Semiconductor Electronics: Materials, Devices and Simple Circuits Test 29

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 29

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Weekly Quiz Competition

Question 1
1 / -0

The width of the depleted region of a p-n junction is of the order of a few tenth of a ________.

A
millimeter

B
micrometer

C
metre

D
nanometre

Solution

The width of the depletion layer is about $$10^{-6}$$ to $$10^{-7}$$ m. i.e, few tenths of a micrometer.
Question 2
1 / -0

p-n junction is said to be forward bias. If

A
the n-side is connected to the -ve side and p-side to the +ve terminal of an external battery.

B
the n-side is connected to the +ve side and p-side to the -ve side of an external battery.

C
n-side is connected to +ve terminal and p-side in neutral.

D
none of these

Solution

A forward biased conductor takes place when the $$p$$-type is connected to the $$+ve$$ terminal and $$n$$-type is connected to the $$-ve$$ terminal of the battery.
Question 3
1 / -0

The delpetion layer in a p-n junction diode is $$10^{-6}m$$ wide and its knee potential is $$0.5V$$, then the inner electric field in the depletion region is:

A
$$5\times 10^{6}V/m$$

B
$$5\times 10^{-7}V/m$$

C
$$5\times 10^{5}V/m$$

D
$$5\times 10^{-1}V/m$$

Solution

In both forward biasing and reverse biasing, applied potential establishes an internal electric field which acts against or towards the potential barrier. This internal electric field is weakened or stronger at the junction. In forward biasing knee voltage is the forwards voltage at which the current through the junction starts to increase rapidly. Once the applied forward voltage exceeds the knee voltage, the current starts increasing rapidly. In forward biasing condition, the inner electric field is given by
$$E = -\dfrac {\triangle V}{\triangle r}$$
or $$|E| = \dfrac {\triangle V}{\triangle r} = \dfrac {5\times 10^{-1}}{10^{-6}}$$
$$= 5\times 10^{5}V/m$$
Question 4
1 / -0

The circuit given represents which of logic operations

A
AND

B
NOT

C
OR

D
NOR

Solution

$$Y=\overline{\overline{(A.B)}.\overline{(A.B)}} $$
$$(A,B)$$ $$Y$$
$$(1,1)$$ $$1$$
  $$(1,0)$$ $$ 0$$
  $$(0,1)$$ $$ 0$$
  $$(0,0)$$ $$0$$
Question 5
1 / -0

Which logic gates is represented by the following combination of logic gates?

A
OR

B
NAND

C
AND

D
NOR

Solution

This is a case of AND gate. Input and output are shown below
$$\therefore$$ $$y=\overline { \overline { A } +\overline { B } } =\overline { \overline { A } .\overline { B } } =AB$$ (since $$\overline { A } +\overline { B } =\overline { A } .\overline { B } $$)
Question 6
1 / -0

Identify the logic operation of the following logic circuit:

A
NAND

B
AND

C
NOR

D
OR

Solution

logic equation for the following circuit can be written as,
$$\overline{A.B}=C\longrightarrow \overline{C.C}=\overline{C}=\overline{\overline{A.B}}=A.B=Y=$$ Output
Hence logic operation of the given circuit is equivalent to AND which is A.B
Question 7
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

From input signals, we have
$$A$$ $$B$$ Output NAND gate
$$0$$ $$0$$ $$1$$
$$1$$ $$0$$ $$1$$

$$1$$ $$1$$ $$0$$
$$0$$ $$1$$ $$1$$
The output signal is shown $$B$$
Question 8
1 / -0

The input to the digital circuit are as shown below. The output Y is:

A
$$A+B+\overline{C}$$

B
$$(A+B)\overline{C}$$

C
$$\overline{A}+\overline{B}+\overline{C}$$

D
$$\overline{A}+\overline{B}+C$$

Solution

The output of digital circuit can be given as, $$T=\overline{AB}+\overline{C}$$
It will be same as, $$Y=\overline {A}+\overline{B}+\overline{C}$$
Question 9
1 / -0

As we are growing in technology, use of wires in an electrical circuit is reducing. This is done by using semiconductor chips or reducing length of wire as much as possible. What could be the possible reason for the above?

A
Using chips allows more compact devices

B
Heat loss in wires is reduced

C
Power requirements are reduced

D
All of the above

Solution

Compactness of a device plays a major role in the success of product.
With reduction in surface area, heat loss is tremendously reduced.
Less power requirements makes the application cheap and reducing electricity bill.
Question 10
1 / -0

The part of a transistor which is heavily dopped to produce a large number of majority carriers is:

A
Base

B
Emitter

C
Collector

D
None of these

Solution

The transistor has three regions, namely emitter, base and collector. The base is much thinner than the emitter, while collector is wider than both as shown in figure.
The emitter is heavily doped, so that it can inject a large number of charge carriers (electrons or holes) into the base. The base is lightly doped and very thin, it passes most of the emitter injected charge carriers to the collector. The collector is moderately doped.