Questions - Answers
Very Short Answer type Questions
1. What is an n - type semiconductor? What are the majority and minority charge carriers in it?
A: When pentavalent impurity is added to a pure semiconductor, it is called 'n - type semiconductor'. Majority charge carriers are electrons and minority charge carriers are holes in it.
2. What are intrinsic and extrinsic semiconductors?
A: Pure form of semiconductors are called 'Intrinsic semiconductors'. Impure form of semiconductors are called 'Extrinsic semiconductors'.
3. What is a p - type semiconductor? What are the majority and minority charge carriers in it?
A: When trivalent impurity is added to a pure semiconductor, it is called 'p - type semiconductor'. Majority charge carriers are holes and minority charge carriers are electrons in it.
4. What is a p - n junction diode? Define depletion layer?
A: When p - type semiconductor is suitably joined to n - type semiconductor, a p - n junction is formed and the arrangement is known as p - n junction diode.
Depletion layer: A narrow region on either side of junction where there are no mobile charge carriers is called depletion layer.
5. How is a battery connected to a junction diode in (i) Forward and (ii) Reverse bias?
A:
In forward bias, battery +ve is connected to p - region and -ve is connected to n - region.
In reverse bias, battery -ve is connected to p - region and +ve is connected to n - region.
6. What is the maximum percentage of rectification in half wave and full wave rectifiers?
A: Percentage of rectification in half wave rectifier is 40.6% and in full wave rectifier is 81.2%.
7. What is Zener Voltage (VZ) and how will a Zener diode be connected in circuits generally?
A: When reverse voltage applied to the p - n junction is increased, at a particular voltage, the current increases enormously. This voltage is called Zener voltage or breakdown voltage. In circuits, Zener diode is connected in reverse bias.
8. Write the expressions for the efficiency of a full wave rectifier and half wave rectifier?
9. What happens to the width of the depletion layer in a p - n junction diode when it is (i) Forward - biased and (ii) Reverse - biased?
A: In forward biased depletion layer width decreases. In reverse bias depletion layer width increases.
10. Draw the circuit symbols for p - n - p and n - p - n transistors?
11. Define amplifier and amplification factor?
A: The process of raising the strength of weak - signal is called amplification and the device used for amplification is called amplifier. The ratio between output voltage to input voltage is called amplification factor.
12. In which bias can a Zenor diode be used as a voltage regulator?
A: In reverse bias, Zener diode is used as voltage regulator.
13. Which gates are called universal gates?
A: 'NAND' and 'NOR' gates are called universal gates.
14. Write the truth tables of NAND gate. How does it differ from AND gate?
A: Truth table for NAND gate Truth table for AND gate
The output of 'NAND' gate is an inversion of output of 'AND' gate.
Short Answer type Questions
15. What are n - type and p - type semiconductors? How is a semiconductor junction formed?
A: When a pure semiconductor is doped with a trivalent impurity atoms, it is called a p - type semiconductor and when it is doped with a pentavalent impurity atoms, it is called as n - type semiconductor.
Formation of p - n junction: If one side of a single crystal of pure semiconductor [Ge (or) Si] is doped with trivalent impurity atoms and the other side is doped with trivalent impurity atoms, a p-n junction is formed.
p - region has high concentration of holes and n - region contains a large number of electrons.
16. Draw and explain the Current - Voltage (I - V) characteristic curves of a junction diode in forward and reverse bias.
A: A graph between the voltage applied across the terminates of p - n junction diode and current flow in the circuit is called V - I characterisitcs of junction diode. Usually voltage is taken along X - axis and current along Y - axis.
From graph it is observed that,
In forward bias:
(a) As forward voltage is zero (Vf = 0), current in the circuit is zero. This is indicated by 'O' in the graph.
(b) From O to A, as forward voltage increases, increase in the current increase is small because forward voltage is less than barrier voltage.
(c) When forward voltage is greater than potential barrier, the current increases linearly from point 'A'. The voltage at point Ais called cut in voltage (or) knee voltage.
In reverse bias:
(d) When reverse voltage is increased from O to 'C' a small reverse current flows due to minority carriers crossing the junction.
(e) When reverse voltage reaches the point 'C' the reverse current suddenly increases to large value, because breaking of covalent bonds and releasing large number of electron - hole pairs. This voltage is called a break - down voltage (or) Zener voltage. Hence resistance of diode changes with applied voltage.
Dynamic resistance (rd):
The ratio of small change in voltage (ΔV) to the small change in current (ΔI) is called dynamic resistance of junction diode.
