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#### Page No 416:

#### Answer:

For sodium, the atomic number is 11. The electronic configuration of sodium is 1s^{2}^{ }2s^{2} 2p^{6} 3s^{1}.

One sodium atom has 11 electrons. Thus, if the sodium crystals consist of *N* atoms, the total number of electrons will be 11 *N*. We know that for each atom, there are two states in the energy level 1s. Thus, the sodium crystal will have 2 *N* states for 1s energy level. Similarly, the number of states in 3s energy level will also be 2 *N. *1s state is filled under normal condition. But the 3s state has only one electron per sodium atom, so the 3s band will be half-filled.

#### Page No 416:

#### Answer:

A solid consists of a combination of closely spaced energy levels. These energy levels are discrete but they have very small energy gap between two consecutive levels so they are reffered as band.However, the energy levels in the band are discrete.

#### Page No 416:

#### Answer:

It will be a conductor. As the elements having partially filled conduction band belong to the category of elements whose outermost subshell consists of an odd number of electrons, they are good conductors of electricity. When an electric field is applied, electrons in the partially filled band gain energy and start drifting. So, the conductor will conduct even at 0 K. A semiconductor behaves like an insulator at 0 K and an insulator conducts poorly only at very high temperatures.

As the given material has free electrons to conduct even at 0 K, it is a conductor.

#### Page No 416:

#### Answer:

An electron jumps from the valence band to the conduction band only when it has gained sufficient energy. The thermal collisions sometimes do not provide sufficient energy to the electron to jump. Also, energy is lost in the form of heat because of the collision of the carriers with other charge carriers and atoms. Because of all these losses, only few electrons are left with sufficient energy to jump from the valence band to the conduction band. So, the population of electron in the conduction band does not keep on increasing with time.

#### Page No 417:

#### Answer:

Fermi level: it is the energy level occupied by the highest energy electron.

In an extrinsic semiconductor for example in n-type semiconductor, fermi level lies close to the conduction band so it needs a very small amount of energy to excite the electron from fermi level to conduction band. This energy is comparable to the thermal excitation energy. So even at room temperature,these semiconductors can conduct.For a p-type semiconductor, fermi level lies close to valence bane because here conduction takes place majorly via holes.So by the thermal excitation,a bond is broken and an electron hole pair is created.Out of this,hole comes to the valence band for conduction or equivalently an electron goes to the conduction band. In an intrinsic semiconductor, no impurity is doped so fermi level lies at the centre of band gap. Here only few electrons get sufficient energy via repeated thermal collisions to jump from the fermi level to the conduction band.Hence the conductivity of intrinsic semiconductor is less as compared to extrinsic semiconductor.

#### Page No 417:

#### Answer:

At 0 K, the valence band is full and the conduction band is empty. As no electron is available for conduction in an intrinsic semiconductor, the intrinsic semiconductor at 0 K acts as an insulator and hence offers infinite resistance.

#### Page No 417:

#### Answer:

Holes do not exist in reality. They exist only virtually. When an electron jumps from the valence band to the conduction band, a vacancy is created at the place from where the electron had jumped. This vacancy is called a hole. So, a valence or conduction hole is a virtual concept only.

#### Page No 417:

#### Answer:

A *p*-type semiconductor is formed by doping a group 13 element with group 14 element (Si or Ge). As the group 13 element has only 3 electrons in its valence shell and the group 14 element has 4 electrons in its valence shell, when the group 13 element, say, Al, replaces one Si in the silicon crystal, only 3 covalent bonds are formed by it. And the fourth covalent bond is left in need of one electron. So, it creates a hole. Since the atom as a whole is electriclly neutral, the *p*-type semiconductor is also neutral.

In a *p*‒*n* junction, when the diffusion of holes takes place across the junction because of the difference in the concentration of charge carriers from *p* to *n* sides, these holes neutralise some of the electrons on the *n* side. So, the atom attached with that electron becomes one electron deficient and hence positively charged. This makes the *n* side of the *p‒n* junction positively charged and the *p* side of the *p*‒*n* junction negatively charged.

#### Page No 417:

#### Answer:

When the temperature of a reverse-biassed *p*‒*n* junction is increased, the breaking of bonds takes place because of the increase in the thermal energy of the charge carriers. Drift current is due to the flow of the minority carriers across the junction. So, when a *p‒n* junction is reverse biassed, the applied voltage supports the flow of minority charge carriers across the junction. Thus, the drift current increases with increase in temperature in a reverse-biassed* p‒n* junction.

