Level 2 — RecallDiodes & Applications

Diodes & Applications

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level 2 — Recall & Standard Problems Time limit: 30 minutes Total marks: 40

Answer all questions. Assume a silicon diode forward drop of $0.7\text{ V}$ unless stated otherwise. Use $\pi \approx 3.1416$ .

Q1. Define the following diode terms as given in a datasheet: (a) forward voltage $V_f$ , (b) reverse (leakage) current $I_r$ , (c) peak inverse voltage (PIV). (3 marks)

Q2. A half-wave rectifier is fed from a transformer secondary of peak voltage $V_p = 12\text{ V}$ . Assuming an ideal diode, calculate: (a) the peak output voltage, (b) the average (DC) output voltage. (4 marks)

Q3. State two advantages of a full-wave bridge rectifier compared to a single-diode half-wave rectifier. (2 marks)

Q4. A Zener diode with $V_Z = 5.1\text{ V}$ is used to regulate a load. The unregulated supply is $12\text{ V}$ and a series resistor limits the current. If the desired Zener current is $20\text{ mA}$ and load current is negligible, calculate the required series resistance $R_S$ . (4 marks)

Q5. An LED with a forward voltage of $2.0\text{ V}$ is to operate at $15\text{ mA}$ from a $5\text{ V}$ supply. Calculate the required series current-limiting resistor. (4 marks)

Q6. Briefly explain how a photodiode differs from an LED in terms of its operating principle and typical biasing. (4 marks)

Q7. State one key advantage of a Schottky diode over a standard silicon PN-junction rectifier diode, and give one typical application. (3 marks)

Q8. (a) What is the function of a varactor diode? (b) State how its junction capacitance varies with reverse bias voltage. (4 marks)

Q9. A positive series clipper uses an ideal diode. A sinusoidal input $v_{in}=10\sin(\omega t)\text{ V}$ is applied. Sketch/describe the output waveform and state the peak output voltage for the half-cycles that pass. (4 marks)

Q10. A two-input diode OR gate has inputs A and B (logic HIGH = 5 V, LOW = 0 V) with a pull-down resistor to ground. Complete the truth table for output Y (ideal diodes) for all four input combinations. (4 marks)

END OF PAPER

Answer keyMark scheme & solutions

Q1. (3 marks)

(a) $V_f$ : the voltage across the diode when conducting a specified forward current (~0.7 V for Si, ~0.3 V for Ge/Schottky). (1)
(b) $I_r$ : the small leakage current flowing when the diode is reverse-biased below breakdown. (1)
(c) PIV: the maximum reverse voltage the diode can withstand without breaking down. (1)

Q2. (4 marks)

(a) Ideal diode → peak output = peak input: $V_{out,peak}=12\text{ V}$ . (2)
(b) Half-wave average: $V_{DC}=\dfrac{V_p}{\pi}=\dfrac{12}{3.1416}=3.82\text{ V}$ . (2) Why: over a full period only one half-cycle passes, so the mean of the rectified sine is $V_p/\pi$ .

Q3. (2 marks) Any two of: higher average DC output (uses both half-cycles); higher ripple frequency (twice line freq) → easier filtering; better transformer utilization; no centre-tapped transformer needed. (1 each)

Q4. (4 marks)

Voltage across $R_S$ : $12 - 5.1 = 6.9\text{ V}$ . (1)
$R_S=\dfrac{V_{in}-V_Z}{I_Z}=\dfrac{6.9}{0.020}=345\ \Omega$ . (3) Why: the series resistor drops the excess voltage at the chosen Zener current.

Q5. (4 marks)

Voltage across resistor: $5-2.0=3.0\text{ V}$ . (1)
$R=\dfrac{V_S-V_{LED}}{I}=\dfrac{3.0}{0.015}=200\ \Omega$ . (3)

Q6. (4 marks)

LED: forward-biased; electron-hole recombination emits photons (electrical → optical). (2)
Photodiode: reverse-biased (or zero-biased in photovoltaic mode); incident light generates carriers producing a current (optical → electrical). (2)

Q7. (3 marks)

Advantage: lower forward voltage drop (~0.2–0.4 V) and very fast switching / low reverse recovery time. (2)
Application: high-frequency rectification, switch-mode power supplies, RF detectors. (1)

Q8. (4 marks)

(a) A varactor is used as a voltage-controlled capacitor (e.g. for tuning oscillators/filters). (2)
(b) Junction capacitance decreases as reverse bias voltage increases (depletion region widens). (2)

Q9. (4 marks)

Positive series clipper with ideal diode passes only the half-cycles that forward-bias the diode. For the conducting half-cycles output follows the input; the other half-cycles are clipped to 0 V. (2)
Peak output of passing half-cycles $=10\text{ V}$ (ideal, no drop). (2)

Q10. (4 marks) Diode OR gate (½ mark each row × 4, plus 2 for correct logic):

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

Output HIGH if either input is HIGH (ideal diodes, no drop). (4)

[
  {"claim":"HW average output = 12/pi ≈ 3.82 V","code":"Vdc=12/pi; result=abs(float(Vdc)-3.8197)<0.01"},
  {"claim":"Zener series resistor = 345 ohm","code":"Rs=(12-5.1)/0.020; result=abs(Rs-345)<1e-6"},
  {"claim":"LED series resistor = 200 ohm","code":"R=(5-2.0)/0.015; result=abs(R-200)<1e-6"},
  {"claim":"Clipper peak output = 10 V ideal","code":"Vpk=10; result=Vpk==10"}
]