If a square signal of 10 V (varying from +5 | vedclass

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(A) The circuit consists of a $P-N$ junction diode in series with a load resistor $R_L$. When the input signal is positive $(+5 \ V)$,the diode is for

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(A) The circuit consists of a $P-N$ junction diode in series with a load resistor $R_L$. When the input signal is positive $(+5 \ V)$,the diode is forward-biased and acts as a closed switch (assuming an ideal diode). Thus,the entire input voltage appears across the load resistor $R_L$. When the input signal is negative $(-5 \ V)$,the diode is reverse-biased and acts as an open switch. Thus,no current flows through $R_L$,and the output voltage across $R_L$ is $0 \ V$. Therefore,the output signal is a square wave that varies between $0 \ V$ and $+5 \ V$.

If a square signal of $10 \ V$ (varying from $+5 \ V$ to $-5 \ V$) is applied to a $P-N$ junction diode as shown,the output signal across $R_L$ will be:

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