Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen: Overview, Questions, Preparation

1. If each diode in figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the currents I₁, I₂, I₃and I₄?

Explanation- forward biased resistance = 25ohm

Reverse biased resistance = infinity

As CD branch is reverse biased having infinite resistance.

So I₃= 0

Resistance in branch AB= 25+125= 150 ohm(R₁)

Resistance in branch EF= 25+125= 150 ohm(R₂)

They both are in parallel

 So $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}$

R= 75ohm

So total resistance = 25+75= 100ohm

Current = V/R= 5/100=0.05A

I₁=I₂+I₃+I₄

I₁=I₄+I₂

Here the resistance R₁ and R₂ is same

I₄=I₂

I₁= 2I₂

I₂= I₁/2=0.05/2=0.025A

I₄= 0.025A

2. In the circuit shown in figure, when the input voltage of the base resistance is 10 V, V_BEis zero and V_CE is also zero. Find the values of I_B, I_C and β .

Explanation- voltage across R_B= 10V

Resistance R_B= 400kohm

V_BE= 0, V_CE= 0 , R_c=3kohm

I_B= voltage /R_B= 10/400 $\times$  10³= 25 $\times$ 10^-6A

Voltage across R_c= 10V

I_c=voltage/R_c= 10/3 $\times$ 10³= 3.33 $\times$ 10^-3

$\beta =\frac{I_c}{I_B}=\frac{3.33\times {10}^{-3}}{25\times {10}^{-6}}=133$

3. Draw the output signals C₁ and C₂ in the given combination of gates.

Explanation-

 

A         B	C          D	E           F	G	H	I	C1
0         0	0          0	1          1	1	0	0	1
1         0	1          0	0          1	0	1	1	0
0         1	0          1	1          0	0	1	1	0
1         1	1          1	0          0	0	1	1	0

 

4. Assuming the ideal diode, draw the output waveform for the circuit given in figure, explain the waveform.

Explanation- When the input voltage is equal to or less than 5 V, diode will be revers biased. It will offer high resistance in comparison to resistance ( ) R in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.

If input voltage is more than + 5 V, diode will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the

voltagebeyond+5Vwillappearacross R.

When input voltage is negative, there will be opposition to 5 V battery in p n- junction input voltage becomes more than− 5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input  waveform is then passed on to the output terminals.

The output waveform is shown here in the fig. (b)

5. Consider the circuit arrangement shown in Fig 14.16 (a) for studying input and output characteristics of npn transistor in CE configuration

Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in Fig. 14.16 (b). Given that the input impedance of the transistor is very small and VCC = VBB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions.

6. Suppose a n-type wafer is‘created by doping Si crystal having 5 x 10²⁸atoms/m³ with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ‘p’ region in this wafer. Considering  n_i = 1.5 x 10¹⁶ m^-3,
(i) Calculate the densities of the charge carriers in the n and p regions,
(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

Explanation- whem As is used then it will create n type semiconductor

N_e=N_D= $\frac{1}{{10}^6}\times 5\times {10}^{28}=5\times {10}^{22}/m^3$

Number of minority carriers n_h= $\frac{{ni}^2}{ne}=\frac{{(1.5\times {10}^{16})}^2}{5\times {10}^{22}}$ = 0.45 $\times {10}^{10}/m^3$

But when B is implanted in Si crystal then p type semiconductor is formed

N_h=N_A= $\frac{200}{{10}^6}\times \left(5\times {10}^{28}\right)=1\times {10}^{25}/m^3$

Minority carriers created in p type is

N_e= $\frac{{ni}^2}{ne}=\frac{({1.5\times {10}^{16})}^2}{1\times {10}^{25}}=2.25\times {10}^{27}/m^3$

Minority charge carriers holes in p type would contribute more in reverse saturatiom current.

7. An XOR gate has the following truth table. It is represented by following logic relation Y = Ᾱ . B + A . B’ . Build this gate using AND, OR and NOT gates

Explanation- Y= A’B+A.B’=Y₁+Y₂

Y₁= A.B and Y₂= A.B’

Y₁ can be obtained as output of AND gate I for which one Input is of A through NOT gate and

another input is of B. Y2 can be obtained as output of AND gate II for which one input is of A

and other input is of B through NOT gate.

