = 2.5 mm² = 2.5 × 10⁻⁶ m²
    R = ρl/A = (1.68 × 10⁻⁸ × 500) / (2.5 × 10⁻⁶)
      = (8.4 × 10⁻⁶) / (2.5 × 10⁻⁶) = 3.36 Ω
  8. Describe THREE advantages of connecting domestic appliances in parallel rather than in series. (6 marks)
    Model answer
    Each appliance receives the full supply voltage of 240 V and so operates at its rated power. Each can be switched on or off independently without interrupting the others. If one appliance fails open-circuit, the remaining appliances continue to work, so the system is fault tolerant. Additional loads can also be added without re-rating existing ones.
  9. A sinusoidal supply has a peak value of 325 V. Determine its r.m.s. value, its average value and the form factor. (6 marks)
    Model answer
    Vᵣᵐᵣ = 0.707 × 325 = 229.8 V  ≈ 230 V
    Vᴡᵛ  = 0.637 × 325 = 207.0 V
    Form factor = Vᵣᵐᵣ/Vᴡᵛ = 229.8/207.0 = 1.11
    This confirms that a 325 V peak waveform is the familiar 230 V mains supply.
2

Identify electronic components

~30 min

By the end of this outcome you should be able to

  • Identify resistors, capacitors, diodes and inductors by symbol and by appearance
  • State the characteristics and applications of each component
  • Read the resistor colour code and interpret tolerance in practical terms
  • Calculate capacitor and inductor combinations, reactance, impedance and resonance
  • Identify integrated circuit characteristics, packages and handling precautions

Identification of electronic components

Components split into two families. Passive components cannot produce power gain — they resist, store or transform energy. Active components can control or amplify current, and need a supply to do it.

Resistor

Opposes current and drops voltage. Measured in ohms. Passive, non-polarised.

Capacitor

Stores charge in an electric field. Measured in farads. Passive; some types polarised.

Inductor

Stores energy in a magnetic field. Measured in henries. Passive.

Diode

Conducts one way only. Active semiconductor device.

Transistor

Switches or amplifies. The building block of all digital logic. Active.

Integrated circuit

Thousands to billions of components on one silicon chip.

Varistor (VDR)

Non-linear: resistance collapses above a threshold voltage. Used for surge protection.

Zener diode

Designed to conduct in reverse at a precise voltage. Used for regulation.

LED

Emits light when forward biased. Needs a series resistor to limit current.

Resistors

A resistor opposes the flow of current and converts electrical energy into heat. Its value in ohms, its tolerance and its power rating are the three things you need before choosing one.

Characteristics

  • Resistance value — from a fraction of an ohm to many megohms
  • Tolerance — how far the actual value may stray from the marked one (±1 %, ±5 %, ±10 %)
  • Power rating — watts it can dissipate without burning, typically 0.125 W to 5 W
  • Temperature coefficient — how much the value drifts as it warms
  • Resistors are non-polarised: they can be fitted either way round
  • Preferred values follow the E-series (E12, E24), which is why you find 4.7 kΩ and 6.8 kΩ but not 5 kΩ

Types of resistor

TypeBehaviourWhere used
Fixed (carbon film, metal film, wirewound)One permanent valueCurrent limiting, biasing, pull-ups
Variable (potentiometer, rheostat)Value adjusted by handVolume controls, calibration, dimmers
Thermistor (NTC / PTC)Value changes with temperatureTemperature sensing, inrush limiting
Light dependent resistor (LDR)Value falls as light increasesStreet lights, camera exposure
Varistor (VDR)Value falls sharply above a threshold voltageSurge and transient protection
The varistor, in the words papers expect

A varistor is a voltage-dependent, non-linear resistor. At normal working voltage its resistance is very high and it draws almost no current, so the circuit behaves as if it were not there. When a surge pushes the voltage above its clamping level, its resistance collapses to a few ohms and it diverts the surge energy safely to the return path, protecting the equipment. The two main types are the metal oxide varistor (MOV) and the silicon carbide varistor.

Applications of resistors

Limiting current through an LED, dividing voltage to a safe level for a sensor input, setting the gain of an amplifier, pulling a logic input to a known state, biasing a transistor, and discharging a capacitor safely after power-off.

An array of axial-lead resistors of different colour bands and resistance values
Real resistors, left to right: 39 kΩ, 47 Ω, 1.5 kΩ, 100 Ω, 2.2 kΩ and 47 kΩ. All carry a gold tolerance band (±5 %) and are rated 0.25 W. Photo: Evan-Amos, public domain · Wikimedia Commons

Reading the resistor colour code

Read from the band nearest an end, keeping the tolerance band — usually gold or silver, and spaced slightly apart — on the right.

A common mnemonic for the digits: Black Brown Red Orange Yellow Green Blue Violet Grey White.
ColourDigitMultiplierTolerance
Black0×1—
Brown1×10±1 %
Red2×100±2 %
Orange3×1 k—
Yellow4×10 k—
Green5×100 k±0.5 %
Blue6×1 M±0.25 %
Violet7×10 M±0.1 %
Grey8—±0.05 %
White9——
Gold—×0.1±5 %
Silver—×0.01±10 %

Worked example — past-paper style

A resistor carries the bands Red, Violet, Brown, Gold.
(a) Determine its value. (b) State the tolerance.
(c) Explain what that tolerance means in practical use.

(a) Red = 2, Violet = 7, Brown = ×10
    Value = 27 × 10 = 270 Ω
(b) Gold = ±5 %
(c) The true resistance may lie anywhere between
    270 − 5%  = 256.5 Ω  and
    270 + 5%  = 283.5 Ω
    The designer must ensure the circuit still works
    anywhere in that band. Where the exact value matters
    — a timing circuit or a precision divider — a
    ±1 % (brown) resistor is specified instead.
Resistor colour-code decoder
1st 2nd × tol

Capacitors

A capacitor stores energy in the electric field between two conducting plates separated by an insulating dielectric. Capacitance is measured in farads (F), though practical values are microfarads, nanofarads and picofarads.

Q = C V  ·  energy W = ½ C V²  ·  time constant τ = R C  ·  C = εA/d

Characteristics

  • Blocks d.c., passes a.c. — once charged, no steady current flows through it
  • Opposition to a.c. is capacitive reactance, XC = 1/(2πfC), which falls as frequency rises
  • Has a working voltage that must never be exceeded
  • Capacitance rises with plate area and with the dielectric constant, and falls as plate separation increases
  • Electrolytic and tantalum types are polarised — fitting them backwards can make them vent or burst
  • Charges to 63 % of the supply in one time constant, and is taken as fully charged after five
TypeTypical rangeWhere used
Ceramic1 pF – 1 µFDecoupling, high-frequency filtering
Electrolytic (polarised)1 µF – 10 000 µFPower supply smoothing, bulk storage
Tantalum (polarised)0.1 µF – 100 µFCompact, stable supply decoupling
Film (polyester, polypropylene)1 nF – 10 µFAudio, timing, a.c. coupling
Variable / trimmer5 pF – 500 pFRadio tuning
Supercapacitor0.1 F – 3000 FMemory backup, energy buffering
A group of aluminium electrolytic capacitors in snap-in, leaded and surface-mount styles
Electrolytic capacitors in snap-in, leaded and surface-mount styles. The polarity stripe and the printed capacitance and voltage rating are visible on each can. Photo: Zhao JingC · Wikimedia Commons

Capacitors in series and parallel — the opposite of resistors

Series: 1/CT = 1/C₁ + 1/C₂ + 1/C₃  ·  same charge Q on each, voltages add
Parallel: CT = C₁ + C₂ + C₃  ·  same voltage across each, charges add

Worked example — capacitors in series

C1 = 4 F and C2 = 2 F are connected in series across 12 V.
Find (i) total capacitance, (ii) total charge,
(iii) the voltage across each capacitor.

(i)  1/Cₜ = 1/4 + 1/2 = 1/4 + 2/4 = 3/4
     Cₜ   = 4/3 = 1.333 F

(ii) Q = Cₜ × V = 1.333 × 12 = 16 C
     (In series the SAME charge flows through both.)

(iii) V₁ = Q/C₁ = 16/4 = 4 V
      V₂ = Q/C₂ = 16/2 = 8 V
      Check: 4 + 8 = 12 V  ✓

Note the pattern: the SMALLER capacitor takes the
LARGER share of the voltage.

Applications of capacitors

Smoothing the ripple after a rectifier, coupling an audio signal while blocking d.c. bias, decoupling supply noise beside an IC, setting the frequency of an oscillator or 555 timer, correcting power factor in industrial plant, and tuning a radio receiver.

Charged capacitors bite

A large electrolytic in a power supply can hold a dangerous charge for minutes after switch-off. Discharge it through a suitable bleed resistor — never by shorting the terminals with a screwdriver.

Inductors, reactance and impedance

An inductor is a coil of wire that stores energy in a magnetic field when current flows. Inductance is measured in henries (H).

A collection of small electronic inductors of various shapes and colours
Small signal inductors. The coloured bands on some of these follow the same colour code system as resistors, but read as microhenries instead of ohms. Photo: Miguel/FDominec, GFDL · Wikimedia Commons
Induced e.m.f. e = −L (dI/dt)  ·  energy W = ½ L I²  ·  XL = 2πfL
Series: LT = L₁ + L₂  ·  Parallel: 1/LT = 1/L₁ + 1/L₂  (same as resistors)

Characteristics

  • Opposes any change in current — current through an inductor cannot jump instantly
  • Passes d.c. easily but opposes a.c. more strongly as frequency rises — the opposite of a capacitor
  • Lenz's law: the induced e.m.f. always acts in the direction that opposes the change producing it. That is the meaning of the minus sign in the formula
  • Switching an inductor off produces a large back-e.m.f. spike, which is why relay and motor coils need a flyback diode
Inductor typeCharacteristicArea of application
Air coreLow inductance, no core losses, no saturationRadio frequency circuits, tuned circuits, transmitters
Iron coreHigh inductance, high losses at high frequencyMains transformers, low-frequency chokes, power filtering
Ferrite coreGood at high frequency, low eddy-current lossSwitch-mode supplies, EMI suppression beads
Variable (tapped or slug-tuned)Inductance adjusted by moving the coreTuning and calibration in radio receivers
frequency → opposition → R — flat X₊ = 1/(2πfC) — falls Xₗ = 2πfL — rises resonance: Xₗ = X₊
Figure 2.1 — Resistance is independent of frequency; capacitive and inductive reactance move in opposite directions. Where they are equal, the circuit is at resonance.

Impedance and resonance

In an a.c. circuit, total opposition is impedance Z, measured in ohms, combining resistance and reactance as a right-angled triangle rather than a simple sum.

Series R-C: Z = √(R² + XC²)  ·  Series R-L: Z = √(R² + XL²)  ·  I = V / Z
Resonant frequency fr = 1 / (2π√(LC))  ·  at fr, XL = XC and Z is purely resistive

Worked example — reactance and impedance

Three resistors of 10 Ω, 20 Ω and 30 Ω in parallel are fed
from 240 V. A 50 µF capacitor is then added in series with
the parallel network, on a 50 Hz supply.

Parallel resistance (from LO1):  R = 60/11 = 5.4545 Ω
Current before the capacitor:    I = 240/5.4545 = 44 A

X₊ = 1 / (2πfC)
   = 1 / (2 × 3.1416 × 50 × 50 × 10⁻⁶)
   = 1 / (0.015708)
   = 63.66 Ω

Z  = √(R² + X₊²)
   = √(5.4545² + 63.66²)
   = √(29.75 + 4052.6)
   = √4082.35  =  63.89 Ω

New current I = V/Z = 240 / 63.89 = 3.76 A

The capacitor dominates: current drops from 44 A to under 4 A.
An industrial use of combining R and C is a snubber network
across contactor contacts, which suppresses arcing and
voltage transients when the contacts open.
Reactance, impedance and resonance calculator

Diodes, integrated circuits and safe handling

A diode is a two-terminal semiconductor that conducts in one direction only. The terminals are the anode (positive) and the cathode (marked with a band). The physics is in LO3; here it is the component itself that matters.

TypeKey characteristicApplication
Rectifier / signal diodeConducts above about 0.7 V (Si) or 0.3 V (Ge)Converting a.c. to d.c., signal detection, flyback protection
Zener diodeConducts in reverse at a precise breakdown voltageVoltage regulation and reference
Light emitting diode (LED)Emits light when forward biasedIndicators, displays, lighting
PhotodiodeGenerates current when light falls on itLight meters, optical receivers, remote-control receivers, solar cells, smoke detectors
Schottky diodeVery low forward drop, very fast recoveryHigh-frequency switching, low-loss rectifiers
VaractorCapacitance varies with reverse voltageElectronic tuning in radio and TV
Transient voltage suppression (TVS)Clamps fast voltage spikesProtecting inputs from ESD and surges
A 1N4001 rectifier diode with its cathode band visible, shown against a centimetre ruler
1N4001 rectifier diode. The black band marks the cathode. Photo: Vonvon · Wikimedia Commons
Several small coloured LEDs of different colours lit up
LEDs of different colours, each needing its own series resistor to limit current. Photo: Ingo Dierking, CC BY-SA · Wikimedia Commons
A disc-shaped metal oxide varistor rated at 385 volts
A metal oxide varistor rated at 385 V. It conducts heavily above this voltage to shunt a surge away from the protected circuit. Photo: Michael Schmid · Wikimedia Commons

Identification of integrated circuit characteristics

An integrated circuit (IC) is a complete circuit — transistors, resistors, diodes and their interconnections — fabricated on a single chip of semiconductor, usually silicon, and sealed in a package with external pins.