#### Page No 417:

#### Answer:

It should be an ideal vacuum diode. When a pn junction diode is reverse biassed then a small current called reverse current flows across the diode.As the the *p‒n* junction diode allows some current in reverse biassed condition also so the given diode can not be a pn junction diode.

#### Page No 417:

#### Answer:

The amplifier takes this energy from the power supply. Amplifiers take the energy from the power supply and control the output to match with the input signal but with greater amplitude.

#### Page No 417:

#### Answer:

(c) both electrons and holes

A hole is created in a semiconductor when a valence electron moves to the conduction band. When potential difference is applied across the semiconductor, the electron drifts opposite to the electric field applied, while the hole moves along the electric field. Therefore, electric conduction takes place in a semiconductor because of both electrons and holes.

#### Page No 417:

#### Answer:

(b) *n* will increase but *v* will decrease

As we increase the temperature, additional electron‒hole pairs are created in a semiconductor. As a result, the number of charge carriers increases.

Now, drift velocity $\left({v}_{d}\right)$ is given by

${v}_{d}=\frac{-eE\tau}{m}$

As the temperature increases, the relaxation time of charge carriers $\left(\tau \right)$ decreases. As a result, *v _{d}* decreases.

#### Page No 417:

#### Answer:

(b) *n _{p}* =

*n*

_{e}As the intrinsic semiconductor is free from all impurities, the number of electrons is equal to the number of holes.

#### Page No 417:

#### Answer:

(d) *n _{p}* ≠

*n*

_{e}Extrinsic semiconductors are formed by doping either an

*n*-type material or

*a*

*p*-type material with a pure semiconductor, Thus, it may have more number of holes (if the doping material is of

*p*type) or more number of electrons (if the doping material is of

*n*type).

#### Page No 417:

#### Answer:

(c) uncharged

A *p*-type semiconductor is formed by doping a pure semiconductor with a *p*-type material. As impurity atoms take the position of the germanium atom in a germanium crystal, three electrons of a *p*-type material form covalent bonds by sharing electrons with three neighbouring germanium atoms. However, the fourth covalent bond is left incomplete, with a want of one electron. This creates a hole. As the atom as a whole is neutral, the *p*-type material is also neutral.

#### Page No 417:

#### Answer:

(a) increases

When an impurity (either a *p*-type atom or an *n*-type atom) is doped into an intrinsic semiconductor, it increases the number of charge carriers in the intrinsic semiconductor. As conductivity is directly related to the number of charge carriers, the conductivity of a semiconductor increases with doping.

#### Page No 417:

#### Answer:

(a) there will not be a steady current in the circuit

In a *p‒n* junction, current flows only if it is connected to the battery. If two ends of a *p‒n* junction are joined by a wire, then there will be diffusion and drift currents in the circuit and they will cancel each other. Hence, no current will flow in the circuit.

#### Page No 417:

#### Answer:

(a) from the *n*-side to the *p*-side

After the diffusion of majority charge carriers across a *p‒n* junction, an electric field is set up because of the accumulation of immobile ions at the junction. These further oppose the motion of majority charge carriers across the junction. As a result, electrons from the *p* region start moving to the *n* region and holes from the *n* region start moving to the *p *region. This constitutes the drift current. As the direction of the current is opposite to the direction of the motion of the electrons, the direction of the drift current is from the *n* side to the *p* side.

In forward biasing, there is no movement of electrons from the *p* region to the *n* region and of holes from the *n* region to the *p *region. Hence, there is not drift current.

#### Page No 417:

#### Answer:

(b) from the *p*-side to the *n*-side

When a *p‒n* junction is formed then because of the difference in the concentration of charge carriers in the two regions, electrons from the *n *region move to the *p* region and holes from the *p *region move to the *n *region. Since the direction of the current is always opposite to the motion of electron, the direction of the current is from the *p* side to the *n *side.

Similarly, when the junction is forward biassed, the positive terminal of the battery is connected to the *p* side of the* p‒n* junction and the negative terminal of the battery is connected to the *n* side of the *p‒n* junction. As a result, electrons in the *n *side of the *p‒n* junction are repelled by the negative terminal of the battery and they move to the *p* side, where the positive terminal of the battery attracts them. Similarly, holes from the *p* side of the *p‒n* junction are repelled by the positive terminal of the battery and they move to the *n* side, where the negative terminal of the battery attracts them. Thus, they give diffusion current from the *p* side to the *n* side across the *p‒n* junction.

