41. Match each of the items A, B and C with and appropriate item from 1, 2, 3, 4 and 5.
    In a JFET
    (A) the pinch-off voltage decreases
    (B) the transconductance increases
    (C) the transit time of the carriers in the channel is reduced
    1. the channel doping is reduced
    2. the channel length is increased
    3. the conductivity of the channel increased
    4. the channel length is reduced
    5. the gate area is reduced

1. 1. A-1, B-4, C-5

   
   
   

   2. A-2, B-1, C-3

   
   
   

   3. A-3, B-2, C-1

   
   
   

   4. A-4, B-3, C-2

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that, pinch-off voltage
   

   | VP | =	q·ND·a2	where VP = pinch-off voltage
   	2∈0

   
   a = channel width
   N_D = doping concentration
   ∈ = relative permeability Hence
   A – 1
   

   ID = b ND. µn		W			. VDS → hence B – 4
   		L

   
   Gate decide the flow of electrons to the drain. Hence C – 5.
   Therefore
   A – 1
   B – 4
   C – 5
   Hence alternative (A) is the correct choice.

   Correct Option: A

   We know that, pinch-off voltage
   

   | VP | =	q·ND·a2	where VP = pinch-off voltage
   	2∈0

   
   a = channel width
   N_D = doping concentration
   ∈ = relative permeability Hence
   A – 1
   

   ID = b ND. µn		W			. VDS → hence B – 4
   		L

   
   Gate decide the flow of electrons to the drain. Hence C – 5.
   Therefore
   A – 1
   B – 4
   C – 5
   Hence alternative (A) is the correct choice.

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42. Match each of the items A, B and C with an appropriate item from 1, 2, 3, 4 and 5. In a bipolar junction transistor:
    (A) the current gain increases
    (B) the collector break-down voltage increases
    (C) the cut-off frequency increases
    1. the base doping is increased and the base width is reduced
    2. the base doping is reduced and the base width is increased
    3. the base doping and the base width are reduced
    4. the emitter area is increased and the collector area is reduced
    5. the base doping and the base width are increased

1. 1. A - 1, B - 2, C - 4

   
   
   

   2. A - 5, B - 4, C - 1

   
   
   

   3. A - 3, B - 1, C - 1

   
   
   

   4. A - 2, B - 4, C - 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   A – 3
   B – 1
   C – 1

   Correct Option: C

   A – 3
   B – 1
   C – 1

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43. An n-channel JFET has I_DSS = 1 mA and V_P = – 5 V. Its maximum transconductance is:

1. 1. 0.1 milli mho

   
   
   

   2. 0.4 milli mho

   
   
   

   3. 1.0 milli mho

   
   
   

   4. 4.0 milli mho

   
   
   

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1. View Hint View Answer Discuss in Forum

   gmo =	2IDSS	=	2 × 1 × 10–3	= 0.4 milli mho.
   	| VP |		5

   Correct Option: B

   gmo =	2IDSS	=	2 × 1 × 10–3	= 0.4 milli mho.
   	| VP |		5

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44. An amplifier has an open-loop gain of 100, and its lower and upper-cut -off frequency of 100 Hz and 100 kHz, respectively. A feedback network with a feedback factor of 0.99 is connected to the amplifier. The new lower and upper-cut-off frequencies are at___and ___

1. 1. f_H = 10 MHz, f_L = 1 Hz

   
   
   

   2. f_H = 25 MHz, f_L = 10 Hz

   
   
   

   3. f_H = 100 MHz, f_L = 100 Hz

   
   
   

   4. f_H =10 MHz, f_L = 10 Hz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Af =	A	=	100	= 1
   	1 + Aβ		1 + 100 × 0·99

   
   f^*_H = f_H. (1 + Aβ) = 100 × 10³ (1 + 0.99 × 100)
   = 10 MHz
   

   f*L =	fL	=	100	= 1 = 1 Hz.
   	1 + Aβ		(1 + 100 × 0·99)

   Correct Option: A

   Af =	A	=	100	= 1
   	1 + Aβ		1 + 100 × 0·99

   
   f^*_H = f_H. (1 + Aβ) = 100 × 10³ (1 + 0.99 × 100)
   = 10 MHz
   

   f*L =	fL	=	100	= 1 = 1 Hz.
   	1 + Aβ		(1 + 100 × 0·99)

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45. A power amplifier delivers 50 W output at 50% efficiency. The ambient temperature is 25° C. If the maximum allowable junction temperature is 150° C, then the maximum thermal resistance Q_jc that can be tolerated is:

1. 1. 25° C/W

   
   
   

   2. 20° C/W

   
   
   

   3. 5° C/W

   
   
   

   4. 1° C/W

   
   
   

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1. View Hint View Answer Discuss in Forum

   P_D = P₀ × efficiency
   = dissipated power at the output
   

   PD =	50 × 50	= 25 W
   	100

   
   Given that, T_i = 150° C
   T_A = 25°C
   Now,
   

   PD =	Ti – TA
   	Qjc

   
   where, T_i = instantaneous temperature
   T_A = ambient temp.
   

   Qjc =	Ti – TA	=	150 – 25	= 5° C/W
   	PD		25

   
   Q_jc = thermal resistance

   Correct Option: C

   P_D = P₀ × efficiency
   = dissipated power at the output
   

   PD =	50 × 50	= 25 W
   	100

   
   Given that, T_i = 150° C
   T_A = 25°C
   Now,
   

   PD =	Ti – TA
   	Qjc

   
   where, T_i = instantaneous temperature
   T_A = ambient temp.
   

   Qjc =	Ti – TA	=	150 – 25	= 5° C/W
   	PD		25

   
   Q_jc = thermal resistance

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