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

$${D_2}$$ is forward biased whereas $${D_1}$$ is reversed biased.
So effective resistance of the circuit
$$\eqalign{ & R = 8 + 4 = 12\Omega \cr & \therefore i = \frac{{10}}{{12}} = 0.8\,A \cr} $$

Graph (a) is for a simple diode.
Graph (b) is showing the $$V$$ Break down used for zener diode.
Graph (c) is for solar cell which shows cut-off voltage and open circuit current.
Graph (d) shows the variation of resistance $$h$$ and hence current with intensity of light.

A $$p$$ - $$n$$ junction is said to be reverse biased, if the positive terminal of the external battery $$B$$ is connected to $$n$$-side and the negative terminal to $$p$$-side of the $$p$$ - $$n$$ junction.
In reverse biasing, the reverse bias voltage supports the potential barrier $${V_B}.$$  Now the majority carriers are pulled away from the junction and the depletion region becomes thick.
There is no conduction across the junction due to majority carriers. However, a few minority carriers (holes in $$n$$-section and electrons in $$p$$-section) of $$p$$ - $$n$$ junction diode cross the junction after being accelerated by high reverse bias voltage. They constitute a current that flows in the opposite direction.

So when $$p$$ - $$n$$ junction is reverse biased, the flow of current is due to drifting of minority charge carries across the junction.

$$E = \frac{V}{d} = \frac{{0.1}}{{{{10}^{ - 6}}}} = {10^5}\,V/m$$

If $$M$$ is the molar mass and $$\rho $$ is the density then volume of one mole
$$V = \frac{M}{\rho }.$$
The number of atoms per unit volume
$$\eqalign{ & = \frac{{{N_A}}}{V} = \frac{{{N_A}}}{{\frac{M}{\rho }}} = \frac{{{N_A}\rho }}{M} = \frac{{\left( {6.02 \times {{10}^{23}}} \right) \times \left( {8.96} \right)}}{{63.54}} \cr & = 8.49 \times {10^{22}}c{m^{ - 3}} \cr & = 8.49 \times {10^{28}}{m^{ - 3}} \cr} $$
As each copper (monovalent) atom has one electron, so number of electrons per unit volume
$$ = 8.49 \times {10^{26}}{m^{ - 3}}$$

For forward biasing of $$p$$ - $$n$$ junction, $$p$$-side should be at higher potential than $$n$$ -side. Now we apply this rule to the four options.
(A) Here, $$p$$-side is at lower potential $$\left( {0\,V} \right)$$  and $$n$$ -side at higher potential $$\left( {2\,V} \right).$$  So, this diode is not forward biased.
(B) Here, $$p$$ -side is at higher potential $$\left( {0\,V} \right)$$  and $$n$$-side at lower potential $$\left( { - 2\,V} \right).$$  So, this diode is forward biased.
(C) Here, $$p$$ -side is at lower potential $$\left( { - 2\,V} \right)$$  and $$n$$-side at higher potential $$\left( {0\,V} \right).$$  So, this diode is not forward biased.
(D) Here $$p$$ -side is at lower potential $$\left( {2\,V} \right)$$  and $$n$$ -side at higher potential $$\left( {5\,V} \right).$$  So, this is not forward biased.
Hence, choice (B) is correct.

Truth table for given network is

A	B	$${Y_1}$$	$${Y_2}$$	Y
0	0	1	0	1
1	0	0	1	0
0	1	0	1	0
1	1	0	1	0

Output $$Y$$ of network matches with that of NOR gate.

In a junction diode, when electron jumps to conduction band from valence band due to thermal agitation or any other circumstances, then a vacancy is created in valence band which has positive charge equal to charge of electron in magnitude. This is called a hole.
Thus, holes in junction diode are due to missing electrons.

When a small amount of pentavalent impurity is added to a pure semiconductor, it is known as $$n$$-type semiconductor. Arsenic (33) is pentavalent impurity. The addition of pentavalent impurity provides a large number of free electrons in the semiconductor crystal.

$$\lambda = \frac{{hc}}{E} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.75 \times 1.6 \times {{10}^{ - 19}}}} = 16500\,\mathop {\text{A}}\limits^ \circ $$