166. An amplifier has a voltage gain of 100. To reduce distortion, 10% negative feedback is employed. The gain of the amplifier with feedback is:

1. 1. 1.01

   
   
   

   2. 9.09

   
   
   

   3. 90.9

   
   
   

   4. 101

   
   
   

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   Given A = 100
   

   β = 10% =	10	= 0.1
   	100

   
   

   Af =	A	=	100	=	100	= 9.09
   	1 + Aβ		1 + 100 × 0.1		11

   Correct Option: B

   Given A = 100
   

   β = 10% =	10	= 0.1
   	100

   
   

   Af =	A	=	100	=	100	= 9.09
   	1 + Aβ		1 + 100 × 0.1		11

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167. The slew rate of an operational amplifier is 0.5 V/micro sec. The maximum frequency of a sinusoidal input of 2 V_rms that can be handled without excessive distortion is:

1. 1. 3 kHz

   
   
   

   2. 30 kHz

   
   
   

   3. 300 kHz

   
   
   

   4. 2 MHz

   
   
   

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   fmax =	Slew rate
   	2π·Vm

   
   

   =	0·5 × 106	Vrms = 2 =	Vm
   	2 × π × 2 √2		√2

   
   ⇒ V_m = 2√2 = 28 kHz
   Hence alternative (B) is the most closet answer.

   Correct Option: B

   fmax =	Slew rate
   	2π·Vm

   
   

   =	0·5 × 106	Vrms = 2 =	Vm
   	2 × π × 2 √2		√2

   
   ⇒ V_m = 2√2 = 28 kHz
   Hence alternative (B) is the most closet answer.

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168. The voltage gain of a device is 6 dB. This means that ratio of the amplitudes of the input to output will be approximately:

1. 1. 6

   
   
   

   2. 1

      6

   
   
   

   3. 2

   
   
   

   4. 1

      2

   
   
   

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   20 log10	V0	= 6
   	Vin

   
   

   or log10	V0	=	3
   	Vin		10

   
   

   or	V0	= 100.3
   	Vin

   
   

   V0	≈ 2
   Vin

   
   

   or	Vin	=	1
   	V0		2

   Correct Option: D

   20 log10	V0	= 6
   	Vin

   
   

   or log10	V0	=	3
   	Vin		10

   
   

   or	V0	= 100.3
   	Vin

   
   

   V0	≈ 2
   Vin

   
   

   or	Vin	=	1
   	V0		2

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169. The figure shows a class B amplifier. Given that V_CC = 6.3 V, R_L = 4 ohms and v₀ is sinusoidal voltage of value 4 V, the average power drawn by the two supply voltage will be approximately:
     
     
     

1. 1. 2 W

   
   
   

   2. 4 W

   
   
   

   3. 6 W

   
   
   

   4. 8 W

   
   
   

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1. View Hint View Answer Discuss in Forum

   P_dc = V_CQ. I_CQ
   In class B push pull amplifier
   V_CQ = V_CC
   

   ICQ =	2Im	=	2	·	Vm
   	π		π		RL

   
   

   Now Pdc = VCC		2Vm		=	2 × 6.3 × 4	≈ 4 watts
   		π·RL			3.14 × 4

   Correct Option: B

   P_dc = V_CQ. I_CQ
   In class B push pull amplifier
   V_CQ = V_CC
   

   ICQ =	2Im	=	2	·	Vm
   	π		π		RL

   
   

   Now Pdc = VCC		2Vm		=	2 × 6.3 × 4	≈ 4 watts
   		π·RL			3.14 × 4

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170. In a transistor push-pull amplifier:

1. 1. there is no d.c. present in the output.

   
   
   

   2. there is no distortion in the output.

   
   
   

   3. there is no even harmonics in the output.

   
   
   

   4. there is no odd harmonics in the output.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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