17. Describe how a semiconductor diode is used as half wave rectifier?
A: A circuit which is rectifies half of the a.c. wave is called half wave rectifier.
Explanation:
* During +ve half cycle of a.c. input, end A becomes +ve and end B becomes -ve. This makes diode forward bias and conducts current. So output is obtained across RL.
* During -ve half cycle, end A becomes -ve, end B becomes +ve. This makes the diode reverse bias and does not conduct current. So no output is obtained across RL. Thus a half wave rectifier gives discontinuous and pulsating d.c. output across load resistance (RL).
* In half wave rectifier only 40.6% of a.c. is converted into d.c.
18. What is rectification? Explain the working of a full wave rectifier?
A: Rectification: The process of converting alternating current (a.c.) into directed current (d.c.) is called rectification.
Full Wave Rectifier (F.W.R.): The circuit which rectifies both half cycles of the a.c. wave is called Full Wave Rectifier.
Working:
(1) During +ve half cycle, end 'A' becomes +ve and end 'B' becomes -ve. This makes diode D1, forward bias and diode D2 reverse bias. So, D1 conducts and D2 does not conduct. So output voltage is obtained through load resistance (RL) due to diode D1.
(2) During -ve half cycle, end A becomes -ve, end B becomes +ve. This makes diode D1 reverse bias and D2 forward bias. So D2 conducts and D1 does not. So output voltage obtained across RL is due to D2.
(3) The full wave rectifier gives continuous and pulsating d.c. output across RL.
(4) In full wave rectifier 81.2% of a.c. is converted into d.c.
Where rf = diode resistance
RL = load resistance
19. Distinguish between half wave and full wave rectifiers.
A:
20. What is a photodiode? Explain the working with a circuit diagram and draw its I - V characteristics.
A: Photodiode: A p - n junction diode made from light sensitive semiconductor is called photodiode.
Working: When light photons each of energy E = hυ fall on the semiconductor the valence electrons absorbs this energy and jump to conduction band living a hole in the valence band. Thus electron - hole pairs are produced. This electron hole pairs are constitute a photocurrent which flows in the circuit. As intensity of light increases photocurrent also increases.
V - I characteristics of photodiode:
(1) When no light falls on the diode, a small reverse current flow due to minority carriers. This current is called dark current.
(2) With increase of intensity of incident light, the value of reverse current also increases.
(3) Measurement of change in reverse current on illumination can give values of light intensity.
21. Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
A: A light emitting diode (LED) is a forward biased p-n junction diode, which emits visible light when energized.
Working: When a junction diode is forward biased, electrons from n - side and holes from p - side move towards the depletion region and recombination takes place. When an electron in the conduction band recombines with a hole in the valency band, energy is released. In the case of semiconducting materials like Gallium Arsenide (GaAs) a greater percentage of energy is given out in the form of light. If the semiconductor material is translucent, light is emitted and the junction becomes a light source.
Advantages of LED over conventional incandescent lamps:
(1) low operational voltage and less power consumption.
(2) Fast action and no warm up time required.
(3) The band width of emitted light is 100 A° to 500 A° i.e. light is nearly monochromatic.
(4) long life and raggedness.
(5) Fast ON/ OFF switching capability.
22. Define NAND and NOR gates. Give their truth tables.
A: NAND gate: NAND is a combination of AND + Not.
The logic gate in which the output of AND gate is connected to the input of NOT gate is called NAND gate.
In NAND gate the output is low (0) when both the inputs are in high (1) otherwise high.
NOR gate: NOR gate is a combination of OR + NOT.
The logic gate in which the output of OR gate is connected to the input of NOT gate is called NOR gate.
The Boolean equation for NOR gate is Y = 
.
Symbol:
Truth Table:
The output of NOR gate is high (1) when both the inputs are low (0) otherwise low.
23. Discuss the behaviour of p-n junction. How does a potential barrier develop at the junction?
A: Behaviour of p - n junction: In forward bias, p - n junction offers only low resistance and conducts current. In reverse bias, p - n junction offers high resistance and does not conduct current.
Formation of depletion layer and potential barrier: When p - n junction is formed, the electrons diffuse from n-side to p-side to combine with holes and becomes neutral. Similarly holes diffuse from p-side to n-side to combine with electrons and becomes neutral. As a result of narrow region is formed on either side of junction where there are no mobile carriers. This region is called depletion layer.
