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P-N Junction Diode and its applications

Practice Questions:

Question 1: easy

What is the current through an ideal PN-junction diode shown in figure below :

1. Zero

2. 10 mA

3. 20 mA

4. 50 mA

View Answer

i = 2/100 = 20 mA

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Question 2: easy

A reverse biased diode is :

View Answer

In Reverse bias mode potential of p-type is less than that of n-type semiconductor

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Question 3: difficult

A zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the zener diode is:

1. 5 mA

2. 10 mA

3. 15 mA

4. 20 mA

View Answer

Voltage across 250 ohm resistor= 5 volt

Current through 250 ohm resistor= 5/250 = 20 mA

Current through 1000 ohm resistor = 15/1000 = 15mA

Current through Zener Diode = 20 mA -15 mA = 5 mA

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Question 4: moderate

In the following circuit of PN junction diodes D1, D2, D3 are ideal then i is :

1. E/R

2. E/2R

3. 2E/3R

4. Zero

View Answer

Diode D3 is forward biased so it will short circuit both diodes D1 and D2. So, equivalent resistance of circuit becomes R. Hence,

i= E/R

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Question 5: moderate

If in a p-n junction, a square input signal of 10 V is applied as shown, then the output across RL will be :

View Answer

Diode conducts in forward biased case and stops conduction in reverse biased mode.

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Question 6: moderate

The breakdown in a reverse biased p–n junction diode is more likely to occur due to

(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strong electric field in a depletion region if the doping concentration is small
(d) strong electric field in the depletion region if the doping concentration is large.

1. a, d

2. c only

3. b only

4. b, c

View Answer

The breakdown in a reverse biased p–n junction diode is more likely to occur due to

a. Avalanche Breakdown - large velocity of the minority charge carriers if the doping concentration is small

b. Zener Breakdown - strong electric field in the depletion region if the doping concentration is large

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Question 7: easy

In the following figure, the diodes which are forward biased, are :

1. (a), (b) and (d)

2. (c) only

3. (a) and (c)

4. (b) and (d)

View Answer

For Forward biasing of PN Junction diode potential of P side should be greater than potential of N-side semiconductor

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Question 8: moderate

Which of the following circuits provides full-wave rectification of an ac input :

View Answer

Theoretical question based on bridge rectifier.

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Question 9: moderate

A sinusoidal voltage of peak value 200 volt is connected to a diode and resistor R in the
circuit shown so that half-wave rectification occurs. If the forward resistance of the diode
is negligible compared to R, the r.m.s. voltage across R is approximately :

1. 200

2. 100

3. 100/√2

4. 283

View Answer

RMS voltage for half wave rectifier is Vmax/2 = 100 Volt

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Question 10: easy

A p–n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly :

1. \[1\times 10_{14} Hz\]

2. \[20\times 10_{14} Hz\]

3. \[10\times 10_{14} Hz\]

4. \[5\times 10_{14} Hz\]

View Answer

Energy of EM wave E = hυ

⇒ 2 eV = 6.626 × 10^-34×υ

⇒ \[v= 5\times 10_{14} Hz\]

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