421. In a PN-junction diode, holes diffuse from the P region to the N region because—

1. 1. The free electrons in the N region attracted them

   
   
   

   2. They are swept across the junction by the potential difference

   
   
   

   3. There is greater concentration of holes in the P region as compared to N region

   
   
   

   4. None of the above

   
   
   

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   NA

   Correct Option: C

   NA

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422. In an unbiased PN-junction, the junction current at equilibrium is—

1. 1. Due to diffusion of minority carriers only

   
   
   

   2. Due to diffusion of majority carriers only

   
   
   

   3. Zero, because equal but opposite carriers crossing the junction

   
   
   

   4. Zero, because no charges are crossing the junction

   
   
   

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   In an unbiased PN-junction the junction current at equilibrium is zero, because equal but opposite carriers crossing the junction.

   Correct Option: C

   In an unbiased PN-junction the junction current at equilibrium is zero, because equal but opposite carriers crossing the junction.

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423. The potential barrier at a PN-junction is due to the charges on either side of the junction. These charges are—

1. 1. Minority carriers

   
   
   

   2. Majority carriers

   
   
   

   3. Both majority and minority carriers

   
   
   

   4. FIxed donor and acceptor ions

   
   
   

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   The potential barrier at a PN-junction is due to the charges of fixed donor and acceptor ions.

   Correct Option: D

   The potential barrier at a PN-junction is due to the charges of fixed donor and acceptor ions.

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424. The output waveform will be—
     
     
     

1. 1. 

   
   
   

   2. 

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   During the forward biased
   

   Vo =		Vi	600		– Vr
   			600 + 100

   
   Where, Vr = .6 V
   

   = 10 ×	600	- 0.6
   	700

   
   = 8.57 – 0.6 ≈ 8.00 V
   

   Correct Option: D

   During the forward biased
   

   Vo =		Vi	600		– Vr
   			600 + 100

   
   Where, Vr = .6 V
   

   = 10 ×	600	- 0.6
   	700

   
   = 8.57 – 0.6 ≈ 8.00 V
   

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425. The circuit in figure shows a zener-regulated d.c. power supply. The zener diode voltage is 15 V. The minimum value of R_L down to which the output voltage remains constant is—
     
     
     

1. 1. 27 ohm

   
   
   

   2. 45 ohm

   
   
   

   3. 15 ohm

   
   
   

   4. 24 ohm

   
   
   

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1. View Hint View Answer Discuss in Forum

   From given figure
   V_Z = 15 V
   

   I =		24 – VZ	=	24 – 15
   		27		27

   
   

   I =	1
   	3

   
   

   RL min =		VZ	=	15	= 45 Ω
   		IL (max)		1 / 3

   
   

   Correct Option: B

   From given figure
   V_Z = 15 V
   

   I =		24 – VZ	=	24 – 15
   		27		27

   
   

   I =	1
   	3

   
   

   RL min =		VZ	=	15	= 45 Ω
   		IL (max)		1 / 3

   
   

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