Potential barrier (Vb): Due to immobile negative ions on p-side and immobile +ve ions or n-side set up a potential difference across the junction. This potential difference is called barrier potential (Vb). This potential barrier opposes further diffusion of electrons and holes across the junction. For Si, Vb = 0.7 V and for Ge Vb = 0.3 V.
24. Distinguish between Zener breakdown and Avalanche breakdown.
A:
25. Explain the working of a solar cell and draw its I - V characteristics.
A: Solar cell: It is a junction diode which converts solar energy into electricity and is based on photovoltaic effect.
Working: When light photons (with energy hυ > Eg) reach the junction, electron - hole pairs, generated in the depletion region and move in opposite direction due to barrier field. Now electrons move towards p-side. Thus p-side becomes +ve and n-side becomes -ve giving rise to a photovoltage. When load resistance RL is connected in the external circuit; a photocurrent (IL) flows. The current is proportional to intesity of light.
V - I characteristics: From graph, Voc depends on intensity of light. Hence output power of an solar cell depends on intensity of incident sunlight.
26. Explain the different transistor configurations with diagrams.
A: In electric circuit transistor can be arranged in three configurations.
(1) Common base configuration (CB)
(2) Common emitter configuration (CE)
(3) Common collector configuration (CC)
Common base configuration (CB): If base is common to both input and output of a transistor is called CB configuration.
Common emitter configuration (CE): If emitter is common to both input and output of a transistor is called CE configuration.
Common collector configuration (CC): If collector is common to both input and output of a transistor is called CC configuration.
Among the three, CE is more advantageous because it gives more amplification in current, voltage and power.
27. Explain the power of a NOT gate and give its truth table.
A: NOT gate: NOT gate is a gate with only one input and one output. It also called inverter because its output is complement to the input.
In NOT gate, when input is low, ouput is high and vice versa.
28. Explain hole conduction in the intrinsic semiconductor.
A: Pure semiconductors are called intrinsic semiconductors. At low temperature valency band is filled with electrons and conduction band is empty. Hence it acts as an insulator at low temperatures.
As temperature increases valence band electrons get energy and jump into the conduction band crossing the forbidden gap. At their places in the valence band a vacancy is created. This vacancy of electron in the valence band is called as 'hole'. 'Hole' has +ve charge and moves only in the valence band giving hold current (Ih). The excited electrons in the conduction band is equal to number of holes in the valence band inp = ne.
Current due to electrons is equal to current due to holes Ie = Ih
but total current I = Ie + Ih.
29. Explain how transistor can be used as switch?
A: Switch is a device for the on or off the current in the circuit. A n-p-n transistor with common emitter circuit is used as a switch.
Using Kirchoff's voltage law:
For the input circuit -VBB + IBRB + VBE = 0
VBB = IBRB + VBE ............. (1)
For Si transistor VBE is 0.6 V to 0.7 V
For the output circuit -VCC + ICRC + VCE = 0 or VCE = VCC - ICRC .......... (2)
If VBB is d.c. input voltage Vi and VCE is output voltage Vo then
from (1) Vi = IBRB + VBE ..............(3)
Vo = VCC - ICRC ............. (4)
for Si transistor if Vi < 0.6 V then collector current (IC = 0) then
from eq (4), Vo = VCC, the transistor will be in cutoff state or switched off state.
When Vi > 0.6 V. IC increases and ouput.
Voltage Vo, decreases and the transistor will be inactive state. If Vi > 1.0 V, Vi and Vo is non linear with increase in Vi, Vo decreases then IC becomes max and transistor is in saturation state or switched on state.
The low input state switches the transistor off and a high input state switches it on. Hence transistor acts as a switch.
30. Explain how transistor can be used as an oscillator?
A: An oscillator is a device for converting energy of dc source into an alternating voltage of high frequency.
The basic oscillating circuit consists of an inductance L connected in parallel to a capacitor 'C' which is called a tank circuit. The frequency of oscillations (υ) is given by
An oscillator generates ac output signal without any input signal. A part of the output is feedback to the input and this feedback signal is the only input to the internal amplifier of oscillator. If the feedback voltage and input voltage are in same phase the input to the amplifier increased. This type of feedback which increases the gain of the amplifier is called +ve feedback. An oscillator works on +ve feed back.
+ve feedback is given by Af = 
here F is called feedback factor.
Af - over all gain of amplifier.
If AB = 1 then Af = ∞ then amplifier becomes an oscillator.
The starting or noise voltage produced by random motion of electrons in resistors used in the circuit contains all the sinusoidal frequencies the ouput of the oscillator will contain only a single sinusoidal frequency (υ).
