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P-N Junction Diode and its applications

Practice Questions:

Question 1: 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 2: 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 3: 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 4: 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 5: 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 6: moderate

In the following circuit the equivalent resistance between A and B is

1. 20/3 Ω

2. 10 Ω

3. 16 Ω

4. 20 Ω

View Answer

Both the diodes are reversed biased so can be replaced by open circuit.

Req= 8+2+6= 16 ohm

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Question 7: moderate

If break down voltage of zener diode is 6 V then value of I2 will be :

1. 18 mA

2. 24 mA

3. 6 mA

4. Zero

View Answer

Voltage across 500 ohm resistor = 12 Volt

I = 12/500 = 24 mA

I 1 = 6/1000= 6mA

I2= I - I1= (24-6 )mA= 18 mA

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Question 8: moderate

Determine V0 for the network

1. 6.6 V

2. 9 V

3. -9 V

4. -6.6 V

View Answer

Both the diodes are forward biased so current will flow through the circuit

i= 10/4.5= 20/9 mA

V0= i R= (20/9)×3.3 Volt= 6.6 volt

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Question 9: moderate

Two PN-junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be :

1. In the circuit (1) and (2)

2. In the circuit (2) and (3)

3. In the circuit (1) and (3)

4. Only in the circuit (1)

View Answer

In circuit 1. First Diode is forward biased and second diode is reverse biased. So, potential across 1st diode is less than 2nd diode.

In circuit 2: Both the diodes are forward biased so potential difference across them is equal

In circuit 3: Both the diodes are reverse biased so potential difference across them is equal to V/2.

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