Semiconductor Electronics: Materials, Devices and Simple Circuits Test 63

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 63

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Question 1
1 / -0

Find the minimum load resistance which can be used for the zener diode as shown in figure. Given, $$V_Z=10 V, R_Z=0 \Omega, R=450 \Omega, I_Z(min)=2 mA $$ and $$I_Z(max)=60 mA$$

A
$$0 \Omega$$

B
$$333.3 \Omega$$

C
$$31.95 \Omega$$

D
$$319.5 \Omega$$

Solution

Apply Kirchoff's voltage law to loop including battery and zener diode.
$$\implies 25-iR-V_Z=0$$
$$\implies i=33.3mA$$
Thus this current can be separated into zener diode branch and load resistance branch.
Hence $$I_{RL}+I_{Z}=33.3mA$$
$$\implies I_{RL_{max}}=33.3mA-I_{Z_{min}}=33.3mA=31.3mA$$
Hence minimum load resistance which can be used=$$\dfrac{V_{RL}}{I_{RL_{max}}}$$
$$=\dfrac{V_Z}{I_{RL_{max}}}=\dfrac{10}{0.0313}\Omega$$
$$=319.5\Omega$$
Question 2
1 / -0

A NOR gate and a NAND gate are connected as shown in the figure. Two different sets of inputs are given to this setup. In the first case, the inputs to the gates are $$A=0,B=0,C=0$$. In the second case, the inputs are $$A=1,B=0,C=1$$. The output $$D$$ in the first case and second case respectively are:

A
$$0$$ and $$0$$

B
$$0$$ and $$1$$

C
$$1$$ and $$0$$

D
$$1$$ and $$1$$

Solution

In first case $$A=0,B=0$$
$$\therefore$$ Output of NOR gate, $$Y=\overline { A+B } =1$$
This output is the input for NAND gate ie, $$Y=1$$ and $$C=0$$
$$\therefore$$ $$D=\overline { Y.C } =1$$
In second case $$A=1, B=0$$
$$\therefore$$ Output of NOR Gate, $$Y=\overline { A+B } =0$$
This output is the input for NAND gate ie $$Y=0$$ and $$C=1$$
$$\therefore$$ $$D=\overline { Y.C } =1$$
Question 3
1 / -0

A zener diode voltage regulator operated in the range $$120-180\ V$$ produces a constant supply of $$110\ V$$ and $$250\ mA$$ to the load. If the maximum current is to be equally shared between the load and zener, then the values of series resistance ($$\displaystyle { R }_{ S }$$) and load resistance ($$\displaystyle { R }_{ L }$$) are:

A
$$\displaystyle { R }_{ L }=70\ \Omega ;{ R }_{ S }=280\ \Omega $$

B
$$\displaystyle { R }_{ L }=440\ \Omega ;{ R }_{ S }=140\ \Omega $$

C
$$\displaystyle { R }_{ L }=140\ \Omega ;{ R }_{ S }=440\ \Omega $$

D
$$\displaystyle { R }_{ L }=280\ \Omega ;{ R }_{ S }=70\ \Omega $$

Solution

The resistance offered by Zener diode will be: $${ R }_{ Z }=\dfrac { 110 }{ 0.25 } \ \Omega =440\ \Omega $$

Hence, for equal current in load and Zener diode, $${ R }_{ L }={ R }_{ Z }=440\ \Omega $$

Now, total resistance will be: $$\dfrac { 440 }{ 2 } +{ R }_{ S }={ R }_{ S }+220$$

Since current through Zener diode is $$0.25A$$, current through combination will be: $$0.25\ A\times 2=0.5\ A$$.

Hence, for current to be maximum, the maximum voltage can be $$180\ V$$.

