101. Match each of the items A, B and C with and appropriate item from 1, 2, 3, 4 and 5.
     (A) Common-collector amplifier.
     (B) Common-emitter amplifier.
     (C) Common-base amplifier.
     1. provides voltage gain but no current gain
     2. provides current gain but no voltage gain
     3. provides neither voltage nor power gain
     4. provides neither current nor power gain
     5. provides both voltage and current gain

1. 1. A-5, B-3, C-1

   
   
   

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

   
   
   

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

   
   
   

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

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A – 2
   B – 5
   C – 1

   Correct Option: D

   A – 2
   B – 5
   C – 1

---

102. Match each of the items A, B and C with and appropriate item from 1, 2, 3, 4 and 5. In an extrinsic semiconductor:
     (A) the resistivity decreases
     (B) the temperature coefficient of resistivity is negative.
     (C) the photo conductivity is low
     1. the doping concentration is low
     2. the length of the semiconductor is reduced
     3. the band gap is high
     4. the area of cross-section of the semiconductor is increased
     5. the doping concentration is increased

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

   
   
   

   2. A-5, B-5, C-3

   
   
   

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

   
   
   

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

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A – 5
   B – 5
   C – 1

   Correct Option: B

   A – 5
   B – 5
   C – 1

---

---

103. 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

   
   
   

---

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.

---

104. 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

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A – 3
   B – 1
   C – 1

   Correct Option: C

   A – 3
   B – 1
   C – 1

---

---

105. 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

   
   
   

---

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

---