Due to the resistance in circuit damped oscillations are formed and the energy loss is balanced by feed back i.e. by supplying a suitable amount of energy from the o/p to i/p in proper phase. The amount of energy feedback is determined by the coupling between L and L' due to mutual induction.
Long Answer type Questions
31. What is a rectifier? Explain the working of halfwave and fullwave rectifiers with diagrams.
A: The process of converting an alternating current (a.c.) into a direct current (d.c.) is called rectification. The device used for this purpose is called rectifier.
Half Wave Rectifier:
A half wave rectifier is constructed with a single diode D as shown in figure.
The a.c. supply to be rectified is applied to the p-n junction diode connected in series with the load resistance RL. The d.c. output is taken across the load resistance RL.
During the +ve half cycle, the diode is forward biased and current flows through the diode. During the negative half cycle, the diode is reverse biased and current does not flow through it. Current flows in RL only in one direction. Thus half wave rectifier gives discontinuous and pulsative d.c. output across RL. Efficiency of half wave rectifier is defined as the ratio of d.c. output power to the a.c. input power.
In half wave rectifier maximum efficiency is 40.6%.
Full wave rectifier:
Full wave rectifier can be constructed with two diodes D1 and D2. The terminals of the secondary coils are connected to the p-sides of the diodes and n-sides are connected with each other. Load resistance RL is connected in between the centre tap and the common terminal of n-sides are of the diodes D1, D2. The voltage at A & B w.r.t. to centre tap C are out of phase with each other.
During the +ve half cycle of a.c. input the diode D1 is in forward bias, it allows current. The diode D2 is reverse biased, it will not allow current. The current flows through RL due to D1 only.
Similarly during the -ve half cycle of a.c. input, the diode D2 is in forward bias, it allows current. The diode D1 is reverse biased, it will not allow currnet. The current flows through RL due to D2 only.
Hence current flows through the RL during both half cycles and in the same direction. Thus the output d.c is continuous but pulsating efficiency of full wave rectifier is defined as the ratio of d.c. output power to the a/c input power.
Where rf is diode forward resistance.
In full wave rectifier maximum efficiency is 81.2%.
32. What is a Zener diode? Explain how it is used as a voltage regulator?
A: Zener diode: A heavily doped p-n junction diode which has sharp breakdown voltage, when operated in the reverse bias condition is called Zener diode. The circuit 
symbol of Zener diode is shown in figure.
Zener diode as voltage regulator:
The irregulated input voltage is applied across Zener diode and resistance R which are in series in such a way to make Zener diode reverse biased. The load resistence RL is connected across Zener diode output is taken across RL.
If I is input current IZ and IL are Zener and load currents respectively, then I = IZ + IL
Vin = IR + VZ, But Vout = Vin - IR
The value of R is selected in such a way that in the absence of load resistance RL, maximum safe current flows through the diode.
During the fluctuations even though the current in the source changes, the voltage across Zener diode remains constant.
The voltage across the Zener diode remains constant even if the load resistance RLvaries. When RL is increased but applied input voltage (Vin) is fixed, the current IL decrease and IZ increases by an equal amount. So, that total current I remains constant. Hence output voltage remains constant. Thus Zener diode works as a voltage regulator.
33. Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
A: OR gate: An OR gate has two inputs and only one output. Figure shows the symbol for an OR gate. The output of OR gate is Y = A + B
When one of the inputs is high (or) when both inputs are high, the output is high.
Implementation of OR gate using diodes:
Let D1 & D2 represent two diodes. A potential of 5 V represents zero (0).
When A = 0, B = 0 both diodes are reverse biased and there is no current through the resistance. So, potential Y is zero (0).
When A = 5 V, B = 0, the diode is forward biased and current flows. So Y = 1.
When A = 0, B = 5 V then Y = 1. When A = 5 V, B = 5 V, then Y = 1.
The diode behaves like a closed switch. The output is same as that of OR gate.
34. Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors?
A: AND gate: An AND gate has two inputs and one output. Figure shows the symbol for AND gate. The output of AND gate is Y = A . B
The output of AND gate is high only when both the inputs are high.
Implementation & AND gate using diodes:
Let D1, D2 represents two diodes. A potential of 5 V represents the logical value 1 V and 0 V represents the logical value 0.
When A= 0, B = 0 both diodes are forward biased and they behave like closed switches. Output Y = 0 when A = 1, B = 1 both diodes are reverse biased and they behave open switches. The output is Y = 1. The output is same as that of an AND gate.
