Semiconductor Electronics: Materials, Devices and Simple Circuits Test 62

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 62

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Weekly Quiz Competition

Question 1
1 / -0

For the given combination of gates, if the logic states of inputs A, B, C are as follows $$A = B = C = 0$$ and $$A= B = 1, C = 0$$ then the logic states of output D are

A
$$0, 0$$

B
$$0, 1$$

C
$$1, 0$$

D
$$1, 1$$

Solution

The output D for the given combination:
$$D=\bar{(A+B).C}=\bar{(A+B )}+C$$
If $$A = B = C = 0$$ then $$D = (\bar{0+0})+\bar{0}=\bar{0}+\bar{0}=1+1=1$$
if $$ = B = 1, C = 0$$ then $$D = (\bar{1+1})+\bar{0}=\bar{1}+\bar{0}=0+1=1$$
Question 2
1 / -0

Which logic gate is represented by the following combination of logic gates:

A
OR

B
NAND

C
AND

D
NOR

Solution

$$Y=\overline {\overline {A}+\overline {B}}$$
According to $$De$$ morgan's theorem
$$Y=\overline {\overline {A}+\overline {B}}=\overline {\bar A . \bar B}=A.B$$
Question 3
1 / -0

Pure silicon crystal of length $$l (0.1m)$$ and area $$A$$($$10^{-4}m^2$$) has the mobility of electrons ($$\mu_e$$) and holes ($$\mu_h$$) as $$0.135 m^2/Vs$$ and $$0.048 m^2/Vs$$ , respectively. If the voltage applied across it is $$2V$$ and the intrinsic charge concentration is $$n_i = 1.5 \times 10^6 m^{-3}$$, then the total current flowing through the crystal is

A
$$8.78\times 10^{-17}A$$

B
$$6.25\times 10^{-17}A$$

C
$$7.89\times 10^{-17}A$$

D
$$2.456\times 10^{-17}A$$

Solution

Given:
$$l = 0.1m$$
$$A = 10^{-4} m^{-2}$$
$$\mu_e = 0.135 m^2 (Vs)^{-1}$$
$$\mu_h = 0.048 m^2 (Vs)^{-1}$$
$$n_i = 1.5 \times 10^{6} m^{-3}$$
$$E = 2 V$$

The conductivity is:
$$\sigma = n_i(\mu_e + \mu_h) e$$
$$\sigma = 1.5 \times 10^{6} (0.135 + 0.048) 1.6 \times 10^{-19}$$
$$\sigma = 1.5 (0.183) 1.6 \times 10^{-13}$$
$$\sigma = 0.4392 \times 10^{-13} Sm^{-1}$$

Now, the resistance is given by,
$$R = \dfrac{l}{\sigma (A)}$$
$$R = \dfrac{0.1}{0.4392 \times 10^{-13} (10^{-4})}$$
$$R = 0.2276 \times 10^{17} \Omega$$

Hence, the current is:
$$I = \dfrac{V}{R} = \dfrac{2}{0.2276 \times 10^{17}}$$
$$I = 8.787 \times 10^{-17} A$$
Question 4
1 / -0

The real time variation of input signals A and B are as shown below. If the inputs are fed into NAND gate, then select the output signal from the following

A

B

C

D

Solution

As the given symbol stands for $$NAND$$ gate and its truth table is shown above.
Hence the output signal shown by option B is the correct answer because it satisfies the truth table.
Question 5
1 / -0

Directions For Questions

Doping changes the fermi energy of a semiconductor. Consider silicon, with a gap of $$1.11 eV$$ between the top of the valence bond and the bottom of the conduction band. At $$300K$$ the Fermi level of the pure material is nearly at the midpoint of the gap. Suppose that silicon is doped with donor atoms, each of which has a state $$0.15 eV$$ below the bottom of the silicon conduction band, and suppose further that doping raises the Fermi level to 0.11 eV, below the bottom of that band.

...view full instructions

For both pure and doped silicon, calculate the probability that a state at the bottom of the silicon conduction band is occupied?

