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Semiconductors and diodes: board exam questions with solutions

On the ECE exam under Electronics Engineering and the REE exam under Electrical Engineering Professional Subjects. Try each question before you open its solution, then practise with fresh numbers below: 4 computed question types on this topic.

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Build it in the lab

Diode forward and reverse bias · Zener shunt regulator · Bridge rectifier and filter · Half-wave rectifier · LED and its series resistor: the same circuits on a live bench, with meters you can move.

Questions

1. Doping intrinsic silicon with phosphorus (a pentavalent element) produces what kind of material?

  1. p-type, with phosphorus acting as an acceptor
  2. n-type, with phosphorus acting as a donor
  3. n-type, with phosphorus acting as an acceptor
  4. Intrinsic silicon with higher resistivity
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Answer: B. n-type, with phosphorus acting as a donor

Phosphorus has five valence electrons; four bond with silicon and the fifth is donated as a free electron, giving n-type material.

2. What is the thermal voltage kT/q of a pn junction at 300 K?

  1. 25.85 mV
  2. 0.70 V
  3. 2.59 mV
  4. 258.5 mV
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Answer: A. 25.85 mV

VT = kT/q = (1.381 × 10⁻²³ × 300)/(1.602 × 10⁻¹⁹) = 25.85 mV.

3. A Zener regulator has a 24 V input, a 220 Ω series resistor, a 12 V Zener diode and a 1 kΩ load. What is the Zener current?

  1. 54.5 mA
  2. 109.1 mA
  3. 12.0 mA
  4. 42.5 mA
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Answer: D. 42.5 mA

Is = (24 − 12)/220 = 54.5 mA, IL = 12/1000 = 12.0 mA, Iz = Is − IL = 42.5 mA.

4. A bridge rectifier with silicon diodes (0.7 V each) is fed from a 12 V rms transformer secondary. What is the peak output voltage?

  1. 16.27 V
  2. 16.97 V
  3. 15.57 V
  4. 10.60 V
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Answer: C. 15.57 V

Vm = 12√2 = 16.97 V; two diodes conduct in series, so Vout(peak) = 16.97 − 1.4 = 15.57 V.

5. A full-wave rectifier on 60 Hz mains supplies 0.5 A to a load through a 2200 µF filter capacitor. What is the approximate peak-to-peak ripple?

  1. 3.79 V
  2. 1.89 V
  3. 0.95 V
  4. 0.55 V
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Answer: B. 1.89 V

Vr(p-p) ≈ I/(fr·C) with ripple frequency fr = 2 × 60 = 120 Hz: 0.5/(120 × 2200 × 10⁻⁶) = 1.89 V.

6. In an ideal full-wave bridge rectifier with peak secondary voltage Vm, what peak inverse voltage must each diode withstand?

  1. Vm
  2. 2Vm
  3. Vm/2
  4. Vm/√2
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Answer: A. Vm

In a bridge, a non-conducting diode is placed across the secondary through a conducting diode, so it sees about Vm (versus 2Vm in a centre-tapped circuit).

7. An ideal diode (n = 1) at 300 K is forward biased well above its knee. If the forward voltage is increased by 60 mV, the current increases by about what factor?

  1. About 2×
  2. About 60×
  3. About 1.06×
  4. About 10×
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Answer: D. About 10×

I ∝ e^(V/VT), so the ratio = e^(0.060/0.02585) = 10.2, i.e. about one decade per 60 mV.

8. A Zener regulator has Vs = 20 V, Rs = 100 Ω and Vz = 10 V. The Zener needs at least 5 mA to stay in regulation. What is the minimum load resistance?

  1. 100.0 Ω
  2. 2000.0 Ω
  3. 105.3 Ω
  4. 95.2 Ω
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Answer: C. 105.3 Ω

Is = (20 − 10)/100 = 100 mA; ILmax = 100 − 5 = 95 mA; RLmin = 10/0.095 = 105.3 Ω.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. An LED with Vf = 2.1 V must run at 15 mA from 24 V. What series resistance is needed?

  1. 1.74 kΩ
  2. 1.46 kΩ
  3. 140 Ω
  4. 1.6 kΩ
Show the answer and solution

Answer: B. 1.46 kΩ

R = (Vs − Vf)/I = (24 − 2.1)/0.015 = 1.46 kΩ.

10. A 9.1 V Zener regulator runs from 24 V and supplies 30 mA to a load, with 10 mA kept in the Zener. What series resistor is needed?

  1. 600 Ω
  2. 373 Ω
  3. 497 Ω
  4. 1.49 kΩ
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Answer: B. 373 Ω

The series resistor carries load plus Zener current: R = (24 − 9.1)/(30 + 10) mA = 373 Ω.

11. A full-wave rectifier on 60 Hz mains feeds 0.5 A into a 4.7 mF reservoir capacitor. Estimate the peak-to-peak ripple.

  1. 887 mV
  2. 282 mV
  3. 1.77 V
  4. 443 mV
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Answer: A. 887 mV

A full-wave rectifier recharges at 2f = 120 Hz: Vr ≈ I/(2fC) = 0.5/(120 × 4.7 mF) = 887 mV peak-to-peak.

12. What is the average (DC) output of an unfiltered full-wave rectifier with a peak of 50 V (ideal diodes)?

  1. 31.8 V
  2. 45 V
  3. 35.4 V
  4. 15.9 V
Show the answer and solution

Answer: A. 31.8 V

Vdc = 2Vm/π = 2 × 50/π = 31.8 V (ideal diodes, no filter).

Other topics in the same subjects

DC circuits · AC circuits · Transients (RC, RL, RLC) · BJT and FET circuits · Operational amplifiers · Oscillators and feedback · Power electronics · Digital electronics · Microprocessors and microcontrollers · Control systems · Instrumentation and measurement · Three-phase systems · Transformers · DC machines · Induction motors · Synchronous machines · Power systems and transmission · Protection: fuses, breakers, relays · Illumination · Power plant engineering · PEC: branch circuits, ampacity and protection

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