CBSE Class 12 Physics Chapter-14 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 14 – Semiconductor Electronic: Material, Devices And Simple Circuits 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter 14 – Semiconductor Electronic: Material, Devices And Simple Circuits
CBSE Class 12 Physics Important Questions Chapter 14 – Semiconductor Electronic: Material, Devices And Simple Circuits
3 Marks Questions
1. What is an ideal diode? Draw the output wave form across the load resistor R, if the
input waveform is as shown in the figure.
Ans.An ideal diode has zero resistance when forward biased and an infinite resistance when it is reverse biased. Output wave from is
2. With the help of a labeled circuit diagram, explain full wave rectification using junction diode. Draw input and output wave forms?
Ans.Full wave rectifier consists of two diodes and a transformer with central tap. For any half cycle of a.c. input only one diode is forward biased where as the other one is reverse biased.
Suppose for positive half of a.c. input diode D1 is forward biased and D2 is reverse biased, then the current will flow across D1 where as for negative half of a.c. input diode D2 is forward biased and the current flows across D2. Thus for both the halves output is obtained and current flows in the same direction across load resistance R2 and thus a.c. is converted into d.c.
3.Name the gate shown in the figure and write its truth table?
Ans.It is AND gate and its truth table is
| A | B | Y |
| 0 1 1 | 0 1 1 | 0 1 |
4. In the following diagrams indicate which of the diodes are forward biased and which are reverse bias?
Ans.(a) Forward Biased
(b) Reverse Biased
(c) forward Biased
5.In the given figure, is
(i) The emitter base
(ii) collector base forward or reverse biased? Justify.
Ans.Figure shows n-p-n transistor
(i) Emitter is reversed biased because n-region is connected to higher potential.
(ii) Collector is also reversed biased because n-region of p-n junction is at higher potential than p-region.
6.Two semiconductor materials A and B shown in the figure are made by doping germanium crystal with arsenic and indium respectively. The two are joined end to end and connected to a battery as shown.
(a) Will the junction be forward biased or reverse biased? Justify
(b) Sketch a V-I graph for this arrangement
Ans.Material A is n-type as it is doped with pentavalent impurity and material B is p-type as it is doped with trivalent impurity. As a result the junction becomes reverse biased because positive terminal of the battery is connected to n-type and negative terminal to the p-type hence it is reversed biased.
V-I graph for the given circuit
7. Calculate emitter current for which 
and
?
Ans.
Using 
8. Draw the circuit diagram for common – emitter transistor characteristics using N-P-N transistor? Draw the input and output characteristic curve ?
Ans.
Input characteristic curve is the variation of base current 
(Input) with base – emitter voltage (VEB) at constant collector emitter voltage 
output characteristics is the variation of the collector current 
with collector emitter voltage 
at constant base current 
is called output characteristics.
9. For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 
is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1
.
Ans.Collector resistance, 
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor,
= 100
Base resistance, 
Input signal voltage = 
Base current = 
We have the amplification relation as:
Voltage amplification 
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation:
Therefore, the base current of the amplifier is 10 
10. Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.
Ans.Voltage gain of the first amplifier, 
= 10
Voltage gain of the second amplifier, 
= 20
Input signal voltage, = 0.01 V
Output AC signal voltage = 
The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,
V =
= 200
We have the relation:
= 2 V
Therefore, the output AC signal of the given amplifier is 2 V.
11. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Ans.Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, 
= 6000 nm =
The energy of a signal is given by the relation:
E = 
Where,
h = Planck’s constant
=
c = Speed of light
=
E
= 
But 
= 1 eV
∴E =
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV – the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
12. The number of silicon atoms per
. This is doped simultaneously with 
atoms per 
of Arsenic and 
atoms of Indium. Calculate the number of electrons and holes. Given that
. Is the material n-type or p-type?
Ans.Number of silicon atoms, N = 
Number of arsenic atoms, 
Number of indium atoms, 
Number of thermally-generated electrons, 
Number of electrons, 
Number of holes = 
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
Therefore, the number of electrons is approximately 
and the number of holes is about
. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.
13.Write the truth table for a NAND gate connected as given in Fig. 14.45.
Hence identify the exact logic operation carried out by this circuit.
Ans.A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.
Hence, the output can be written as:
……………(i)
The truth table for equation (i) can be drawn as:
| A | Y |
| 0 | 1 |
| 1 | 0 |
This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:
14. You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify the logic operation carried out by the two circuits.
Ans.In both the given circuits, A and B are the inputs and Y is the output.
(a) The output of the left NAND gate will be
, as shown in the following figure.
Hence, the output of the combination of the two NAND gates is given as:
Hence, this circuit functions as an AND gate.
(b) 
is the output of the upper left of the NAND gate and 
is the output of the lower half of the NAND gate, as shown in the following figure.
Hence, the output of the combination of the NAND gates will be given as:
Hence, this circuit functions as an OR gate.
15.Write the truth table for circuit given in Fig. 14.47 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.
(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Ans.A and B are the inputs of the given circuit. The output of the first NOR gate is
. It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.
Hence, the output of the combination is given as:
The truth table for this operation is given as:
| A | B | Y(=A+B) |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.