Explain the use of the junction diode as a half-wave rectifier by drawing a circuit and draw input and output waves.

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(N/A) half-wave rectifier consists of a transformer,a junction diode,and a load resistance $R_{L}$.
The primary coil of the transformer is connected to the $AC$ mains voltage. The secondary coil of the transformer is connected in series with the junction diode and the load resistance $R_{L}$. This circuit is called a half-wave rectifier.
The required $AC$ voltage is obtained between ends $A$ and $B$ of the secondary coil of the transformer.
During the first positive half-cycle of the $AC$ voltage,end $A$ is positive with respect to $B$. As a result,the $p-n$ junction diode is in forward bias,and current flows through the load resistance $R_{L}$ from direction $X$ to $Y$.
During the second half-cycle,$A$ becomes negative with respect to $B$. As a result,the $p-n$ junction diode is in reverse bias,and no current flows through the load resistance $R_{L}$.
Thus,during the positive half-cycle of incoming cycles,the current flows in $R_{L}$ in the $X$ to $Y$ direction.
The reverse saturated current for the diode is negligible and is ignored.
The reverse breakdown voltage of the diode must be sufficiently higher than the peak $AC$ voltage at the secondary of the transformer to protect the diode from reverse breakdown.
The $AC$ voltage across the ends of $R_{L}$ and the waveform of the rectified voltage are shown in the figure.

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In a half-wave rectifier using a $PN$ junction diode,the input voltage has an amplitude of $25 \ V$ and a frequency of $50 \ Hz$. No filter is used,and the load resistance is $1000 \ \Omega$. The forward resistance of the diode is $10 \ \Omega$. Calculate the peak,average,and $rms$ values of the load current,respectively.

For a half-wave rectifier,if the maximum output voltage is $10 \ V$,what will be the $D.C.$ output voltage?

The output across the diode in the given circuit is:

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