Test: Electronic Devices - 3 - Electronics and Communication Engineering (ECE) MCQ

For the p-n junction diode, the depletion region width is given by

W = 10 μm

For practical diode

When

V_i > 12 + 0.7 = 12.7 diode conducts

i.e

24 sin θ >12.7

θ > 31.93

Conduction angle = 180 – 2 (31.93)

= 116.12°

The circuit shown in a half-wave voltage double During Positive half cycle C₁ and D₁ Conducts and C is charged to V_max During Negative half cycle C₂ and D₂conducts and

 V_out = V_max + V_Cl

= 2 V_max

If r is the internal resistance of LED

Total current = 

As the drop across LED = 1.5 V

Internal resistance r = 
The external resistance to be connected

R = 400 – 100 = 300Ω

Power dissipated in the resistor

I²R = 67.5 mW

I_Z = (125 - I_Z)mA

Case A:

I_Z = 5 mA

I_Z = 120 mA

= 0.12 A

Case B:

I_Z = 100 mA

I_Z = (125 – 100) mA

= 25 mA

= 0.025 A

For P^±n junction the current density is given by

Substituting

D_P = 0.025 × 450

= 11.25

τ_p = 4.32 × 10^-9 sec

For voltages -5V < V_I < 5 V both the diodes are off since the diodes are reverse biased.

The output voltage V_o = V_I

For V_I ≤ -5V

The diode D₁ is forward biased, while D₂ is reverse biased

For V_I ≥ 5 V

The diode D₁ is reverse biased, while D₂ is forward biased


Drawing sketch of V_o Vs V_I

The built in potential is the integral of election field at zero bais:

⇒ 75 × 10^-2 V

⇒ 0.75 V

For positive cycle when V_I = 0.7 V, the 6 V Zener diode is forward biased while 4 V is reverse biased

For negative cycle V_I <-0.7 V, the 6 V Zener diode is reverse biased 4 V Zener diode is forward biased

For -0.7 < V_I < 0.7 both the diodes are reverse biased hence V_out = V_I

The output waveform will be of form

Diffusion capacitance is due to transfer of minority carries during forward bias. The minority carries diffuse from one end of junction to other, causing variation of charge with applied voltage. This leads to capacitance, it is present only in forward bias and is significantly higher than depletion capacitance in forward bias.

Depletion capacitance is due to storage of charges in reverse bias junction, which acts like parallel plate capacitance with forward voltage, the depletion width decreases increasing depletion capacitance. It is less than diffusion capacitance in forward bias mode.