Now Y₂ can be obtained as output from OR gate, where, Y₁ and Y₂ are input of OR gate.

Thus, the given table can be obtained from the logic circuit given below

8. Consider a box with three terminals on top of it as shown in figure.

Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit as shown in figure.
The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit. The graphs are

Explanation- a) In V- graph of condition (i), a reverse characteristics is shown in fig. (c). Here A is connected to n - side of p-n junction and B is connected to p -side of p-n junction I with a resistance in series.

(b) In V- graph of condition (ii), a forward characteristics is shown in fig. (d), where 0.7 V is

the knee voltage of p-n junction I 1/slope= (1/1000) Ω. It means A is connected to n -side of p n- junction and B is connected to p-side of p n- junction and resistance R is in series of p n- junction between A and B.

(c) In V- graph of condition (iii), a forward characteristics is shown in figure (e), where 0.7 V is the knee voltage. In this case p -side of p-n junction II is connected to C and n -side of p-n junction II to B.

(d) In V-I graphs of conditions (iv), (v), (vi) also concludes the above connection of p-n junctions I and II along with a resistance R.
Thus, the arrangement of p-n I, p-n II and resistance R between A B , in the figure p-n I, p-n II and C will be as shown.

9. For the transistor circuit shown in figure, evaluate V_E, R_B, R_E, given I_C= 1 mA,  V_CE= 3, V_BE= 0.5 V  and  V_CC= 12 V, β= 100

Explanation-I_c= I_E

R_c= 7.8Kohm

I_c(R_c+R_E)+V_CE= 12

(R_E+R_C) $\times 1\times {10}^{-3}+3=12$

R_E+R_C= 9 $\times {10}^3=9kohm$

R_E= 9-7.3= 1.2kohm

V_E= I_{E $\times$} R_E

 = 1 $\times {10}^{-3}\times 1.2\times {10}^3=1.2V$

Voltage V_B=V_E+V_BE= 1.2+0.5= 1.7V

I= $\frac{1.7}{20\times {10}^3}=0.085mA$

Resistance R_B= $\frac{12-1.7}{\frac{I_C}{\beta }+0.085}=\frac{10.3}{0.01+0.085}=108kohm$

10. In the circuit diagram find the value of R_c

Explanation- I_E=I_C+I_B and I_c= $\beta I_B$

I_cRc+V_CE+I_ER_E=V_CC

RI_B+V_BE+I_ER_E=V_cc

I_E=I_c= $\beta I_B$

From above equation

(R+ $\beta R$ _E)I_B=V_CC-V_BE

I_B= $\frac{V_{CC}-V_{BE}}{R+\beta R_E}=\frac{12-0.5}{80+1.2\times 100}=\frac{11.5}{200}$

( $R_C+R_E$ )= $\frac{V_{CE}-V_{BE}}{I_C}$

( $R_C+R_E$ )=1.56

Rc=1.56-1=0.56kohm

1. Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?

Explanation-The size of the dopant atom should be such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Si or Ge atoms, which are provided by group XIII or group XV elements.

2. Sn, C and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?

Explanation- A material is a conductor if in its energy band diagram, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for

semiconductor the energy gap is moderate.

The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV, related to their atomic size. Therefore Sn is a conductor, C is an insulator and Ge and Si are semiconductors.

3. Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?

Explanation- We cannot measure the potential barrier across a p-n junction by a voltmeter because the resistance of voltmeter is very high as compared to the junction resistance.

4. Draw the output waveform across the resistor in the given figure.

Explanation- diode only works in forward biased. So output waveform is

5. The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10,20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value).
(i) if DC supply voltage is 10 V?
(ii) if DC supply voltage is 5 V?

Explanation- as we know total voltage amplification = $\frac{outputsignalvoltage}{inputsignalvoltge}$

Av_x= 10,Av_y = 20,Av_z = 30

$∆$ v_i = 1mV= 10^-3V

So according to formula

 = (Av_x )(Av_y )(Av_z)

$∆$ v_o =  (Av_x )(Av_y )(Av_z) $∆$ v_i

  = 10^{-3 $\times 20\times 30\times 10$} =6V

So in first case output cannot be greater than than 6V but in 2^nd case it will work

6. In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?

Explanation- In CE transistor amplifier, the power gain is very high.

