If in a p-n junction, a square input signal of 10 V is applied, as shown, [figure] then the output across RL will be
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Answer
(2) [figure]
Solution
During the +5 V part of the input, the diode is forward biased and conducts (ideal diode, no drop), so the output across RL is +5 V.
During the −5 V part, the diode is reverse biased and does not conduct, so no current flows through RL and the output is 0 V.
The output is a pulse of +5 V with zero output in between (half-wave rectification).
Answer: (2)