In reverse biassing, there is no flow of majority carriers across the junction; hence, there is not diffusion current. Here, the flow of majority carriers is opposed by the applied voltage.

#### Page No 417:

#### Answer:

(a) if the junction is forward-biassed

In the* *forward biassing of a *p*−*n* junction, the positive terminal of the battery is connected to the *p* side of the* **p*−*n* junction and the negative terminal of the battery is connected to the *n* side of the *p*−*n* junction. As a result, electrons in the *n *side of the *p*−*n* junction are repelled by the negative terminal of the battery and move to the* p* side, where the positive terminal of the battery attracts the electrons. Similarly, holes from the *p* side of the *p*−*n* junction are repelled by the positive terminal of the battery and move to the *n* side, where the negative terminal of the battery attracts the holes. Thus, they give diffusion current across the *p*−*n* junction.

In case of reverse biassing, no conduction takes place across the junction because of the diffusion of majority carriers. Hence, there is no diffusion current.

If the junction is unbiased, then diffusion current is initially maximum. But at equilibrium, diffusion current becomes equal to drift current.

#### Page No 417:

#### Answer:

(b) circuit 2 and circuit 3

In circuit 1, one diode is forward biassed and the other diode is reverse biassed.

The forward-biassed diode offers zero resistance (ideally) to the current flow, so it can be replaced by a short circuit. The voltage drop across the first diode will be zero. The second diode is reverse biassed, so it can be replaced by an open circuit; hence, the voltage drop across this diode will be maximum.

In circuit 2, both the diodes are forward biassed, so they can be replaced by short circuits; hence, the voltage drop across both of them will be minimum and equal.

In circuit 3, both the diodes are reverse biassed, so both can be replaced by open circuits; hence, the voltage drop across both of them will be maximum and equal.

#### Page No 417:

#### Answer:

In circuit (a), the diode is forward biassed. So, it offers negligible resistance to the flow of current and can thus be replaced by a short circuit. Now, the capacitor charge will leak through the resistance and decay exponentially with time.

Capacitor charge = $\frac{VC}{\mathrm{e}}$

In circuit (b), the diode is reverse biassed. So, it offers infinite resistance to the current flow and can thus be replaced by an open circuit. As the circuit is open now, no current can flow across the resistance. So, the charge in the capacitor cannot leak through the resistor.

Capacitor charge = *VC*

#### Page No 418:

#### Answer:

(c) a bond is broken on the *n*-side and the electron freed from the bond jumps to a broken bond on the *p*-side to complete it

A hole diffuses from the *p* side to the *n* side in a *p*−*n* junction; that is, an electron moves from the *n* side to the *p* side. This implies that a bond is broken on the *n* side. As the electron travels towards the *p* side, which is rich in holes, it combines with a hole. A hole is created because of the deficiency of one electron. So, when an electron combines with a hole, it completes that bond.

#### Page No 418:

#### Answer:

(c) the base has the least concentration of impurity

When the emitter of a transistor is forward biassed, the majority carriers move from the emitter to the collector through the base. As the base is thin and lightly doped, only small amount of a combination of electrons and holes takes place, leading to weak base current. This makes the collector current equal to the emitter current.

If we make any other choice for least concentration of impurity, we will have a low value of collector current. Thus, the purpose of a transistor will not be fulfilled.

#### Page No 418:

#### Answer:

(d) 'base' and 'base'

In transistors, the emitter−base junction is forward biassed and the collector−base junction is reverse biassed. This is done to make majority carriers cross the emitter−base junction; this helps the electric current to flow through the transistor.

#### Page No 418:

#### Answer:

(a) there are no free electrons at 0 K

(c) the number of free electrons increases with temperature

(d) the number of free electrons is less than that in a conductor

In semiconductors, the valence band is full at 0 K, but the conduction band is empty. So, no free electron is available for conduction at 0 K.

As the temperature increases, covalent bonds that provide free charge carriers for conduction in a semiconductor break.

As the conduction band in metals is already partially filled at 0 K, many free electrons below the Fermi level acquire energy from an external source or temperature, jump to the conduction band and start behaving like free electrons. Hence, metals contain more free electrons than semiconductors.

