Semiconductor Electronics: Materials, Devices and Simple Circuits Test 50

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 50

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Weekly Quiz Competition

Question 1
1 / -0

Which of the PN-junction is forward biased?

A

B

C

D
None of these

Solution
Question 2
1 / -0

A PN-junction has a thickness of the order of :

A
$$1\, gm$$

B
$$1\, mm$$

C
$$1\, \mu m$$

D
$$1\, pm$$

Solution

$$P-N$$ Junction or depletion region is the area between $$P-$$ type and $$N-$$ type semiconductors; which is free from charge carriers.
This region between the two arts as a barrier and has a very small width $$\approx\ 10^{-6}m$$
or
$$1\mu\ m$$
Question 3
1 / -0

The output of an AND gate is connected to both the inputs of a NOR gate, then this circuit will act as a

A
OR gate

B
NOR gate

C
AND gate

D
NOT gate

E
NAND gate

Solution

The correct option is E.
Question 4
1 / -0

In a p-type semiconductor, the acceptor level lies

A
near the conduction band

B
half way between conduction and valence bands

C
within conduction band

D
near the valence band

Solution

In a p-type semiconductor, the acceptor level lies near the valence band.
The correct option is D.
Question 5
1 / -0

In the following common emitter circuit if $$\beta = 100, V _ { C E } = 7 V , V _ { B E } = \text { negligible, } R _ { C } = 2 \mathrm { k } \Omega $$ then $$I _ { B }$$ is

A
$$0.01 \mathrm { mA }$$

B
$$0.04 \mathrm { mA }$$

C
$$0.02 \mathrm { mA }$$

D
$$0.03 \mathrm { mA }$$

Solution
Question 6
1 / -0

The equivalent circuit is :

A
NAND gates

B
OR gates

C
AND gates

D
NOR gates

Solution

From the above,
The output of the above circuit is,
$$Y=\overline {A+B}$$
Which is equivalent to $$NOR$$ gate.
So, the correct option is $$(D)$$.
Question 7
1 / -0

When PN junction os forward biased , the current in junction

A
Increase

B
decrease

C
remains same

D
None of these

Solution

Reverse Bias of P-N junction When the voltage is applied this way round it tends to pull the free electrons and holes apart, and increases the height of the energy barrier between the two sides of the diode. ... A few lucky electrons and holes may happen to pick up a lot of thermal (kinetic) energy
Question 8
1 / -0

Energy bands in solids are a consequence of

A
Ohm's Law

B
Pauli's exclusion principle

C
Bohr's theory

D
Heisenberg's uncertainty principle

Solution

According to Pauli's exclusion principle, two electrons cannot have the same quantum numbers in a given molecule.
Therefore, when two identical atoms combine to give rise to a diatomic molecule, each atomic orbital breaks up into two different molecular orbitals( different energy).
Question 9
1 / -0

Find the value of $${ V }_{ 0 }$$

A
5.7 V

B
5.3 V

C
6 V

D
5 V

Solution

From the given diagram,
$$V = 6V$$
$${V_{Ge}} = 0.3V$$
$${V_{Si}} = 0.7V$$
Germanium starts conducting at 0.3 V and silicon starts conducting at 0.7 V. As both the diodes are in forward biased condition, the current will flow through the diode having less potential difference i.e. through Germanium
Therefore, the value $${V_0}$$ will be
$${V_0} = 6 - 0.3 = 5.7V$$
Hence Option (A) is the correct answer
Question 10
1 / -0

For the circuit show below, the current through the zener diode is :

A
5 mA

B
Zero

C
14 mA

D
9 mA

Solution

Assuming zener diode does not undergo breakdown, current in circuit = $$\dfrac{120}{15000} =8mA$$
Voltage drop across diode = 80 V > 50 V. The diode undergo breakdown.
Current is $$R_1 = \dfrac{70}{5000} = 14 mA$$
Current is $$R_2 = \dfrac{50}{10000} =5A$$
current through diode = 9 mA