GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 24 of 65

Marks: +1, -0

In the circuit shown,

{V}_{ {s}}

is a square wave of period

T

with maximum and minimum values of

8 {V}

and

-10 {V}

, respectively. Assume that the diode is ideal and

{R}_{1}= {R}_{2}

=50 Ω

The average value of

V_{L}

is ____________ volts (rounded off to 1 decimal place).

Solution:

When

{V}_{ {s}}

is +ve
Diode will be reserve biased

{V}_{ {L}}=\;{ {R}_{2}}/{ {R}_{1}+ {R}_{2}} {V}_{ {S}}

∴ {V}_{ {L}}=\;{50}/{50+50} × 8

∴ V_{L}=4 V

.....(1)
When

V_{5}

is -ve
Diode will be forward biased

∴ {VL}= {VS}=-10 {V}

.....(ii)
From (i) and (ii)
Average value

=\;{4+(-10)}/{2}=-3

∴

Average value

=-3

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