Semiconductor Electronics: Materials, Devices and Simple Circuits Test 30

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 30

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Question 1
1 / -0

The output $$Y$$ of the logic circuit given above is:

A
$$\bar { A } +B$$

B
$$\bar { A } $$

C
$$\overline { \left( \bar { A } +B \right) } \cdot \bar { A } $$

D
$$\overline { \left( \bar { A } +B \right) } \cdot A$$

Solution

Using the Boolean expression, we write the output for each logic gate as shown.
The output of signal $$A$$ after passing through $$NOT$$ gate is $$\bar{A}$$
The output of signals $$\bar{A}$$ and $$B$$ after passing through $$AND$$ gate is $$\bar{A}.B$$
The output of signals $$\bar{A}.B$$ and $$\bar{A}$$ after passing through $$OR$$ gate is $$\bar{A} + \bar{A}.B$$

Thus the output signal of the logic circuit $$Y = \bar{A} + \bar{A}.B$$
$$\therefore$$ $$Y = \bar{A} (1 + B)$$
Using $$1+ B = 1$$ , we get $$Y = \bar{A}.1$$
$$\implies Y = \bar{A}$$
Question 2
1 / -0

The follwoing combined logic gate diagram is eqivalent to

A
NOR gate

B
OR gate

C
AND gate

D
NAND gate

Solution

Its clear from the diagram that
A=1, B=1, Y=0
A=1, B=0, Y=1
A=0, B=1, Y=1
A=0, B=0, Y=1
So this is a NAND gate.
Question 3
1 / -0

Which logic gate produced 'LOW' output when any of the inputs is 'HIGH'?

A
AND

B
OR

C
NAND

D
NOR

Solution
Question 4
1 / -0

If the band-gap between valence band and conduction band in a material is $$5.0\ eV$$, then the material is a/an

A
semiconductor

B
good conductor

C
superconductor

D
insulator

Solution

The band gap in a conductor is Zero.
The band gap in semiconductor is nearly equal to $$1 eV$$
For example : For silicon, band gap is $$1.11eV$$

Insulators generally have very large band gap and thus they do not allow the electrons to jump from valance band to conduction band.
The band gap in an insulator $$\geq 5eV$$
For example : Diamond has a band gap of $$5.5eV$$

Hence the above given material with band gap of $$5.0eV$$ must be an insulator.
Question 5
1 / -0

In a $$p-n$$ junction photocell, the value of photoelectromotive force produced by monochromatic light is proportional to

A
The voltage applied at $$p-n$$ junction

B
The barrier voltage at $$p-n$$ junction

C
The intensity of light falling on cell

D
The frequency of light falling on cell

Solution

When a light ( wavelength sufficient to break the covalent bond ) falls on the junction, new hole electron pairs are created. No. of produced electron hole pair deponed upon no. of photons. So photo emf or current proportional to intensity of light. Hence, in a p−n junction photocell, the value of photoelectromotive force produced by monochromatic light is proportional to the intensity of light falling on cell.
Question 6
1 / -0

How many NAND gates are required to form an AND gate?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Two NAND gates arranged as shown, would give an AND gate.
Here output from first NAND gate is $$\overline{A.B}$$
This, as input, is again fed to a NAND gate, giving
$$\overline{\overline{A.B}.\overline{A.B}}=\overline{\overline{A.B}}+\overline{\overline{A.B}}=A.B$$
Question 7
1 / -0

Which type of gate the following truth table represents?
Input Input Output
$$A$$ $$B$$ $$Q$$
$$0$$ $$0$$ $$1$$
$$0$$ $$1$$ $$1$$
$$1$$ $$0$$ $$1$$
$$1$$ $$1$$ $$0$$

A
NOT

B
AND

C
OR

D
NAND

Solution

$$NAND$$ gate is a logic gate which produces high output $$(1)$$ when either of the input signals is low and produces low output when both the input signals are high.
$$NAND$$ gate is a combination of $$AND$$ gate followed by $$NOT$$ gate.
The above truth table represents $$NAND$$ gate.
Question 8
1 / -0

Of the following NAND gate is :

A

B

C

D

Solution

NAND gate is AND gate followed by NOT gate and is represented as shown in the figure.

Hence, answer is option D.
Question 9
1 / -0

If a rod has resistance 4 ohm and if rod is turned as half circle, then the resistance along diameter is

A
1.56 ohm

B
2.44 ohm

C
4 ohm

D
2 ohm

Solution

$$R=\rho\dfrac{L}{A}$$

Here $$L$$ is calculated along the direction in which the current flow.

And in the circular conductor also , the current will flow along its length,
Here the distance traveled by current is $$L$$ and not the distance between start and end point.

Hence on bending the conductor its resistance will not change.

$$R=4\Omega$$

Answer-(C)
Question 10
1 / -0

In the following circuit the output Y becomes zero for the input combinations:

A
A=1, B=0, C=0

B
A=0, B=1, C=1

C
A=0, B=0, C=0

D
A=1, B=1, C=0

Solution

$$Y=\overline{(A.B).\overline{C}} $$
So, to get Y as 0, the inputs of the final NAND gate should be 1, 1.
This means, $${A.B}$$ = 1 and $$\overline{C}$$ = 1
This is possible, when A = 1, B = 1 and C = 0