A
5.20 $$\times$$ 10$$^{-2}$$

B
1.40 $$\times$$ 10$$^{-2}$$

C
10.5 $$\times$$ 10$$^{-2}$$

D
14 $$\times$$ 10$$^{-2}$$

Solution

The probability that a state with energy E is occupied is given by
$$\displaystyle P(E) = \frac{1}{e^{(E-E_F)/K_T+1}}$$, where $$E_F$$ is the Fermi energy, T is the temperature on the Kelvin scale, and K is the Boltzmann constant. If energies are measured from the top of the valence band, then the energy associated with a state at the bottom of the conduction band is E = 1.11 eV.
Furthermore, KT = (8.62$$\times$$ 10$$^{-5}$$ eV/K) (300K) = 0.02586 eV. For pure silicon, $$E_F$$= 0.555 eV and (E -$$E_F$$)/kT =(0.555eV) / (0.02586eV) = 21.46. Thus, $$\displaystyle P(E) = \frac{1}{e^{21.46}+1} = 4.79 \times 10^{-10}$$
For the doped semi-conductor, (E -EF)/ = (0.11 eV)/ (0.02586 eV) = 4.254
and $$\displaystyle P(E) = \frac{1}{e^{4.254}+1} = 1.40 \times 10^{-2}$$
Question 6
1 / -0

Directions For Questions

Doping changes the fermi energy of a semiconductor. Consider silicon, with a gap of $$1.11 eV$$ between the top of the valence bond and the bottom of the conduction band. At $$300K$$ the Fermi level of the pure material is nearly at the midpoint of the gap. Suppose that silicon is doped with donor atoms, each of which has a state $$0.15 eV$$ below the bottom of the silicon conduction band, and suppose further that doping raises the Fermi level to 0.11 eV, below the bottom of that band.

...view full instructions

Calculate the probability that a donor state in the doped material is occupied?

A
0.824

B
0.08

C
0.008

D
8.2

Solution

The energy of the donor state, relative to the top of the valence bond, is 1.11 eV - 0.15 eV= 0.96 eV. The Fermi energy is 1.11 eV- 0.11 eV= 1.00 eV. Hence, (E- $$E_F$$)/kT = (0.96eV -1.00eV)/(0.02586eV) = -1.547 and $$\displaystyle p(E) = \frac{1} {e^{-1.547}+1} = 0.824$$
Question 7
1 / -0

The following circut represents

A
OR gate

B
XOR gate

C
AND gate

D
NAND gate

Solution

Output of upper AND gate =$$\bar{A}B$$
Output of lower AND gate = $$A\bar{B}$$
$$\therefore$$ Output of OR gate, $$Y = A\bar{B} + B\bar{A}$$
This is boolean expression for XOR gate.
Question 8
1 / -0

The following configuration of gate is equivalent to

A
NAND gate

B
XOR gate

C
OR gate

D
NOR gate

Solution

$$Y_1 = A + B, Y_2 = \overline{A.B}$$
$$Y = (A+B)\cdot \overline{AB} = A\cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$$
$$ = 0 + A \cdot \bar{B} + B \cdot \bar{A} + 0 = A \cdot \bar{B} + B \cdot \bar{A}$$
This expression is for XOR
Question 9
1 / -0

The diagram of a logic circuit is given below. The output $$F$$ of the circuit is represented by

A
$$W. (X +Y)$$

B
$$W. (X .Y)$$

C
$$W + (X.Y)$$

D
$$W + (X +Y)$$

Solution

Hint:
By law of distribution of Boolean Algebra, we have
$$A + (B.C) = (A+B).(A+C)$$

Step 1: Write outputs for the OR gates with inputs W and X, and W and Y.
Input for first OR gate are $$W$$ and $$X$$. The output for this OR gate is,
$$Y_1 = W + X$$
Input for second OR gate are $$W$$ and $$Y$$. The output for this OR gate is,
$$Y_2 = W + Y$$

Step 2: Write output for AND gate whose inputs are the output of OR gate.
Output of AND gate, $$F$$, whose inputs are $$Y_1$$ and $$Y_2$$ is given as,
$$ F = Y_1 . Y_2$$
$$\Rightarrow F = (W+X).(W+Y)$$
By using Law of distribution of Boolean Algebra, $$A + (B.C) = (A+B).(A+C)$$

Therefore,
$$F = W + (X.Y)$$

Therefore, output $$F$$ for the given circuit is given by, $$F = W + (X.Y)$$
Option C is correct
Question 10
1 / -0

The equivalent resistance of the network shown across the point A and O is R and the resistance of each branch of the octagon is r. Find the value of $$\displaystyle \frac { 210R }{ 47r } $$

A
5

B
3

C
2

D
4

Solution

see image 1
$$\displaystyle { V }_{ C }={ V }_{ C }^{ ' },\quad { V }_{ D }={ V }_{ D }^{ ' }$$
On folding (see image 2)
$$\displaystyle R=\frac { 47 }{ 105 } r$$
So, $$\displaystyle \frac { 210 }{ 47 } \frac { R }{ r } =2$$