351. The emitter current in a junction transistor with normal bias—

1. 1. May be greatly increased by a small change in collector bias

   
   
   

   2. Is equal to the sum of the base current and collector current

   
   
   

   3. Is approximately equal to the base current

   
   
   

   4. Is designated as I_CO

   
   
   

---

1. View Hint View Answer Discuss in Forum

   In normal bias I_E = I_B + I_C
   Hence alternative (B) is the correct answer.

   Correct Option: B

   In normal bias I_E = I_B + I_C
   Hence alternative (B) is the correct answer.

---

352. One way in which the operation of an NPN transistor differes from that of a PNP transistor is that—

1. 1. The emitter junction is reverse-biased in the NPN

   
   
   

   2. The emitter injects minority carriers into the base region of the PNP and majority carriers in the base region of the NPN

   
   
   

   3. The emitter injects holes into the base region of the PNP and electrons into the base region of the NPN

   
   
   

   4. The emitter injects electrons into the base region of the PNP and holes into the base region of the NPN

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---

---

353. A small increase in the collector reverse-bias will cause—

1. 1. A large increase in emitter current

   
   
   

   2. A large increase in collector current

   
   
   

   3. A large decrease in collector current

   
   
   

   4. Very small change in collector reverse saturation current
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

---

354. The p-n-p transistor circuit as shown in the above figure is in saturation. Given that V_BE = 0.7 V and V_{CE (sat)} = 0.3 V and β_min. = 20, the value of I_C will be—
     
     
     

1. 1. 4.7 mA

   
   
   

   2. 5.3 mA

   
   
   

   3. 8.6 mA

   
   
   

   4. 10 mA

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given V_BE = 0.7 V, V_{CE (sat)} = 0.3 V, β_min. = 20,
   R_C = 1 kΩ, I_C =?
   

   IC =	VCC – VCE (sat)
   	RC

   
   

   or IC =	5 – 0.3	= 4.7 mA
   	1 kΩ

   
   Hence alternative (A) is the correct answer.

   Correct Option: A

   Given V_BE = 0.7 V, V_{CE (sat)} = 0.3 V, β_min. = 20,
   R_C = 1 kΩ, I_C =?
   

   IC =	VCC – VCE (sat)
   	RC

   
   

   or IC =	5 – 0.3	= 4.7 mA
   	1 kΩ

   
   Hence alternative (A) is the correct answer.

---

---

355. Consider the following semiconductor diodes—
     1. Germanium diode
     2. Silicon diode
     3. Tunnel diode
     4. Schottky diode
     The correct increasing order of forward voltage drop of diodes is—

1. 1. 1, 2, 3, 4

   
   
   

   2. 3, 4, 1, 2

   
   
   

   3. 3, 1, 4, 2

   
   
   

   4. 1, 3, 2, 4

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---