66. Mirror current I of the given circuit is _____mA .
    

1. 1. 10

   
   
   

   2. 5.27

   
   
   

   3. 27

   
   
   

   4. 10.27

   
   
   

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   I = Ix =	VCC - VBE	=	12 - 0.7	=10.27 mA
   	RE		1.1 kΩ

   
   

   Correct Option: D

   I = Ix =	VCC - VBE	=	12 - 0.7	=10.27 mA
   	RE		1.1 kΩ

   
   

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67. In the circuit shown in the given figure, assuming that the capacitor C is almost shorted for the frequency range of interest of the input signal, voltage gain of the amplifier will be approximately __________ .
    

1. 1. 1

   
   
   

   2. 2

   
   
   

   3. 3

   
   
   

   4. 4

   
   
   

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   Av =	AI RL	=	(1 + hfe) RL
   	Ri		Ri

   
   

   =	(1 + hfe) RL	=	100 × 0.66 K	= 66
   	hie		1 K

   
   

   Avs =	AV . Ri	= 66 ×	1	= 0.985 ≈ 1
   	Ri + Rs		66 + 1

   Correct Option: A

   Av =	AI RL	=	(1 + hfe) RL
   	Ri		Ri

   
   

   =	(1 + hfe) RL	=	100 × 0.66 K	= 66
   	hie		1 K

   
   

   Avs =	AV . Ri	= 66 ×	1	= 0.985 ≈ 1
   	Ri + Rs		66 + 1

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68. A CE amplifier has a resistor R_F connected between collector an d base R_F = 40 k. R_C = 4k. I f h_fe = 50, r_π = 1 k, then output resistance R₀ is _______ kΩ .
    

1. 1. 33

   
   
   

   2. 0.66

   
   
   

   3. 6.6

   
   
   

   4. 3.3

   
   
   

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1. View Hint View Answer Discuss in Forum

   R0 = RL' = Rc ∥ Rp =	4 × 40	= 3.64 kΩ
   	44

   
   

   1 + β =	β RC	= 50 ×	4K	=	50
   	RF + RC		44K		11

   
   

   1 + βA = D =	61
   	11

   
   

   Rof =	R0	=	116	×	11	= 0.66 kΩ
   	D		44		61

   
   

   Correct Option: B

   R0 = RL' = Rc ∥ Rp =	4 × 40	= 3.64 kΩ
   	44

   
   

   1 + β =	β RC	= 50 ×	4K	=	50
   	RF + RC		44K		11

   
   

   1 + βA = D =	61
   	11

   
   

   Rof =	R0	=	116	×	11	= 0.66 kΩ
   	D		44		61

   
   

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69. For the n-channel enhancement MOSFET shown in the given figure, threshold voltage V_tn = 2V. The drain current I_D of the MOSFET is 4 mA when drain resistance R_D is 1 kΩ. If value of R_D is increased to 4 kΩ, then drain current I_D will become _______ mA .
    

1. 1. 2.8

   
   
   

   2. 8

   
   
   

   3. 28

   
   
   

   4. All of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   I_D = K(V_GS – [V_E(TH)]²
   ⇒ 4mA = K (V_GS – 2)²
   Here, V_GS = 10 – 4 × 1 = 6
   ∴ 4 = K(6 – 2)²
   

   ⇒ K =	4	=	1
   	16		4

   
   When R_D is increased to 4 kΩ,
   V_GS = 10 – 4I_D
   

   Now , ID =	4	(10 – 4ID - 2)2
   	16

   
   ⇒ 4I_D = (8 – 4I_D)² = 64 + 16 I_D² – 64 I_D
   ⇒ 16I_D² – 68 I_D + 64 = 0
   ⇒ 4I_D² – 17 I_D + 16 = 0
   ⇒ I_D = 2.8 mA
   

   Correct Option: A

   I_D = K(V_GS – [V_E(TH)]²
   ⇒ 4mA = K (V_GS – 2)²
   Here, V_GS = 10 – 4 × 1 = 6
   ∴ 4 = K(6 – 2)²
   

   ⇒ K =	4	=	1
   	16		4

   
   When R_D is increased to 4 kΩ,
   V_GS = 10 – 4I_D
   

   Now , ID =	4	(10 – 4ID - 2)2
   	16

   
   ⇒ 4I_D = (8 – 4I_D)² = 64 + 16 I_D² – 64 I_D
   ⇒ 16I_D² – 68 I_D + 64 = 0
   ⇒ 4I_D² – 17 I_D + 16 = 0
   ⇒ I_D = 2.8 mA
   

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70. If α = 0.98, I_CO = 6 µA and I_β = 100 A for a transistor, then value of I_C will be _______ mA

1. 1. 15.2

   
   
   

   2. 5.2

   
   
   

   3. 2.2

   
   
   

   4. 8.4

   
   
   

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1. View Hint View Answer Discuss in Forum

   IC =	ICO	+	α	IB =	6	+	0.98	× 100 = 5.2 mA
   	1 - α		1 - α		1 - 0.98		1 - 0.98

   
   

   Correct Option: B

   IC =	ICO	+	α	IB =	6	+	0.98	× 100 = 5.2 mA
   	1 - α		1 - α		1 - 0.98		1 - 0.98

   
   

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