451. The base width in a junction transistor is deliberately chosen small so that—

1. 1. The electric field is large

   
   
   

   2. The concentration gradient of injected carrier increases

   
   
   

   3. To reduce recombination of injected minority carriers

   
   
   

   4. The majority carriers easily reach the collector
      

   
   
   

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   NA

   Correct Option: C

   NA

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452. In a p-n-p transistor biased to operate in the active region, the current in the base region consists of—

1. 1. Only holes

   
   
   

   2. Only electrons

   
   
   

   3. Predominantly holes

   
   
   

   4. Predominantly electrons

   
   
   

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   NA

   Correct Option: C

   NA

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453. p-n-p transistor biased the active region of operation. The base current is due to—

1. 1. Holes entering the base from the terminal for recombination

   
   
   

   2. Holes injected in to the emitter from base

   
   
   

   3. Only saturation current due to reverse bias of collector junction

   
   
   

   4. Electrons recombining with a part of the holes diffusing in the base region

   
   
   

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   NA

   Correct Option: D

   NA

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454. In a half-wave rectifier, the peak value of the ac voltage across the secondary of the transformer is 20 V. If no filter circuit is used, the maximum dc voltage across the load will be—
     

1. 1. 28.28 V

   
   
   

   2. 20 V

   
   
   

   3. 14.14 V

   
   
   

   4. 9 V

   
   
   

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   Vrms =	Vm
   	√2

   
   

   Vrms =	20 √2	= 20 V
   	√2

   Correct Option: C

   Vrms =	Vm
   	√2

   
   

   Vrms =	20 √2	= 20 V
   	√2

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455. Given that C_D at forward current of 1 mA is 0.8 µF, its value for a current of 4 mA is—

1. 1. 0.8 µF

   
   
   

   2. 0.2 µF

   
   
   

   3. 1.6 µF

   
   
   

   4. 3.2 µF

   
   
   

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   We know that diffusion capacitance C_D is given by relation
   

   CD =	τ·1
   	η – VT

   
   where, I is the forward current.
   i.e. CD ∝ I
   

   or		CD1	=	I1
   		CD2		I2

   
   

   CD2 =		I2·CD1	=	4 × 8µF	3.2 µF
   		I1		1

   Correct Option: D

   We know that diffusion capacitance C_D is given by relation
   

   CD =	τ·1
   	η – VT

   
   where, I is the forward current.
   i.e. CD ∝ I
   

   or		CD1	=	I1
   		CD2		I2

   
   

   CD2 =		I2·CD1	=	4 × 8µF	3.2 µF
   		I1		1

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