Semiconductor Electronics: Materials, Devices and Simple Circuits Test 54

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 54

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Weekly Quiz Competition

Question 1
1 / -0

Determine the current through zener diode for the circuit shown in figure is: (Given: zener diode break down voltage $$V_z=5.6V$$)

A
$$7mA$$

B
$$17mA$$

C
$$10mA$$

D
$$15mA$$

Solution

For zener brak down potential difference across $$800\Omega $$ resistor will be $$5.6V$$4
$$V_z=5.6V$$
$$i_2=\dfrac{V_z}{800}=\dfrac{5.6}{800}=7mA$$
$$\Delta V$$ across $$200\Omega =9-5.6=3.4V$$
$$i_1=\dfrac{3.4}{200}=17mA$$
$$i_1=1_2+i_z$$
$$i_z=17mA-7mA=10mA$$
Question 2
1 / -0

The output of the given combination of gates is equivalent to:

A
NAND

B
OR

C
AND

D
NOR

Solution

$$\overline{\bar{A} \cdot \bar{B}} = A + B$$
Question 3
1 / -0

A device or power strip designed to protect electronic equipment from power surges and spikes is known as.....

A
Surge Suppressors

B
Ping

C
Gateway

D
None of these

Solution
Question 4
1 / -0

Logic gate for inpute A an B is given in figure .
Which table is correct for given gates system:

A

B

C

D

Solution

Output of OR gate $$C=A+B$$
$$y=(A+B).A$$
Question 5
1 / -0

The term 'Pentium' is related to what?

A
Mouse

B
Hard Disk

C
Microprocessor

D
DVD
Question 6
1 / -0

The logic circuit in the figure represents characteristics of which logic gate?

A
NOR

B
OR

C
NAND

D
NOT

Solution

Input Output
$$0$$ $$1$$
$$1$$ $$0$$
Question 7
1 / -0

The reverse breakdown voltage of a Zener diode is $$5.6\ V$$ in the given circuit.
The current $$I_{Z}$$ through the Zener is

A
$$7\ mA$$

B
$$17\ mA$$

C
$$10\ mA$$

D
$$15\ mA$$

Solution

$$9 = V_{Z} + V_{R_{1}}$$
$$V_{Z} = 5.6\ V$$
$$V_{R_{1}} = 9 - 5.6$$
$$V_{R_{1}} = 3.4$$
$$I_{R_{1}} = \dfrac {V_{R_{1}}}{R} = \dfrac {3.4}{200}$$
$$I_{R_{1}} = 17\ mA$$
$$V_{Z} = V_{R_{2}} = I_{R_{2}} (R_{2})$$
$$\dfrac {5.6}{800} = I_{R_{2}}$$
$$I_{R_{2}} = 7\ mA$$
$$I_{Z} = (17 - 7) mA$$
$$= 10\ mA$$.
Question 8
1 / -0

The truth table for the circuit given in the fig. is:

A
$$\begin{vmatrix} A&B&Y \\0&0&1\\0&1&1\\1&0&0\\1&1&0 \end{vmatrix}$$

B
$$\begin{vmatrix} A&B&Y \\0&0&0\\0&1&0\\1&0&1\\1&1&1 \end{vmatrix}$$

C
$$\begin{vmatrix} A&B&Y \\0&0&1\\0&1&0\\1&0&0\\1&1&0 \end{vmatrix}$$

D
$$\begin{vmatrix} A&B&Y \\0&0&1\\0&1&1\\1&0&1\\1&1&1 \end{vmatrix}$$

Solution

$$C = A + B$$
and $$y = \overline{A.C}$$
$$A$$ $$B$$ $$C = (A + B)$$ $$A.C$$. $$y = \overline{A.C}$$
0 0 0 0 1
0 1 1 0 1
1 0 1 1 0
1 1 1 1 0
Question 9
1 / -0

For the given circuit shown in figure, the potential of the battery is varied from $$10V$$ to $$16V$$. If by zener diode breakdown voltage is $$6V$$, find maximum current through zener diode.

A
$$1.5\ mA$$

B
$$3.5\ mA$$

C
$$5\ mA$$

D
$$6.5\ mA$$

Solution

$$i_{i} = \dfrac {6}{4} = 1.5\ mA$$
Maximum current will be obtained for battery voltage $$16\ V$$
$$i = \dfrac {16 - 6}{2} = 5mA\ i_{z}(max) = 5 - 1.5 = 3.5\ mA$$.
Question 10
1 / -0

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V and the load resistance is $$R_L = 4 k \Omega$$. The series resistance of the circuit is $$R_i = 1 k \Omega$$. If the battery voltage $$V_B$$ varies from $$8V$$ to $$16V$$, what are the minimum and maximum values of the current through Zener diode ?

A
$$0.5 \, mA ; \, 6 \, mA$$

B
$$0.5 \, mA ; \, 8.5 \, mA$$

C
$$1.5 \, mA ; \, 8.5 \, mA$$

D
$$1 \, mA ; \, 8.5 \, mA$$

Solution

At $$V_B = 8V$$
$$i_L = \dfrac{6 \times 10^{-3}}{4} = 1.5 \times 10^{-3} A$$
$$i_R = \dfrac{8 - 6 \times 10^{-3}}{1} = 2 \times 10^{-3} A$$
$$\therefore i_{zener \, diode} = i_R - i_{load}$$
$$= 0.5 \times 10^{-3} A$$
At $$V_B = 16 V$$
$$i_L = 1.5 \times 10^{-3} A$$
$$i_R = \dfrac{(16 - 6) \times 10^{-3}}{1} = 10 \times 10^{-3} A$$
$$\therefore i_{zener \, diode} = i_R - i_L$$
$$= 8.5 \times 10^{-3} A$$