Vγ = 0V Vγ = 0.7V, Where Vγ is cutin voltage diode.

Mumbai University > Electronics ana telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: Dec 14

(i) Clamper with γ = 0V (Ideal diode)

The required clamper circuit is biases clamper with negative DC shift is as shown in fig 1.6(a).

The expression for the output voltage of this configuration is as follows: $V_o= V_i- (V_m- V)$ … Positive half cycle of $V_i$

2.7=10-(10-V)

V = 2.7Volts

The battery voltage is equal to 2.7 volts.

In the negative half cycle the output voltage is given by,

$V_o= -V_i- (V_m- V)$

= -10 – (10 – 2.7) = - 17.3 Volts

(ii)Clamper with γ = 0.7 V (Practical silicon diode):

The clamper circuit configuration with the practical silicon diode is same as that shown in fig (a). However the DC source voltage will be different. It will be V = 2 Volts rather than 2.7 Volts.