Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 14 - Semiconductor Electronics

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Page No 87:

Question 1:

The conductivity of a semiconductor increases with increase in temperature because

(a) number density of free current carriers increases.
(b) relaxation time increases.
(c) both number density of carriers and relaxation time increase.
(d) number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density.

Answer:

The conductivity of a semiconductor increases with an increase in temperature this is because there is an increase in the number density of current carries due to which relaxation time decreases and the effect of a decrease in relaxation is much less than the increase in number density.
Hence, the correct answer is option (d).

Page No 87:

Question 2:

In Figure, V_o is the potential barrier across a p−n junction, when no battery is connected across the junction

(a) 1 and 3 both correspond to forward bias of junction
(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction
(c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction.
(d) 3 and 1 both correspond to reverse bias of junction.

Answer:

When p-n junction is forward biased, it opposes the potential barrier junction. But when p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction. This means in forward bias, the potential barrier reduces, whereas in reverse bias, the potential barrier increases. So in the graph shown, 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction.
Hence, the correct answer is option (b).

Page No 88:

Question 3:

In given figure 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:

In the given figure, the p-side of p-n junction D₁ is connected to lower voltage and the n-side of D₁ to higher voltage. Thus, D₁ is reverse biased.
The p -side of p-n junction D₂ is at higher potential and the n-side of D₂ is at lower potential. Thus, D₂ is forward biased.
Hence, the correct answer is option (b).

Page No 88:

Question 4:

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

(a) 220 V.
(b) 110 V.
(c) 0 V.
(d) $220 \sqrt{2} V$

Answer:

Given: v_rms= 220 V
Potential difference across the capacitor will be equal to the peak voltage of the given AC voltage.
$\begin{array}{rcl} \Rightarrow & v_{0} = v_{rms} \sqrt{2} \\ = & 220 \sqrt{2} V \end{array}$

Hence, the correct answer is option (d).

Page No 88:

Question 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 artifically created particle.

Answer:

Hole is basically a vacancy created when an electron leaves a covalent bond. For example, when Si or Ge is doped with a trivalent impurity like Al, B, In, etc. The dopant has one valence electron less than Si or Ge and, therefore, this atom can form covalent bonds with neighbouring three Si atoms but does not have any electron to offer to the fourth Si atom. So the bond between the fourth neighbour and the trivalent atom has a vacancy or hole. Since the neighbouring Si atom in the lattice wants an electron in place of a hole, an electron in the outer orbit of an atom in the neighbourhood may jump to fill this vacancy, leaving a vacancy or hole at its own site. Thus the hole is available for conduction.
Hence, the correct answer is option (b).

Page No 88:

Question 6:

The output of the given circuit in given figure

(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:

When positive cycle is at A, diode will be forward biased and resistance due to v_msinωt diode will be approximately zero which means the diode is shorted. So, potential difference will be zero.
However, when negative cycle is at A, diode will be reverse biased and resistance will be maximum. So, potential difference across diode is v_msinωt with negative at A.

So, we get only negative output across the diode which means it behaves like half wave rectifier with negative cycle in output.

Hence, the correct answer is option (c).

Page No 89:

Question 7:

In the circuit shown in given figure if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

(a) 1.3 V
(b) 2.3 V
(c) 0
(d) 0.5 V

Answer:

R₁ = 5 kΩ
R₂ = 5 kΩ
Applying Kirchhoff's loop law,
V_AB = (5000 × 0.2 × 10^–3) + 0.3 + (5000 × 0.2 × 10^–3)
⇒ V_AB = 1 + 0.3 + 1
⇒ V_AB= 2.3 V

Hence, the correct answer is option (b).

