Semiconductors and Electronic Devices MCQ Questions & Answers in Modern Physics | Physics

For conduction in a $$p-n$$  junction, it should be forward biased because it offers minimum resistance to the flow of current. For this $$p$$-side must be connected to positive terminal (higher potential) and $$n$$-side must be connected to negative terminal (lower potential). Figure below shows the $$p-n$$  junction in a conducting state {forward biased condition)

The truth table for the above logic gate is :

A	B	C
1	1	1
1	0	1
0	1	1
0	0	0

This truth table follows the boolean algebra $$C = A + B$$   which is for OR gate

$$\eqalign{ & {i_c} = \beta \cdot \frac{{{V_s}}}{{{R_{{\text{in}}}}}} = 50 \times \frac{{0.01}}{{1000}} = 500 \times {10^{ - 6}}A \cr & = 500\,\mu A \cr} $$

The gate circuit can be solved by giving two inputs $$A$$ and $$B.$$
Output of NOR gate, $${Y_1} = \overline {A + B} $$
Output of NAND gate, $${Y_2} = \overline {{Y_1} \cdot {Y_1}} $$
$$\eqalign{ & = \overline {\overline {A + B} \cdot \overline {A + B} } \cr & = \overline{\overline {A + B}} + \overline{\overline {A + B}} \cr & = A + B + A + B \cr & = A + B \cr} $$
Output of NOT gate, $$Y = \overline {{Y_2}} = \overline {A + B} $$
which is the output of NOR gate.

In the forward biasing of $$P-N$$  junction, $$p$$ side of junction diode is connected to higher potential and $$n$$ side of junction diode is connected to lower potential. Hence, the option (A) is correct answer.

KEY CONCEPT: For a semi conductor $$n = {n_0}{e^{\frac{{ - Eg}}{{kT}}}}$$   where $${n_0}$$ = no. of free electrons at absolute zero, $$n$$ = no. of free electrons at $$T$$ kelvin, $${E_g}$$ = Energy gap, $$k$$ = Boltzmann constant.
As $${E_g}$$ increases, $$n$$ decreases exponentially.

$$\eqalign{ & {\text{For}}\,\,A = 0,B = 0,\overline {A + B} = 1\,\,{\text{and}}\,\,\overline {A.B} = 1 \cr & {\text{So}}\,\,\overline {\left( {A + B} \right)} .\overline {\left( {A.B} \right)} = 1. \cr} $$

Barrier potential does not depend on diode design while it depends on temperature, doping density and forward biasing.

When pentavalent impurity is added to silicon then $$n$$-type semiconductor is formed. Out of given options, only phosphorus ($$P$$) is pentavalent, so it should be doped to silicon to make it $$n$$-type semiconductor.

Interatomic spacing for a fcc lattice is given by
$$r = {\left[ {{{\left( {\frac{a}{2}} \right)}^2} + {{\left( {\frac{a}{2}} \right)}^2} + {{\left( 0 \right)}^2}} \right]^{\frac{1}{2}}} = \frac{a}{{\sqrt 2 }}$$
where, a being lattice constant.
$$\therefore a = \sqrt 2 r = \sqrt 2 \times 2.54 = 3.59\,\mathop {\text{A}}\limits^ \circ $$
NOTE
Interatomic spacing is just the nearest neighbours distance.