Reasons for using ICs in electronic design
  1. Very small size and light weight compared with the discrete circuit they replace
  2. Low cost per function, because thousands of chips are produced on a single wafer
  3. High reliability — far fewer soldered joints and interconnections to fail
  4. Low power consumption and higher operating speed, since signals travel shorter distances
  5. Easier design and maintenance — one chip replaces a board of parts and is simply swapped if faulty
  6. Consistent performance, because all components on the chip are made together and age together
Limitations
  • Not repairable — a faulty IC is replaced, never fixed internally
  • Sensitive to electrostatic discharge and to exceeding the supply voltage
  • Limited power handling, so high-power stages still use discrete devices
  • Inductors and large-value capacitors cannot be practically fabricated on chip

Classification by scale of integration

ScaleComponents per chipExample
SSI — small scaleup to 100Logic gate packages (7400 series)
MSI — medium scale100 – 1 000Counters, decoders, multiplexers
LSI — large scale1 000 – 100 000Early microprocessors, small memories
VLSI — very large scale105 – 107Modern CPUs, GPUs, RAM chips
ULSI — ultra large scaleabove 107System-on-chip in a smartphone

ICs are also classified by function: analogue or linear ICs (operational amplifiers, voltage regulators such as the 7805, audio amplifiers), digital ICs (logic gates, counters, memories, microprocessors), and mixed-signal ICs containing both, such as analogue-to-digital converters.

12 34 8765 DIP-8 · notch marks pin 1 QFP — surface mount BGA — solder balls
Figure 2.2 — Three common IC packages. Pin 1 is found by the notch or dot, and numbering runs anticlockwise from it.
A PIC16F84 microcontroller in an 18-pin DIP package on a workbench
A microcontroller IC in a DIP package — the notch at the left end marks pin 1. Photo: Wollschaf, GFDL · Wikimedia Commons
Several assorted integrated circuit chips of different sizes and pin counts
Assorted ICs — the same black epoxy package hides anything from a single logic gate to a whole microprocessor. Photo: Mataresephotos · Wikimedia Commons
An everyday example: the diode and IC inside a computer mouse

An optical mouse is a compact case study in this outcome. It shines an infrared LED onto the desk surface, and a matching photodiode sensor — itself a small IC — captures thousands of images of that surface every second to work out how far and which way the mouse has moved. The photo below shows the actual LED, lens and sensor package taken from inside one.

The infrared LED, lens and optical sensor chip removed from inside a computer mouse
Inside an optical mouse: an infrared LED, a small lens, and the optical sensor IC (an Avago ADNS-5090) that images the desk surface and reports movement to the host computer. Photo: Raimond Spekking, CC BY-SA 4.0 · Wikimedia Commons
Safety precautions when handling and testing ICs
  • Guard against electrostatic discharge. Wear an earthed wrist strap, work on an anti-static mat, and store chips in conductive foam or anti-static bags. A static spark you cannot even feel can destroy a CMOS input.
  • Power down and discharge before inserting or removing. Hot-plugging an IC can latch it up and destroy it instantly.
  • Observe correct orientation and pin numbering. Locate pin 1 by the notch or dot; reversing the chip usually applies the supply backwards and kills it.
  • Never exceed the rated supply voltage shown in the datasheet, and check polarity with a meter before connecting.
  • Use a temperature-controlled, earthed soldering iron and work quickly; excessive heat damages the die and lifts PCB pads.
  • Use an IC extractor tool rather than a screwdriver, to avoid bending pins or slipping across adjacent tracks.

Quick check — LO2

1. A resistor is banded brown, black, red, gold. What is its value?

Brown = 1, black = 0, red = ×100, so 10 × 100 = 1000 Ω = 1 kΩ. Gold is ±5 %.

2. Which component blocks d.c. but passes a.c.?

Once a capacitor charges, no steady current flows; a changing voltage keeps charging and discharging it.

3. Two capacitors of 4 µF and 4 µF are connected in parallel. Total capacitance is

Capacitors in parallel add: 4 + 4 = 8 µF. In series they would give 2 µF.

4. What happens to capacitive reactance when frequency increases?

X₊ = 1/(2πfC), so reactance is inversely proportional to frequency. Inductive reactance does the opposite.

5. Which application belongs to an air-core inductor?

Air cores have no core loss and cannot saturate, which suits high frequencies. Iron cores serve low-frequency, high-inductance roles.

Practice questions — LO2 42 marks
  1. Briefly explain the role of each of the following in an electronic circuit: (a) inductor (b) resistor (c) transistor (d) diode. (8 marks)
    Model answer
    Inductor: stores energy in a magnetic field and opposes changes in current; used in filters, chokes, transformers and tuned circuits.
    Resistor: opposes current flow and drops voltage, converting electrical energy to heat; used to limit current, divide voltage and set bias.
    Transistor: an active three-terminal device that amplifies a small signal or switches a larger current on and off; the basis of logic circuits.
    Diode: allows current to flow in one direction only; used to rectify a.c. to d.c., protect against reverse polarity and suppress spikes.
  2. A student finds a resistor with the bands Yellow, Violet, Orange, Silver. Determine the value, the tolerance, and the range within which the true value must lie. (5 marks)
    Model answer
    Yellow = 4, Violet = 7, Orange = ×1 000
    Value = 47 × 1 000 = 47 000 Ω = 47 kΩ
    Silver = ±10 %
    Range: 47 k − 4.7 k = 42.3 kΩ
           47 k + 4.7 k = 51.7 kΩ
  3. Describe the operation of a varistor and state TWO types of this component. (4 marks)
    Model answer
    A varistor is a voltage-dependent non-linear resistor. Below its clamping voltage its resistance is extremely high and it draws negligible current. When the applied voltage rises above that threshold, its resistance falls sharply to a few ohms, diverting the surge current away from the protected equipment and clamping the voltage.
    Two types: the metal oxide varistor (MOV) and the silicon carbide varistor.
  4. Describe step by step how you would use a multimeter to measure (a) the resistance of a resistor and (b) the voltage across a lamp in a live circuit. (4 marks)
    Model answer
    (a) Resistance: switch off and isolate the supply, and discharge any capacitors. Select an ohms range above the expected value. Touch the probes together to confirm the meter reads near zero. Lift one leg of the resistor out of circuit so parallel paths do not affect the reading. Connect a probe to each lead and read the display, multiplying by the range factor if the meter is analogue.
    (b) Voltage: select a.c. or d.c. volts on a range higher than the expected value. Leaving the circuit energised, connect the probes in parallel across the lamp terminals, observing polarity on d.c. Read the display without touching the metal probe tips.
  5. Two capacitors, 6 µF and 3 µF, are connected in series across a 9 V supply. Calculate the total capacitance, the charge stored, and the p.d. across each capacitor. (6 marks)
    Model answer
    1/Cₜ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6
    Cₜ = 2 µF
    Q  = CₜV = 2 × 10⁻⁶ × 9 = 18 µC
    V₁ = Q/C₁ = 18/6 = 3 V
    V₂ = Q/C₂ = 18/3 = 6 V
    Check 3 + 6 = 9 V  ✓
  6. Calculate the capacitive reactance of a 100 µF capacitor at 50 Hz, and state what happens to this value if the frequency is doubled. (5 marks)
    Model answer
    X₊ = 1/(2πfC)
       = 1/(2 × 3.1416 × 50 × 100 × 10⁻⁶)
       = 1/0.031416
       = 31.83 Ω
    At 100 Hz the reactance halves to 15.92 Ω, because XC is inversely proportional to frequency.
  7. List FOUR reasons for using integrated circuits when designing electronic systems. (4 marks)
    Model answer
    Very small size and light weight; low cost per function due to mass production on a single wafer; high reliability because there are fewer interconnections and solder joints to fail; low power consumption and higher operating speed; easier design, assembly and maintenance since one chip replaces a whole board of discrete parts.
  8. Describe THREE safety precautions to observe when handling and testing integrated circuits in the laboratory. (6 marks)
    Model answer
    Guard against electrostatic discharge by wearing an earthed wrist strap, using an anti-static mat and storing chips in conductive packaging, because static can destroy a CMOS input without any visible sign.
    Always power down and discharge the circuit before inserting or removing a chip, since hot-plugging can cause latch-up and permanent damage.
    Identify pin 1 from the notch or dot and observe correct orientation and the rated supply voltage, because reversing the chip or over-volting it destroys it immediately.
3

Understand semi-conductor theory

~32 min

By the end of this outcome you should be able to

  • Define a semiconductor and the related terms
  • Describe the structure of matter and electron behaviour in conductors and semiconductors
  • Distinguish intrinsic from extrinsic semiconductors and state the advantages of semiconductor devices
  • Explain how P-type and N-type materials are formed and name their charge carriers
  • Describe P-N junction operation under forward and reverse bias, including breakdown
  • Explain PNP and NPN transistor operation and calculate collector current and VCE

Definition of semiconductor and related terms

A semiconductor is a material whose electrical conductivity lies between that of a conductor and an insulator, and which can be controlled by adding impurities, by temperature, by light or by an applied voltage. That controllability is the entire reason modern electronics exists.

Atom
The smallest particle of an element that still has the properties of that element.
Nucleus
The dense centre of an atom, containing protons (positive) and neutrons (neutral).
Electron
A negatively charged particle orbiting the nucleus in shells.
Neutron
A neutral nuclear particle that adds mass, binds protons together against repulsion, and gives rise to isotopes.
Valence shell
The outermost electron shell; it determines chemical and electrical behaviour.
Valence electron
An electron in the valence shell. Semiconductors have four of them.
Free electron
An electron with enough energy to escape its atom and drift through the material.
Hole
The vacancy left by a departed electron; behaves as a mobile positive charge carrier.
Covalent bond
A bond formed when neighbouring atoms share valence electrons.
Doping
Deliberately adding a small amount of impurity to change conductivity.
Intrinsic
A pure, undoped semiconductor. Electrons and holes exist in equal numbers.
Extrinsic
A doped semiconductor — either P-type or N-type.
Majority carrier
The carrier present in larger numbers: electrons in N-type, holes in P-type.
Minority carrier
The carrier present in smaller numbers: holes in N-type, electrons in P-type.
Biasing
Applying an external d.c. voltage to set a device's operating condition.
Depletion region
The carrier-free zone either side of a P-N junction.
Barrier potential
The voltage across the depletion region: about 0.7 V for silicon, 0.3 V for germanium.

Intrinsic versus extrinsic semiconductors

PointIntrinsicExtrinsic
PurityPure, no impurity addedDeliberately doped
CarriersElectrons and holes in equal numbersOne carrier type dominates
ConductivityVery low at room temperatureMuch higher and controllable
Depends onTemperature onlyDoping concentration mainly
ExamplePure silicon or germanium crystalSilicon doped with boron or phosphorus

Advantages of semiconductor devices over valves and other devices

  1. Very small size and light weight, allowing miniaturised equipment
  2. Low power consumption, with no heater or filament required
  3. No warm-up time — they operate the instant power is applied
  4. Long operating life and high mechanical ruggedness, with no fragile glass envelope
  5. Low operating voltage, so they are safer and cheaper to supply
  6. Cheap to mass produce and easily integrated into ICs
  7. Silent operation with no vibration

Disadvantages: sensitive to heat and to electrostatic discharge; limited power handling compared with valves; performance drifts with temperature; permanently damaged by voltage or current overload; and they cannot be repaired once failed.

Description of the structure of matter

All matter is made of atoms. An atom has a nucleus of protons and neutrons, surrounded by electrons in shells labelled K, L, M, N outward from the centre. Each shell holds at most 2n² electrons, where n is the shell number: 2, 8, 18, 32.

An atom is electrically neutral because the number of protons equals the number of electrons. Only the valence electrons in the outermost shell take part in conduction — inner-shell electrons are bound too tightly to matter.

+14 4 valence electrons Silicon (Si), atomic number 14 Shell K: 2 electrons Shell L: 8 electrons Shell M: 4 electrons — the valence shell Four valence electrons is the signature of a semiconductor. Conductors have one to three; insulators have five to eight. Germanium (Ge), atomic number 32, has shells 2, 8, 18, 4 — also four in the valence shell.
Figure 3.1 — Shell structure of a silicon atom. Papers ask you to sketch and label exactly this, so practise drawing it with the shell counts written in.

