Semiconductors(12th Grade > Physics ) Questions and answers for exam Preparation

Question 1. A sinusoidal voltage of peak value 200 volt is connected to a diode and resistor R in the circuit figure, so that halfwave rectification occurs. If the forward resistance of the diode is negligible compared to R, the RMS voltage (in volt) across R is approximately

200
100
200√2
280

Explanation:-

Answer: Option C. -> 200√2
:
C
If half wave rectifier the output voltage is the RMS voltage

= \frac{V_{o}}{\sqrt{2}} = \frac{200}{\sqrt{2}}

Question 2. In the half wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be

25 Hz
50 Hz
70.7 Hz
100 Hz

Explanation:-

Answer: Option B. -> 50 Hz
:
B
In half wave rectifier, we get the output only in one half cycle of input AC therefore, the frequency of the ripple of the output is same as that of input AC ie, 50 Hz.

Question 3. Consider the junction diode is idea. The value of current in the figure is

zero
10−2 A
10−1 A
10−3 A

Explanation:-

Answer: Option B. -> 10−2 A
:
B

I_{f} = \frac{4 - 1}{300} = \frac{1}{100} = 10^{- 2} A

Question 4. P-type semiconductor are

positively charged
produced when boron is added as an impurity
produced when phosphorus is added as an impurity to silicon
produced when carbon is added as an impurity to germanium.

Explanation:-

Answer: Option B. -> produced when boron is added as an impurity
:
B
Boron has valency three. When boron is doped in a pure semiconductor, then p-type semiconductor is formed.

Question 5. The correct curve between potential (V) and distance (d) near p-n junction is

a
b
c
d

Explanation:-

Answer: Option A. -> a
:
A
V-d curve near the junction will be as shown by curve (a).

Question 6. Ge and Si diodes conduct at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the valve of

V_{0}

changes by

Ge And Si Diodes Conduct At 0.3 V And 0.7 V Respectively. In...

0.2 V
0.4 V
0.6 V
0.8 V

Explanation:-

Answer: Option B. -> 0.4 V
:
B
Consider the case when Ge and Si diodes are connected as show in the given figure.
Equivalent voltage drop across the combination Ge and Si diode = 0.3 V

\Rightarrow

Current

i = \frac{12 - 0.3}{5 k Ω} = 2.34 m A

∴

Out put voltage

V_{0} = R i = 5 k Ω \times 2.34 m A = 11.7 V

Now consider the case when diode connection are reversed. In this case voltage drop across the diode's combination = 0.7 V

\Rightarrow

current

i = \frac{12 - 0.7}{5 k Ω} = 2.26 m A

∴ V_{0} = i R = 2.26 m A \times 5 k Ω = 11.3 V

Hence charge in the value of

v_{0}

=11.7 - 11.3=0.4V

Question 7. In the case of p-n junction diode at a high value of reverse bias voltage, the current rises sharply. The value of reverse bias voltage is known as

cut-off voltage
zener voltage
inverse voltage
critical voltage

Explanation:-

Answer: Option B. -> zener voltage
:
B
In reverse bias of p-n junction when a high voltage is applied, electric breakdown of junction takes place, resulting in a large increase in reverse current. This high voltage applied is called zener voltage.

Question 8. A working transistor with its three legs marked P, Q and R is tested using a multi-meter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multi-meter. Which of the following is true for the transistor?

It is an n-p-n transistor with R as collector
It is an n-p-n transistor with R as base
It is an p-n-p transistor with R as collector
It is an p-n-p transistor with R as emitter
35 mA

Explanation:-

Answer: Option B. -> It is an n-p-n transistor with R as base
:
B
When a multimeter is connected between P and Q there is no conduction between P and Q, then P and Q are of same type of semiconductor, ie, either both are on n-type or of p-type. It means emitter and collector of a transistor are P and Q. When R is connected to negative terminal of multimeter and positive terminal to P or Q, then emitter-base junction will conduct to some extent due to reverse biasing.
The similar is the case of collector – base junction. Thus the transistor is n-p-n transistor with R as base.

Question 9. A transistor has a base current of 1 mA and emitter current 90 mA. The collector current will be

90 mA
1 mA
89 mA
91 mA

Explanation:-

Answer: Option C. -> 89 mA
:
C

I_{c} = I_{e} - I_{b}

= 90 - 1 = 89mA.

Question 10. In an unbiased p-n junction,

potential at p is equal to that at n
potential at p is +ve and that at n is –ve
potential at p is more than that at n
potential at p is less than that at n

Explanation:-

Answer: Option D. -> potential at p is less than that at n
:
D
Option (d)
At the barrier, there are electron concentration on p side.