446. The forward bias applied to a PN-junction diode is increased from zero to higher values. Rapid increase in the current flow for a relatively, small increase in voltage occurs—

1. 1. Immediately

   
   
   

   2. Only after the forward bias exceeds the potential barrier

   
   
   

   3. When the flow of minority carriers is cause an avalanche breakdown

   
   
   

   4. When the depletion area becomes large than the space-charge area
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

447. When a pn-junction is biased in the forwad direction—
     

1. 1. Only holes in the p region are injected into the n region

   
   
   

   2. Only electrons in the n region are injected into the p region

   
   
   

   3. Majority carriers in each region are injected into the other region

   
   
   

   4. No carriers move

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---

---

448. Avalanche breakdown in a semiconductor diode occurs when—

1. 1. Forward current exceeds a certain value

   
   
   

   2. Reverse bias exceeds a certain value

   
   
   

   3. Forward bias exceeds a certain value

   
   
   

   4. The potential barrier is reduced to zero
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---

449. The depletion or space-charge region in a junction diode contains charges that are—
     

1. 1. Mostly majority carriers

   
   
   

   2. Mostly minority carriers

   
   
   

   3. Mobile donor and acceptor ions

   
   
   

   4. FIxed donor and acceptor ions

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

---

---

450. A junction transistor with β = 49 and I_CO = I_CBO = 1 µA has I_B = 10 µA. The value of I_C is given in µA by—

1. 1. 441

   
   
   

   2. 490

   
   
   

   3. 539

   
   
   

   4. 540

   
   
   

---

1. View Hint View Answer Discuss in Forum

   We know that the collector current is given by the relation
   I_C = βI_B + (1 + β) I_CO
   = 49 × 10. 10^{– 6} + (1 + 49) 1 × 10^{– 6}
   = (490 + 50) × 10^{– 6} = 540 µA.

   Correct Option: D

   We know that the collector current is given by the relation
   I_C = βI_B + (1 + β) I_CO
   = 49 × 10. 10^{– 6} + (1 + 49) 1 × 10^{– 6}
   = (490 + 50) × 10^{– 6} = 540 µA.

---