Electrons in conductors and semiconductors

MaterialValence electronsEnergy gapConductivityExamples and uses
Conductor1 – 3None — bands overlapVery highCopper, silver, aluminium — cables and PCB tracks
Semiconductor4Small (Si ≈ 1.1 eV, Ge ≈ 0.7 eV)Intermediate, controllableSilicon, germanium, GaAs — diodes, transistors, ICs, solar cells
Insulator5 – 8Large (above about 5 eV)Practically zeroRubber, PVC, mica, glass — cable sheathing, tool handles, PCB substrate
Conductor bands overlap no gap to cross Semiconductor small gap ≈ 1.1 eV heat or light lifts electrons Insulator wide gap above 5 eV electrons stay put conduction valence
Figure 3.2 — Energy band model. A semiconductor conducts only when electrons gain enough energy to cross the forbidden gap into the conduction band, leaving holes behind.
Why semiconductors behave oddly with heat

In a metal, heating increases resistance because the lattice vibrates and obstructs electrons. In a semiconductor, heating frees more carriers, so resistance falls. A semiconductor has a negative temperature coefficient of resistance. This is why power transistors need heatsinks: hot means more conductive, which means hotter still — a runaway that must be designed against.

A thin, mirror-polished circular silicon wafer catching the light
A silicon wafer — a thin slice cut from a single crystal of ultra-pure silicon. Hundreds of individual chips are patterned onto one wafer like this before being cut apart and packaged. Photo: Inductiveload, public domain · Wikimedia Commons

Types of semiconductor materials

Silicon

  • Atomic number 14, four valence electrons, energy gap about 1.1 eV
  • Extremely abundant — sand is silicon dioxide, so raw material cost is low
  • Tolerates higher temperature, up to about 150 °C, before leakage becomes a problem
  • Low reverse leakage current
  • Forms a stable oxide, SiO₂, which makes reliable ICs possible
  • Forward voltage drop about 0.7 V

Germanium

  • Atomic number 32, four valence electrons, energy gap about 0.7 eV
  • Lower forward drop of about 0.3 V, which suits very small signal detection
  • Higher leakage current and a low temperature limit of roughly 70 °C
  • Scarcer and more expensive than silicon, so now largely replaced by it

Compound semiconductors also matter: gallium arsenide (GaAs) for high-frequency and microwave devices, gallium nitride (GaN) for efficient power conversion and blue LEDs, and silicon carbide (SiC) for high-voltage, high-temperature power electronics.

P-type and N-type materials

Pure silicon barely conducts at room temperature because each atom's four valence electrons are locked into covalent bonds with four neighbours. Doping adds a controlled impurity — roughly one atom in ten million — and transforms conductivity.

N-type — pentavalent donor Si Si Si Si P e⁻ the 5th electron has no bond to join → free electron, majority carrier P-type — trivalent acceptor Si Si Si Si B one bond is left empty → hole, majority carrier
Figure 3.3 — Doping. A donor atom brings a spare electron; an acceptor atom leaves a hole. Both materials remain electrically neutral overall.
FeatureN-typeP-type
Impurity addedPentavalent (5 valence electrons)Trivalent (3 valence electrons)
Doping elementsPhosphorus, arsenic, antimony, bismuthBoron, aluminium, gallium, indium
Impurity calledDonor — it donates an electronAcceptor — it accepts an electron
Majority carriersElectronsHoles
Minority carriersHolesElectrons
Conduction mechanismElectron drift towards the positive terminalHole movement towards the negative terminal, as electrons hop from bond to bond
Overall chargeNeutralNeutral
Common exam trap

N-type is not negatively charged and P-type is not positively charged. Both are electrically neutral — the added atom brings its own protons with it. "N" and "P" describe which carrier is in the majority, nothing more.

P-N junction diode operation

Join P-type to N-type and you form a P-N junction. At the boundary, free electrons from the N side diffuse across and recombine with holes on the P side. This leaves exposed ions and creates a thin depletion region containing no free carriers, with a potential barrier across it — about 0.7 V for silicon and 0.3 V for germanium.

a  Unbiased P N depletion region barrier ≈ 0.7 V (Si) b  Forward biased — conducts P N + − region narrows → large current flows c  Reverse biased — blocks P N − + region widens → only tiny leakage Symbol & polarity anode (+) cathode (−) The bar matches the band on the body.
Figure 3.4 — The same junction under three conditions. The arrow in the symbol points the way conventional current can flow.

Forward biasing

The positive terminal of the supply is connected to the P side and the negative to the N side.

  • The applied voltage opposes the barrier potential, so the depletion region narrows
  • Once the supply exceeds the barrier (0.7 V Si, 0.3 V Ge), majority carriers cross freely
  • A large forward current flows and the diode acts almost like a closed switch
  • Forward resistance is low, typically a few ohms
  • A series resistor is essential, or the current will destroy the diode

Reverse biasing

The positive terminal is connected to the N side and the negative to the P side.

  • The applied voltage reinforces the barrier, so the depletion region widens — and the wider it is, the more the junction behaves like a capacitor, which is what a varactor exploits
  • Majority carriers are pulled away from the junction and cannot cross
  • Only a small reverse leakage current flows, carried by minority carriers
  • Reverse resistance is very high, so the diode acts like an open switch

Breakdown: avalanche and Zener

If reverse voltage keeps rising, the junction eventually breaks down and conducts heavily. Two distinct mechanisms produce this, and papers ask about both.

  • Avalanche breakdown occurs in lightly doped junctions at higher voltages, typically above about 6 V. The wide depletion region and strong electric field accelerate the few minority carriers to high velocity. They collide with atoms in the lattice and knock out further electron-hole pairs by impact ionisation. Those new carriers are accelerated in turn and collide again, so the carrier population multiplies like an avalanche and reverse current rises sharply.
  • Zener breakdown occurs in heavily doped junctions at lower voltages, below about 5 V. The depletion region is so narrow that the field rips electrons directly out of covalent bonds without any collision.

Practical applications of reverse breakdown: Zener diodes used as voltage regulators and reference sources; avalanche photodiodes used in optical receivers because the multiplication amplifies weak light signals; and transient voltage suppression diodes that clamp surges to protect circuits.

Breakdown is only safe when the current is limited by a series resistor. Beyond the rated peak inverse voltage (PIV), an ordinary rectifier diode suffers permanent thermal damage.

V I (forward) I (reverse) 0.7 V knee forward conduction small leakage current breakdown / PIV avalanche or Zener reverse bias ←
Figure 3.5 — V-I characteristic of a silicon diode. Note the knee at 0.7 V, the almost flat leakage region, and the near-vertical breakdown that a Zener diode is designed to sit in.
Testing a diode with a multimeter

On the diode range, the meter should read roughly 0.5–0.7 V one way (red probe on the anode) and show open-circuit the other way. Two identical readings mean a fault: near zero both ways is a short circuit, open both ways is an open circuit.

Operation of transistors

A bipolar junction transistor (BJT) is a three-layer, two-junction semiconductor device with terminals named emitter (E), base (B) and collector (C). It can amplify a small signal or switch a large current.

Three construction facts explain everything the device does:

  • The emitter is heavily doped — its job is to inject carriers
  • The base is very thin and lightly doped — most carriers pass straight through it
  • The collector is moderately doped and physically largest — it collects carriers and dissipates the heat
NPN N P N emitter base collector B C E arrow points out "Not Pointing iN" PNP P N P emitter base collector B C E arrow points in towards the base
Figure 3.6 — Construction and symbols. The arrow is always on the emitter and always shows conventional current direction.
Assorted discrete transistors in TO-3, TO-126, TO-92 and SOT-23 packages, largest to smallest
Real transistor packages, largest to smallest: TO-3 (bolted to a heatsink for power use), TO-126, the common small-signal TO-92, and the surface-mount SOT-23. CC BY-SA 3.0 · Wikimedia Commons

How an NPN transistor works

For normal amplifying operation the emitter-base junction is forward biased and the collector-base junction is reverse biased.

  1. The forward-biased emitter-base junction lets the heavily doped emitter inject a large flood of electrons into the base
  2. The base is thin and lightly doped, so only about 1–5 % of those electrons recombine with holes there and leave as base current IB
  3. The remaining 95–99 % are swept across into the reverse-biased collector by its positive potential, forming the collector current IC
  4. A small change in IB therefore produces a large change in IC — that is current amplification

A PNP transistor works identically but with every polarity reversed and holes as the carriers. Its emitter is held positive with respect to the collector, and it is turned on by making the base more negative than the emitter. Holes injected from the emitter cross the thin N base and are collected by the negative collector.

IE = IB + IC  ·  β = IC / IB (50–400)  ·  α = IC / IE  ·  β = α/(1−α)
Common emitter: VCE = VCC − ICRC  ·  voltage across load VRC = ICRC

Worked example — common emitter calculation

A transistor has Iᵝ = 200 µA, β = 120, Vᶜᶜ = 12 V, Rᶜ = 600 Ω.
Find (i) Iᶜ, (ii) the voltage across Rᶜ, (iii) VᶜᲩ.

(i)   Iᶜ = β × Iᵝ
         = 120 × 200 × 10⁻⁶
         = 0.024 A = 24 mA

(ii)  Vᵣᶜ = Iᶜ × Rᶜ
         = 0.024 × 600
         = 14.4 V

(iii) VᶜᲩ = Vᶜᶜ − IᶜRᶜ
         = 12 − 14.4
         = −2.4 V

A negative VᶜᲩ is impossible, so the transistor is
SATURATED. It cannot supply 24 mA into that load from
a 12 V rail. The actual collector current is limited to
Iᶜ(sat) = Vᶜᶜ/Rᶜ = 12/600 = 20 mA, and VᶜᲩ falls to
about 0.2 V. This is exactly what you want from a
transistor used as a SWITCH.

Also find IᲩ:  IᲩ = Iᵝ + Iᶜ = 0.0002 + 0.020 = 20.2 mA

Regions of operation

RegionE-B junctionC-B junctionUsed for
Cut-offReverseReverseSwitch fully OFF — no collector current, VCE ≈ VCC
ActiveForwardReverseLinear amplification
SaturationForwardForwardSwitch fully ON — maximum current, VCE ≈ 0.2 V

The transistor as a switch

Driven only between cut-off and saturation, a transistor behaves like a relay with no moving parts. With no base current it is off and the load is dead. With enough base current to saturate it, the collector-emitter path drops only about 0.2 V, so almost the full supply appears across the load and power wasted in the transistor is small. This is how a microcontroller output, which can supply only a few milliamps, controls a motor, a relay coil or a wireless transmitter drawing hundreds of milliamps.

To guarantee saturation, designers supply a base current several times larger than IC/β — an overdrive factor of 2 to 10 — because β varies between individual devices and falls with temperature.

PointNPNPNP
Layer orderN – P – NP – N – P
Majority carriersElectronsHoles
Arrow on emitterPoints outwardPoints inward
Collector supplyPositive relative to emitterNegative relative to emitter
Turned on byBase more positive than emitterBase more negative than emitter
Switching speedFaster — electrons are more mobileSlower
Common useDefault choice, low-side switchingHigh-side switching, complementary output pairs
Common-emitter transistor calculator

Quick check — LO3

1. How many valence electrons does a semiconductor atom have?

Four. Conductors have one to three; insulators have five to eight.

2. Adding phosphorus to silicon produces which material?

Phosphorus is pentavalent; its fifth electron has no bond to join, so it is free. Boron, being trivalent, gives P-type.

3. When a diode is reverse biased, the depletion region

The applied voltage adds to the barrier potential, pulling carriers away from the junction. Only tiny leakage remains.

4. In an NPN transistor operating as an amplifier

That is the active region. Both reverse is cut-off; both forward is saturation.

5. Avalanche breakdown is caused by

High-velocity minority carriers collide with lattice atoms, freeing more electron-hole pairs, which collide in turn.

6. A transistor with β = 100 and IB = 50 µA has a collector current of

IC = βIB = 100 × 50 µA = 5000 µA = 5 mA.

