96. The nominal quiescent collector current of a transistor is 1.2 mA. If the range of β for this transistor is 80 ≤ β ≤ 120 and if the quiescent collector current changes by ± 10 per cent, the range in value for r_π is:
    

1. 1. 1.73 kΩ < r_π < 2.59 kΩ

   
   
   

   2. 1.93 kΩ < r_π < 2.59 kΩ

   
   
   

   3. 1.73 kΩ < r_π < 2.59 kΩ

   
   
   

   4. 1.56 kΩ < r_π < 2.88 kΩ

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that, I_CQ = 1.2 mA ± 10%
   

   i.e., ICQ min =		1.2 –	1.2 × 10		= 1.08 mA
   			100

   
   

   80 ≤ β ≤ 120 ICQ max =		1.2 +	1.2 × 10		= 1.32 mA
   			100

   
   i.e. β_max = 120 and β_min = 80
   we know that,
   

   rπ =	β	=	β·VT
   	gm		ICQ

   
   

   rπ max =	βmax · VT
   	ICQ min

   
   

   =	120 × 26 × 10–3	= 2.88 kΩ
   	1.08 × 10–3

   
   

   rπ min =	βmin ·VT
   	ICQ max

   
   

   =	80 × 26 × 10–3	= 1.56 kΩ
   	1.32 × 10–3

   
   Hence alternative (D) is the correct choice.

   Correct Option: D

   Given that, I_CQ = 1.2 mA ± 10%
   

   i.e., ICQ min =		1.2 –	1.2 × 10		= 1.08 mA
   			100

   
   

   80 ≤ β ≤ 120 ICQ max =		1.2 +	1.2 × 10		= 1.32 mA
   			100

   
   i.e. β_max = 120 and β_min = 80
   we know that,
   

   rπ =	β	=	β·VT
   	gm		ICQ

   
   

   rπ max =	βmax · VT
   	ICQ min

   
   

   =	120 × 26 × 10–3	= 2.88 kΩ
   	1.08 × 10–3

   
   

   rπ min =	βmin ·VT
   	ICQ max

   
   

   =	80 × 26 × 10–3	= 1.56 kΩ
   	1.32 × 10–3

   
   Hence alternative (D) is the correct choice.

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97. The small signal voltage gain AV =	V0	is:
    	VS

1. 1. – 4.38

   
   
   

   2. 4.38

   
   
   

   3. – 1.88

   
   
   

   4. 1.88

   
   
   

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1. View Hint View Answer Discuss in Forum

   AV =	V0	= – gm. Rc.		rπ
   	VS			rπ + RB

   
   

   = – 24 × 10– 3. 4 × 103		5 × 103
   		5 × 103 + 250 × 103

   
   

   = – 120 ×	5	= – 1.88
   	255

   Correct Option: C

   AV =	V0	= – gm. Rc.		rπ
   	VS			rπ + RB

   
   

   = – 24 × 10– 3. 4 × 103		5 × 103
   		5 × 103 + 250 × 103

   
   

   = – 120 ×	5	= – 1.88
   	255

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98. The hybrid-π parameter values of g_m, r_π and r₀ are:

1. 1. 24 mA/V, ∞, 5 kΩ

   
   
   

   2. 24 mA/V, 5 kΩ, ∞

   
   
   

   3. 48 mA/V, 10 kΩ, 18.4 kΩ

   
   
   

   4. 48 mA/V, 18.4 kΩ, 10 kΩ

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given that, β = 120
   V_A = ∞
   

   IB =	2 – VBE	=	2 – 7	= 5.2 µA
   	250 kΩ		250 kΩ

   
   I_CQ = β I_B = 120 × 5.2 × 10^–6 = 0.642 mA
   Now,

   gm =	ICQ	=	·642 × 10–3	= 24 mA/V
   	VT		26 × 10–3

   
   

   rπ =	β	=	120	= 5 kΩ
   	gm		24 × 10–3

   
   

   r0 =	| VA |	=	∞	= ∞
   	ICQ		24 × 10–3

   Correct Option: B

   Given that, β = 120
   V_A = ∞
   

   IB =	2 – VBE	=	2 – 7	= 5.2 µA
   	250 kΩ		250 kΩ

   
   I_CQ = β I_B = 120 × 5.2 × 10^–6 = 0.642 mA
   Now,

   gm =	ICQ	=	·642 × 10–3	= 24 mA/V
   	VT		26 × 10–3

   
   

   rπ =	β	=	120	= 5 kΩ
   	gm		24 × 10–3

   
   

   r0 =	| VA |	=	∞	= ∞
   	ICQ		24 × 10–3

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99. If the transistor parameter are β = 180 and early voltage V_A = 140 V and it is biased at I_CQ = 2 mA, the values of hybrid-π parameter g_m, r_x and r₀ are respectively:

1. 1. 14 A/V, 2.33 kΩ, 90 kΩ

   
   
   

   2. 14 A/V, 90 kΩ, 2.33 kΩ

   
   
   

   3. 77 mA/V, 2.33 kΩ, 70 kΩ

   
   
   

   4. 77.2 A/V, 70 kΩ, 2.33 kΩ

   
   
   

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1. View Hint View Answer Discuss in Forum

   gm =	IC Q	=	2 × 10–3	= 0.0769 ≈ 77 mA/V
   	VT		26 × 10–3

   
   

   rπ =	β	=	180	= 2.33 kΩ
   	gm		77 × 10–3

   
   

   r0 =	| VA |	=	140	= 70 kΩ
   	ICQ		2 × 10–3

   Correct Option: C

   gm =	IC Q	=	2 × 10–3	= 0.0769 ≈ 77 mA/V
   	VT		26 × 10–3

   
   

   rπ =	β	=	180	= 2.33 kΩ
   	gm		77 × 10–3

   
   

   r0 =	| VA |	=	140	= 70 kΩ
   	ICQ		2 × 10–3

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100. The bandwidth of an RF tuned amplier is dependent on:

1. 1. Q-factor of the tuned output circuit

   
   
   

   2. Q-factor of the tuned input circuit

   
   
   

   3. Quiescent operating point

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Bandwidth =	f0
   	Q

   
   where, f₀ = resonant frequency & Q = quality factor

   Correct Option: A

   Bandwidth =	f0
   	Q

   
   where, f₀ = resonant frequency & Q = quality factor

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