Page No 89:

Question 8:

Truth table for the given circuit (in given figure) is

(a)

A	B	E
0 0 1 1	0 1 0 1	1 0 1 0

(b)

A	B	E
0 0 1 1	0 1 0 1	1 0 0 1

(c)

A	B	E
0 0 1 1	0 1 0 1	0 1 0 1

(d)

A	B	E
0 0 1 1	0 1 0 1	0 1 1 0

Answer:

From the given figure, E = C . D
C = A · B
$and D = \bar{A} \cdot B$
$\Rightarrow E = (A \cdot B) . (\bar{A} \cdot B)$

A	B	$\bar{A}$	C = A · B	$D = \bar{A} \cdot B$	E = C . D
0	0	1	0	0	0
0	1	1	0	1	0
1	0	0	0	0	0
1	1	0	1	0	0

Disclaimer: None of the given options are correct

Page No 89:

Question 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:

When electric field is applied across a semi-conductor, electrons in conduction band gets accelerated and acquires an energy. This means they move from lower energy level to higher energy level. However, holes move from higher energy level to lower energy level in valence band.

Hence, the correct options are (a) and (c).

Page No 90:

Question 10:

Consider an npn 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:

In a npn transistor with its base-emitter junction forward biased and collector base junction reverse biased, electrons crossover from emitter to collector and also electrons move from emitter to base.
Hence, the correct answers are option (a) and (c).

Page No 90:

Question 11:

In given figure shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true?

(a) At V_i = 0.4V, transistor is in active state.
(b) At V_i = 1V , it can be used as an amplifier.
(c) At V_i= 0.5V, it can be used as a switch turned off.
(d) At V_i = 2.5V, it can be used as a switch turned on.

Answer:

(a) At V_i = 0.4 V, I_c = 0, so only output voltage is due to V_cε which means 0.4 V transistor is not in active mode.
(b) When V_i = 1 V transistor is in between active region so it can be used as an amplifier.
(c) At V_i = 0.5 V, transistor is in cut off state can be used as a switch turned off.
(d) At V_i = 2.5 V, transistor is beyond the active region. I_c is in saturated state, so it can be used as a switch turned on.

Hence, the correct options are (b), (c) and (d).

Page No 90:

Question 12:

In a npn 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:

Let I_e, I_b and I_c be the emitter current, collector current and the base current.
Collector current, I_c= 10 mA
I_c= 95% of I_{e
$\Rightarrow I_{c} = \frac{95}{100} I_{e} \Rightarrow I_{e} = \frac{10 \times 100}{95} mA = 10.53 mA$}
_Also,
I_e= I_b+ I_c
⇒ I_b= 10.53 − 10 = 0.53 mA

= 0.53 mA

Hence, the correct options are (b) and (c).

Page No 90:

Question 13:

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

Answer:

Depletion zone is formed due to diffusion of electron from n to p side. Due to diffusion positive immobile ions exists on n side and negative ions on p- side .Recombination of electrons and holes takes place on p-side. This results in (n-side more positive than p-side) formation of a junction field and potential barrier across the depletion zone.
Hence, the correct answers are options (a), (b) and (c).

Page No 90:

Question 14:

What happens during regulation action of a Zener diode?

(a) The current in and voltage across the Zenor 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:

During the reverse biased mode, the zener diode offers constant voltage drop across the terminals. And during the regulation action of the zener diode, the current across the series resistance R_s is changed, so, the resistance offered by the zener diode also changes. The current in the zener diode changes, but the voltage drop remains constant.
Hence, the correct options are (b) and (d).

Page No 90:

Question 15:

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

Answer:

Here R_L is
Ripple factors of a full wave rectifier using capacitor filter is given by
$r = \frac{1}{\sqrt[4]{3} R_{L} C v}$
$\Rightarrow r \propto \frac{1}{R_{L}}, r \propto \frac{1}{C}, r \propto \frac{1}{v},$
This means in order to reduce r, the value of R_{L ,}v and C should be increased.

Hence, the correct answers are options (a), (c) and (d).

Page No 91:

Question 16:

The breakdown in a reverse biased p–n junction diode 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) strong electric field in a depletion region if the doping concentration is small.
(d) strong electric field in the depletion region if the doping concentration is large.

Answer:

In case of reverse bias, minority charge carriers will be accelerated, which on striking with atoms can cause ionization that results in more charge carriers because of secondary electrons and hence breakdown occurs.
Now, when doping is large, more number of ions will be there in the depletion region which results in strong electric field and hence breakdown occurs.

Hence, the correct options are (a) and (d).

