Semiconductor Electronics: Materials, Devices and Simple Circuits Test 14

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 14

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Question 1
1 / -0

The thickness of the depletion region is of the order of ___________ of a micrometre.

A
one-hundredth

B
one-tenth

C
one-thousandth

D
None of these

Solution

Depletion region is a region near the p-n junction where flow of charge carriers (free electrons and holes) is reduced over a given period and finally results in zero charge carriers.The width of depletion region which is generally $1\mu m$ , depends on the amount of impurities added to the semiconductor. Impurities are the atoms (pentavalent and trivalent atoms) added to the semiconductor to improve its conductivity.
hence answer is $1\mu m$ and correct option is $D$ - none of these.
Question 2
1 / -0

The zone in a semiconductor diode where no free charge carriers exists is known as the:

A
anode region

B
cathode region

C
depletion region

D
None of these

Solution

In depletion region negative charge carrier electrons are attached with their positive charge carrier holes. hence there is no free charge carriers exists.
Question 3
1 / -0

Two important processes that occur during the formation of a p-n junction.

A
Diffusion

B
Drift

C
Both drift and diffusion

D
Drift in p region and diffusion in n region

Solution

$Answer:-$ C
Three important phenomena occurs during formation of pn junction:-
1) Diffusion 2) Formation of space charge 3) Drift
Question 4
1 / -0

In a semiconductor diode, the depletion region is removed when:

A
the diode is in its forward conducting state

B
the diode is in its reverse non-conducting state

C
there is no potential difference between the anode and cathode

D
None of these

Solution

In a semiconductor diode, when diode in its forward conducting state current starts in diode and depletion region starts to deformed. After sometime depletion region is removed.
Question 5
1 / -0

The leakage current across a pn junction is due to _________

A
Minority carriers

B
Majority carriers

C
Junction capacitance

D
None of these

Solution

When a diode is reverse biased, its depletion region increases in width. Thus depletion region would correspond to an insulator. However still some current actually flows through the depletion region. This is due to the pushing of minority carriers across the depletion region causing very small reverse 'leakage current'.
Question 6
1 / -0

The battery connections required to forward bias a pn junction are ____________

A
+ve terminal to p and -ve terminal to n

B
-ve terminal to p and +ve terminal to n

C
-ve terminal to p and -ve terminal to n

D
None of these

Solution

$Answer:-$ A
To forward bias the p-n junction, the p side is made more positive, so that it is "downhill" for electron motion across the junction. An electron can move across the junction and fill a vacancy or "hole" near the junction. It can then move from vacancy to vacancy leftward toward the positive terminal, which could be described as the hole moving right. The conduction direction for electrons in the diagram is right to left, and the upward direction represents increasing electron energy.
Question 7
1 / -0

Which of these is the best description of a Zener diode ?

A
It is a constant voltage device.

B
It operates in the reverse region.

C
It is a constant current device.

D
It works in forward region.

Solution

Zener diode is a p-n junction diode working in the breakdown region. It is used as a voltage regulator/stabilizer to provide a constant voltage from a source whose voltage may fluctuate over a wide range.
Question 8
1 / -0

The inputs to the digital circuit are shown below. The output Y is.

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

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

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

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

Solution

A and B are inputs to the NAND gate. The output of that gate is $\overline{(A.B)}=\overline{A}+\overline{B}$ .
The output of C being input to the NOT gate is $\overline{C}$ .
These two are inputs of OR gate.
Hence $Y=\overline{(A.B)}+\overline{C}$
$=\overline{A}+\overline{B}+\overline{C}$
Question 9
1 / -0

If the bandgap between valence band and conduction band in a material is $0 eV$ , then the material is

A
Semiconductor

B
Good conductor

C
Superconductor

D
Insulator

Solution

In good conductors, the valance band and the conduction band overlap each other resulting in a zero band gap energy. Whereas, bandgap in insulator is nearly $5$ eV and that in semiconductors is nearly $1.1$ eV.
Hence the given material is a good conductor.
Question 10
1 / -0

In an LED device, the intensity of emitted light

A
increases continuously with the forward current of the diode.

B
increases continuously with the forward current of the diode, reaches a maximum and then decreases.

C
decreases continuously with the forward current of the diode.

D
decreases continuously with the forward current of the diode, reaches a minimum and then increases.

Solution

The intensity of emitted light increase with the foreword current more the light intensity.

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 14

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 14

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SHARING IS CARING

Question 1

one-hundredth

one-tenth

one-thousandth

None of these

Solution

Question 2

anode region

cathode region

depletion region

None of these

Solution

Question 3

Diffusion

Drift

Both drift and diffusion

Drift in p region and diffusion in n region

Solution

Question 4

the diode is in its forward conducting state

the diode is in its reverse non-conducting state

there is no potential difference between the anode and cathode

None of these

Solution

Question 5

Minority carriers

Majority carriers

Junction capacitance

None of these

Solution

Question 6

+ve terminal to p and -ve terminal to n

-ve terminal to p and +ve terminal to n

-ve terminal to p and -ve terminal to n

None of these

Solution

Question 7

It is a constant voltage device.

It operates in the reverse region.

It is a constant current device.

It works in forward region.

Solution

Question 8

A+B+CˉA+B+\bar{C}A+B+Cˉ

(A+B)Cˉ(A+B)\bar{C}(A+B)Cˉ

Aˉ+Bˉ+Cˉ\bar{A}+\bar{B}+\bar{C}Aˉ+Bˉ+Cˉ

Aˉ+Bˉ+C\bar{A}+\bar{B}+{C}Aˉ+Bˉ+C

Solution

Question 9

Semiconductor

Good conductor

Superconductor

Insulator

Solution

Question 10

increases continuously with the forward current of the diode.

increases continuously with the forward current of the diode, reaches a maximum and then decreases.

decreases continuously with the forward current of the diode.

decreases continuously with the forward current of the diode, reaches a minimum and then increases.

Solution

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$A+B+\bar{C}$

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

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

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