35. What is a junction diode? Explain the formation of depletion region at the junction. Explain the variation of depletion region in forward and reverse biased conditions.
A: Junction diode: A p - n junction diode is a two terminal device. When a semiconductor material of Silicon or Germanium is doped with Ni such a way that one side of it becomes a
p-type and other side becomes n-type, we get p - n junction diode.
Circuit symbol of p-n junction diode is shown in figure. 
Formation of depletion layer:
When p - n junction is formed the free electrons on n-side diffuse over to p-side, combine with holes and becomes neutral. Similarly the holes on p-side diffuse over to n-side, combine with electrons and become neutral. "This results in the formation of a narrow region on either side of the junction, which becomes free from charge carriers. This region is called "depletion layer".
Near the n-type the junction becomes positively charged and near the p-type, the junction becomes negatively charged. This creates an electric field about the junction and a potential difference exists about the junction and repels further flow of charge carriers to cross the junction. This potential difference is called barrier potential (VB).
Forward bias:
"When the battery +ve terminal is connected to p-side and negative terminal of the battery to n-side and move towards the junction is called forward bias".
The holes on p-side repelleld by the +ve terminal of the battery and move towards the junction. At the same time the electrons on n-side are attracted by the +ve terminal of the battery and move away from the junction. As a result the thickness of depletion layer increases. The diode will not allow current through it. In reverse bias the resistance is high.
36. Describe a transistor and explain its working.
A: Transistor consists of two p - n junctions joined back to back. The word transistor means transfer of resistance. A transistor has three regions.
They are: 1) Emitter (E) 2) Base (B) 3) Collector (C).
Emitter (E): The section at one end of transistor is called emitter. It is heavily doped region. It contains large number of charge carriers.
Base (B): The middle section of transistor is called base. This is lightly doped and very thin. Most of the charge carriers injected in it to flow into collector without neutralised.
Collector (C): The section at the other end is called collector, it is moderately doped. Physically it is the largest. It collects the charge carriers from the base. Transistors are of two types: 1) p-n-p transistor
2) n-p-n transistor
1. p-n-p transistor: In a p-n-p transistor the base part is made of n-type, emitter and collector are made of p-type. The circuit symbol of p-n-p transistor is shown in figure. 
In p-n-p transistor emitter junction is forward biased by connecting +ve terminal of a battery to the emitter (E) and negative to the base. In collector junction is reverse biased by connecting battery +ve to the base and -ve to the collector.
The holes in the emitter are repelled by the +ve terminal of battery and cross the emitter junction enters in the base causing emitter current IE. A base is lightly doped, a few number of holes combined with electrons causing a base current IB. Majority of holes enters in the collector region. The collector terminal connected to the -ve of a battery. It rapidly sweep the holes in the collector, causing a collector current IC.
IE = IB + IC
A continuous supply of holes injected into the emitter flows across the base to the collector.
In p-n-p transistor majority charge carriers inside the circuit are holes and outside the circuit charge carriers are electrons.
n-p-n transistor: In n-p-n transistor base part is made of p-type, emitter and collector parts are made of n-type. The circuit symbol of n-p-n transistor is shown in figure. 
In n-p-n transistor the emitter junction is forward biased with -ve terminal of the battery connected to the emitter and +ve to the base. The collector junction is reverse biased with +ve terminals to the collector and -ve to the base.
The electrons in the emittor are repelled by the -ve terminal of the battery and cross the base region causing the emitter current IE. As base is lightly doped, few electrons combine with holes causing base current IB and the majority of electrons enters in the collector. These electrons rapidly swept out by the +ve terminal of the battery, causing collector current IC.
IE = IB + IC
A continuous supply of electrons injected into the emitter flows across the base of the collector.
In n-p-n transistor charge carriers inside and outside the circuit are electrons.
37. What is amplification? Explain the working of a common emitter amplifier with necessary diagram.
A: Amplification: The process of raising the strength of a weak signal is known as amplification.
Amplifier: The devise which is used the process of raising the strength of a weak signal is called amplifier.
Uses: Amplifiers have wide applications in communication systems, radioreceivers, television etc.
The block diagram of amplifier is shown in the following figure.
Types of amplifiers:
a) Power amplifiers: The amplifier which is used to raise the power level is known as "power amplifier".
b) Voltage amplifiers: The amplifier that is used to raise the voltage level is known as "voltage amplifier".
Amplification factor: The ratio between output voltage to input voltage is called amplification factor.