Page No 91:

Question 17:

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

Answer:

The elemental dopants for Silicon and Germanium are usually chosen from group XIII or group XV. The size of dopant should be such that it should not distort the pure semi-conductor lattice and should able to easily contribute a charge carrier during the formation of covalent bond with the host atom. The elements of group XIII or XV fulfills these requirements, hence they are used as a dopants.

Page No 91:

Question 18:

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?

Answer:

The conduction level of any element depends upon the energy gap in between the conduction band and valence band.
In conductors, no energy gap is in between the conduction band and valence band.
For insulator, energy gap is large.
For semiconductor, energy gap is moderate.
The energy gap for Sn is 0 eV, C is 5.4 eV, Si is 1.1 eV and Ge is 0.7 eV.
That is why Sn is conductor, C is insulator, Ge and Si are semiconductors.

Page No 91:

Question 19:

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

Answer:

We cannot measure the potential barrier across p-n junction by voltmeter because the voltmeter must have a resistance very high as compared to the junction resistance, the latter being nearly infinite.

Page No 91:

Question 20:

Draw the output waveform across the resistor (in given figure).

Answer:

In the given circuit, waveform is connected at A and waveform obtained when the diode only works in forward biased. So the output obtained only when positive input is given.
So, the output waveform will be as shown below:

Page No 91:

Question 21:

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 10V?
(ii) if dc supply voltage is 5V?

Answer:

The voltage gains of X, Y and Z are
Av_x= 10
Av_y= 20
Av_z= 30
Input signal peak value, Δv_i= 1 mV

= 10^–3 V

Total voltage amplification = \frac{Output Signal Voltage}{Input Signal Voltage}

\begin{array}{rcl} \Rightarrow & Output voltage, ∆ v_{0} = A v_{x} A v_{y} A v_{z} \times ∆ v_{i} \\ = & 10 \times 20 \times 30 \times 10^{- 3} \\ = & 6 V \end{array}

(i) It is given that dc voltage is 10 V, but the output is 6V and as the theoretical gain is equal to practical gain so output can never be greater than 6V.
So, V_o = 6 V

(ii) It is given that dc voltage is 5 V, so output cannot exceed 5 V.
So, V_o = 5 V

Page No 91:

Question 22:

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 voilate conservation of energy?

Answer:

In CE transistor amplifier, DC supply is connected to give energy to signal due to which the power gain is very high. The extra power required for amplified output is obtained from DC source and due to which the circuit does not violet the law of conservation.

Page No 92:

Question 23:

(i) Name the type of a diode whose characteristics are shown in given figure (A) and Fig. 14.9(B).
(ii) What does the point P in figure (A) represent?
(iii) What does the points P and Q in Fig. (B) represent?

Answer:

The I-V characteristics of a Zener diode is shown in graph (a). It is seen that when the applied reverse bias voltage (V) reaches the breakdown voltage at Vz (at point P) of the Zener diode, there is a large change in the current. After the breakdown voltage Vz , a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant, even though current through the Zener diode varies over a wide range. This property of the Zener diode is used for regulating supply voltages so that they are constant.
Now, the I-V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes in graph (b). This is because a solar cell does not draw current but supplies the same to the load.
(i) Characteristics graph (a) shows Zener junction diode and graph (b) shows solar cell.
(ii) Point P in figure (a) represents Zener breakdown voltage.
(iii) Point P represents open circuit voltage while point Q represents short circuit current on solar cell characteristic curve (b).

Page No 92:

Question 24:

Three photo diodes D₁, D₂ and 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 ?

Answer:

Photo-diodes D₁, D₂, D₃ have band gaps of 2.5 eV, 2 eV and 3 eV.
Wavelength, λ = 6000 Å
Energy of the incident photon, $E = \frac{h c}{λ}$
$\Rightarrow E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6000 \times 10^{- 10}} J = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{6000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} eV = 2.06 eV$
It means it is only valid for D₂ as the incident radiation can be detected by photo-diode if energy of the incident radiation is greater than the band gap. So, D₂ will detect these radiations.

Page No 92:

Question 25:

If the resistance R₁ is increased (in given figure), how will the readings of the ammeter and voltmeter change?

