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

Forbidden energy gap $$ = \frac{{12400\left( {eV - \mathop {\text{A}}\limits^ \circ } \right)}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}$$
$$ = \frac{{12400}}{{24750}}eV = 0.5\,eV$$

The power gain in case of $$CE$$  amplifier,
$$\eqalign{ & {\text{Power gain}} = {\beta ^2} \times {\text{Resistance gain}} \cr & = {\beta ^2} \times \frac{{{R_o}}}{{{R_i}}} = {\left( {10} \right)^2} \times 5 = 500. \cr} $$

Semiconductor have negative temperature coefficient of resistance i.e. the resistance of a semiconductor decreases with the increase in temperature as no. of electrons increases in conduction bond. Copper has positive temperature coefficient of resistance, i.e. the resistance of copper increases with the increase in temperature due to the collision of electrons. Hence, at $$80\,K,$$  the resistance of copper will decrease while that of germanium will increase.

We can see from the truth table that output is 1 only when its both inputs are 1.
This is possible only for AND gate. The Boolean expression for AND gate is $$Y = A \cdot B$$
which satisfies the truth table as below
$$\eqalign{ & 0 \cdot 0 = 0 \cr & 1 \cdot 0 = 0 \cr & 0 \cdot 1 = 0 \cr & 1 \cdot 1 = 1 \cr} $$
Here symbol $$\left( \cdot \right)$$ represents AND operation.

Consider the given figure,
The resultant boolean expression of the above logic circuit will be
$$Y = (A + B).C$$
Now, let us try with inputs $$A,B$$  and $$C$$ given in the options and lets see, which one of them will give output 1 at $$Y.$$
$$\eqalign{ & {\text{If}}\,\,A = 0,B = 0,C = 0 \cr & \Rightarrow Y = \left( {0 + 0} \right).0 \Rightarrow Y = 0 \cr & {\text{If}}\,\,A = 1,B = 1,C = 0 \cr & \Rightarrow Y = \left( {1 + 1} \right).0 \cr & \Rightarrow Y = 1.0 \Rightarrow Y = 0 \cr & {\text{If}}\,\,A = 1,B = 0,C = 1 \cr & \Rightarrow Y = \left( {1 + 0} \right).1 \Rightarrow Y = 1.1 \Rightarrow Y = 1 \cr & {\text{If}}\,\,A = 0,B = 1,C = 0 \cr & \Rightarrow Y = \left( {0 + 1} \right).0 \cr & \Rightarrow Y = 1.0 \Rightarrow Y = 0 \cr} $$
So, we have seen that among the given options, only option (C) is the correct choice, i.e.
Output $$Y = 1$$  only when inputs $$A = 1,B = 0$$   and $$C = 1.$$

In common emitter configuration for $$n-p-n$$   transistor input and output signals are $${180^ \circ }$$ out of phase i.e., phase difference between output and input voltage is $${180^ \circ }.$$

Forbidden energy gap $$ = \frac{{12400\left( {eV - \mathop {\text{A}}\limits^ \circ } \right)}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}$$
$$ = \frac{{12400}}{{38950}}eV = 0.3\,eV.$$

Here$$y = \overline {\left( {\overline A + \overline B } \right)} = \overline{\overline A} .\overline{\overline B} = A \cdot B.$$       Thus it is an $$AND$$  gate for which truth table is

A	B	y
0	0	0
0	1	0
1	0	0
1	1	1

Current gain in common-base configuration is,
$$\alpha = {\left( {\frac{{\Delta {i_c}}}{{\Delta {i_e}}}} \right)_{{V_{cb}}}}\,\,\left[ {{\text{at constant voltage across}}\,{\text{collector base junction}}} \right]$$
Current gain in common-emitter configuration is,
$$\beta = {\left( {\frac{{\Delta {i_c}}}{{\Delta {i_b}}}} \right)_{{V_{ce}}}}\,\,\left[ {{\text{at constant voltage across}}\,{\text{collector emitter junction}}} \right]$$
$$\eqalign{ & {\text{Also}}\,\,{i_b} = {i_e} - {i_c} \cr & {\text{or}}\,\,\Delta {i_b} = \Delta {i_e} - \Delta {i_c} \cr & {\text{As}}\,\,\beta = \frac{{\Delta {i_c}}}{{\Delta {i_b}}} = \frac{{\Delta {i_c}}}{{\Delta {i_e}}} \times \frac{{\Delta {i_e}}}{{\Delta {i_b}}} \cr & {\text{or}}\,\,\beta = \alpha \times \frac{{\Delta {i_e}}}{{\Delta {i_e} - \Delta {i_c}}} \cr & {\text{or}}\,\,\beta = \alpha \times \frac{1}{{1 - \frac{{\Delta {i_c}}}{{\Delta {i_e}}}}} \cr & {\text{or}}\,\,\beta = \frac{\alpha }{{1 - \alpha }} \cr} $$
Alternative
As we know that relation between current gain in common base and common emitter configuration is given by
$$\eqalign{ & \alpha = \frac{\beta }{{1 + \beta }} \cr & \Rightarrow \alpha + \alpha \beta = \beta \cr & \Rightarrow \beta - \alpha \beta = \alpha \cr & \Rightarrow \beta \left( {1 - \alpha } \right) = \alpha \cr & \Rightarrow \beta = \frac{\alpha }{{1 - \alpha }} \cr} $$
NOTE
$$\beta $$ is always greater than $$\alpha .$$ Also $$\alpha < 1$$  and $$\beta > 1.$$

$${\text{NOR}} + \left( {{\text{NAND}} = {\text{NOT}}} \right) + {\text{NOT}} \Rightarrow {\text{NOR}}.$$