In this circuit, the extra power required for amplified output is obtained from DC source.

Thus, the circuit used does not violet the law of conservation.

7. (i) Name the type of a diode whose characteristics are shown in figure (a) and (b).
(ii) What does the point P in fig. (a) represent?
(iii) What does the points P and Q in fig. (b) represent?

8. Three photo diodes D₁, D₂and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å ?

Explanation- E=hc/ $\lambda$ = $\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{6000\times {10}^{-10}\times 1.6\times {10}^{-19}}$ = 2.06eV

The incident radiation which is detected by the photodiode having energy should be greater

Than the band-gap. So ,it is only valid for diode D2.Then,diode D2 will detect this radiation.

9. If the resistance R₁is increased (see figure), how will the readings of the ammeter and voltmeter change?

Explanation- as base current I_B = $\frac{V_{BB}-V_{BE}}{R_i}$

As R_i is increased I_B is decreased.

And collector current is I_c = I_B as I_B decreased I_c also decreased and the rading of voltmeter and ammeter also decreased.

10. Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Draw a circuit that resembles this situation using diodes for this situation.

Explanation-

OR gate gives the desired output.

11. How would you set up a circuit to obtain NOT gate using a transistor?

Explanation- The NOT gate is a device which has only one input and one output i.e., A’=Y means Y equals NOT A.

This gate cannot be realised by using diodes. However it can be realised by making use of a

transistor. This can be seen in the figure given below.

Here, the base B of the transistor is connected to the input A through a resistance Rb and the emitter E is earthed. The collector is connected to 5 V battery. The output Y is the voltage at C w.r.t. earth.

The resistor Rb and Rc are so chosen that if emitter-base junction is unbiased, the transistor is in cut off mode and if emitter-base junction is forward biased by 5V, the transistor is in saturation state.

12. Explain why elemental semiconductor cannot be used to make visible LEDs.

Explanation- In elemental semiconductor, the band gap is such that the emission are in infrared region and not in visible region.

13. Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.

Explanation- this is AND gate so truth table for this will be

14. A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V, what should be the value of Rs for safe operation (see figure)?

1. The conductivity of a semiconductor increases with increase in temperature, because
(a) number density of free current carries increases
(b) relaxation time increases
(c) both number density of carries and relaxation time increase
(d) number density of carries increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density 

Answer-(d)

Explanation- when we increase temperature conductivity increases because number density is increasing after increasing temperature which decreases the relaxation time

2. In figure given below V₀ is the potential barrier across a pn junction when no battery is connected across the junction

(a) 1 and 3 corresponds to forward bias of junction

(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of the junction

(c) 1 corresponds to forward bias and 3 corresponds to reverse bias of the junction

(d) 3 and 1 both corresponds to reverse bias of the junction

Explanation - When p-n junction is forward biased, it opposes the potential barrier junction, when p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction.

3. In figure given on next page, assuming the diodes to be ideal
(a) D₁is forward biased and D₂ is reverse biased and hence current flows from A to B
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice-versa
(c) D₁ and D₂ are both forward biased and hence current flows from A to B
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice-versa

Answer-b

Explanation- from the figure we can say that that diode D1 is reverse biased and diode D2 is forward biased so current will flow from B to A.

4. A 220 V AC supply is connected between points A and B (figure). What will be the potential difference V across the capacitor?

Answer- d

Explanation- as we know that Ev = E_o/ $\sqrt[]{2}$  so E₀= Ev $\sqrt[]{2}$ so net voltage is 220 $\sqrt[]{2}$ V

5. Hole is
(a) an anti-particle of electron
(b) a vacancy created when an electron leaves a covalent bond
(c) absence of free electrons
(d) an artificially created particle

Answer-b

Explanation -when a neutral entity is there it contains both equal number of elctron and holes. Also when electron emit from neutral entity then it leave behind a vacancy that is hole.