Practice questions — LO3 44 marks
  1. Sketch and label the atomic structure of a silicon atom, then describe briefly what happens to this structure when a pentavalent impurity is introduced. (8 marks)
    Model answer
    Sketch as Figure 3.1: a nucleus labelled +14, with three shells holding 2, 8 and 4 electrons, the outer four labelled as valence electrons.
    When a pentavalent atom such as phosphorus is introduced, it takes the place of a silicon atom in the lattice. Four of its five valence electrons form covalent bonds with the four neighbouring silicon atoms. The fifth has no bond to join and is only loosely held, so at room temperature it becomes a free electron available for conduction. The material becomes N-type, with electrons as majority carriers and holes as minority carriers, while remaining electrically neutral overall.
  2. Outline any FIVE advantages and FOUR disadvantages of semiconductor devices over other devices. (9 marks)
    Model answer
    Advantages: very small size and light weight; low power consumption with no heater or filament; instant operation with no warm-up time; long life and mechanical ruggedness with no fragile glass envelope; low operating voltage, making them safer and cheaper to supply; cheap to mass produce; silent, vibration-free operation.
    Disadvantages: very sensitive to heat, needing heatsinks and derating; easily destroyed by electrostatic discharge; limited power handling compared with valves; permanently damaged by voltage or current overload, and not repairable once failed; characteristics drift with temperature.
  3. Outline TWO differences between intrinsic and extrinsic semiconductors. (4 marks)
    Model answer
    An intrinsic semiconductor is pure with no impurity added, while an extrinsic semiconductor has been deliberately doped. In an intrinsic semiconductor electrons and holes exist in equal numbers and conductivity is very low and depends only on temperature; in an extrinsic semiconductor one carrier type dominates and conductivity is much higher and is set mainly by the doping concentration.
  4. Explain how the magnitude of the applied voltage affects the depletion layer of a P-N junction. (4 marks)
    Model answer
    Under forward bias the applied voltage opposes the barrier potential, so as its magnitude increases the depletion layer becomes progressively narrower; once it exceeds the barrier of about 0.7 V for silicon the layer is effectively collapsed and heavy current flows.
    Under reverse bias the applied voltage reinforces the barrier, so as its magnitude increases the depletion layer becomes progressively wider, the junction capacitance falls and only a tiny minority-carrier leakage current flows — until the breakdown voltage is reached.
  5. In a reverse-biased P-N junction, increasing the applied voltage produces avalanche breakdown. (a) Explain how this process occurs. (b) Give TWO practical applications of this phenomenon. (4 marks)
    Model answer
    (a) As reverse voltage rises, the electric field across the wide depletion region becomes very strong. The few minority carriers crossing it are accelerated to high velocity and collide with atoms in the crystal lattice, knocking valence electrons out of their covalent bonds by impact ionisation. Each new electron-hole pair is accelerated in turn and produces further collisions, so the carrier population multiplies rapidly like an avalanche and reverse current rises sharply.
    (b) Voltage regulation and reference using Zener and avalanche diodes; avalanche photodiodes in optical communication receivers, where the multiplication amplifies weak light signals. Transient voltage suppression diodes are also acceptable.
  6. Draw a well-labelled circuit symbol of a Zener diode, label its terminals and indicate the direction of current flow in normal use. (3 marks)
    Model answer
    Draw the standard diode triangle pointing at a bar, with the bar bent at both ends into a Z or flag shape. Label the triangle side as the anode and the barred side as the cathode. In normal regulating use the Zener is reverse biased, so the arrow for conventional current runs from cathode to anode, that is against the triangle — opposite to an ordinary diode. Marks are lost for drawing a plain diode bar without the bent ends.
  7. Describe how P-type and N-type semiconductors are formed, stating the charge-carrier mechanism in each. (4 marks)
    Model answer
    P-type: formed by doping pure silicon with a trivalent impurity such as boron, aluminium, gallium or indium. Each impurity atom has only three valence electrons, so one covalent bond is left incomplete, creating a hole. Conduction occurs by holes moving towards the negative terminal as electrons hop from bond to bond in the opposite direction. Holes are the majority carriers.
    N-type: formed by doping with a pentavalent impurity such as phosphorus, arsenic or antimony. The fifth valence electron has no bond to join and becomes free. Conduction occurs by these free electrons drifting towards the positive terminal. Electrons are the majority carriers.
  8. Demonstrate, with the aid of a diagram, the operation of an NPN transistor as a switch. (4 marks)
    Model answer
    Draw an NPN transistor with emitter grounded, the load (lamp or relay coil) between VCC and the collector, and a base resistor from the control input to the base.
    With no base current the transistor is in cut-off: both junctions are reverse biased, no collector current flows, the load is off and VCE is approximately equal to VCC. When sufficient base current is applied, the transistor saturates: both junctions are forward biased, collector current is limited only by the load, VCE falls to about 0.2 V, and almost the whole supply appears across the load, switching it on.
  9. A transistor in common-emitter configuration has β = 150, IB = 40 µA, VCC = 9 V and RC = 1 kΩ. Calculate IC, IE, the voltage across RC and VCE. (4 marks)
    Model answer
    Iᶜ = βIᵝ  = 150 × 40 × 10⁻⁶ = 6 mA
    IᲩ = Iᵝ + Iᶜ = 0.04 mA + 6 mA  = 6.04 mA
    Vᵣᶜ = IᶜRᶜ = 0.006 × 1000       = 6 V
    VᶜᲩ = Vᶜᶜ − IᶜRᶜ = 9 − 6      = 3 V
    VCE is positive and well away from both rails, so the transistor is operating in the active region as intended.
4

Identify and classify memory

~24 min

By the end of this outcome you should be able to

  • Define memory and state the units in which it is measured
  • Classify memories as RAM, ROM and DAM, and state the functions of ROM
  • Distinguish SRAM from DRAM, and main memory from cache
  • Distinguish semiconductor memories from magnetic memories
  • Describe the structure of a magnetic disk and perform simple disk calculations
  • Select suitable memory for a given embedded application and justify the choice

Definition of memory

Memory is the part of an electronic system that stores data, instructions and results, either temporarily during processing or permanently for later use. Every memory is an array of cells, each holding one bit — a single 0 or 1 — with a unique address so the processor can find it.

Storage units. 1 byte = 8 bits; a nibble is half a byte.
UnitSizeHolds roughly
Bit (b)1 binary digitOne on/off state
Nibble4 bitsOne hexadecimal digit
Byte (B)8 bitsOne character
Kilobyte (KB)1 024 BA short page of text
Megabyte (MB)1 024 KBA photograph or a minute of music
Gigabyte (GB)1 024 MBA couple of hours of video
Terabyte (TB)1 024 GBA large hard disk

Key memory characteristics

  • Volatility — does the content survive a power cut?
  • Access time — how long from request to data, in nanoseconds or milliseconds
  • Capacity — how much it holds
  • Cost per bit — fast memory is always expensive memory
  • Access method — random, direct or sequential
  • Read/write ability — read-write, read-mostly, or read-only

Classification of memories

MEMORY Primary (main) semiconductor, fast Secondary (backing) bulk, non-volatile RAM volatile · read-write ROM non-volatile · read SRAM DRAM PROM EPROM EEPROM · Flash Magnetic Optical Solid-state HDD · tape CD · DVD · Blu-ray SSD · flash drive
Figure 4.1 — Memory classification. Primary memory is what the CPU reaches directly; secondary memory holds everything else.
A 4GB DDR3 SO-DIMM RAM stick showing its memory chips and gold edge connector
A 4 GB DDR3 RAM module (SO-DIMM form factor, for laptops). Each black chip on the board is a DRAM package; the gold edge plugs directly into a slot on the motherboard. Photo: Tobias b Köhler · Wikimedia Commons

RAM — Random Access Memory

RAM is the computer's working memory. Any location can be reached in the same time regardless of its address, which is what "random access" means. It is volatile: switch off and the contents are lost. RAM holds the operating system, the programs currently running and the data being worked on.

FeatureSRAM (static)DRAM (dynamic)
Storage cellFlip-flop, 4–6 transistorsOne transistor and one capacitor
RefreshingNot requiredMust be refreshed thousands of times a second
SpeedVery fast, 1–10 nsSlower, 50–100 ns
DensityLow — takes more chip area per bitHigh — many more bits per chip
Cost per bitHighLow
PowerHigher when idleLower, but refresh costs power
Used asCPU cache and registersMain system memory (DDR modules)

Main memory versus cache memory

PointMain memoryCache memory
TechnologyDRAMSRAM
CapacityGigabytesKilobytes to a few megabytes
SpeedSlower than the CPUClose to CPU speed
LocationOn modules in slots on the motherboardInside or beside the CPU die
PurposeHolds all running programs and dataHolds recently and frequently used data to reduce waiting
Cost per bitLowerMuch higher

Cache is organised in levels: L1 is smallest and fastest and sits inside each core, L2 is larger, and L3 is shared across cores. Hit rates above 90 % are normal, which is why cache has such a large effect on real performance.

ROM — Read Only Memory

ROM holds data permanently and is non-volatile. It stores firmware — the instructions a device needs the instant it powers on.

Functions of ROM
  1. Stores the bootstrap loader / BIOS or UEFI that starts the computer and locates the operating system
  2. Holds the power-on self-test (POST) routines that check hardware at start-up
  3. Stores permanent firmware for embedded devices such as routers, washing machines and microwave controllers
  4. Holds fixed data tables such as character generators, look-up tables and calibration constants
  5. Provides basic input/output routines that let the processor communicate with attached hardware before the operating system loads
TypeWrittenErasedTypical use
MROM Mask ROMAt manufactureNeverMass-produced fixed firmware
PROMOnce, by the user with a programmerNeverOne-off production runs
EPROMElectricallyBy ultraviolet light through a quartz windowDevelopment and prototyping
EEPROMElectrically, byte by byteElectrically, byte by byteSettings, calibration data, small permanent stores
FlashElectrically, in blocksElectrically, in blocks — fastSSDs, memory cards, USB drives, BIOS

DAM — Direct Access Memory

Direct access memory is storage in which the read head moves straight to the region holding the data and then searches a short distance sequentially. It sits between pure random access (RAM) and pure sequential access (tape). Hard disk drives, optical discs and floppy disks all use direct access: the head steps to the right track, then waits for the right sector to rotate underneath it.

Access methodHow data is foundSpeedExample
Random accessAddress supplied; any cell reached in equal timeNanosecondsRAM, ROM
Direct accessJump near the target, then scanMillisecondsHard disk, CD, DVD
Sequential accessRead from the start until you reach itSeconds to minutesMagnetic tape

Types of memories

Semiconductor memories

These store data as charge or as the state of transistors on a silicon chip. They have no moving parts, which makes them fast, silent, shock-resistant and power-efficient.

  • Registers — a handful of the fastest stores, inside the CPU itself
  • SRAM — CPU cache at levels L1, L2 and L3
  • DRAM — main memory modules
  • ROM, PROM, EPROM, EEPROM — firmware and configuration
  • Flash (NAND and NOR) — SSDs, microSD cards, USB flash drives, phone storage

Magnetic memories

These store data as the direction of magnetisation of tiny regions on a coated surface. A read/write head magnetises a spot one way for 1 and the other way for 0.

  • Hard disk drive (HDD) — rigid spinning platters, high capacity, low cost per gigabyte, direct access
  • Magnetic tape — sequential access, very cheap per terabyte, still used for archival backup
  • Floppy disk — obsolete, but a classic example of removable magnetic direct-access storage
  • Magnetic stripe cards — bank and access control cards
A 2.5 inch SATA solid-state drive with its metal casing
A 2.5-inch SSD. No motor, no platters — just flash memory chips and a controller inside a sealed metal case. via Wikimedia Commons
A Travan magnetic tape cartridge used for computer data backup
A magnetic tape cartridge, still used for cheap, high-capacity archival backup. Data is read sequentially from the start. via Wikimedia Commons

Optical and solid-state secondary memory

  • CD-ROM, DVD, Blu-ray — data stored as pits and lands read by a laser; direct access; cheap to distribute, slow to write
  • Solid-state drive (SSD) — NAND flash; no moving parts, very fast, shock-proof, more expensive per gigabyte and with a finite write endurance
  • USB flash drive and memory card — small, removable NAND flash for transfer and expansion
  • Cloud storage — remote disk arrays reached over a network; unlimited scaling but dependent on connectivity
Close-up of the underside of a CD and DVD showing their reflective data surface
CD and DVD undersides, where a laser reads the pattern of microscopic pits and lands moulded into the reflective layer. via Wikimedia Commons
Point of comparisonSemiconductor memoryMagnetic memory
Storage principleElectric charge / transistor stateDirection of magnetisation
Moving partsNoneSpinning platters, moving heads
Access timeNanoseconds to microsecondsMilliseconds
Cost per gigabyteHigherLower
Shock resistanceHighLow — a drop can wreck a running drive
Power use and noiseLow, silentHigher, audible
Data retentionVolatile (RAM) or limited-cycle (flash)Long-term, many rewrites
Typical roleWorking memory, firmware, SSDBulk storage and archive

Structure of a magnetic disk

A hard disk contains one or more rigid platters coated with magnetic material, spinning on a common spindle. Read/write heads on an actuator arm float microscopically above each surface.

Close-up of a hard disk drive's read write head positioned above its mirror-shine magnetic platters
Inside a real hard disk drive (Seagate Medalist ST33232A, 1998): the read/write head sits at the end of the actuator arm, a fraction of a millimetre above the spinning platter. Photo: Sting · Wikimedia Commons
Track one concentric ring Sector a pie-slice of the whole platter Disk sector (block) track × sector — the smallest addressable unit, usually 512 B or 4 KB Cylinder same track on every platter actuator arm and read/write head
Figure 4.2 — Magnetic disk structure. Tracks are the rings, sectors are the pie slices, a disk sector or block is where the two intersect, and a cylinder is the same track across every platter.

Disk performance calculations

Transfer time = data size ÷ transfer rate  ·  Revolutions per second = rpm ÷ 60
Average rotational latency = ½ × (60 ÷ rpm)  ·  Access time = seek time + latency + transfer time

Worked example — past-paper style

An HDD has: capacity 1 TB, rotational speed 7200 rpm,
transfer rate 150 MB/s.

(i)  Time to transfer 600 MB
     t = 600 / 150 = 4 seconds

(ii) Revolutions per second
     n = 7200 / 60 = 120 rev/s

(iii) Average rotational latency
     One revolution takes 1/120 = 0.00833 s = 8.33 ms
     Average latency = half a revolution
                     = 8.33 / 2 = 4.17 ms

The memory hierarchy

Systems combine several memory technologies. Moving down the hierarchy, capacity and cost-effectiveness rise while speed falls.

