Test: Breakdown Diodes


10 Questions MCQ Test Electronic Devices | Test: Breakdown Diodes


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This mock test of Test: Breakdown Diodes for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Breakdown Diodes (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Breakdown Diodes quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Breakdown Diodes exercise for a better result in the exam. You can find other Test: Breakdown Diodes extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

 A zener diode works on the principle of_________

Solution:

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.

QUESTION: 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

Solution:

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.

QUESTION: 3

When the voltage across the zener diode increases_________

Solution:

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.

QUESTION: 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?

Solution:

 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.

QUESTION: 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?

Solution:

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

QUESTION: 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?

Solution:

 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.

QUESTION: 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 _______

Solution:

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.

QUESTION: 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?

Solution:

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

QUESTION: 9

Avalanche breakdown in zener diode is ______

Solution:

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.

QUESTION: 10

The zener diode is heavily doped because______

Solution:

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.

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