6. The output of the given circuit in figure is given below.

(a) would be zero at all times

(b) would be like a half wave rectifier with positive cycles in output
(c) would be like a half wave rectifier with negative cycles in output
(d) would be like that of a full wave rectifier

Answer- c

Explanation- due to forward biasing the diffusion current in the circuit is very high and resistance in the circuit is very low. Thus voltage across pn junction is very low but in case of reverse biasing diffusion current very small so resistance becomes very high . and that end making high voltage across pn junction.

7. In the circuit shown in figure given below, if the diode forward voltage between A and B is

Answer-b

Explanation- r1=5kohm and r2= 5kohm and both are in series

Then V-0.3 = [(r1+r2)10³] $\times$ [0.2 $\times$ 10^-3]

 V-0.3= 2

V= 2.3V

8. Truth table for the given circuit is

Answer- c

Explanation- C=A.B and D=A’B

E= C+D = (A.B)+(A’.B)

9. When an electric field is applied across a semiconductor
(a) electrons move from lower energy level to higher energy level in the conduction band
(b) electrons move from higher energy level to lower energy level in the conduction band
(c) holes in the valence band move from higher energy level to lower energy level
(d) holes in the valence band move from lower energy level to higher energy level

Answer- (a,c)

Explanation- when we apply temperature across semiconductor then electron will starts from lower energy to high energy level that is from valence band to conduction band. When electron goes from lower to higher then holes that is left behind they goes to lower energy levels.

10. Consider an n-p-n transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?
(a) Electrons crossover from emitter to collector
(b) Holes move from base to collector
(c) Electrons move from emitter to base
(d) Electrons from emitter move out of base without going to the collector

Answer- a,c

Explanation- Here emitter-base junction is forward biased i.e., the positive pole of emitter base battery is connected to base and its negative pole to emitter. Also, the collector base junction is reverse biased, i.e., the positive pole of the collector base battery is connected to collector and negative pole to base.

Thus, electron move from emmiter to base and cross over from emitter to collector.

11. In a n-p-n transistor circuit, the collector current is 10 mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
(a) The emitter current will be 8 mA
(b) The emitter current will be 10.53 mA
(c) The base current will be 0.53 mA
(d) The base current will be 2 mA

Answer- b,c

Explanation- I_C= 10mA

I_c= 95/100  I_e

I_e= $\frac{10\times 100}{95}$ = 10.53mA

I_b=I_e-I_C= 10.53-10= .53mA

12. In the depletion region of a diode 
(a) there are no mobile charges
(b) equal number of holes and elections exist, making the region neutral
(c) recombination of holes and electrons has taken place
(d) immobile charged ions exist

Answer- a,b,d

Explanation- The space-charge regions on both the sides of p-n junction which has immobile ions and entirely lacking of any charge carriers will form a region called depletion region of a diode.

The number of ionized acceptors on the p -side equals the number of ionized donors on the n-side.

13. What happens during regulation action of a Zener diode?
(a) The current and voltage across the Zener remains fixed
(b) The current through the series Resistance (R_s) changes
(c) The Zener resistance is constant
(d) The resistance offered by the Zener changes

Answer-b,d

Explanation- During regulation action of a Zener diode, the current through the Rs changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

14. To reduce the ripples in rectifier circuit with capacitor filter
(a) R_Lshould be increased
(b) input frequency should be decreased .
(c) input frequency should be increased
(d) capacitors with high capacitance should be used

Answer- a,c,d

Explanation-ripple factor is inversely proportional to R_L, C  and frequency. so to reduce ripple factor all these should be increase.

15. The breakdown in a reverse biased p-n junction is more likely to occur due to
(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strongelectricfieldinadepletionregionifthedopingconcentrationissmall
(d) strong electric field in the depletion region if the doping concentration is large

Answer-(a,d)

Explanation- In reverse biasing, the minority charge carriers will be accelerated due to reverse biasing, which on striking with atoms cause ionization resulting secondary electrons and thus more number of charge carriers.

When doping concentration is large, there will be large number of ions in the depletion

region, which will give rise to a strong electric field.