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If in a p–n junction, a square input signal of 10 V is applied, as shown
Then the output across $R_L$ will be
Then the output across $R_L$ will be
A
-
B
-
C
-
D
-
Detailed Solution
The input square wave swings between +5 V and −5 V.
The diode conducts only when forward biased, so only the positive half (+5 V) appears across $R_L$; the circuit acts as a half-wave rectifier.
So the output is a 5 V positive pulse.
The diode conducts only when forward biased, so only the positive half (+5 V) appears across $R_L$; the circuit acts as a half-wave rectifier.
So the output is a 5 V positive pulse.
