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

The final boolean expression is,
$$X = \overline {\left( {\bar A.\bar B} \right)} = \overline {\bar A} + \overline {\bar B} = A + B \Rightarrow {\text{OR}}\,{\text{gate}}$$

$$\eqalign{ & {\text{Voltage gain}} \cr & {A_v} = \beta \frac{{{R_C}}}{{{R_B}}} = 60 \times \frac{{5 \times {{10}^3}}}{{2 \times {{10}^3}}} = 150 \cr & {\text{Power gain}} = {\beta ^2}\frac{{{R_C}}}{{{R_B}}} = 60 \times 60 \times 2.5 = 9000 \cr & {\text{Base current}} = {I_B} = \frac{{12 \times {{10}^{ - 3}}}}{{2 \times {{10}^{ - 3}}}} = 6 \times {10^{ - 6}}A \cr & {I_C} = \beta {I_B} = 60 \times 6 \times {10^{ - 6}} = 3.6 \times {10^{ - 4}}A \cr & {\text{Output}} = {I_C}{R_L} = 3.6 \times {10^{ - 4}} \times 5 \times {10^3}\Omega = 1.8\,V \cr} $$

For ‘fcc’ structure, $$r = \frac{a}{{2\sqrt 2 }}$$
where $$r =$$  radius of the atom in the packing
$$a =$$  edge length of the cube
Here, $$a = 3.6\,\mathop {\text{A}}\limits^ \circ $$
$$\therefore r = \frac{{3.6}}{{2 \times 1.4}} = 1.27\,\mathop {\text{A}}\limits^ \circ $$

When a battery is connected to junction diode with $$p$$-side connected to negative terminal or lower potential and $$n$$-side to the positive terminal or higher potential, the junction diode is reverse biased.
The circuit shown in figure can be redrawn as

In the option (C), the $$p$$-end of the diode is connected to negative terminal of the battery, so the diode has been reverse biased.

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

OR gate

By expanding this Boolen expression
$$\Upsilon = A.\overline B + B.\overline A $$
Thus the truth table for this expression should be (A).

Truth table of the gate is as follows :

A	B	C	A+B	$$Y = \left( {A + B} \right).C$$	$${\bar Y}$$
0	0	0	0	0	1
1	1	0	1	0	1

In a $$p$$ - $$n$$ junction diode, majority carriers are holes on $$p$$-side and electrons on $$n$$-side. Holes, thus diffuse to $$n$$-side and electrons to $$p$$-side. This diffusion causes a layer of positive charge in the $$n$$-region and layer of negative charge in the $$p$$-region near the junction.
This double layer of opposite charge creates an electric field which exerts a force on the electrons and holes, against their diffusion. This electric field becomes strong enough as diffusion proceeds to stop it. In the equilibrium position, there is a barrier, for charge motion with the $$n$$-side at a higher potential than the $$p$$-side.
The junction region has a very low density of either $$p$$ or $$n$$-type carriers, because of inter diffusion. It is called depletion region.

It is the circuit of full wave rectifier.

At $$0\,K,$$  the motion of electrons cease and so electric current becomes zero.