Answer:

From the circuit diagram. Base current, $I_{B} = \frac{V_{B B} - V_{B E}}{R_{1}}$

When R₁ is increased, base current will decrease. Current in ammeter is collector current.
Collector current, I_c = $β$ I_B ,
So, when base current I_Bdecreases, collector current I_cwill also decreases.
Hence, the reading of voltmeter and ammeter also decreases.

Page No 92:

Question 26:

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

Answer:

When a car enters the gate, any one or both are opened. This means OR gate gives the desired output.
The truth table for the OR gate is:

A	B	Output
0 0 1 1	0 1 0 1	0 1 1 1

The corresponding circuit is as drawn below:

Page No 92:

Question 27:

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

Answer:

NOT gate consist of one input and one output. This gate cannot be realized by using diodes whereas it can be realized by using transistors.
For making it, bast B of the transistor is connected to input A through base resistor whereas emitter is earthed and the collector is to be connected with a 5V battery. So, the output y is the voltage at C w.r.t. earth. In this base and collector emitter are chosen such 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 the saturation state.

A	Output
0	0
1	1

Page No 92:

Question 28:

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

Answer:

In case of elemental semiconductor, the band gap lies in infrared region that is why it cannot be used to make visible LEDs.

Page No 93:

Question 29:

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

Answer:

Boolean expression of the given circuit, Y = A·B. Which means the given circuit resemble AND gate the truth table will be

A	B	Y = A·B
0 0 1 0	0 1 0 1	0 0 0 1

Page No 93:

Question 30:

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

Answer:

Power, P = 1 W
Zener breakdown voltage, V_z = 5 V
Minimum voltage, V_min = 3 V
Maximum voltage, V_max = 7 V
Power, P = VI
So, the maximum current,
$\Rightarrow {I_{z}}_{\max} = \frac{P}{V_{z}} = \frac{1}{5} = 0.2 A So, R_{s} = \frac{V_{\max} - V_{z}}{I_{z_{\max}}} = \frac{(7 - 5)}{0.2} = \frac{2}{0.2} = 10 Ω$
Hence, the value of R_s for safe operation is 10 Ω.

Page No 93:

Question 31:

If each diode in given figure has a forward bias resistance of 25Ω and infinite resistance in reverse bias, what will be thevalues of the current I₁, I₂, I₃ and I₄?

Answer:

Given → Forward biased resistance = 25 Ω
Reverse biased resistance = ∞
Resistance in branch AB, R₁ = 25 + 125 = 150 Ω
Resistance in branch EF, R₂ = 25 + 125 = 150 Ω
Also, AB is parallel to EF, resultant resistance
$\Rightarrow \frac{1}{R'} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{150} + \frac{1}{150} = \frac{2}{150} = \frac{1}{75} \Rightarrow R' = 75 Ω$
Total resistance, R = 75 + 25 = 100 Ω
Current, $I_{1} = \frac{V}{R} = \frac{5}{100} = 0.05 A$
The diode in branch CD is in reverse biased having resistance infinite.
$\Rightarrow I_{3} = 0 Also, I_{1} = I_{2} + I_{3} + I_{4} \Rightarrow I_{1} = I_{2} + I_{4}$
Resistance across R₁ and R₂ are same, so current I₄ = I_{2
$\Rightarrow I_{1} = I_{2} + I_{2} \Rightarrow I_{1} = 2 I_{2} \Rightarrow I_{2} = 0.025 A$

Hence, the values of current $I_{1} = 0.05 A, I_{2} = 0.025 A, I_{3} = 0 and I_{4} = 0.025 A$ .}

Page No 93:

Question 32:

In the circuit shown in given figure, when the input voltage of the base resistance is 10V, V_be is zero and V_ce is also zero. Find the values of I_b, I_c and β.

Answer:

Given → Voltage across R_B, V_B= 10 V
Base resistance, R_B = 400 kΩ
Collector resistance, R_C = 3 kΩ
Voltage across R_C , V_C = 10 V
$Base current, I_{B} = \frac{V_{B}}{R_{B}} = \frac{10}{400 \times 1000} = 25 \times 10^{- 6} A Collector current, I_{C} = \frac{V_{C}}{R_{C}} = \frac{10}{3 \times 1000} = 3.3 \times 10^{- 3} A Current gain, β = \frac{I_{C}}{I_{B}} = \frac{3.3 \times 10^{- 3}}{25 \times 10^{- 6}} = 132$
Hence, the value of I_B= 25 × 10^–6 A, I_C= 3.3 × 10^–3 A, β = 132.

