481. Find I₁ and I₂ in diagram shown below in terms of V_S assume V_T = 0.7 volt for all diode—
     
     
     

1. 1. I1 =	5 – Vx	if 0 < VS < 0.7 V and I1 = I2 = 1.8 mA if VS > .7 V
      	2K

   
   
   

   2. I2 = I1 =	5 – Vx	if VS > .7 V I2 = 1.8 mA if 0 < VS < 0.7 V
      	2K

   
   
   

   3. I2 = 1.8 mA if VS > 0.7 V I1 =	5 – Vx	if VS > 0.7 V
      	2K

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   From figure we see that here everything depends upon voltage. V_x
   

   I1 =	5 – Vx
   	2K

   
   if 0 < V_S < .7 V
   Also V_x = V_S + .7 V
   
   If V_S > 0.7 V then
   The diode D₁ will be in reverse biased and
   

   I1 = I2 =	5 – 1.4	= 1.8 mA
   	2K

   
   Where, 1.4 V is the voltage drop across diode D₂ and D₃

   Correct Option: A

   From figure we see that here everything depends upon voltage. V_x
   

   I1 =	5 – Vx
   	2K

   
   if 0 < V_S < .7 V
   Also V_x = V_S + .7 V
   
   If V_S > 0.7 V then
   The diode D₁ will be in reverse biased and
   

   I1 = I2 =	5 – 1.4	= 1.8 mA
   	2K

   
   Where, 1.4 V is the voltage drop across diode D₂ and D₃

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482. The emitter resistor R_E bypassed by a capacitor C_E—

1. 1. Reduces the voltage gain

   
   
   

   2. Increases the voltage gain

   
   
   

   3. Causes thermal runaway

   
   
   

   4. Stabilizes the Q point

   
   
   

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1. View Hint View Answer Discuss in Forum

   The main function of the emitter resistor R_E bypassed by a capacitor is to improve the stability of the transistor.

   Correct Option: D

   The main function of the emitter resistor R_E bypassed by a capacitor is to improve the stability of the transistor.

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483. The Q point in a voltage amplifier is selected in the middle of the active region because—

1. 1. It gives a distortionless output

   
   
   

   2. The operating point then becomes very stable

   
   
   

   3. The circuit then requires less number of resistors

   
   
   

   4. It then requires a small dc voltage

   
   
   

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1. View Hint View Answer Discuss in Forum

   Because if we select the Q-point other than the middle of the active region then there will be the more chances of clipping the output signal (i.e. either voltage or current or both).

   Correct Option: A

   Because if we select the Q-point other than the middle of the active region then there will be the more chances of clipping the output signal (i.e. either voltage or current or both).

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484. The operating point of an NPN transistor amplifier should not be selected in the saturation region as—

1. 1. It may drive the transistor to thermal runaway

   
   
   

   2. It may cause output to be clipped in the negative half of the input signal

   
   
   

   3. It may cause output to be clipped in the positive half of the input signal

   
   
   

   4. It may requird high dc collector supply.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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485. The potential-divider method of biasing is used in amplifiers to—

1. 1. Limit the input ac signal going to the base

   
   
   

   2. Make the operating point almost independent of β

   
   
   

   3. Reduce the base current

   
   
   

   4. Reduce the cost of the circuit

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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