Electronic Devices and Circuits Multiple Choice Questions on “Breakdown Diodes”.
1. A zener diode works on the principle of_________
A. tunneling of charge carriers across the junction
B. thermionic emission
C. diffusion of charge carriers across the junction
D. hopping of charge carriers across the junction
Answer: A
Clarification: Due to zener effect in reverse bias under high electric field strength, electron quantum tunneling occurs. It’s a mechanical effect in which a tunneling current occurs through a barrier. They usually cannot move through that barrier.
2. Which of the following are true about a zener diode?
1) it allows current flow in reverse direction also
2) it’s used as a shunt regulator
3) it operates in forward bias condition
A. 3 only
B. 1 and 2
C. 2 and 3
D. 2 only
Answer: B
Clarification: The operation of a zener diode is made in reverse bias when breakdown occurs. So, it allows currnt in reverse direction. The most important application of a zener diode is voltage or shunt regulator.
3. When the voltage across the zener diode increases_________
A. temperature remains constant and crystal ions vibrate with large amplitudes
B. temperature increases and crystal ions vibrate with large amplitudes
C. temperature remains constant and crystal ions vibrate with smaller amplitudes
D. temperature decreases and crystal ions vibrate with large amplitudes
Answer: B
Clarification: When voltage is increased, the tunnelling at reverse bias increases. The voltage rises temperature. The crystal ions with greater thermal energy tend to vibrate with larger amplitudes.
4. For the zener diode shown in the figure, the zener voltage at knee is 7V, the knee current is negligible and the zener dynamic resistance is 10Ω. If the input voltage (Vi) ranges from 10 to 16 volts, the output voltage (Vo) ranges from?
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A. 7 to 7.29V
B. 6 to 7V
C. 7.14 to 7.43V
D. 7.2 to 8V
Answer: C
Clarification: If i is the current flowing, then V0=10i+7
i=(VI-7)/210. By substituting, if VI=10V then i=1/70 and V0=(1/7)+7=7.14V
if VI =16V then i=3/70 and V0=(3/7)+7=7.43V.
5. In the circuit below, the knee current of ideal zener diode is 10mA. To maintain 5V across the RL, the minimum value of RL is?
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A. 120
B. 125
C. 250
D. 100
Answer: B
Clarification: Here, IKNEE=10mA, VZ=5V. I=IL+IZ. I= (10-5)/100=50mA
Now, 50=10+ILMAX .
ILMAX=40mA. RLMIN=5/40mA=125 Ω.
6. The zener diode in the circuit has a zener voltage of 5.8V and knee current of 0.5mA. The maximum load current drawn with proper function over input voltage range between 20 and 30V is?
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A. 23.7mA
B. 20mA
C. 26mA
D. 48.3mA
Answer: A
Clarification: Here, I1MAX=IZMIN+ILMAX.
IZMIN =0.5mA, I1MAX =(V1MAX-VZ )/RS . Putting the values we get , I1MAX =24.2mA.
So, 24.2-0.5=23.7mA.
7. In the given limiter circuit, an input voltage Vi=10sin100πt is applieC. Assume that the diode drop is 0.7V when it’s forward biaseC. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______
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A. 6.1V,-0.7V
B. 0.7V,-7.5V
C. 7.5V,-0.7V
D. 7.5V,-7.5V
Answer: C
Clarification: With VI= 10V when maximum, D1 is forward biased, D2 is reverse biaseC. Zener is in breakdown region. VOMAX=sum of breakdown voltage and diode drop=6.8+0.7=7.5V. VOMIN=negative of voltage drop=-0.7V. There will be no breakdown voltage here.
8. The 6V Zener diode shown has zener resistance and a knee current of 5mA. The minimum value of R so that the voltage does not drop below 6V is?
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A. 1.2Ω
B. 80 Ω
C. 50 Ω
D. 70 Ω