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 5 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
5 Marks Questions
1.Distinguish between conductors, insulators and semiconductors on the basis of energy band diagrams?
Ans.Conductor – Conduction band in a conductor is either partially filled or conduction and valence band overlaps each other. There is no energy gap in a conductor.
Insulators – conduction band and valence band of all insulator are widely separated by and energy gap of the order 6 to 9eV Also conduction band of an insulator is almost empty.
Semiconductor – In semiconductors the energy gap is very small i.e. about 1ev only.
2.The following truth table gives the output of a 2-input logic gate.
A B output
0 0 1
01 0
10 0
11 0
Identify the logic gate used and draw its logic symbol. If the output of this gate is fed
as input to a NOT gate, name the new logic gate so formed?
Ans.The gate is NOR gate. If the output of NOR gate is connected to a NOT gate then the figure will be
New truth table is
A B Y
000
011
101
111
It is the truth table of OR gate
3.With the help of a diagram, show the biasing of a light emitting diode (LED). Give its two advantages over conventional incandescent lamps?
Ans. Light emitting diode is forward biased i.e. energy is released at the junction.
Advantages of LED
(1) They are used in numerical displays as compact in size.
(2) It works at low voltage and has longer life than incandescent bulbs.
4.The input resistance of a silicon transistor is 665
. Its base current is changed by 15
A, which results in the change in collector current by 2mA. This transistor is used as a common emitter amplifier with a load resistance of 5k. Calculate current gain
.
Ans.(1) Trans conductance (gm) (2) voltage gain (Av) of the amplifier.
Here 
(1) Trans conductance, gm = 
(2) Voltage gain (Av) = 
5.Draw the symbol for zener diode? Zener diodes have higher dopant densities as
compared to ordinary p-n junction diodes. How dos it affect the (i) width of the
depletion layer (i) junction field?
Ans.Symbol for zener diode
(i) Width of the depletion layer of zener diode becomes very small due to heavy doping of p and n-regions
(ii) Junction field will be high.
6.A P-N-P transistor is used in common – emitter mode in an amplifier circuit. A change of 
in the base current brings a change of 2mA in collector current and 0.04V in base – emitter voltage. Find (i) input resistance (ii) current amplification factor
. If a load resistance of 6k is used, then find voltage gain?
Ans.
Voltage gain 
7.A semiconductor has equal electron and whole concentration of 
.
On doping with certain impurity, electron concentration increases to
.
(i) Identify the new semiconductor
(ii) Calculate the new whole concentration.
(iii) How does the energy gap vary with doping?
Ans.(i) New semiconductor obtained is N-type because
(ii) 
(iii) Energy gap decreases due to creation of donor level in between the valence band and the conduction band.
8.Draw a labeled circuit diagram of a common emitter transistor amplifier. Draw the input and the output wave forms and also state the relation between input and output signal?
Ans.
Input wave from
Output wave form
Relation – output waveform has 
phase reversal as compared to input and also the output is being amplified.
9.In an intrinsic semiconductor the energy gap 
is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration 
is given by
where 
is a constant.
Ans.Energy gap of the given intrinsic semiconductor, = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Where
= Boltzmann constant =
T = Temperature
= Constant
Initial temperature,
= 300 K
The intrinsic carrier-concentration at this temperature can be written as:
… (1)
Final temperature, 
= 600 K
The intrinsic carrier-concentration at this temperature can be written as:
… (2)
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.
Therefore, the ratio between the conductivities is 
10.In a p-n junction diode, the current I can be expressed as
where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, k Bis the Boltzmann constant (
) and T is the absolute temperature. If for a given diode 
and T = 300 K, then
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Ans.In a p-n junction diode, the expression for current is given as:
Where,
= Reverse saturation current = 
T = Absolute temperature = 300 K
= Boltzmann constant =
V = Voltage across the diode
(a) Forward voltage, V = 0.6 V
∴Current, I
Therefore, the forward current is about 0.0256 A.
(b) For forward voltage, V‘ = 0.7 V, we can write:
Hence, the increase in current, ΔI = I‘ – I
= 1.257 – 0.0256 = 1.23 A
(c) Dynamic resistance 
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.
11. You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
Ans.(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure.
Hence, the output of the NOR Gate =
This will be the input for the NOT Gate. Its output will be
= A + B
∴Y = A + B
Hence, this circuit functions as an OR Gate.
(b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.
Hence, the output of the given circuit can be written as:
Hence, this circuit functions as an AND Gate.
12.Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
Ans.(a)A acts as the two inputs of the NOR gate and Y is the output, as shown in the following figure. Hence, the output of the circuit is
.
Output, 
The truth table for the same is given as:
| A | Y |
| 0 | 1 |
| 1 | 0 |
This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.
(b) A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are
,as shown in the following figure.
are the inputs for the last NOR gate. Hence, the output for the circuit can be written as:
The truth table for the same can be written as:
| A | B | Y( ) |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.