Page No 93:

Question 33:

Draw the output signals C₁ and C₂in the given combination of gates (in given figure).

Answer:

Combination of gates C₁ and C₂:
Truth table of C₁:

Time slot	A	_B	C	_D	E	F	G	H	I	C₁
0-1	1	0	1	0	0	1	0	1	1	0
1-2	1	1	1	1	0	0	0	1	1	0
2-3	0	1	0	1	1	0	0	1	1	0
3-4	1	0	1	0	0	1	0	1	1	0
4-5	0	0	0	0	1	1	1	0	0	1

Truth table of C₂:

Time slot	A	_B	C	_D	E	F	G	C₂
0-1	1	0	1	0	0	1	1	0
1-2	1	1	1	1	0	0	0	1
2-3	0	1	0	1	1	0	1	0
3-4	1	0	1	0	0	1	1	0
4-5	0	0	0	0	1	1	1	0

Hence, the corresponding output signal for C₁ and C₂ are shown below:

Page No 94:

Question 34:

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 R_B and R_C for a transistor whose V_BE = 0.7 V,so that the transistor is operating at point Q as shown in the
characteristics shown in in given figure (b).

Given that the input impedance of the transistor is very small and V_CC = V_BB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions.

Answer:

Given,
V_BE= 0.7 V
V_CC= V_BB= 16 A
V_CE= 8V
I_C= 4 mA = 4 × 10^–3 A
I_B= 30µA = 30 × 10^–6 A
Output characteristic,
$V_{C C} = I_{C} R_{C} + V_{C E} \Rightarrow R_{C} = \frac{V_{C C} - V_{C E}}{I_{C}} = \frac{16 - 8}{4 \times 10^{- 3}} = 2 kΩ$
$Also, V_{B B} = T_{B} R_{B} + V_{B E} \Rightarrow R_{B} = \frac{16 - 0.7}{30 \times 10^{- 6}} = 510 \times 10^{3} Ω = 510 kΩ$
$Current gain, β = \frac{I_{C}}{I_{B}} = \frac{4 \times 10^{3}}{30 \times 10^{- 6}} = 133 Voltage gain = \frac{β R_{C}}{R_{B}} = \frac{133 \times 2 \times 10^{3}}{510 \times 10^{3}} = 0.52$
$\begin{array}{rcl} Power gain & = & β \times Voltage gain \\ = & 133 \times 0.52 \\ = & 69 \end{array}$

Page No 94:

Question 35:

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

Answer:

When 20 sinωt signal gives input voltage less than 5 V then diode will be in reverse bias because after 5 V diode will get a positive voltage at its p-junction. So, the resistance of diode remain infinity and the input signal will not pass through diode and battery. So, the output increase from 0-5 V. However, when voltage decreases the diode will be in reverse bias and the output will again fall from 5 V to 0 V when input changes. When input voltage becomes negative (there is opposite of 5 V battery in p-junction input voltage becomes more than 5 V) the diode is in reverse biased and will not conduct current through CD and in output across AB will get same as input AC for negative cycle diode offer infinite resistance as compared to R. So, the waveform will be

Page No 94:

Question 36:

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

Answer:

(i) When a pentavalent AS is added to Si, n-type wafer is produced.
Number of majority carriers electrons due to doping of AS,
$\begin{array}{rcl} n_{e} & = & N_{D} = \frac{1}{10^{6}} \times 5 \times 10^{28} \\ = & 5 \times 10^{22} / m^{3} \end{array}$
Number of minority carriers in n-type wafer is,
$\begin{array}{rcl} n_{h} & = & \frac{n i^{2}}{n_{e}} = \frac{{(1.5 \times 10^{16})}^{2}}{5 \times 10^{22}} \\ = & 0.45 \times 10^{10} / m^{3} \end{array}$
When boron is implanted in silicon crystal p-type wafer is created with number of holes.
$\Rightarrow n_{h} = N_{A} = \frac{200}{10^{6}} (5 \times 10^{28}) = 1 \times 10^{25} / m^{3}$
Number of minority carriers in p-type wafer is $n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{1 \times 10^{25}} = 2.25 \times 10^{27} / m^{3}$

(ii) The minority carrier holes of n-region wafer will contribute more to reverse saturation current when diode is reverse biased than minority carrier electrons of p-region wafer.