The n-p-n transistor common emitter amplifier circuit is shown in figure.
The emitter junction is in forward biased with Bb, and the collector junction is reverse biased with BC.
The input signal to be amplified is connected in series with Bb and output voltage is taken at load resistance RL.
According to the input signal, the base current changes (ΔIb), which results in a large change in collector current (ΔIC).
Current gain: The change in collector current to change in base currnet is called current gain.
Voltage gain: The ratio of change in output voltage to change in input voltage.
Power gain: The product of current gain and voltage gain is called power gain.
Power gain = current gain × voltage gain.
PROBLEMS
1. In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 20 ohm is used in the circuit then find the efficiency of this half wave rectifier.
Sol: Given rf = 20 Ω, RL = 2 Ω
% of efficiency (η) = 0.036 × 100 = 3.69%.
2. A fullwave p-n junction diode rectifier used a bad resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Sol: Given rf = 9 Ω, RL = 1300 Ω
% of efficiency (η) = 0.8064 × 100 = 80.64%.
3. Calculate the current amplification factor β when change in collector current is 1 mA and change in base current is 20 μA.
Sol: ΔIC = 1 mA = 10-3 A, ΔIB = 20 μA = 20 × 10-6 A
4. For transistor amplifier, the collector load resistance RL = 2 k ohm and the input resistance Ri = 1 k ohm. If the current gain is 50, calculate the voltage, gain of the amplifier.
Sol: Given RL = 2 kΩ = 2 × 103 Ω, Ri = 1 kΩ = 1 × 103 Ω
β = 50
AV = 100
5. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Sol: Eg = 2.8 eV, λ = 6000 nm = 6000 × 10-9 m
As E < Eg, the p-n junction cannot detect the rediation of wavelength 6000 nm.
6. For a CE - transisttor amplifier, the audio signal voltage across the collected resistance of 2 kΩ 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 kΩ.
Sol: Here Vo = 2 V, β = 100
RC (collector resistance) = 2 kΩ, Rβ = 1 kΩ
7. 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 voltage. Calculate the output ac signal.
Thus Vo = 200 Vi = 200 × 0.01 V = 2 V
8. The number of Silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that nt = 1.5 × 1018 m-3. Is the material n-type or p-type?
Sol: Here Nd = 5 × 1022 / m3
Nd = 5 × 1020 / m3 = 0.05 × 1022 / m3
ne = 1.5 × 1016 / m3
For the semiconductor to remain neutral (nd - na) = (ne - nh) ........... (1)
and nenh = ni2 ............. (3)
from equations (1), (2) & (3)
As nd - na = 5 × 1022 - 0.05 × 1022 = 4.95 × 1022
Since ne >> nh, the material is n - type.
9. In an intrinsic semiconductor, the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration nt is given by ni = no exp 
where no is a constant.
Sol: Here Eg = 1.2 eV, T1 = 300 K, T2 = 600 K,
KB = 8.62 × 10-5 eV/K
Conductivity of an intrinsic semiconductor
σ = e(neμe + nhμh) = eni(μe + μh) = eniμe
10. You are given the two circuits as shown in figure. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
A: The circuit consists of NOR gate followed by a NOT gate. The truth table is as follows.
Since Y = A + B, the circuit acts as a OR gate.
(a) The circuit consists two NOT gates followed by a NOR gate. The truth table is as follows.
Since Y = A + B, the circuit acts OR gate.
(b) The circuit consists of two NOT gates followed by a NOR gate. The truth table is as follows
11. Write the truth table for a NAND gates connected as given in figure. Hence identify the exact logic operation carried out by the cicuit.
A:
12. You are given two circuits as shown in figure which consist of NAND gates. Identify the logic operation carried out by two circuits.
13. Write the truth table for the circuit given in figure below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.
A: The first gate is a NOR gate. The second gate is also a NOR gate, with both the input terminals connected together. The truth table for the above circuit is as follows.
Since in this case Y = 
= A + B, the given circuit performs the function of an OR gate.
14. Write the truth table for the circuits given in figure consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
A: (a) The circuit of fig (a) is a NOR gate with its input terminals connected together. The truth table is for this circuit is as follows.
Since Y = = 
, a NOR gate with both its terminals connected together, performs the NOT operation.
(b) The inputs A & B are inverted by the two NOT gates (obtained from NOR gates as detailed in (a)]. These outputs & 
are fed to a NOR gate and the truth table is as follows.
Since in the case Y = = A. B, the fig (b) performs the function of an AND gate.
