76. An amplifier has a gain of 1000 ± 10. Negative feedback is provided such that the gain variation is within 0.1%. Then the amount of feedback β_f is:

1. 1. 1

      10

   
   
   

   2. 1

      9

   
   
   

   3. 9

      100

   
   
   

   4. 9

      1000

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that A = 1000 ± 10
   

   	dA	=	10	=	1	= 0.01
   	A		1000		100

   
   

   	dAf	= 0.1% =	.1	= 0.001
   	Af		100

   
   Now, from relation
   

   	dAf	=	1	x	dA
   	dA		(1 + Aβ)		A

   
   

   0.001 =	1	× 0.01
   	(1 + 1000 β)

   
   or 1 + 1000 β = 10
   or 1000 β = 9
   

   or β =	9
   	1000

   
   Hence alternative (D) is the correct choice.

   Correct Option: D

   Given that A = 1000 ± 10
   

   	dA	=	10	=	1	= 0.01
   	A		1000		100

   
   

   	dAf	= 0.1% =	.1	= 0.001
   	Af		100

   
   Now, from relation
   

   	dAf	=	1	x	dA
   	dA		(1 + Aβ)		A

   
   

   0.001 =	1	× 0.01
   	(1 + 1000 β)

   
   or 1 + 1000 β = 10
   or 1000 β = 9
   

   or β =	9
   	1000

   
   Hence alternative (D) is the correct choice.

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77. The high frequency response of an amplifier is given by
    

    A =	A0
    		1 + j	ω
    			ω2

    
    where A₀ = 1000 and ω₂ = 10⁴. Negative feedback with β = 9/1000 is now given. The value of ω₂ now is:

1. 1. 10⁴

   
   
   

   2. 10³

   
   
   

   3. 10⁵

   
   
   

   4. none of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   From given high frequency response
   

   A =	A0
   		1 + j	ω
   			ω2

   
   where, ω2 is the higher cut-off frequency.
   and due to negative feedback ω2 increases, so the new,
   ω₂* = ω₂ (1 + Aβ)
   

   = 104		1 + 1000 ×	9		= 104. 10 = 105
   			1000

   
   Hence alternative (C) is the correct choice.

   Correct Option: C

   From given high frequency response
   

   A =	A0
   		1 + j	ω
   			ω2

   
   where, ω2 is the higher cut-off frequency.
   and due to negative feedback ω2 increases, so the new,
   ω₂* = ω₂ (1 + Aβ)
   

   = 104		1 + 1000 ×	9		= 104. 10 = 105
   			1000

   
   Hence alternative (C) is the correct choice.

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78. An amplifier with mid band gain A = 500 has negative feedback applied of value β = 1/100. Given the upper cut off frequency without feedback was 60 kHz. With feedback it becomes:

1. 1. 10 kHz

   
   
   

   2. 12 kHz

   
   
   

   3. 300 kHz

   
   
   

   4. 360 kHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that
   

   β =	1
   	100

   
   A = 500
   f_H = 60 kHz
   After feedback. New upper cut-off frequency increases as bandwidth increases after feedback.
   f_H* = f_H (1 + Aβ)
   

   = 60		1 +	500		= 60 × 6 = 360 kHz
   			100

   Correct Option: D

   Given that
   

   β =	1
   	100

   
   A = 500
   f_H = 60 kHz
   After feedback. New upper cut-off frequency increases as bandwidth increases after feedback.
   f_H* = f_H (1 + Aβ)
   

   = 60		1 +	500		= 60 × 6 = 360 kHz
   			100

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79. The amount of feedback applied to an amplifier reduces the gain by a factor of 10. The bandwidth:

1. 1. decreases by a factor of 10

   
   
   

   2. increases by a factor of 10

   
   
   

   3. remains the same

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since the product of gain and bandwidth remains constant. Therefore if gain reduces by factor of 10, then surely bandwidth will increase by a factor of 10.

   Correct Option: B

   Since the product of gain and bandwidth remains constant. Therefore if gain reduces by factor of 10, then surely bandwidth will increase by a factor of 10.

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80. The feedback amplifier shown in figure has:
    
    
    

1. 1. current-series feedback

   
   
   

   2. voltage-series feedback

   
   
   

   3. voltage-shunt feedback

   
   
   

   4. current-shunt feedback

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here the output current is feedback in the node. Hence the type of feedback is current-shunt.

   Correct Option: D

   Here the output current is feedback in the node. Hence the type of feedback is current-shunt.

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