#### Page No 418:

#### Answer:

(b) holes and conduction electrons systematically go from the *p*-side to *n*-side and from the *n*-side to *p*-side, respectively

(c) there is no net charge transfer between the two sides

(d) there is a constant electric field near the junction

Because of the difference in the concentration of charge carriers in the *p*−*n* junction, holes from the *p* side move to the *n *side and electrons from the *n *side move to the *p* side. This motion of charge carriers gives rise to diffusion current.

Because of this, a negative space charge region is formed in the *p* region and a positive space region is formed in the* **n* region. This sets up an electric field across the junction. Thus, there is a constant electric field near the junction.

This electric field further opposes the diffusion of majority charge carriers across the junction. As a result, an electron from the* p* region starts moving to the *n* region and a hole from the* n* region starts moving to the *p* region. This sets up drift current. Thus, there is a systematic flow of charge carriers across the junction. Also, there is no net charge transfer between the two sides.

#### Page No 418:

#### Answer:

(a) new holes and conduction electrons are produced continuously throughout the material

(d) holes and conduction electrons recombine continuously throughout the material except in the depletion region

In a *p‒n* junction diode, diffusion current flows because of the diffusion of holes from the *p* side to the * n* side and of electrons from the

*n*side to the

*p*side. The current flowing in the diode due to the diffusion of charge carriers across the junction is called the diffusion current. The current flowing in the diode due to the movement of minority carriers across the junction due to their thermal energy is called the drift current. In an unbiased diode, the net current flowing across the junction is zero due to the cancellation of the drift current by the diffusion current. For the flow of diffusion and drift currents, holes and electrons are produced continuously throughout the material. When a hole crosses the junction, it combines with an electron on the

*n*side. As the depletion region is devoid of free charge carriers, this recombination never takes place inside the depletion region.

#### Page No 418:

#### Answer:

(b) boron

(d) aluminium

A *p-*type semiconductor is formed by doping an intrinsic semiconductor with a trivalent atom (atom having valency 3). As phosphorous and boron have three valence electrons, they can be doped with silicon to make a *p*-type semiconductor.

#### Page No 418:

#### Answer:

(a) increasing the temperature

(b) doping acceptor impurities

(c) doping donor impurities

(d) irradiating ultraviolet light on it

We know that the conductivity of any semiconductor can be increased by increasing the number of charge carriers. All the given methods are effective in increasing the number of free charge carriers. Hence, all options are correct.

#### Page No 418:

#### Answer:

(d) a *p*−*n* junction

As a *p*−*n* junction allows the flow of current in forward bias and stops the current in reverse bias (almost negligible reverse leakage current flows in the reverse-biassed *p−n* junction), the device should be a *p*−*n* junction. Other options are examples of semiconductors that allow moderate current to flow and that do not have any effect of changing the polarity of the battery.

#### Page No 418:

#### Answer:

(c) The electron concentration increases.

When a semiconductor is doped with a donor type such as arsenic or phosphorous, which has five valence electrons, the donor atom replaces the Si or Ge atom. As a result, four out of the five electrons of the donor atom form a covalent bond by sharing an electron with four atoms of silicon. However, the fifth electron is free to move. Also, due to the breaking up of covalent bonds at room temperature, equal number of electrons and holes are produced. Thus, the total number of holes in the *n-*type semiconductor is less compared to the number of free electrons.

#### Page No 418:

#### Answer:

(a) *i _{C}* is slightly smaller than

*i*

_{E}(c)

*i*is much smaller than

_{B}*i*

_{E}The highlighted parts could not be edited, as the meaning could not be understood. Also, please check the last line for logical accuracy. Only one option is given as correct, while the solution gives two correct options.

We know that in the transistor base is slightly doped, therefore when the majority carriers due to forward biasing of emitter base junction, feel the repulsive force from the battery and pass over to the base region. This gives the emitter current

*i*

_{E}.

As the base is thin and lightly doped, only few majority carriers of the emitter are neutralised at the base. This gives the base current. Hence, base current $\left({i}_{B}\right)$ is low.

The remaining majority carriers of the emitter pass to the collector and give collector current

*i*

_{C}.

Thus, we get the relation given below:

*i*

_{E}

*= i*

_{B}+ i_{C}Thus, because of the base, current

*i*is slightly smaller than

_{C}*i*.

_{E}Hence, option (a) and (c) are correct.

