Semiconductor Electronics: Materials, Devices and Simple Circuits Test 27

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 27

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Weekly Quiz Competition

Question 1
1 / -0

The cause of the potential barrier in a p-n diode is

A
depletion of positive charges near the junction

B
concentration of positive charges near the junction

C
depletion of negative charges near the junction

D
concentration of positive and negative charges near the junction

Solution

During the formation of a junction diode, holes from p- region diffuse into n-region and electrons from n-region diffuse into p-region. In both cases, when an electrons meets a hole, they cancel the effect at each other and as a result, a thin layer at the junction becomes free from any of charges carriers. This is called depletion layer. There is a potential gradient in the depletion layer, negative on the p-side, and positive on the n-side. The potential difference thus developed across the junction is called potential barrier.
Question 2
1 / -0

To which logic gate does the truth table given below correspond?
A B Y
0 0 0
1 0 0
0 0 0
1 1 1

A
OR gate

B
AND gate

C
NAND gate

D
NOR gate

Solution

For this truth table, Boolean expression is $Y = A.B$ which is for AND gate
Question 3
1 / -0

Which of the following represents NAND gate ?

A

B

C

D

Solution
Question 4
1 / -0

What is the value of $A. \bar{A}$ in Boolean algebra?

A
$0$

B
$1$

C
$A$

D
$\bar{A}$

Solution
Question 5
1 / -0

To which logic gate does the truth table given below correspond?
A B Y
0 0 1
1 0 1
0 1 1
1 1 0

A
OR gate

B
AND gate

C
NAND gate

D
NOR gate

Solution

For this truth table, Boolean expression is $Y = \overline{A.B}$ which is for NAND gate
Question 6
1 / -0

If the ratio of the concentration of electrons to that of holes in a semiconductor is $\displaystyle \frac{7}{5}$ and the ratio of current is $\displaystyle \frac{7}{4}$ , then what is the ratio of their drift velocities?

A
$\displaystyle \frac{5}{8}$

B
$\displaystyle \frac{4}{5}$

C
$\displaystyle \frac{5}{4}$

D
$\displaystyle \frac{4}{7}$

Solution

$\displaystyle \dfrac{I_e}{I_h} = \dfrac{n_eeAv_e}{n_heAv_h} \Rightarrow \frac{7}{4} = \dfrac{7}{5} \times \dfrac{V_e}{V_h}$

$\displaystyle \Rightarrow \dfrac{V_e}{V_h} = \dfrac{5}{4}$
Question 7
1 / -0

Which logic gate is represented by following diagram?

A
XOR

B
NOR

C
OR

D
AND

Solution

This represents AND gate. Output Y=AB
Question 8
1 / -0

Directions For Questions

A student performs an experiment for drawing the static characteristic curve of a triode value in the laboratory. The following data were obtained from the linear portion of the curves:
Grid voltage V $_g$ (volt) $-2.0$ $-3.5$ $-2.0$
Plate voltage V $_p$ (volt) $180$ $180$ $120$
Plate current I $_p$ (mA) $15$ $7$ $10$

...view full instructions

Calculate the plate resistance r $_p$ of the triode valve?

A
$0.12$ $\times$ $10$ $^4$ ohm

B
$1.2$ $\times$ $10$ $^4$ ohm

C
$1.3$ $\times$ $10$ $^4$ ohm

D
$1.4$ $\times$ $10$ $^4$ ohm

Solution

$\displaystyle r_p = \dfrac{\Delta V_p}{\Delta I_p} = \dfrac{(180-120)}{(15-10) \times 10^{-3}} = 1.2 \times 10^{-4}$ ohm.
Question 9
1 / -0

Diodes are used to ___________.

A
convert DC into AC

B
convert AC into DC

C
resistance control

D
current regulation

Solution

Diodes are used as a rectifier. A rectifier is a device made from heavily doped p-n junction diode that is used to convert a.c to d.c.
Question 10
1 / -0

n-type silicon is obtained by:

A
doping with tetravalent element

B
doping with pentavalent element

C
doping with trivalent element

D
doping with a mixture of trivalent and tetravalent current

Solution

Doping the intrinsic semiconductor with a pentavalent impurity produces n-type semiconductor. This is because for each pentavalent dopant (or donor impurity) added, an electron is produced in the intrinsic semiconductor. As a result, electrons are more than that of holes in such doped semiconductor. Since electrons are majority charge carriers, it is an n-type semiconductor.