Semiconductor Electronics: Materials, Devices and Simple Circuits Test 24

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Statement -1: NAND and NOR gates are called digital building blocks.
Statement-2: The repeated use of NAND (or NOR) gates can produce all the basis or complicated gates.

A
Statement -1 is false, Statement -2 is true

B
Statement 1-is true, Statement -2 is true and Statement -2 is correct explanation of Statement-1

C
Statement 1-is true, Statement -2 is true and statement -2 is not correct explanation of statement-1

D
Statement 1-is true, Statement -2 is false

Solution

NAND or NOR gates are called universal gates or digital building blocks. Also, the repeated use of NAND (or NOR) gates can produce all the basis or any complicated gates can be formed using NAND (or NOR) gates. Hence, statement 1-is true statement -2 is true statement -2 is correct explanation of statement-1.
Question 2
1 / -0

Carbon, silicon and germanium have four valence electrons each. At room temperature the appropriate statement is

A
The number of free electrons for conduction is significant only in Si and Ge but small in C

B
The number of free conduction electrons is significant in C but small Si and Ge.

C
The number of free conduction electrons is negligibly small in all the three.

D
The number of free conduction electrons is significant in all the three.

Solution

Option A is correct.
In Si and Ge at room temperature($$300K$$) ; the energy band gap is low as the result of which electrons in the covalent bonds gain kinetic energy, they break the bond and move to conduction band. A hole is also created in the valence band. So the number of free electrons for conduction is significant in Si and Ge. The energy band gap in case of carbon is high as the result of which there are not significant number of electrons in the conduction band even at room temperature.
Question 3
1 / -0

Carbon, Silicon, and Germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy band gaps represented by $$(Eg)_{C}, (Eg)_{Si}$$, and $$ (Eg)_{Ge}$$ respectively. Which one of the following relationships is true in their case:

A
$$(Eg)_{C} < (Eg)_{Ge}$$

B
$$(Eg)_{C} > (Eg)_{Si}$$

C
$$(Eg)_{C} = (Eg)_{Si}$$

D
$$(Eg)_{C} < (Eg)_{Si}$$

Solution

The energy band gap in maximum for carbon , less for silicon and learnt for germanium
(Eg)c>(Eg)Si>(Eg)
Hence the correct option is B
Question 4
1 / -0

In a semiconductor, which of the following statement is correct?

A
At $$0 K$$, Si is a super conductor.

B
In a p-type semiconductor the acceptor level lies near the conduction band.

C
Each donor atom contributes one hole.

D
p - n junction is electrically neutral.

Solution

Generally Si is a semiconductor. In p type semiconductor acceptor level lies near the valence band and each donor atom contribute one electron. p-n junction electrically neutral because total charge is zero.
Question 5
1 / -0

A zener diode, having breakdown voltage equal to $$15 V,$$ is used in a voltage regulator circuit shown in figure. The current through the diode is :

A
$$5 mA$$

B
$$10 mA$$

C
$$15 mA$$

D
$$20 mA$$

Solution

Break down Voltage$$=15V$$
To find, current through the diode$$?$$
For $$1k\Omega,$$
$$i_1=\cfrac{V}{R}=\cfrac{15V}{1K\Omega}=\cfrac{15}{1\times 10^{3}}=15 mA$$
For $$250 \Omega$$
$$i=\cfrac{V}{R}=\cfrac{20-15}{250}=\cfrac{5}{250}=\cfrac{1}{50}A$$
$$\left(\cfrac{1}{50}\times 1000\right)mA=20 mA$$
$$\therefore i_2=$$current through the diode$$=i-i_1$$
$$=\left(20-15\right) mA$$
$$= 5 mA$$
Question 6
1 / -0

A pn photo diode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength :

A
$$496A^o$$

B
$$6000A^o$$

C
4000 nm

D
6000 nm

Solution

$$\lambda_{max}=\frac{hc}{eV}$$

$$=\frac{1242eVA^o}{2.5eV}=4960A^o$$
Question 7
1 / -0

When a P - N junction is unbiased, the junction current at equilibrium is :

A
zero as no charges cross the junctions

B
zero as equal number of charge carriers cross the barrier in opposite directions

C
mainly due to diffusion of majority charge carriers

D
mainly due to diffusion of minority charge carriers

Solution

When a diode is connected in a Zero Bias condition, no external potential energy is applied to the PN junction. However if the diodes terminals are shorted together, a few holes (majority carriers) in the P-type material with enough energy to overcome the potential barrier will move across the junction against this barrier potential. This is known as the Forward Current.
Likewise, holes generated in the N-type material (minority carriers), find this situation favourable and move across the junction in the opposite direction. This is known as the Reverse Current and is referenced as IR. This transfer of electrons and holes back and forth across the PN junction is known as diffusion.
Then an Equilibrium or balance will be established when the majority carriers are equal and both moving in opposite directions so that the net result is zero current flowing in the circuit. When this occurs the junction is said to be in a state of Dynamic Equilibrium.
Question 8
1 / -0

Following diagram performs the logic function of :

A
AND gate

B
NAND gate

C
OR gate

D
XOR gate

Solution
Question 9
1 / -0

Following truth table represent which logic gate :
A B C
1 1 0
0 1 1
1 0 1
0 0 1

A
XOR

B
NOT

C
NAND

D
AND

Solution

The given logic gate represents NAND gate because the results are the complement of the AND gate .
Question 10
1 / -0

Absence of electron in a valence band of semiconductor is called

A
Electron

B
Positron

C
Holes

D
Valence

Solution