# Electronic Devices and Circuits Questions and Answers – Breakdown Diodes

This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses 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

Explanation: 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

Explanation: 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

Explanation: 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?

a) 7 to 7.29V
b) 6 to 7V
c) 7.14 to 7.43V
d) 7.2 to 8V

Explanation: 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?

a) 120
b) 125
c) 250
d) 100

Explanation: Here, IKNEE=10mA, VZ=5V. I=IL+IZ. I= (10-5)/100=50mA
Now, 50=10+ILMAX .
ILMAX=40mA. RLMIN=5/40mA=125 Ω.
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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?

a) 23.7mA
b) 20mA
c) 26mA
d) 48.3mA

Explanation: 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 applied. Assume that the diode drop is 0.7V when it’s forward biased. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______

a) 6.1V,-0.7V
b) 0.7V,-7.5V
c) 7.5V,-0.7V
d) 7.5V,-7.5V

Explanation: With VI= 10V when maximum, D1 is forward biased, D2 is reverse biased. 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?

a) 1.2Ω
b) 80 Ω
c) 50 Ω
d) 70 Ω

Explanation: Here, Vz =6V, IZMIN=5mA.IS=IZMIN+ILMAX.
80=5+ILMAX . ILMAX=75Ma.RLMIN=VI/ILMAX=6/75mA
=80 Ω.

9. Avalanche breakdown in zener diode is ______
a) electric current multiplication takes place
b) phenomenon of voltage multiplication takes place
c) electrons are decelerated for a period of time
d) sudden rise in voltage takes place.

Explanation: The carriers in transition region are accelerated by electric field to energies. That energies are sufficient to create electron current multiplication. A single carrier that is energized will collide with another by gaining energy. Thus an avalanche multiplication takes place.

10. The zener diode is heavily doped because______
a) to have low breakdown voltage
b) to have high breakdown voltage
c) to have high current variations
d) to maintain perfect quiescent point

Explanation: The value of reverse breakdown voltage at which zener breakdown occurs is controlled by amount of doping. If the amount of doping is high, the value of voltage at which breakdown occurs will decrease. Better doping gives a sooner breakdown voltage.

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

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