Q 1 :    

Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of photodiode is found to be (x8)eV. The value of X is (Given, h=6.6×10-34Js,e=1.6×10-19C)               [2024]

  • 15

     

  • 11

     

  • 13

     

  • 21

     

(A)         Energy bond gap Eg is

               Eg=hcλ=1240λ(in nm)

                     =1240660=1.87eV

               or Eg=(158)eV

               x=15

 



Q 2 :    

The value of net resistance of the network as shown in the given figure is                   [2024]

  • (5/2)Ω

     

  • (30/11)Ω

     

  • (15/4)Ω

     

  • 6Ω

     

(4)

Diode 2 is in reverse bias, so current will not flow in branch of 2nd diode, So we can assume it to be broken wire.
Diode 1 is in forward bias, so it will behave like conducting wire. The circuit is equivalent to

Req=15×1015+10=15×1025=6Ω



Q 3 :    

Which of the diode circuit shows correct biasing used for the measurement of dynamic resistance of p-n junction diode?           [2024]

  •  

  •  

  •  

(2)

For conduction of current, diode must be in forward biased mode.
So, option (2) is correct.

 



Q 4 :    

The I-V characteristics of an electronic device shown in the figure. The device is:                   [2024]

  • a solar cell

     

  • a transistor which can be used as an amplifier

     

  • a Zener diode which can be used as a voltage regulator

     

  • a diode which can be used as a rectifier

     

(3)

It is a theoretical conceptual question. Zener diode used as voltage regulator

 



Q 5 :    

Which of the following circuits is reverse-biased?              [2024]

  •  

  •  

  •  

  •  

(4)

 



Q 6 :    

In the given circuit, the breakdown voltage of the Zener diode is 3.0 V. What is the value of IZ?            [2024]

  • 3.3 mA

     

  • 5.5 mA

     

  • 10 mA

     

  • 7 mA

     

(2)

Breakdown voltage of the Zener diode, Vz=3V

Let potential at B=0V

Potential at E(VP)=10V and VC=VA=3V

I=10-31000=71000A and I1=32000A

Also, IZ+I1=I

Therefore, IZ=7-1.51000=5.5 mA

 



Q 7 :    

A Zener diode of breakdown voltage 10V is used as a voltage regulator as shown in the figure. The current through the Zener diode is               [2024]

  • 50 mA

     

  • 0

     

  • 30 mA

     

  • 20 mA

     

(3)

i1=20-10200=120

i2=10-0500=150

iz=i1-i2=120-150

=3100 A=30 mA



Q 8 :    

In the given circuit, the voltage across load resistance (RL) is          [2024]

  • 8.75 V

     

  • 9.00 V

     

  • 8.50 V

     

  • 14.00 V

     

(1)

(VB)Si=0.7V

(VB)Ge=0.3V

Voltage drop across both diodes =0.7+0.3=1V

So remaining 14 volts will drop across RL

VL=2.5(1.5+2.5)×14=2.54×14 volt=8.75 volt



Q 9 :    

In the given circuit if the power rating of Zener diode is 10 mW, the value of series resistance Rs to regulate the input unregulated supply is             [2024]

  • 0.5 

     

  • 10 Ω

     

  •  

  • 10 

     

(1)

Potential difference across Rs

V1=8-5=3V

Current through the load resistor

I=51×103=5 mA

Maximum current through Zener diode

Izmax.=105=2 mA

and minimum current through Zener diode

Izmin.=0

 Ismax.=5+2=7 mA

and Rsmin=V1Ismax.=37 kΩ

Similarly, Ismin.=5 mA

and Rsmax.=V1Ismin.=35 kΩ

 37 kΩ<Rs<35 kΩ



Q 10 :    

A potential divider circuit is connected with a DC source of 20 V, a light-emitting diode of glow in voltage 1.8 V and a Zener diode of breakdown voltage of 3.2 V. The length (PR) of the resistive wire is 20 cm. The minimum length of PQ to just glow the LED is ____ cm.           [2024]



(5)

VPQ=3.2+1.8=5 V

And, VPR=20 V  (given)

Now 20 V20 cm×=5 V