Therefore, $$180=0.5({ R }_{ S }+220)$$

$${ R }_{ S }=140\ \Omega $$
Question 4
1 / -0

A 5 V zener diode is used to regulate the voltage across load resistor $$R_L$$ and the input voltage varies in between 10 V to 15 V. The load current also varies from 5 mA to 50 mA. Find the value of series resistance R.
Given, $$I_Z(min)=20 mA$$

A
$$70 \Omega$$

B
$$90.91 \Omega$$

C
$$142.86 \Omega$$

D
$$71.43 \Omega$$

Solution

The zener current $$I_Z$$ will minimum when the load resistance is maximum and the input voltage will also minimum.
Thus, $$I_T=I_Z(min)+I_L(max)$$
By KVL, $$V_{in}(min)-V_Z=I_TR=[I_Z(min)+I_L(max)]R$$
so, $$R=\dfrac{10-5}{(20+50)\times 10^{-3}}=71.43 \Omega$$
Question 5
1 / -0

Find the minimum and maximum load currents for which the zener diode as shown in figure will maintain regulation. Given, $$V_Z=10 V, R_Z=0 \Omega, R=450 \Omega, I_Z(min)=2 mA $$ and $$I_Z(max)=60 mA$$

A
$$0 mA, 33.3 mA$$

B
$$31.3 mA, 0 mA$$

C
$$2 mA, 31.3 mA$$

D
$$31.3 mA, 31.3 mA$$

Solution
Question 6
1 / -0

Which of the following is NOT true about a zener diode?

A
It is used as a voltage regulator.

B
It is forward biased.

C
It is reverse biased.

D
It has a breakdown voltage.

Solution

Zener Diode is a heavily doped p-n junction diode that when in the Breakdown region (i.e when reverse bias is applied to the diode). It acts as a voltage regulator.
Question 7
1 / -0

Consider an open-circuited p-n junction in thermal equilibrium. Let $$J_{p,drift}$$ and $$J_{p,diff}$$ be the hole drift and diffusion current densities, respectively, and let $$J_{n,drift}$$ and $$J_{n,diff}$$ be the corresponding electron current densities. Of the following equalities, mark false,at all points in the p-n junction structure.

A
$$J_{n,drift}$$ + $$J_{p,diff}$$ = -$$J_{n,diff} - J_{p,diff}$$

B
$$J_{n,drift}$$ + $$J_{n,diff}$$ = $$J_{p,drift} + J_{p,diff}$$

C
$$J_{n,drift}$$ = $$J_{n,diff}$$ and $$J_{p,drift} = J_{p,diff}$$

D
$$J_{n,drift}$$ + $$J_{p,diff}$$ = - $$J_{n,drift} - J_{p,diff}$$

Solution

electron diffusion and drift are not equal in pn junction.same holds for holes drift and diffusuion velocities
Question 8
1 / -0

In the circuit diagram given A and B are switches. The logic operation, which the switches can perform, is.....?

A
NAND

B
OR

C
AND

D
None of these

Solution
Question 9
1 / -0

When a semiconductor is heated, its resistance.

A
Remains the same

B
Decreases

C
Increases

D
May increases or decreases depending on the semiconductor

Solution
Question 10
1 / -0

The waveforms A and B given below are given as input to a NAND gate. Then, its logic output $$y$$ is

A
for $$t_1$$ to $$t_2 ; y=0$$

B
for $$t_2$$ to $$t_3 ; y=1$$

C
for $$t_3$$ to $$t_4 ; y=1$$

D
for $$t_4$$ to $$t_5 ; y=0$$

E
for $$t_5$$ to $$t_6 ; y=0$$

Solution

Time A B $$Y=\overline{AB}$$
$$t_1$$ 1 0 1
$$t_2$$ 1 1 0
$$t_3$$ 1 1 0
$$t_4$$ 0 1 1
$$t_5$$ 1 0 1
$$t_6$$ 1 1 0
NAND gate is the combination of AND and NOT gate and the output of the two inputs (A and B) NAND gate is $$Y=\overline{AB}$$
Here, the truth table for the NAND gate will be the following table,
(note 1 for high signal and 0 for low signal)