Diode valve is discovered by Fleming.

N-type semiconductor is having excess of free electrons (unbound electrons) for conduction. The total number of electrons in an atom is equal to total number of protons in the nucleus. Therefore, n –type semiconductor is neutral.

Boron has valency three. When boron is doped in a pure semiconductor, then p-type semiconductor is formed.

For semiconductor, $n=A{T}^{\frac{3}{2}}{e}^{\frac{-E_g}{2kT}};$
So, $n\alpha T^{\frac{3}{2}}$

V-d curve near the junction will be as shown by curve (a).

In circuit 1, N is connected with N, which is not a series combination of p-n junction. In circuit 2, each p-n junction is forward biased, hence same current flows, giving same potential difference across p-n junction. In circuit 2, each p-n junction is reverse biased; same leakage current will flow, giving equal potential difference across each p-n junction diode.

If half wave rectifier the output voltage is the RMS voltage
$=\frac{V_o}{\sqrt{2}}=\frac{200}{\sqrt{2}}$

The average value of output direct current in a half wave rectifier is

            =$\frac{\left(\text{average value of current over a cycle}\right)}{2}$

            = $\frac{\left(\frac{2I_0}{\pi }\right)}{2}=\frac{I_0}{\pi }$

In half wave rectifier, we get the output only in one half cycle of input AC therefore, the frequency of the ripple of the output is same as that of input AC ie, 50 Hz.

Here p-n junction as forward biased with voltage
= 5 - 3 = 2 V.
$\therefore$ Current $I=\frac{2}{200}=\frac{1}{100}={10}^{-2}A$