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Unit notes and practice questions

Grade 12 Physics unit 5: Basics of electronics

Everything here is for unit 5 of the Grade 12 Physics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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5
Cards
10
Practice questions on unit 5

What unit 5 covers

  1. 1.Semiconductors
  2. 2.Diodes and their Functions
  3. 3.Rectification
  4. 4.Transistors and their application
  5. 5.Integrated Circuits
  6. 6.Logic gates and logic circuits
  7. 7.Application of electronics

Short notes for unit 5

The cards filed under this chapter on the Grade 12 Physics page. Open one at a time.

TypeDopant groupDopant examplesMajority carrierMinority carrier
N-typeGroup V (pentavalent)As, P, Sb, BiElectronsHoles
P-typeGroup III (trivalent)B, Al, Ga, InHolesElectrons

When you use it

Use to distinguish between N-type and P-type semiconductors, their impurities, and their charge carriers.

Watch out

Doped semiconductors remain electrically neutral overall even though one carrier type outnumbers the other.

Drafted from Grade 12 Physics, page 143 onwards, then checked twice before it went up

GateLogic operationOutput is 1 whenUniversal gate
ORAddition: y=A+By = A + BAt least one input is 1No
ANDMultiplication: y=A⋅By = A \cdot BAll inputs are 1No
NOTInversion: single inputInput is 0No
NORInverted ORAll inputs are 0Yes
NANDInverted ANDAt least one input is 0Yes

When you use it

Use to evaluate Boolean logic operations and outputs for digital logic circuits.

Watch out

In Boolean logic addition, 1 + 1 equals 1, representing TRUE OR TRUE, not arithmetic 2.

Drafted from Grade 12 Physics, page 143 onwards, then checked twice before it went up

When you use it

Use when determining the bias voltage polarities required across the junctions of a bipolar transistor for proper amplifier operation.

Watch out

The emitter-base junction must be forward biased and the collector-base junction must be reverse biased. Setting both junctions to forward bias is incorrect.

Drafted from Grade 12 Physics, page 143 onwards, then checked twice before it went up

β=ICIB,IE=IB+IC\beta = \frac{I_C}{I_B},\quad I_E = I_B + I_C
β\beta
Current gaindimensionlessdimensionless
ICI_C
Collector currentAA
IBI_B
Base currentAA
IEI_E
Emitter currentAA

When you use it

Use when calculating terminal currents or common-emitter current amplification in a bipolar junction transistor.

Watch out

Base current is given in microamperes while collector and emitter currents are in milliamperes. Convert all currents to amperes before computing.

Drafted from Grade 12 Physics, page 143 onwards, then checked twice before it went up

RegionDoping levelPhysical sizeFunction
EmitterHeavily dopedMedium sizeSupplies charge carriers
BaseLightly dopedVery thinPasses carriers to collector
CollectorModerately dopedLargest sizeCollects charge carriers

When you use it

Use to compare the doping levels, physical dimensions, and functions of the three regions of a bipolar junction transistor.

Watch out

The base is both the thinnest and the most lightly doped layer, while the collector is the largest in physical size.

Drafted from Grade 12 Physics, page 143 onwards, then checked twice before it went up

Practice questions on unit 5

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

How does a pure semiconductor behave at absolute zero temperature (0 K)?

  1. A.As a superconductor
  2. B.As an insulator
  3. C.As a good conductor
  4. D.As a magnetic material
Show the answer and why

Answer: B As an insulator

Why

At absolute zero (0 K), a semiconductor acts as an insulator because valence electrons remain tightly held in their covalent bonds and cannot move freely.

Question 2

Which of the following Group III elements is added to pure silicon to create a P-type semiconductor?

  1. A.Arsenic
  2. B.Phosphorus
  3. C.Boron
  4. D.Antimony
Show the answer and why

Answer: C Boron

Why

Boron is a Group III element with three valence electrons. Adding it creates holes, which makes a P-type semiconductor.

Question 3

What is the approximate barrier potential across a silicon P-N junction diode?

  1. A.0.3 V
  2. B.0.7 V
  3. C.1.5 V
  4. D.5.0 V
Show the answer and why

Answer: B 0.7 V

Why

The barrier potential of a P-N junction depends on the material; it is approximately 0.7 V for silicon and 0.3 V for germanium.

Question 4

In a bipolar junction transistor, which section is lightly doped and very thin?

  1. A.Emitter
  2. B.Collector
  3. C.Base
  4. D.Anode
Show the answer and why

Answer: C Base

Why

The base is the middle layer that is lightly doped and very thin compared to the emitter and collector.

Question 5

A student in Hawassa tests a transistor circuit. The base current is IB = 0.05 mA and the collector current is IC = 4.95 mA. What is the emitter current IE?

  1. A.4.90 mA
  2. B.5.00 mA
  3. C.0.05 mA
  4. D.99.0 mA
Show the answer and why

Answer: B 5.00 mA

Why

The emitter current is the sum of the base current and collector current: IE = IB + IC = 0.05 mA + 4.95 mA = 5.00 mA.

Question 6

What happens to the depletion region when a P-N junction diode is forward biased?

  1. A.It becomes very thin and narrow, allowing high current to flow.
  2. B.It becomes wider, blocking all current flow.
  3. C.It remains unchanged while the barrier voltage doubles.
  4. D.It fills with positive ions on the P-side only.
Show the answer and why

Answer: A It becomes very thin and narrow, allowing high current to flow.

Why

In forward bias, charges are repelled toward the junction, making the depletion region very thin and creating a low-resistance path for electric current.

Question 7

In a basic full-wave bridge rectifier circuit containing four diodes (D1 to D4), how many diodes conduct current during each half-cycle of the AC supply?

  1. A.Only one diode
  2. B.Only two diodes
  3. C.Three diodes
  4. D.All four diodes
Show the answer and why

Answer: B Only two diodes

Why

During each half-cycle of an AC signal, two diodes conduct in series while the other two diodes are reverse biased and switch off.

Question 8

An integrated circuit (IC), invented by Jack Kilby and Robert Noyce, combines which four components on a small semiconductor chip?

  1. A.Diodes, transistors, resistors, and capacitors
  2. B.Diodes, transformers, inductors, and batteries
  3. C.Transistors, electric motors, switches, and cells
  4. D.Resistors, capacitors, generators, and coils
Show the answer and why

Answer: A Diodes, transistors, resistors, and capacitors

Why

An integrated circuit consists of diodes, transistors, resistors, and capacitors formed together on a single small piece of semiconductor material.

Question 9

A technician in Addis Ababa measures a common-emitter transistor amplifier with a current gain of β = 100 and a collector current of IC = 10 mA. What is the emitter current IE?

  1. A.0.1 mA
  2. B.9.9 mA
  3. C.10.1 mA
  4. D.1000 mA
Show the answer and why

Answer: C 10.1 mA

Why

First calculate the base current: IB = IC / β = 10 mA / 100 = 0.1 mA. Then calculate the emitter current: IE = IB + IC = 0.1 mA + 10 mA = 10.1 mA.

Question 10

Which transistor configuration provides both current gain and voltage gain, making it the most commonly used circuit arrangement?

  1. A.Common-base (CB)
  2. B.Common-collector (CC)
  3. C.Common-emitter (CE)
  4. D.Common-gate (CG)
Show the answer and why

Answer: C Common-emitter (CE)

Why

The common-emitter (CE) configuration provides both current and voltage gain, giving high overall power gain.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 5

The whole Grade 12 Physics textbook, free as a PDF.

Ministry of Education edition, English, 186 pages

Download the PDF

All Grade 12 Physics short notes

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