296. Three identical single tuned amplifiers are connected in cascade. The 3 dB bandwidth of each amplifier is 100 kHz. The overall 3 dB bandwidth will be approximately:

1. 1. 300 kHz

   
   
   

   2. 100 kHz

   
   
   

   3. 50 kHz

   
   
   

   4. 33 kHz

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given n = 3
   f_H = 100 kHz
   f_H^* = 3 dB
   bandwidth of 3-cascaded stage and is given by the relation
   f_H^* = f_H √2² – 1 kHz
   = 100 √2³ – 1 kHz
   = 100 √1.2570 – 1 kHz
   ≈ 100 × .5 kHz
   ≈ 50 kHz

   Correct Option: C

   Given n = 3
   f_H = 100 kHz
   f_H^* = 3 dB
   bandwidth of 3-cascaded stage and is given by the relation
   f_H^* = f_H √2² – 1 kHz
   = 100 √2³ – 1 kHz
   = 100 √1.2570 – 1 kHz
   ≈ 100 × .5 kHz
   ≈ 50 kHz

---

297. The configuration of a cascade amplifier is:

1. 1. CE-CE

   
   
   

   2. CE-CB

   
   
   

   3. CC-CB

   
   
   

   4. CC-CC

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

---

298. If the input to the circuit given below is a sine wave the output will be a:
     
     
     

1. 1. half wave rectified sine wave

   
   
   

   2. full wave rectified sine wave

   
   
   

   3. triangular wave

   
   
   

   4. square wave

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Since the given circuit arrangement represent open loop system, so the output will be ± V_sat, which resembles a square wave.

   Correct Option: D

   Since the given circuit arrangement represent open loop system, so the output will be ± V_sat, which resembles a square wave.

---

299. The op-amp of figure has a very poor open loop voltage gain of 45 but is otherwise ideal.
     The gain of the amplifier equals:
     
     
     

1. 1. 5

   
   
   

   2. 20

   
   
   

   3. 4

   
   
   

   4. 4.5

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A closed loop gain,
   

   ACL =	V0	=	AOL
   	Vi		1 + AOL

   
   where, A_OL = 45 (given)
   

   β =	2k	(from given figure)
   	2k + 8k

   
   or β = 0.2
   

   Now, ACL =	45
   	1 + 45 × ·2

   
   

   =	45	= 4.5
   	10

   
   

   Correct Option: D

   A closed loop gain,
   

   ACL =	V0	=	AOL
   	Vi		1 + AOL

   
   where, A_OL = 45 (given)
   

   β =	2k	(from given figure)
   	2k + 8k

   
   or β = 0.2
   

   Now, ACL =	45
   	1 + 45 × ·2

   
   

   =	45	= 4.5
   	10

   
   

---

---

300. For the op-amp circuit shown below the voltage gain A_ν = ν₀/ν_i is:
     
     
     

1. 1. – 8

   
   
   

   2. 8

   
   
   

   3. – 10

   
   
   

   4. 10

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Apply KCL at node A
   

   	VA – VC	+	VA	+	VA – VB	= 0 ............(i)
   	R		R		R

   
   KCL at node B
   

   	VB – VA	+	VB – V0	+	VB	= 0 ............(ii)
   	R		R		R

   
   KCL at node C
   

   	VC – Vi	+	VC – VA	= 0 .............(iii)
   	R		R

   
   on putting V_C = 0 (due to virtual ground) and eliminating V_A and V_B from equation (i), (ii) and (iii)
   

   	V0	= – 8
   	Vi

   
   

   Correct Option: A

   Apply KCL at node A
   

   	VA – VC	+	VA	+	VA – VB	= 0 ............(i)
   	R		R		R

   
   KCL at node B
   

   	VB – VA	+	VB – V0	+	VB	= 0 ............(ii)
   	R		R		R

   
   KCL at node C
   

   	VC – Vi	+	VC – VA	= 0 .............(iii)
   	R		R

   
   on putting V_C = 0 (due to virtual ground) and eliminating V_A and V_B from equation (i), (ii) and (iii)
   

   	V0	= – 8
   	Vi

   
   

---