Registers Cache (SRAM) Main memory (DRAM) SSD / flash Hard disk · optical · tape fastercostliersmaller slowercheaperlarger < 1 ns1–10 ns50–100 ns 50–100 µs5–15 ms
Figure 4.3 — The memory hierarchy with rough access times. No single technology is fast, large and cheap at once, so systems use all of them together.
Choosing memory for an embedded system

A design brief that says "16 KB for temporary sensor data and 8 KB for permanent instructions" is asking you to pick RAM (SRAM) for the temporary data — because it is read-write, very fast, and unlimited in write cycles, and the data does not need to survive a power cut — and ROM, specifically EEPROM or flash, for the instructions — because it is non-volatile so the program survives power loss, it cannot be corrupted by ordinary program writes, and the device can boot and run the instant power is applied.

Quick check — LO4

1. Which memory loses its contents when power is removed?

RAM is volatile. ROM, flash and magnetic storage all retain data without power.

2. Which type of RAM needs constant refreshing?

DRAM stores each bit as charge on a tiny capacitor, which leaks and must be topped up thousands of times a second.

3. Magnetic tape uses which access method?

You must wind through everything before the target. A hard disk is direct access; RAM is random access.

4. An EPROM is erased by

EPROMs have a quartz window for UV erasure. EEPROM and flash are erased electrically.

5. The same track position across every platter of a hard disk is called a

Cylinder. Reading a whole cylinder needs no head movement, so it is the fastest set of blocks to access together.

Practice questions — LO4 36 marks
  1. Differentiate between the following as used in computers: (i) DRAM and SRAM (ii) main memory and cache memory. (8 marks)
    Model answer
    (i) SRAM stores each bit in a flip-flop of four to six transistors and needs no refreshing, making it very fast but low in density and expensive per bit; it is used for cache. DRAM stores each bit as charge on a capacitor with a single transistor, so it must be refreshed thousands of times a second; it is slower but far denser and cheaper, and is used for main memory.
    (ii) Main memory is large-capacity DRAM on modules on the motherboard, holding all running programs and data. Cache is a much smaller block of fast SRAM located inside or beside the CPU, holding recently and frequently used items so the processor does not have to wait for main memory. Cache is far faster and far more expensive per bit.
  2. With the aid of a diagram, outline the structure of a magnetic disk, showing tracks, sectors and disk sectors. (6 marks)
    Model answer
    Draw a circular platter as in Figure 4.2. Label: tracks as the concentric rings on the surface, numbered from the outside inwards; sectors as the pie-shaped wedges that divide the whole platter radially; the disk sector or block as the shaded area where one track crosses one sector, which is the smallest addressable unit, usually 512 bytes or 4 KB; the spindle at the centre; and the read/write head on its actuator arm. Mention that the same track across all platters in a stack forms a cylinder.
  3. State FOUR functions of Read Only Memory. (4 marks)
    Model answer
    It stores the bootstrap loader or BIOS/UEFI that starts the computer and locates the operating system. It holds the power-on self-test routines that check hardware at start-up. It stores permanent firmware for embedded devices such as routers and appliances. It holds fixed data such as character generator tables and calibration constants. It provides basic input/output routines allowing the processor to talk to hardware before the operating system loads.
  4. A digital system uses Module A, an 8 GB DDR4 RAM stick, and Module B, a 256 MB EEPROM. (a) Classify each as volatile or non-volatile. (b) State TWO key differences between them. (4 marks)
    Model answer
    (a) Module A (DDR4 RAM) is volatile; Module B (EEPROM) is non-volatile.
    (b) RAM loses its contents when power is removed, whereas EEPROM retains them indefinitely. RAM is far faster, with access times in tens of nanoseconds, while EEPROM writes are much slower. RAM has effectively unlimited write cycles, whereas EEPROM has a limited write endurance. RAM offers much larger capacity at lower cost per bit.
  5. An HDD has a capacity of 1 TB, an average rotational speed of 7200 rpm and a data transfer rate of 150 MB/s. Calculate (i) the time taken to transfer 600 MB of data and (ii) the number of revolutions per second. (4 marks)
    Model answer
    (i)  t = data / rate = 600 / 150 = 4 seconds
    (ii) n = 7200 / 60 = 120 revolutions per second
  6. An embedded system needs 16 KB of memory for temporary data storage and 8 KB for permanent instructions. (a) Identify a suitable memory type for each. (b) Justify each choice with TWO practical reasons. (6 marks)
    Model answer
    (a) Temporary data: RAM, specifically SRAM. Permanent instructions: ROM, specifically flash or EEPROM.
    (b) RAM: it is read-write, so sensor readings can be overwritten continuously; it is very fast, so it keeps up with the sampling rate; and it has effectively unlimited write endurance, unlike flash. Volatility does not matter because the data is transmitted and then discarded.
    Flash/EEPROM: it is non-volatile, so the program survives power loss and the device boots unattended in the field; it cannot be corrupted by ordinary data writes; and it can still be reprogrammed electrically when firmware needs updating.
  7. Explain FOUR types of secondary memory devices. (8 marks)
    Model answer
    Hard disk drive: rigid magnetic platters spun at high speed with floating read/write heads; high capacity at low cost per gigabyte, direct access, but slow relative to solid state and vulnerable to shock.
    Solid-state drive: NAND flash memory with no moving parts; very fast access, silent, shock-resistant and low power, but more expensive per gigabyte with finite write endurance.
    Optical disc (CD, DVD, Blu-ray): data stored as pits and lands read by a laser; cheap to duplicate and distribute, removable and long-lived, but slow and of limited capacity.
    Magnetic tape: data written on a long flexible coated strip; extremely cheap per terabyte and durable for archives, but sequential access makes retrieval very slow.
    (USB flash drives and memory cards are also acceptable.)
5

Apply number systems and binary coding

~35 min

By the end of this outcome you should be able to

  • Define a number system and a binary code, and state the advantages of binary coding
  • Work confidently in decimal, binary, octal and hexadecimal, and convert between them
  • Perform binary addition, subtraction, multiplication and division
  • Use 1's and 2's complement for subtraction
  • Represent decimal numbers in 8421 BCD, 2421 BCD and Excess-3, and perform BCD arithmetic
  • Name and describe Gray code, ASCII and parity codes

Definition of number system and binary code

A number system is a method of representing quantities using a defined set of symbols (digits) and a base or radix — the number of distinct digits available. The value of a digit depends on its position, which is why these are called positional systems.

A binary code is a scheme for representing numbers, letters or other information as patterns of 0s and 1s so that digital circuits can process them.

Advantages of binary coding

  1. Only two states are needed — a transistor either conducts or it does not, so circuits are simple and cheap
  2. High noise immunity — a signal has to be corrupted a long way before a 0 is mistaken for a 1
  3. Reliable storage — two magnetic directions, two charge states or two light levels are easy to distinguish
  4. Suits Boolean logic directly, so arithmetic, comparison and decision-making use the same gates
  5. Error detection and correction are straightforward using parity and Hamming codes
  6. Easy compression, encryption and transmission, since everything is a uniform stream of bits
SystemBaseDigits usedExampleWhy it matters
Decimal100–929₁₀Everyday human counting
Binary20, 111101₂What the hardware actually stores
Octal80–735₈Shorthand for 3-bit groups; Unix file permissions
Hexadecimal160–9, A–F1D₁₆Shorthand for 4-bit groups; memory addresses, colour codes, MAC addresses
Place value: 11101₂ 111 01 2⁴2³2²2¹2⁰ 168421 16 + 8 + 4 + 0 + 1 = 29₁₀ Add up only the columns holding a 1. That is the whole method.
Figure 5.1 — Positional notation. Memorise the powers of two up to 256 and most conversions become mental arithmetic.
Counting table. Learning the first sixteen rows by heart makes hexadecimal effortless.
DecBinaryOctHex8421 BCDExcess-3Gray
0000000000000110000
1000111000101000001
2001022001001010011
3001133001101100010
4010044010001110110
5010155010110000111
6011066011010010101
7011177011110100100
81000108100010111100
91001119100111001101
10101012A0001 0000—1111
11101113B0001 0001—1110
12110014C0001 0010—1010
13110115D0001 0011—1011
14111016E0001 0100—1001
15111117F0001 0101—1000

Base conversion

Any base to decimal — multiply by place value and add

Convert 1101₂, 257₈ and 2AF₁₆ to decimal

1101₂  = 1×2³ + 1×2² + 0×2¹ + 1×2⁰
       = 8 + 4 + 0 + 1                 = 13₁₀

257₈   = 2×8² + 5×8¹ + 7×8⁰
       = 128 + 40 + 7                  = 175₁₀

2AF₁₆ = 2×16² + 10×16¹ + 15×16⁰
       = 512 + 160 + 15                = 687₁₀

Decimal to any base — repeated division, read remainders upward

Convert 45₁₀ to binary, octal and hexadecimal

45 ÷ 2 = 22 r 1   ↑        45 ÷ 8 = 5 r 5   ↑
22 ÷ 2 = 11 r 0   ↑         5 ÷ 8 = 0 r 5   ↑
11 ÷ 2 =  5 r 1   ↑        45₁₀ = 55₈
 5 ÷ 2 =  2 r 1   ↑
 2 ÷ 2 =  1 r 0   ↑        45 ÷ 16 = 2 r 13 (D)  ↑
 1 ÷ 2 =  0 r 1   ↑         2 ÷ 16 = 0 r 2       ↑
45₁₀ = 101101₂              45₁₀ = 2D₁₆

Decimal fractions — repeated multiplication, read downward

Convert 0.625₁₀ to binary

0.625 × 2 = 1.25   → 1   ↓
0.250 × 2 = 0.50   → 0   ↓
0.500 × 2 = 1.00   → 1   ↓   stop when the fraction reaches 0
0.625₁₀ = 0.101₂

Combined: 45.625₁₀ = 101101.101₂

Grouping shortcuts — binary to octal and hexadecimal

Because 8 = 2³ and 16 = 2⁴, you never need to go through decimal. Group the binary digits from the binary point outward: threes for octal, fours for hexadecimal, padding with leading zeros.

Convert 11010110₂, and convert the hex sequence 3F, 2A, 5C, 7D to binary

11010110₂
Octal :  011 | 010 | 110   →  3  2  6   =  326₈
Hex   :  1101 | 0110       →  D  6      =  D6₁₆
Check :  128+64+16+4+2     =  214₁₀

Hex to binary — one nibble per digit:
3F → 0011 1111
2A → 0010 1010
5C → 0101 1100
7D → 0111 1101

Hexadecimal arithmetic

Add 3F₁₆ and 2A₁₆, giving the answer in hex and in binary

Method 1 — column addition in hex:
      3 F
    + 2 A
    -----
  F + A = 15 + 10 = 25 = 16 + 9  → write 9, carry 1
  3 + 2 + 1(carry) = 6           → write 6
  Result = 69₁₆

Method 2 — via decimal (use this to check):
  3F₁₆ = 3×16 + 15 = 63
  2A₁₆ = 2×16 + 10 = 42
  63 + 42 = 105
  105 ÷ 16 = 6 r 9  →  69₁₆  ✓

In binary: 6 → 0110, 9 → 1001
  69₁₆ = 0110 1001₂
Number base converter

Binary arithmetic

Addition

0+0 = 0 · 0+1 = 1 · 1+0 = 1 · 1+1 = 10 (write 0, carry 1) · 1+1+1 = 11 (write 1, carry 1)

1011₂ + 1101₂

  carries  1111
           1 0 1 1      (11)
        +  1 1 0 1      (13)
        -----------
         1 1 0 0 0      (24)  ✓ 11 + 13 = 24

Subtraction by direct borrowing

0−0 = 0 · 1−0 = 1 · 1−1 = 0 · 0−1 = 1 with a borrow of 1 from the next column

1101₂ − 1010₂

           1 1 0 1      (13)
        −  1 0 1 0      (10)
        -----------
           0 0 1 1      (3)   ✓

Subtraction using 1's complement

Form the 1's complement of the subtrahend by inverting every bit, add it to the minuend, then apply the end-around carry: if a carry comes out of the most significant bit, add it back to the least significant bit and the answer is positive. If there is no carry, the answer is negative and you must take the 1's complement of the result.

11011001₂ − 10100111₂ using 1's complement

10100111  invert →  01011000     (1's complement)

    1101 1001      (217)
  + 0101 1000      (1's comp of 167)
  -------------
  1 0011 0001      carry out of MSB = 1
          → end-around carry: add the 1 back
    0011 0001
  +         1
  -------------
    0011 0010      = 50₁₀

Check: 217 − 167 = 50  ✓

Subtraction using 2's complement

Computers do not really subtract — they add the 2's complement of the second number. Form it by inverting every bit and adding 1. Any final carry out is simply discarded.

13 − 10 using 8-bit 2's complement

10          = 0000 1010
invert      = 1111 0101        (1's complement)
add 1       = 1111 0110        (2's complement of 10)

  0000 1101      (13)
+ 1111 0110      (−10)
-----------
1 0000 0011      discard the carry → 0000 0011 = 3  ✓
Why 2's complement wins

It lets one adder circuit handle both addition and subtraction, needs no end-around carry, and has only one representation of zero. In 8 bits it covers −128 to +127, with the leftmost bit acting as the sign: 0 positive, 1 negative.

