Zener Diode as Voltage Regulator

ZENER DIODE AS VOLTAGE REGULATOR

When reverse bias is applied, the voltage across the diode remains constant and, the current through the diode increases.

The voltage across the Zener diode acts as reference voltage and the diode can be used as a voltage regulator.

In the Fig.1.32 shown, the load resistance should be provided with constant voltage. Zener diode is reverse biased and if the input voltage is not less than Zener breakdown voltage (V_z) then the voltage across the diode will be constant and thus the load voltage is also constant.

Applications of Zener Diode

i. Voltage regulators.

ii. Zener limiters to clip the unwanted portion of the voltage waveform.

iii. Over voltage protection.

SOLVED PROBLEMS

Problem 1.5

The forward current of a silicon PN diode is 5 mA at T = 300 K. Determine the forward resistance of a PN junction diode.

Given:

Solution

Problem 1.6

The voltage across a silicon diode at room temperature is 0.7 V when 2 mA current flows through it. If the voltage is increased to 0.75 V, calculate the diode current

Given

Solution

Problem 1.7

A silicon diode has a saturation current of 7.5 μA at room temperature 300 K. Calculate the saturation current at 400 K.

Given:

I₁ = 7.5 x 10^-6 A

T₁ = 300 K

T₂ = 400 K

To find :

I₂ = ?

Solution:

Problem 1.8

Determine the germanium PN diode current for the forward bias voltage of 0.22 V at room temperature 25 °C with reverse saturation current of 1 mA.

Given:

V = 0.22 V

T = 25°C = 273 + 25 = 298 K

I_o = 1 mA = 1 x 10^-3 A

For Germanium diode η = 1

To find :

Diode Current I

Solution:

 

Problem 1.9

A Germanium diode has a saturation current of 10μA at room temperature (300 K). Find the saturation current at 450 K.

Given:

To Find:

Current I₂ at 400 K

Solution:

Problem 1.10

An ideal Germanium diode at a temperature of 125°C has a reverse saturation current of 30 μA. At a temperature of 127 °C, find the dynamic resistance for a 0.2 V bias in (a) forward direction. (b) reverse direction.

Given:

To Find:

Dynamic resistance in forward and reverse direction

Solution:

(a) Dynamic resistance in forward direction

Differentiate with respect to V

(b) Dynamic resistance in reverse direction

In reverse direction, V = -0.2

Substitute in equation,

Problem 1.11

In an N-type semiconductor, the Fermi-level lies 0.3 eV below the conduction band at 27°C. If the temperature is increased to 55°C, find the new position of Fermi level. (AU/ECE - Dec 2007)

Given:

Solution:

The new position of Fermi level lies 0.32 eV below the conduction level.

Problem 1.12

A pn junction diode has a reverse saturation current of 10μA at the room temperature of 27 °C when the room temperature is increased, the reverse saturation current is 30 μA, Calculate the new room temperature germanium.

Given:

Problem 1.13

A full wave diode rectifier has V_i = 100 sin ω t, R_L = 900 Ω and R_f = 100 Ω  

Calculate

(a) Peak load current I_m

(b) dc load current I_dc

(c) AC load current I_rms

(d) dc voltage V_dc

(e) peak instantaneous diode current

(f) PIV of the diode

(g) efficiency

Given:

V_i = 100 sin ω t

V_m = 100 V

R_L = 900 Ω

R_f = 100 Ω

Solution:

Problem 1.14

In a full wave rectifier, a signal of 300 V is applied at 50 Hz frequency. Each diode has an internal resistance of 800 Ω. If the load is 2000 Ω, then calculate

(a) peak value of current in the output

(b) output dc current

(c) efficiency of power transfer

Given:

V_rms = 300 V

f = 50 Hz

R_f = 800 Ω & R_L = 2000 Ω

Solution:

(a) Peak value of current in output

(b) Output dc current

(c) Efficiency of power transfer