Page No 95:

Question 37:

An X-OR gate has following truth table:

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	0

It is represented by following logic relation
Y =

\bar{A}

.B+ A.

\bar{B}

Build this gate using AND, OR and NOT gates.

Answer:

Given:
Y = $\bar{A} . B$ + $A \bar{B}$
= y₁ + y₂
Here, y₁ = $\bar{A} . B$ and y₂ = $A \bar{B}$
y₂ can be obtained by using AND gate of which A is connected to a NOT gate.
y₂ can be obtained using AND gate of which B is connected to a NOT gate.
The logic circuit will be

Page No 95:

Question 38:

Consider a box with three terminals on top of it as shown in given figure (a):

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 shown in Fig. 14.18 (b).

The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit.
The graphs are
(i) when A is positive and B is negative

(ii) when A is negative and B is positive

(iii) When B is negative and C is positive

(iv) When B is positive and C is negative

(v) When A is positive and C is negative

(vi) When A is negative and C is positive

From these graphs of current – voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B and C.

Answer:

(i) In case (i), D₁ is reverse biased as n₁ is positive and p₁ is negative.

The resistance will be linear/ohmic.
(ii) In case (ii), D₁ is forward bias having knee voltage of 0.7 V.
$\Rightarrow Slope of I - V curve = \frac{∆ I}{∆ V} \Rightarrow \frac{1}{Slope} = \frac{∆ V}{∆ I} = resistance = 1000 Ω$

(iii) In this case D₂ is forward bias, as n₂ is negative and p₂ is positive

(iv) In this case D₂ is reverse bias as n₂ is positive and p₂ is negative

(v) In this case both diodes are reverse biased n₁ is positive and p₂ is negative.

(vi) In this case both diodes are forward biased n₁ is negative and p₂ is positive.

Page No 97:

Question 39:

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

Answer:

Given:
Collector Current, I_C = 1 mA,
Collector emitter voltage, V_CE = 3 V
Base emitter voltage, V_BE = 0.5 V
V_CC = 12 V
$β$ = 100
R_C = 7.8 kΩ
Since base current is very small
⇒ I_C ≈ I_E
Also, I_C(R_C + R_E) + V_CE = 12
⇒ (R_E+ R_C) × 1 × 10^–3 + 3 = 12
⇒ R_E + R_C = 9 kΩ
$and R_{B} = \frac{12 - 1.7}{\frac{I_{C}}{β} + 0.085} = \frac{10.3}{0.01 + 0.085} \Rightarrow R_{B} = 108 kΩ$

Page No 97:

Question 40:

In the circuit shown in given figure, find the value of R_C.

Answer:

Firstly image to be added.
$Given \to R_{B} = 100 kΩ, R_{E} = 1 kΩ V_{BE} = 0.5 V, V_{CE} = 3 V, V_{CC} = 12 V$

Since base current is very small
$I_{E} \approx I_{C} = β I_{B} \Rightarrow I_{B} (R_{B} + {βR}_{E}) = V_{CC} - V_{BE} \Rightarrow I_{B} = \frac{V_{CC} - V_{BE}}{R_{B} + β R_{E}} = \frac{12 - 0.5}{100 \times 10^{3} + 100 \times 1 \times 10^{3}} = \frac{11.5}{200} mA$
Similarly, I_C(R_C + R_E) = V_CC – V_CE
$\Rightarrow R_{C} + R_{E} = \frac{V_{CC} - V_{CE}}{I_{C}} = \frac{V_{CC} - V_{CE}}{β I_{B}} \Rightarrow R_{C} + R_{E} = \frac{200}{11.5} \frac{(12 - 13)}{100} = 1.56 kΩ \Rightarrow R_{C} = 1.56 - 1 = 0.56 kΩ$
Hence, the value of R_C is 0.56 kΩ.

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