Multiplication

0×0 = 0 · 0×1 = 0 · 1×0 = 0 · 1×1 = 1

1011₂ × 101₂

       1 0 1 1        (11)
   ×     1 0 1        (5)
   ---------------
       1 0 1 1        × 1
     0 0 0 0          × 0, shifted one place
   1 0 1 1            × 1, shifted two places
   ---------------
   1 1 0 1 1 1        (55)   ✓ 11 × 5 = 55

Division

1100₂ ÷ 10₂  (12 ÷ 2)

        1 1 0        ← quotient = 6
      _______
 1 0 ) 1 1 0 0
       1 0
       ---
        1 0
        1 0
        ---
         0 0
           0
         ---
           0          remainder = 0   ✓ 12 ÷ 2 = 6
Shortcut worth knowing

Shifting a binary number one place left multiplies it by 2; shifting one place right divides it by 2. Processors use shifts instead of multiply instructions wherever they can, because a shift takes a single clock cycle.

Binary codes

A binary code assigns a fixed bit pattern to each symbol. Codes exist because raw binary is not always convenient — digital clocks, calculators and meters need to drive decimal displays directly.

8421 BCD — Binary Coded Decimal

In 8421 BCD each decimal digit is replaced by its own 4-bit binary equivalent. The name comes from the place values of the four bits: 8, 4, 2, 1. It is a weighted code.

Representing decimal numbers in BCD

   5  6  9                2  5  9
   |  |  |                |  |  |
0101 0110 1001        0010 0101 1001

Compare 259 in the two forms:
  Pure binary : 259 = 1 0000 0011   (9 bits)
  8421 BCD    : 259 = 0010 0101 1001 (12 bits)
  • Only the patterns 0000 to 1001 are valid; 1010 to 1111 are illegal in BCD
  • Conversion between BCD and decimal is trivial, which suits displays and keypads
  • It wastes storage — six of the sixteen possible patterns are never used, roughly 20 % less efficient than pure binary
  • Arithmetic circuits are more complex than for pure binary

2421 BCD

2421 BCD is another weighted code, but the bit weights are 2, 4, 2, 1 rather than 8, 4, 2, 1. It is self-complementing: the 9's complement of any digit is found simply by inverting all four bits.

Decimal2421 codeDecimal2421 code
0000051011
1000161100
2001071101
3001181110
4010091111

Advantages of 2421 BCD: it is self-complementing, so subtraction by 9's complement needs only an inverter rather than a subtractor; and because it is weighted, the decimal value can still be recovered by simple addition of the bit weights, unlike a purely unweighted code.

Excess-3 code

Excess-3 is formed by adding 3 (0011₂) to each BCD digit. It is an unweighted, self-complementing code.

Decimal 5 and 2 in Excess-3, and the self-complementing property

Decimal 5:  BCD 0101 + 0011 = 1000   Excess-3
Decimal 2:  BCD 0010 + 0011 = 0101   Excess-3

Self-complementing check:
Excess-3 of 4 = 0111.  Invert → 1000 = Excess-3 of 5.
And 4 + 5 = 9. ✓
Feature8421 BCDExcess-3
Weighted?Yes — 8, 4, 2, 1No
Code for 000000011
Code for 910011100
Self-complementing?NoYes
All-zero pattern used?Yes, for 0No — useful for detecting a dead line
Main useDisplays, calculators, digital metersOlder arithmetic and data transmission circuits

Gray code

Gray code is an unweighted code in which only one bit changes between consecutive values. This eliminates the momentary wrong readings that occur when several bits change at once in a mechanical encoder, so it is used in shaft encoders, position sensors and Karnaugh maps.

Binary → Gray: keep the MSB, then XOR each bit with the one to its left

Convert 1011₂ to Gray code

Binary :  1   0   1   1
          |  / |  / |  /
MSB kept: 1
1 XOR 0 = 1
0 XOR 1 = 1
1 XOR 1 = 0
Gray   :  1 1 1 0

ASCII and parity

ASCII (American Standard Code for Information Interchange) is a 7-bit code representing 128 characters: letters, digits, punctuation and control codes. 'A' is 65 (100 0001₂), 'a' is 97, and '0' is 48. Unicode is its modern multi-byte successor, covering every writing system.

Parity adds one extra bit so that the total number of 1s is even (even parity) or odd (odd parity). It detects any single-bit error but cannot correct it. Hamming code adds several parity bits and can both detect and correct a single-bit error.

BCD arithmetic

BCD arithmetic follows binary rules but with a correction step, because the four-bit groups must never stray outside 0000–1001.

BCD addition — the add-six rule

Add the groups as plain binary. Then, for any group whose result exceeds 9 or which generates a carry out, add 0110 (6) to that group and propagate any resulting carry to the next group.

Case 1 — no correction needed: 25 + 32

   0010 0101      (25)
 + 0011 0010      (32)
 -------------
   0101 0111      (57)   both groups ≤ 9, already valid BCD

Case 2 — correction needed: 28 + 35

   0010 1000      (28)
 + 0011 0101      (35)
 -------------
   0101 1101      units group = 1101 = 13, illegal
 +      0110      add 6 to that group
 -------------
   0110 0011      (63)   ✓ 28 + 35 = 63

Case 3 — three digits with carries: 575 + 895

575 = 0101 0111 0101
895 = 1000 1001 0101

Units :  0101 + 0101 = 1010  → illegal, add 0110
         1010 + 0110 = 1 0000  → write 0000, carry 1
Tens  :  0111 + 1001 + 1 = 1 0001 → carry out, so add 0110
         0001 + 0110 = 0111  → write 0111, carry 1
Hundreds: 0101 + 1000 + 1 = 1110 → illegal, add 0110
         1110 + 0110 = 1 0100 → write 0100, carry 1
Thousands: the final carry = 0001

Result = 0001 0100 0111 0000  =  1470₁₀
Check: 575 + 895 = 1470  ✓

BCD subtraction

Subtract as binary. Wherever a group required a borrow, subtract 0110 (6) from that group to correct it. BCD hardware normally uses 9's or 10's complement addition instead.

52 − 27

   0101 0010      (52)
 − 0010 0111      (27)
 -------------
   0010 1011      units group borrowed and is illegal (1011)
 −      0110      subtract 6 from that group
 -------------
   0010 0101      (25)   ✓ 52 − 27 = 25

BCD multiplication and division

These are rarely performed directly in BCD hardware. The standard method, and the one to state in an assessment, is:

  1. Convert each BCD operand to its decimal or pure binary value
  2. Perform the multiplication or division
  3. Convert the result back into BCD, digit by digit

BCD examples

0010 0011 (23) × 0011 (3)
  23 × 3 = 69
  6 → 0110,  9 → 1001
  Result = 0110 1001

0100 1000 (48) ÷ 0110 (6)
  48 ÷ 6 = 8
  Result = 1000
BCD, Excess-3 and Gray code encoder

Quick check — LO5

1. What is 1101₂ in decimal?

8 + 4 + 0 + 1 = 13.

2. How many binary digits does one hexadecimal digit replace?

16 = 2⁴, so four bits. One octal digit replaces three bits because 8 = 2³.

3. Which of these is an illegal 8421 BCD group?

1100 is 12, above the maximum of 9. Only 0000 to 1001 are valid.

4. In BCD addition, when is 0110 added to a group?

Adding 6 pushes illegal patterns back into valid range and generates the correct carry.

5. Excess-3 code for decimal 6 is

6 in BCD is 0110; add 0011 to get 1001.

6. The 2's complement of 0000 1010 in eight bits is

Invert to 1111 0101, then add 1 to get 1111 0110, which represents −10.

7. What makes Gray code useful in shaft encoders?

If several bits changed at once, a momentary misalignment could be read as a wildly wrong position.

Practice questions — LO5 42 marks
  1. Explain ONE difference between the decimal and the hexadecimal number system, and outline THREE advantages of binary coding. (5 marks)
    Model answer
    Decimal has a base of 10 and uses the digits 0 to 9, whereas hexadecimal has a base of 16 and uses 0 to 9 followed by the letters A to F to represent values 10 to 15. Each hexadecimal digit therefore represents exactly four binary bits, which decimal cannot do.
    Advantages of binary coding: only two states are needed so circuits are simple and cheap; it has high noise immunity because the two levels are far apart; it maps directly onto Boolean logic for arithmetic and decision making; and error detection with parity or Hamming codes is straightforward.
  2. Convert 217₁₀ into (i) binary (ii) octal (iii) hexadecimal. (6 marks)
    Model answer
    217 ÷ 2 = 108 r 1        217₁₀ = 1101 1001₂
    108 ÷ 2 =  54 r 0
     54 ÷ 2 =  27 r 0        Group in 3s: 011 011 001 = 331₈
     27 ÷ 2 =  13 r 1        Group in 4s: 1101 1001   = D9₁₆
     13 ÷ 2 =   6 r 1
      6 ÷ 2 =   3 r 0
      3 ÷ 2 =   1 r 1
      1 ÷ 2 =   0 r 1   read upward
  3. A digital circuit uses 8421 BCD representation. Write the BCD code for the decimal number 569. (2 marks)
    Model answer
    5 → 0101,  6 → 0110,  9 → 1001
    569₁₀ = 0101 0110 1001 in 8421 BCD
  4. Convert the decimal value 259 into (i) pure binary and (ii) 8421 BCD, and comment on the difference in length. (5 marks)
    Model answer
    (i) 259₁₀ = 1 0000 0011₂   (9 bits)
    (ii) 2 → 0010, 5 → 0101, 9 → 1001
         259₁₀ = 0010 0101 1001  (12 bits)
    BCD needs three more bits for the same value because six of the sixteen possible four-bit patterns are never used. BCD is chosen where the number must drive a decimal display directly, and pure binary where storage efficiency and arithmetic speed matter.
  5. A technician stores instructions in hexadecimal as 3F, 2A, 5C, 7D. (a) Convert each into binary. (b) Add 3F and 2A, giving the result in hexadecimal and in binary. (8 marks)
    Model answer
    (a) 3F → 0011 1111      5C → 0101 1100
        2A → 0010 1010      7D → 0111 1101
    
    (b) 3F = 63,  2A = 42,  63 + 42 = 105
        105 ÷ 16 = 6 r 9  →  69₁₆
        69₁₆ = 0110 1001₂
  6. Perform 575 + 895 using BCD representation, showing every correction step. (6 marks)
    Model answer
    575 = 0101 0111 0101
    895 = 1000 1001 0101
    
    Units   : 0101+0101 = 1010 illegal, +0110 = 1 0000
              write 0000, carry 1
    Tens    : 0111+1001+1 = 1 0001 carry out, +0110
              write 0111, carry 1
    Hundreds: 0101+1000+1 = 1110 illegal, +0110 = 1 0100
              write 0100, carry 1
    Thousands: 0001
    
    Answer = 0001 0100 0111 0000 = 1470  ✓
  7. Perform 11011001₂ − 10100111₂ using 1's complement. (5 marks)
    Model answer
    1's complement of 10100111 = 01011000
    
      1101 1001
    + 0101 1000
    -----------
    1 0011 0001    carry out = 1, so answer is positive
    End-around carry:  0011 0001 + 1 = 0011 0010
    
    Answer = 0011 0010₂ = 50₁₀
    Check: 217 − 167 = 50  ✓
  8. Describe the 2421 BCD code and outline TWO of its advantages. (5 marks)
    Model answer
    2421 BCD is a four-bit weighted code in which the bit weights are 2, 4, 2 and 1 rather than 8, 4, 2 and 1. Decimal 0 to 4 use the patterns 0000 to 0100 and decimal 5 to 9 use 1011 to 1111.
    Advantages: it is self-complementing, so the 9's complement of any digit is obtained simply by inverting all four bits, which means subtraction circuits need only inverters rather than full subtractors. It is also weighted, so the decimal value can be recovered by adding the bit weights, which an unweighted code such as Excess-3 does not allow.
6

Identify emerging trends in electronics

~20 min

By the end of this outcome you should be able to

  • Describe current emerging trends in electronics
  • Explain the working principle of IoT sensors and the promise of quantum computing
  • Explain the challenges these trends bring
  • Describe practical ways of coping with emerging trends
  • Perform simple solar PV power calculations

Description of emerging trends

An emerging trend is a new technology, technique or practice that is gaining ground and beginning to change how work is done in a field. In electronics the direction of travel is consistent: smaller, faster, cheaper, more connected and more intelligent.

Internet of Things

Everyday objects fitted with sensors and network connections — smart meters, farm sensors, vehicles, health monitors.

Artificial intelligence chips

Processors designed for machine learning — NPUs, TPUs and edge accelerators running models on-device.

5G and beyond

Very high bandwidth and very low latency wireless, enabling remote control and real-time data at scale.

Nanoelectronics

Transistors measured in a few nanometres, 3D stacking, graphene and carbon nanotube research.

Green and renewable electronics

Solar PV, efficient GaN and SiC power devices, better batteries, energy harvesting and recyclable design.

Flexible and wearable electronics

Printed and bendable circuits, e-textiles, smart watches and patch-based medical monitors.

Quantum computing

Qubits that hold superpositions of states, promising problem-solving far beyond classical machines.

Robotics and automation

Embedded controllers, drones, industrial robots and mechatronic systems replacing manual processes.

Miniaturisation and SMT

Surface-mount assembly, system-on-chip and micro-electromechanical sensors packed into tiny devices.

