121. The voltage V_CB and current I_B for the CB-configuration of the figure shown below will be
     
     

1. 1. 3.4 V and 45.8 A

   
   
   

   2. 6V, and 2.75 mA

   
   
   

   3. 34 V and 4.5 mA

   
   
   

   4. None of these
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   IE =	VEE - VBE	=	4 - 0.7	= 2.75 mA
   	RE		1.2 K

   
   and VCB = V_CC - I_CR_C ..... (assuming I_C ≈ I_E)
   = 10 – (2.75) × 2.4 = 3.4 V
   

   ∴ IB =	IC	=	2.75	= 45.8 A
   	β		60

   
   

   Correct Option: A

   IE =	VEE - VBE	=	4 - 0.7	= 2.75 mA
   	RE		1.2 K

   
   and VCB = V_CC - I_CR_C ..... (assuming I_C ≈ I_E)
   = 10 – (2.75) × 2.4 = 3.4 V
   

   ∴ IB =	IC	=	2.75	= 45.8 A
   	β		60

   
   

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122. If an npn transistor (with C = 0.3 pF) has a unity gain cut-off frequency f_T of 400 MHz at a d.c. bias current I_c = 1 mA, then value of its C is approximately (V_T = 26 mV)
     

1. 1. 15

   
   
   

   2. 30

   
   
   

   3. 50

   
   
   

   4. 96

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : C = 0.3 pF
   

   ∴ gm =	|IC|(mA)	=	1	A/V
   	26 mV		26

   
   

   ⇒ C =	gm	=	1 / 26	= 15 pF
   	2π fT		2π × 4 × 108

   
   

   Correct Option: A

   Given : C = 0.3 pF
   

   ∴ gm =	|IC|(mA)	=	1	A/V
   	26 mV		26

   
   

   ⇒ C =	gm	=	1 / 26	= 15 pF
   	2π fT		2π × 4 × 108

   
   

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123. If an npn transistor has a beta cut-off frequency f_B of 1MHz, and emitter short circuit low frequency current gain β₀ of 200, then unity gain frequency f_T and alpha cut off frequency f_α respectively are
     

1. 1. 200 MHz, 201 MHz

   
   
   

   2. 200 MHz, 199 MHz

   
   
   

   3. 199 MHz, 200 MHz

   
   
   

   4. 201 MHz, 200 MHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   f _T = β₀ × f_β = (200) × 1 MHz = 200 MHz
   

   and fα =	fβ	=	fβ
   	1 - α		1 -	β
   				1 + β

   
   = f _β (β+ 1) = 10⁶ (200 + 1) 10⁶ × 201 = 201 MHz
   

   Correct Option: A

   f _T = β₀ × f_β = (200) × 1 MHz = 200 MHz
   

   and fα =	fβ	=	fβ
   	1 - α		1 -	β
   				1 + β

   
   = f _β (β+ 1) = 10⁶ (200 + 1) 10⁶ × 201 = 201 MHz
   

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124. What will be the input independence Z_i for the network shown below?
     
     

1. 1. 560.5 Ω

   
   
   

   2. 56.05 Ω

   
   
   

   3. 937 Ω

   
   
   

   4. 617.7 Ω

   
   
   

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1. View Hint View Answer Discuss in Forum

   IB =	VCC - VBE
   	RF + βRC

   
   

   =	9 - 0.7	= 11.53 A
   	180 kΩ + 200 × 2.7 kΩ

   
   I_E = (β + 1) I_B = 201 × 11.53 A = 2.32 mA
   

   re =	26 mV	=	26 mV	= 11.21 Ω
   	IE		2.32 mA

   
   

   ∴ Zi =	re				=	11.21	= 560.5 Ω
   	1	+	RC			0.005 + 0.015
   	β		RF

   
   

   Correct Option: A

   IB =	VCC - VBE
   	RF + βRC

   
   

   =	9 - 0.7	= 11.53 A
   	180 kΩ + 200 × 2.7 kΩ

   
   I_E = (β + 1) I_B = 201 × 11.53 A = 2.32 mA
   

   re =	26 mV	=	26 mV	= 11.21 Ω
   	IE		2.32 mA

   
   

   ∴ Zi =	re				=	11.21	= 560.5 Ω
   	1	+	RC			0.005 + 0.015
   	β		RF

   
   

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125. In the following two non-linear transistor biasing circuits, the resistors
     
     

1. 1. R_A and R_B, both have negative temperature coefficient.

   
   
   

   2. R_A and R_B, both have positive temperature coefficient.

   
   
   

   3. R_A has negative temperature coefficient and R_B has positive temperature coefficient.

   
   
   

   4. R_A has positive temperature coefficient and R_B has negative temperature coefficient.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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