Semiconductor Devices and Logic Gates

Given, base current in $n-p-n$ transistor,

$ I_B=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A} $

Since, $95 \%$ of emitted electrons reach the collector, hence Collector current,

$ \begin{equation*} I_C=95 \% \text { of } I_E \Rightarrow I_C=0.95 I_E \Rightarrow I_E=\frac{I_C}{0.95} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

But we know that, $I_E=I_C+I_B$

$ \Rightarrow \quad \frac{I_C}{0.95}=I_C+2 \times 10^{-3} \Rightarrow \frac{I_C}{0.95}-I_C=2 \times 10^{-3} $

The circuit given is as shown

Potential drop across

$ R_L=V_L=V_Z=10 \mathrm{~V} $

Potential drop across $4 \mathrm{k} \Omega$ resistance is

$ V_R=60-10=50 \mathrm{~V} $

Current through $4 \mathrm{k} \Omega$ resistance is

$ I=\frac{V_R}{R}=\frac{50}{4 \times 10^3}=12.5 \times 10^{-3} $

$\mathrm{A}=12.5 \mathrm{~mA}$

Current through load resistance $=I_L$

$ =\frac{V_L}{R_L}=\frac{10}{2 \times 10^3}=5 \mathrm{~mA} $

Also, $I_Z=I-I_L=12.5-5=7.5 \mathrm{~mA}$

Phosphorous is a pentavalent impurity.

So, resulting semiconductor is a $n$-type semiconductor. The energy band diagram of a $n$-type semiconductor is shown below

So, the energy levels of impurity atom created in the semiconductor are close to the bottom of the conduction band.

The output of full wave rectifier with capacitor filter for the given AC input is given as

The given logic circuit is

∴ Boolean expression of the given logic circuit is given as

$ Y=\bar{A}+B $

The given combination of gates is shown below

The output of this combination can be determined by Boolean algebra.

$\therefore$ Output, $X = \overline {\overline A \,.\,\overline B } = \overline{\overline A} + \overline{\overline B} $ [Using De Morgan's theorem]

$ \Rightarrow X = A + B$

This is the output of OR gate. Hence, the given combination of gates yield OR gate.

In this case, the diode is forward biased. Potential drop in a germanium diode in forward bias is around 0.3V.

Hence, current through the given circuit,

$I = {{{V_{net}}} \over {{R_{eq}}}} = {{5 - 0.3} \over {2 + 3}} = {{4.7} \over 5}$

I = 0.94A