An Arduino Uno R3 microcontroller board with its USB port, power jack and pin headers
A microcontroller board (Arduino Uno). Boards like this are the cheap, hobby-accessible way to prototype the sensing, processing and communication stages of an IoT device. via Wikimedia Commons

How IoT smart sensors work

An IoT-enabled smart sensor in a household device follows the same five stages every time, and describing them in order is what earns the marks:

  1. Sensing. A transducer converts a physical quantity — temperature, humidity, soil moisture, current, motion — into an electrical signal, usually a small analogue voltage.
  2. Signal conditioning and conversion. The signal is amplified and filtered, then an analogue-to-digital converter turns it into a digital value the processor can use.
  3. Local processing. An embedded microcontroller reads the value, timestamps it, compares it against thresholds and may act immediately, for example switching a relay.
  4. Communication. A wireless module — Wi-Fi, Bluetooth, Zigbee, LoRa or a cellular modem — transmits the reading to a gateway and on to a cloud server.
  5. Cloud analysis and feedback. The server stores and analyses the data, shows it on a dashboard or phone app, raises alerts, and can send commands back to actuators in the device, closing the control loop.

Quantum computing and electronics

Classical electronics stores information in bits that are strictly 0 or 1. A quantum computer uses qubits, which can exist in a superposition of both states at once and can be entangled with one another, so n qubits explore 2n combinations simultaneously. For certain problems — factoring large numbers, searching unsorted data, simulating molecules and optimising large systems — this offers speed-ups no classical machine can match.

For the electronics trade the consequences are: new demand for cryogenic and microwave control hardware; the need for post-quantum cryptography, because current public-key encryption would be breakable; hybrid systems where a classical processor offloads specific problems to a quantum co-processor; and faster materials and chip design through quantum simulation. Practical machines remain expensive, error-prone and limited to specialised tasks, so classical electronics is not being replaced.

An array of solar photovoltaic panels mounted outdoors
Solar photovoltaic panels, the renewable-electronics application behind the worked example below. via Wikimedia Commons

Solar PV and energy calculations

Output power P = V × I  ·  Energy E = P × t  ·  Operating hours t = Edemand ÷ Ppanel

Worked example — past-paper style

A solar panel delivers 24 V at 3 A. The system consumes
288 Wh per day.

(i)  Output power
     P = V × I = 24 × 3 = 72 W

(ii) Hours of operation required
     t = energy demand / panel power
       = 288 / 72
       = 4 hours of full sunshine per day

In practice you would oversize the panel, because
irradiance varies, panels lose efficiency when hot,
and the battery and charge controller are not 100 %
efficient. A common rule is to add 25–30 % margin,
which here means about a 90–95 W panel.

Challenges of emerging trends

ChallengeWhat it looks like in practice
Rapid obsolescenceEquipment and skills date quickly; a device bought today may be unsupported within three years
High cost of entryNew tools, test equipment and licensed software are expensive, especially for small workshops
Skills gapTraining lags the technology; technicians trained on through-hole work meet 0402 surface-mount parts and BGA packages
Difficult repairabilitySealed, glued and BGA-packaged devices cannot be repaired at component level without specialised rework equipment
Electronic wasteShort product lives create toxic waste containing lead, mercury, cadmium and brominated flame retardants
Security and privacyBillions of connected devices widen the attack surface; many ship with weak default credentials and are never patched
Health and safetyNew materials, high voltages in EV systems, lithium battery fires and possible nanomaterial exposure
Unreliable infrastructurePower interruptions and patchy internet limit what connected systems can achieve in some regions
Job displacementAutomation removes routine roles while creating roles that demand different, higher skills
Standards and interoperabilityCompeting protocols mean devices from different makers often refuse to work together
Counterfeit componentsFake or recycled parts entering the supply chain cause early and sometimes dangerous failures
Manufacturing complexityNanoscale fabrication needs cleanrooms and equipment costing billions, concentrating production in a few countries

Nanotechnology: two advantages and two challenges

AdvantagesChallenges
Far higher component density, so more processing power fits in a smaller, lighter deviceExtremely high fabrication cost, requiring cleanrooms and equipment only a few companies can afford
Lower power consumption and faster switching, because charge travels shorter distancesQuantum effects such as electron tunnelling and heat dissipation become severe as features shrink
New materials such as graphene and carbon nanotubes enable flexible and transparent electronicsUnknown long-term health and environmental effects of nanomaterials, and difficult recycling
Better sensors and medical devices, small enough to be implanted or wornDevices cannot be inspected or repaired by conventional means, so faults mean full replacement

Coping with emerging trends

Coping is not about resisting change; it is about staying employable and safe while it happens.

Keep learning continuously

Short courses, vendor certifications, online tutorials and manufacturer datasheets. Treat learning as part of the job, not an interruption to it.

Upgrade tools and methods

Invest gradually in hot-air rework, a decent oscilloscope, ESD protection and up-to-date test gear rather than all at once.

Join professional networks

Trade associations, maker groups, technician forums and supplier communities spread knowledge fast and cheaply.

Follow standards and safety

Observe ESD handling, RoHS restrictions, correct e-waste disposal and recognised wiring regulations.

Specialise deliberately

Pick a growing niche — embedded systems, solar installation, industrial automation, EV servicing — and go deep in it.

Practise and prototype

Build with Arduino, Raspberry Pi and ESP32 boards. Hands-on projects turn reading into competence and build a portfolio.

Repair, reuse, recycle

Extend equipment life, harvest usable parts, and route dead boards to licensed e-waste handlers rather than landfill.

Build security in

Change default passwords, keep firmware patched, segment IoT devices onto their own network and encrypt data in transit.

A practical stance

The fundamentals in this unit — Ohm's law, the behaviour of a P-N junction, binary arithmetic — have not changed in decades and will not. Every new technology is built on them. Learn them properly and emerging trends become things you can pick up, rather than things that leave you behind.

Quick check — LO6

1. Which of these is a challenge rather than a benefit of emerging electronics?

E-waste is one of the biggest problems created by short product lifecycles, alongside security risk and the skills gap.

2. The Internet of Things refers to

Smart meters, connected farm sensors and wearable health monitors are all IoT devices.

3. The best single response to rapid obsolescence in the trade is to

Skills date faster than the fundamentals. Continuous learning, professional networks and deliberate specialisation are the practical defences.

4. A qubit differs from a classical bit because it

Superposition and entanglement let n qubits explore 2ⁿ combinations at once, which is where the speed-up comes from.

Practice questions — LO6 30 marks
  1. Define the term nanoelectronics and state one way it has affected the design and manufacture of electronic devices. (4 marks)
    Model answer
    Nanoelectronics is the branch of electronics concerned with devices and components whose critical dimensions are measured in nanometres, typically below 100 nm, where quantum effects begin to influence behaviour.
    Effect: it has allowed billions of transistors to be packed onto a single chip, so devices such as smartphones are far smaller, faster and more power-efficient than earlier equipment. It has also pushed manufacturing into ultra-clean, extremely costly fabrication plants, concentrating production among a handful of global manufacturers.
  2. Explain the working principle of IoT-enabled smart sensors in monitoring household devices. (4 marks)
    Model answer
    A transducer converts a physical quantity such as temperature, humidity or current into an electrical signal. The signal is conditioned and converted to digital form by an ADC. An embedded microcontroller processes the reading, timestamps it and compares it with set thresholds, acting locally where necessary. A wireless module then transmits the data over Wi-Fi, Zigbee, LoRa or cellular to a gateway and on to a cloud server, where it is stored, analysed and displayed on a phone app. The server can also send commands back to actuators in the device, closing the control loop and allowing automatic or remote control.
  3. Discuss TWO advantages and TWO challenges of integrating nanotechnology in modern electronics. (8 marks)
    Model answer
    Advantages: far higher component density, so much greater processing power fits into a smaller and lighter device; and lower power consumption with faster switching, because charge carriers travel shorter distances, which extends battery life and raises clock speeds.
    Challenges: extremely high fabrication cost, since nanoscale manufacture demands cleanrooms and lithography equipment costing billions, restricting production to a few firms; and severe physical problems as features shrink, including electron tunnelling through ultra-thin insulating layers, heat dissipation in a tiny volume, and uncertain long-term health and environmental effects of nanomaterials.
  4. A company powers a smart-home system using solar PV. A panel delivers 24 V at 3 A. (i) Calculate the output power. (ii) If the system consumes 288 Wh daily, determine how many hours the panel must operate to meet the demand. (6 marks)
    Model answer
    (i)  P = V × I = 24 × 3 = 72 W
    (ii) t = E / P = 288 / 72 = 4 hours
    In practice the panel should be oversized by about 25 to 30 per cent to allow for variable irradiance, temperature losses and charge-controller and battery inefficiency.
  5. Explain briefly how quantum computing may affect the future of electronics and data processing. (4 marks)
    Model answer
    Quantum computers use qubits, which can exist in superposition and be entangled, so n qubits explore 2n states simultaneously. This offers enormous speed-ups for factoring, searching, optimisation and molecular simulation. For electronics this means new demand for cryogenic and microwave control hardware; an urgent move to post-quantum cryptography, because current public-key encryption would become breakable; hybrid architectures where classical processors offload specific problems to quantum co-processors; and faster materials and chip design through quantum simulation. Machines remain costly and error-prone, so classical electronics will continue alongside them.
  6. State ONE real-world application of IoT in everyday life and TWO challenges associated with IoT implementation. (3 marks)
    Model answer
    Application: a smart electricity meter that reports consumption to the utility automatically, or a smart-farm soil moisture sensor that triggers irrigation.
    Challenges: weak security and privacy, since many devices ship with default credentials and are never patched, creating a large attack surface; and poor interoperability, because competing standards mean devices from different manufacturers often cannot work together. Dependence on reliable power and internet connectivity is also acceptable.
  7. Discuss any SIX challenges of emerging trends in electronics manufacturing. (12 marks)
    Model answer
    Rapid obsolescence: products and production lines are superseded quickly, so capital equipment must be replaced before it has paid for itself.
    High capital cost: nanoscale fabrication requires cleanrooms and lithography tools costing billions, which few countries or firms can fund.
    Skills shortage: the workforce must be retrained continuously for surface-mount, BGA and automated assembly techniques.
    Electronic waste and environmental compliance: short product lives generate toxic waste, and manufacturers must meet RoHS and WEEE-type regulations.
    Supply-chain fragility and counterfeiting: components come from a few global sources, and counterfeit or recycled parts entering the chain cause early failures.
    Quality control at nanoscale: defects invisible to conventional inspection reduce yield, and the finished devices cannot be repaired, only scrapped.
    Security requirements: connected products must be designed with secure firmware and update paths, adding cost and complexity.
★

Mock CDACC written assessment

3 hours
Instructions to candidate

1. This paper consists of TWO sections: A and B.
2. Answer ALL questions in Section A and ANY THREE questions in Section B.
3. Marks for each question are indicated in the brackets.
4. Time allowed: THREE hours. Use a separate answer booklet.
5. Attempt the whole paper before opening any model answers.

Section A — answer ALL questions 40 marks
  1. Name one application of an insulator and one application of a semiconductor material in real life. (2 marks)
    Model answer
    Insulator: PVC sheathing on domestic wiring cable, or the rubber grip on a screwdriver handle. Semiconductor: silicon used to make the transistors inside a microprocessor, or a solar photovoltaic cell.
  2. A diode is connected to a battery in two different ways. (a) Sketch and label the connection when forward biased. (b) Sketch and label the connection when reverse biased. (4 marks)
    Model answer
    (a) Battery positive terminal to the anode (triangle side), negative to the cathode (bar side), with a series resistor. Label the narrow depletion region and show current flowing. (b) Battery positive to the cathode and negative to the anode. Label the widened depletion region and note that only a tiny leakage current flows. See Figure 3.4.
  3. State Ohm's law. A resistor of 120 Ω carries a current of 250 mA. Calculate the voltage across it and the power dissipated. (5 marks)
    Model answer
    Ohm's law: current is directly proportional to p.d.,
    provided temperature and other conditions are constant.
    