#### Page No 418:

#### Answer:

(a) the base−emitter junction is forward-biassed

(d) the base−collector junction is reverse-biassed

In the normal operation of a transistor, the base−emitter junction is forward biassed and the base−collector junction is reverse biassed. This is done so that the conduction of majority carriers can take place across the emitter−base junction and the free electrons can reach the collector to give the output current.

#### Page No 418:

#### Answer:

(c) NAND gate

(d) NOR gate

i)

ii)

#### Page No 419:

#### Answer:

Density of sodium,* **d* = 1013 $\mathrm{kg}/{\mathrm{m}}^{3}$

Volume, *V* = 1 ${\mathrm{m}}^{3}$

Mass of sodium, *m* = *d$\times $V* = 1013 × 1 = 1013 kg

Molecular mass of sodium, *M* = 23

We know that 23 g sodium contains 6 atoms, so the number of atoms in 1023 kg sodium will be

$\frac{1013\times {10}^{3}\times 6\times {10}^{23}}{23}\phantom{\rule{0ex}{0ex}}=\left(\frac{1013\times 6}{23}\right)\times {10}^{26}\phantom{\rule{0ex}{0ex}}=264.26\times {10}^{26}$

(a) As the number of maximum possible electrons that can occupy the 3s band is 2, the total number of states in the 3s band will be

$N=2\times 264.26\times {10}^{26}\phantom{\rule{0ex}{0ex}}=528.52\times {10}^{26}\approx 5.3\times {10}^{28}$

(b) As the atomic number of sodium is 11, its electronic configuration is $1{\mathrm{s}}^{2},2{\mathrm{s}}^{2},2{\mathrm{p}}^{6},3{\mathrm{s}}^{1}$.

This implies that the 3s band is half-filled in case of sodium, so the total number of unoccupied states is 2.65 × 10^{28}.

#### Page No 419:

#### Answer:

Pure semiconductors are intrinsic semiconductors or semiconductors without any doping.

We know that for pure semiconductors, the number of conduction electrons is equal to the number of holes.

Number of electrons in volume 1 m^{3} = $6\times {10}^{19}$

Number of holes in volume 1 m^{3} = $6\times {10}^{19}$

Given volume:

*V* = 1 cm × 1 cm × 1 mm

*$\Rightarrow $V *= 1 × 10^{−2}^{ }× 1 × 10^{−2}^{ }× 10^{−3}

*$\Rightarrow $V* = 10^{−7} m^{3}

Now,

Number of holes in volume ${10}^{-7}{\mathrm{m}}^{3}$:

*N* = 6 × 10^{19} × 10^{−7} = 6 × 10^{12}

#### Page No 419:

#### Answer:

Given:

Band gap between the conduction band and the valence band, *E* = 0.23 eV

Boltzmann's constant, *k* = 1.38 × 10^{−23} J/K

We need to find the temperature at which thermal energy *kT* becomes equal to the band gap of indium antimonide.

∴ *kT = E*

$\Rightarrow 1.38\times {10}^{-23}\times T=0.23\times 1.6\times {10}^{-19}\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{0.23\times 1.6\times {10}^{-19}}{1.38\times {10}^{-23}}\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{0.23\times 1.6\times {10}^{4}}{1.38}\phantom{\rule{0ex}{0ex}}\Rightarrow T=0.2666\times {10}^{4}\approx 2670\mathrm{K}$

#### Page No 419:

#### Answer:

Given:

Band gap of silicon,* E* = 1.1 eV

Temperature, *T* = 300 K

Boltzmann's constant, *k *= _{$8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}$}

(a) We need to find out the ratio of the band gap to *kT*.

Ratio$=\frac{1.1}{kT}$

$=\frac{1.1}{8.62\times {10}^{-5}\times 3\times {10}^{2}}\phantom{\rule{0ex}{0ex}}=42.53=43\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(b) The new ratio is $\frac{1}{10}$th of the earlier ratio.

i.e. New ratio = 4.253

We know,

Ratio = (Band gap)/(*kT*)

$\Rightarrow 4.253=\frac{1.1}{8.62\times {10}^{-5}\times T}\phantom{\rule{0ex}{0ex}}\Rightarrow T\mathit{=}\frac{1.1}{8.62\times {10}^{-5}\times 4.253}\phantom{\rule{0ex}{0ex}}\Rightarrow T\mathit{=}3000.4\mathrm{K}\approx 3000\mathrm{K}$

#### Page No 419:

#### