241. The amplifier shown below is being used to amplify an input signal to a peak output voltage of 100 mV. What is the maximum operating frequency of the amplifier. Given slew rate is 0.5 V/µs
     
     
     

1. 1. 796 Hz

   
   
   

   2. 796 kHz

   
   
   

   3. 50 kHz

   
   
   

   4. 50 Hz

   
   
   

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1. View Hint View Answer Discuss in Forum

   fmax =	Slew rate
   	2πVm

   
   

   =	0.5 × 106	= 796 kHz
   	2 × 3.14 × 100 × 10–3

   Correct Option: B

   fmax =	Slew rate
   	2πVm

   
   

   =	0.5 × 106	= 796 kHz
   	2 × 3.14 × 100 × 10–3

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242. An op-amp has a CMRR of 90 dB. If its differential voltage gain is 30,000.Then its common mode gain is:

1. 1. 1

   
   
   

   2. .95

   
   
   

   3. .85

   
   
   

   4. None of these.

   
   
   

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1. View Hint View Answer Discuss in Forum

   CMRR = 20 log10		Ad
   		Ac

   
   

   90 = 20 log10		30,000
   		Ac

   
   

   4.5 = log10		30,000
   		Ac

   
   

   or Ac =	30‚000	=	30‚000	= 0.9459 = 0.95
   	104.5		31622.77

   Correct Option: B

   CMRR = 20 log10		Ad
   		Ac

   
   

   90 = 20 log10		30,000
   		Ac

   
   

   4.5 = log10		30,000
   		Ac

   
   

   or Ac =	30‚000	=	30‚000	= 0.9459 = 0.95
   	104.5		31622.77

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243. A common-mode rejection ratio CMRR of an-amp, that has a differential gain of 2,00,000 and a common mode gain 6.33:

1. 1. 90 dB

   
   
   

   2. 70 dB

   
   
   

   3. 80 dB

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   CMRR in dB= 20 log10		Ad
   		Ac

   
   

   = 20 log10		2‚00‚000
   		6·33

   
   = 89.99
   = 90 dB

   Correct Option: A

   CMRR in dB= 20 log10		Ad
   		Ac

   
   

   = 20 log10		2‚00‚000
   		6·33

   
   = 89.99
   = 90 dB

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244. In the OP amp. circuit with Zener diodes connected as shown, where the Zener voltage is V_Z. V₁ is the input signal and V_R a reference voltage. The output V_o is given thus:
     
     
     

1. 1. For V₁ > V_R V_o = + V_z

   
   
   

   2. For V₁ < V_R V_o = + V_z

   
   
   

   3. For V₁ > V_R V_o = – V_z

   
   
   

   4. For all V₁ V_o = + V_z

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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245. In the op-amp circuit given below the load current i_L is
     
     
     

1. 1. –	Vs
      	R2

   
   
   

   2. Vs

      R2

   
   
   

   3. Vs

      RL

   
   
   

   4. V_s R_L

   
   
   

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1. View Hint View Answer Discuss in Forum

   V_A = V_B (due to virtual ground)
   KCL at node B
   

   	Vβ	+	VB – Vo	+ iL = 0 .............(i)
   	R2		R2

   
   KCL at node A
   

   	VA – Vs	+	VA – Vo	= 0 .............(ii)
   	R1		R1

   
   

   VA = Vs ·	R1	+ Vo ·	R1
   	R1 + R1		R1 + R1

   
   

   VA =	Vs	+	Vo	= VB
   	2		2

   
   

   	Vs + Vo		Vs + Vo	– Vo
   =	2	+	2		= – io
   	R2		R2

   
   

   Vs	= – io
   R2

   
   

   – io =	Vs
   	R2

   
   

   Correct Option: A

   V_A = V_B (due to virtual ground)
   KCL at node B
   

   	Vβ	+	VB – Vo	+ iL = 0 .............(i)
   	R2		R2

   
   KCL at node A
   

   	VA – Vs	+	VA – Vo	= 0 .............(ii)
   	R1		R1

   
   

   VA = Vs ·	R1	+ Vo ·	R1
   	R1 + R1		R1 + R1

   
   

   VA =	Vs	+	Vo	= VB
   	2		2

   
   

   	Vs + Vo		Vs + Vo	– Vo
   =	2	+	2		= – io
   	R2		R2

   
   

   Vs	= – io
   R2

   
   

   – io =	Vs
   	R2

   
   

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