    V = IR = 0.25 × 120 = 30 V
    P = I²R = 0.25² × 120 = 0.0625 × 120 = 7.5 W
  4. A resistor carries the bands Orange, White, Red, Gold. Determine its value and state its tolerance. (3 marks)
    Model answer
    Orange = 3, White = 9, Red = ×100, so the value is 39 × 100 = 3 900 Ω = 3.9 kΩ. Gold gives a tolerance of ±5 %, so the true value lies between 3.705 kΩ and 4.095 kΩ.
  5. Distinguish between N-type and P-type semiconductor material. (4 marks)
    Model answer
    N-type is formed by doping with a pentavalent impurity such as phosphorus, producing free electrons as majority carriers and holes as minority carriers. P-type is formed by doping with a trivalent impurity such as boron, producing holes as majority carriers and electrons as minority carriers. Both remain electrically neutral overall.
  6. A sinusoidal voltage has a maximum value of 200 V. Calculate its r.m.s. and average values. (4 marks)
    Model answer
    Vᵣᵐᵣ = 0.707 × 200 = 141.4 V
    Vᴡᵛ  = 0.637 × 200 = 127.4 V
  7. Outline FOUR characteristics of integrated circuits. (4 marks)
    Model answer
    Very small size and light weight; low power consumption and high operating speed; high reliability due to few interconnections; low cost per function through mass production; not repairable once faulty; sensitive to electrostatic discharge and over-voltage.
  8. Differentiate between volatile and non-volatile memory, giving one example of each. (4 marks)
    Model answer
    Volatile memory loses its stored contents as soon as power is removed and is used for temporary working storage; an example is RAM, whether SRAM or DRAM. Non-volatile memory retains its contents indefinitely without power and is used for firmware and permanent storage; examples are ROM, EEPROM, flash and hard disks.
  9. Write the 8421 BCD representation of the decimal number 407, and state why 1010 is not a valid BCD group. (4 marks)
    Model answer
    4 → 0100,  0 → 0000,  7 → 0111
    407 = 0100 0000 0111
    1010 represents decimal 10, which is not a single decimal digit. Only 0000 to 1001 are valid in 8421 BCD; the remaining six patterns are illegal and trigger the add-six correction in BCD arithmetic.
  10. Describe THREE safety precautions to observe when handling integrated circuits. (6 marks)
    Model answer
    Use an earthed wrist strap and anti-static mat and store chips in conductive packaging, because electrostatic discharge destroys CMOS inputs invisibly. Power down and discharge the circuit before inserting or removing a chip, since hot-plugging causes latch-up. Identify pin 1 from the notch or dot and never exceed the rated supply voltage, as reversing or over-volting a chip destroys it instantly.
Section B — answer ANY THREE questions 60 marks
  1. A control panel is being wired in a manufacturing plant. Three resistors of 12 Ω, 24 Ω and 8 Ω are connected in parallel across a 240 V supply. A 40 µF capacitor is then added in series with the parallel network, on a 50 Hz supply.
    1. Draw the circuit diagram of the arrangement. (3 marks)
    2. Calculate the total resistance of the parallel network. (3 marks)
    3. Calculate the total circuit current before the capacitor is added. (2 marks)
    4. Calculate the capacitive reactance of the 40 µF capacitor at 50 Hz. (3 marks)
    5. Determine the new circuit impedance and the new total current. (6 marks)
    6. Explain ONE industrial application of combining resistors and capacitors in control circuits. (3 marks)
    Model answer
    (b) 1/R = 1/12 + 1/24 + 1/8
           = 2/24 + 1/24 + 3/24 = 6/24
        R  = 4 Ω
    
    (c) I  = V/R = 240/4 = 60 A
    
    (d) X₊ = 1/(2πfC)
           = 1/(2 × 3.1416 × 50 × 40 × 10⁻⁶)
           = 1/0.012566 = 79.58 Ω
    
    (e) Z  = √(R² + X₊²) = √(16 + 6333)
           = √6349 = 79.68 Ω
        I  = V/Z = 240/79.68 = 3.01 A
    
    (f) An RC snubber network across contactor or relay
        contacts. When the contacts open, the capacitor
        absorbs the inductive energy and the resistor
        damps the resulting oscillation, suppressing
        arcing and voltage transients. This extends
        contact life and prevents interference with
        nearby electronic controls.
    For (a), draw the three resistors side by side between two common nodes, with the capacitor in series between one node and the supply.
  2. A research lab is testing transistors for switching. A transistor has IB = 150 µA, β = 100, VCC = 10 V and RC = 470 Ω.
    1. Draw the circuit diagram of the transistor in common-emitter configuration. (3 marks)
    2. Calculate IC, IE, the voltage across RC and VCE. (8 marks)
    3. State the region in which the transistor is operating and justify your answer. (3 marks)
    4. Convert the hexadecimal values 4B and 3D into binary, then add them and express the result in hexadecimal. (6 marks)
    Model answer
    (b) Iᶜ = βIᵝ = 100 × 150 × 10⁻⁶ = 15 mA
        IᲩ = Iᵝ + Iᶜ = 0.15 + 15 = 15.15 mA
        Vᵣᶜ = IᶜRᶜ = 0.015 × 470 = 7.05 V
        VᶜᲩ = Vᶜᶜ − IᶜRᶜ = 10 − 7.05 = 2.95 V
    
    (c) Active region. The emitter-base junction is
        forward biased and the collector-base junction
        reverse biased, and VᶜᲩ of 2.95 V sits between
        saturation (about 0.2 V) and cut-off (10 V),
        so the device can amplify without clipping.
    
    (d) 4B → 0100 1011      3D → 0011 1101
        4B = 75,  3D = 61,  75 + 61 = 136
        136 ÷ 16 = 8 r 8  →  88₁₆
        88₁₆ = 1000 1000₂
    For (a), draw the transistor with emitter grounded, RC from VCC to the collector, and a base resistor from the input to the base, with input and output taken with respect to the emitter.
  3. An agricultural sensor network must measure soil moisture, store the data and transmit it wirelessly.
    1. The system requires 32 KB of memory for temporary data and 16 KB for permanent instructions. Identify a suitable memory type for each and justify each choice with TWO reasons. (6 marks)
    2. Convert the decimal sensor reading 342 into pure binary and into 8421 BCD. (4 marks)
    3. The wireless transmitter is switched by a PNP transistor. With the aid of a neat diagram, explain how the transistor operates in saturation to allow current to flow. (6 marks)
    4. State TWO challenges the designers will face in deploying this network in a rural area, and one way of addressing each. (4 marks)
    Model answer
    (b) 342 ÷ 2 = 171 r 0
        171 ÷ 2 =  85 r 1
         85 ÷ 2 =  42 r 1
         42 ÷ 2 =  21 r 0
         21 ÷ 2 =  10 r 1
         10 ÷ 2 =   5 r 0
          5 ÷ 2 =   2 r 1
          2 ÷ 2 =   1 r 0
          1 ÷ 2 =   0 r 1
        342₁₀ = 1 0101 0110₂
    
        BCD: 3 → 0011, 4 → 0100, 2 → 0010
        342₁₀ = 0011 0100 0010
    (a) Temporary data: SRAM, because it is read-write with unlimited write cycles for continuous sensor sampling, and fast enough to keep up with the sampling rate; volatility does not matter since readings are transmitted then discarded. Permanent instructions: flash or EEPROM, because it is non-volatile so the firmware survives power loss and the node reboots unattended in the field, and it can still be reprogrammed electrically for firmware updates.
    (c) Draw a PNP transistor with the emitter connected to the positive supply, the load (transmitter module) between the collector and ground, and a base resistor from the control line to the base. To saturate the device the base is pulled sufficiently negative with respect to the emitter, forward biasing the emitter-base junction. Holes are injected heavily from the emitter into the thin base; base current is made large enough that the collector-base junction also becomes forward biased. The transistor is then in saturation: collector current is limited only by the load, VEC falls to about 0.2 V, and almost the whole supply voltage appears across the transmitter, switching it fully on.
    (d) Unreliable mains power — address with solar panels, a charge controller and battery backup sized for several days of autonomy. Patchy network coverage — address with a long-range low-power protocol such as LoRaWAN, plus local buffering in memory so readings are stored and forwarded when the link returns. Harsh environment or theft are also acceptable, addressed by sealed IP-rated enclosures and secure mounting.
  4. Emerging trends and energy.
    1. Explain the working principle of an IoT-enabled smart sensor. (4 marks)
    2. Discuss TWO advantages and TWO challenges of nanotechnology in modern electronics. (8 marks)
    3. A solar panel delivers 18 V at 4.5 A. Calculate the output power, and the hours of operation needed to supply a daily demand of 405 Wh. (5 marks)
    4. State THREE ways a technician can cope with rapid change in the electronics industry. (3 marks)
    Model answer
    (c) P = V × I = 18 × 4.5 = 81 W
        t = E / P = 405 / 81 = 5 hours
    (a) and (b) are covered in full in the LO6 notes above. (d) Commit to continuous learning through short courses and certifications; upgrade tools and test equipment gradually, including ESD protection and rework gear; join professional networks and supplier communities to share knowledge; specialise in a growing niche such as embedded systems, solar installation or industrial automation; and build practical projects to keep skills current.
☰

Formula sheet, checklist & glossary

revision

Every formula in this unit

Circuits and power

V = I R  ·  P = V I = I²R = V²/R  ·  W = P t
E = W/Q  ·  I = Q/t  ·  V = E − I r
R = ρl/A  ·  1 mm² = 10−6 m², 1 µΩm = 10−6 Ωm
Series R = R₁+R₂+R₃  ·  Parallel 1/R = 1/R₁+1/R₂+1/R₃
KCL ΣIin = ΣIout  ·  KVL ΣE = ΣIR

Alternating current

Vrms = 0.707 Vpk  ·  Vav = 0.637 Vpk
form factor 1.11  ·  peak factor 1.414  ·  f = 1/T

Components

Q = C V  ·  W = ½CV²  ·  τ = RC
Capacitors series 1/C = 1/C₁+1/C₂  ·  parallel C = C₁+C₂
Inductors series L = L₁+L₂  ·  W = ½LI²  ·  e = −L(dI/dt)
XC = 1/(2πfC)  ·  XL = 2πfL
Z = √(R² + X²)  ·  I = V/Z  ·  fr = 1/(2π√(LC))

Semiconductors

Barrier: Si 0.7 V  ·  Ge 0.3 V
IE = IB + IC  ·  β = IC/IB  ·  α = IC/IE  ·  β = α/(1−α)
VCE = VCC − ICRC  ·  saturation IC = VCC/RC

Memory and disks

transfer time = size ÷ rate  ·  rev/s = rpm ÷ 60
average latency = ½ × (60 ÷ rpm)  ·  access = seek + latency + transfer

Number systems

BCD add-six rule: correct any group > 9 or producing a carry by adding 0110
Excess-3 = BCD + 0011  ·  2's complement = invert + 1
Binary → Gray: keep MSB, then XOR with the bit to the left

Energy and solar

P = V I  ·  E = P t  ·  t = E ÷ P  ·  1 kWh = 3.6 MJ

Competence checklist

Tick each statement you could do unaided in an assessment. Your ticks are saved in this browser.

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Glossary

Ampere
Unit of current; one coulomb per second.
Avalanche breakdown
Reverse breakdown by impact ionisation multiplying carriers, typically above 6 V.
BCD
Binary Coded Decimal; each decimal digit replaced by its own four-bit code.
Biasing
Applying d.c. voltage to set a device's operating condition.
Cache
Small fast SRAM near the CPU holding recently used data.
Capacitance
Ability to store charge, measured in farads.
Cylinder
The same track position across every platter of a disk stack.
DAM
Direct access memory; jump near the target then scan, as on a hard disk.
Depletion region
Carrier-free zone either side of a P-N junction.
Doping
Adding a controlled impurity to change a semiconductor's conductivity.
DRAM
Dynamic RAM; one transistor and one capacitor per bit, needs refreshing.
E.m.f.
Energy supplied per coulomb by a source, in volts.
Excess-3
Unweighted self-complementing code formed by adding 0011 to BCD.
Extrinsic
A doped semiconductor, either N-type or P-type.
Form factor
Ratio of r.m.s. to average value; 1.11 for a sine wave.
Gray code
Code in which only one bit changes between consecutive values.
Hole
Vacancy left by a missing electron, acting as a positive carrier.
Impedance
Total a.c. opposition combining resistance and reactance, in ohms.
Integrated circuit
A complete circuit fabricated on a single chip of silicon.
Intrinsic
A pure, undoped semiconductor.
Kirchhoff's laws
Current into a junction equals current out; loop e.m.f.s equal loop p.d.s.
Lenz's law
An induced e.m.f. always opposes the change producing it.
Majority carrier
Electrons in N-type, holes in P-type.
Nanoelectronics
Electronics with critical dimensions below about 100 nanometres.
Parity
An extra bit making the count of 1s even or odd, for error detection.
PIV
Peak inverse voltage; the maximum reverse voltage a diode can withstand.
Reactance
Opposition to a.c. from capacitance or inductance, in ohms.
Resistivity
A material property, in ohm metres, relating R to length and area.
R.m.s.
The equivalent d.c. value of an a.c. waveform; 0.707 of the peak.
ROM
Read-only, non-volatile memory holding firmware and boot code.
Saturation
Transistor fully on; both junctions forward biased, VCE about 0.2 V.
Sector
A radial wedge of a disk platter; with a track it defines a block.
Semiconductor
Material with conductivity between a conductor and an insulator.
SRAM
Static RAM; flip-flop cells, fast, no refresh, used for cache.
Time constant
τ = RC; 63 % of full charge in one, effectively full in five.
Tolerance
Permitted deviation of a component from its marked value.
Varistor
Voltage-dependent resistor whose resistance collapses above a threshold.
Volt
Unit of potential difference; one joule per coulomb.
Zener breakdown
Reverse breakdown by direct field emission in heavily doped junctions, below about 5 V.
Final revision routine

Work the practice questions at the end of each outcome without notes, then sit the mock paper under timed conditions: three hours, all of Section A, any three from Section B. Mark yourself against the model answers and go back to any topic where you lost more than a third of the marks.

These notes follow the TVET CDACC course outline for Apply Basic Electronic Skills, unit code ICT/CU/CS/CC/01/6/B, and are written for trainee use. Worked examples and practice questions are modelled on the format of recent written assessments. They are a study aid and not an official CDACC publication — always check the current curriculum and your trainer's guidance.

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