The document Previous year Questions (2016-20) - Semiconductor devices, Communications systems (Part - 1) Notes | EduRev is a part of the JEE Course Physics For JEE.

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**Q.1. Which of the following gives a reversible operation? (2020)(i)(ii)(iii)(iv)Ans.** (4)

A logic gate is reversible if it’s input can be recovered from it’s output

Option (1)

This is NOR gate, it is not reversible

Option (2)

This is OR gate, it is not reversible

Option (3)

This is NAND gate, it is not reversible

Option (4)

This is NOT gate, it is reversible.

(1) 10 V

(2) 5 V

(3) 15 V

(4) zero

Ans.

Solution.

We have

Since, the diode is forward bias so diode behave as short circuit and it will conduct so equivalent circuit will be

Potential difference between A and B is

Now,

Therefore,

Q.3. Boolean relation at the output stage-Y for the following circuit is (2020)

(2) A + B

(3) A.B

(4)

Solution.

Truth table of input output relation is

From truth table, we have

(1) 0

(2) toggles between 0 and 1

(3) will not execute

Ans.

Let two input of the given circuit be A and B

So, output of the circuit is given by

Given that A = 1, B = 0

So, Y = 0+0 = 0

Q.5. Both the diodes used in the circuit shown are assumed to be ideal and have negligible resistance when these are forward biased. Built in potential in each diode is 0.7 V. For the input voltage shown in the figure, the voltage (in Volts) at point A is __________. (2020)

Ans.

Solution.

From given circuit we have

First diode is forward biased and second diode is reversed biased. So, voltage at point A is given by

(1) 0.2 A

(2) 0.6 A

(3) 0.3 A

(4) 0

Ans.

Solution

In the given circuit both diodes are reversed biased. So, they are open circuited.

Redraw the given circuit we have following new figure

So, current I in the circuit is given by

Here R

Given that V = 9V

Therefore,

Ans.

Given that V

So, voltage across resistance is V = V

Current across resistance is

Total power dissipated across Zener diodes is

P = VI = 8 x 1 x 10

= 80 x 10

Power dissipated across each Zener diode is

P

(1) 2 Ω·m

(2) 4 Ω·m

(3) 0.4 Ω·m

(4) 0.2 Ω·m

Ans.

Current in the semiconductor is,

We know that

(1) 0.8 V

(2) 0.6 V

(3) 0.2 V

(4) 0.4 V

Ans.

In the given circuit,

voltage, V = 12 V

Resistance, R = 5 kΩ

So,

Output voltage

V

= (5 × 10

= 11.7 V

When the connection of Ge diode are reversed then the current will be through Si. Current

Therefore, output voltage

16 = 1R

= (2.26 × 10

= 11.3

The value of V

= 0.4 V

(1) X = 0, Y = 1

(2) X = 1, Y = 1

(3) X = 1, Y = 0

(4) X = 0, Y = 0

Ans.

P + Q must be 0

Therefore, Y = 0, X = 1

(1) 10

(2) 10

(3) 10

(4) 10

Ans.

Energy incident on plate per second = IA

= 1.6 × 10−3 × 1 × 10−4

= 1.6 × 10−7 W

Kinetic energy

K = hv− ϕ

= 10 – 5 = 5 eV

Now,

⇒ N = 10

Therefore, number of emitted electrons per second

(1) 9 mA

(2) 5 mA

(3) Zero

(4) 14 mA

Ans.

If Zener diode does not undergo breakdown

V

= 70

⇒ V = IR

70 = I × 5 × 10

= 14 mA

V

V

50 = 10 × 10

Current through diode is

I

= 14 – 5= 9 mA

(1) 0.0 mA

(2) 6.0 mA

(3) 6.7 mA

(4) 4.0 mA

Ans.

When potential drop across 1500 Ω is 10 V then current flowing through it is

[V = IR]

= 6.61 mA

Now, 2 V will be the potential difference across 500 Ω.

Thus, electric current flowing through it is

= 4 mA

So, I

Therefore, voltage across Zener diode must be less than 10 V therefore it will not work in break down region and its resistance will be infinite and current through it is equal to 0.

(1) 0.036

(2) 0.020

(3) 0.027

(4) 0.030

The second diode is reverse biased. Thus, current only flow through first diode and its value is

Ans.

We have,

(1) 25 μA and 3.5 V.

(2) 20 μA and 3.5 V.

(3) 25 μA and 2.8 V.

(4) 20 μA and 2.8 V.

Ans.

For output, at saturation, V

In CE circuit we have

Now,

At input side in BE circuit

V

= 25 × 10

⇒ VBB = 3.5 V

(1) 3.75 × 10

(2) 3.86 × 10

(3) 6.25 × 10

(4) 4.87 × 10

Ans.

= 3.75 × 10

Usable frequency = 1% of f

Therefore, required number of channel

= 6.25 × 10

(1) 65 km

(2) 48 km

(3) 80 km

(4) 40 km

Ans.

Let h

Maximum distance up to which signal can be broadcasted is

(1) 2750 kHz

(2) 2900 kHz

(3) 2250 kHz

(4) 2000 kHz

Ans.

Side band frequency -

AM wave contains three frequency

- wherein,

is carrier frequency,

are side band frequency.

Range of signal 2250KHz to 2750KHz from option for 2000KHz , from = 200 Hz

range = 1800KHz to 2200KHz.

So, only option (4) lies in the given range.

**Q.20. An amplitude modulated signal is given by V(t) = 10[1 + 0.3 cos (2.2 × 10 ^{4}t)] sin (5.5 × 10^{5}t). Here 't' is in seconds. **

(1) 1785 and 1715.

(2) 892.5 and 857.5.

(3) 178.5 and 171.5.

(4) 89.25 and 85.75.

Ans.

We have,

Which one of the following best describes the above signal? [2019]

(1) (9 + sin (2.5π × 10

(2) (1 + 9sin (2π × 10

(3) (9 + sin(2π × 10

(4) (9 + sin(4π × 10

Ans.

(1) 0.3

(2) 0.5

(3) 0.6

(4) 0.4

Ans.

(2) 2

(3) 4

Ans.

Let d be the cover range of TV tower

(1) 0

(2) 11.5 mA

(3) 15 mA

(4) 13.5 mA

Ans:

Ans:

No. of telephonic channels that can be transmitted simultaneously

**Q.26. In a common emitter configuration with suitable bias, it is given that R _{L} is the load resistance and R_{BE} is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given respectively by: **

(1)

(2)

(3)

(4)

Ans:

Voltage gain

Current gain

Power gain = voltage gain x current gain

(1) 10

(2) 15

(3) 20

(4) 8

Ans:

The station will require a band width of 30 kHz

So,

No. of stations = 300/30

= 10

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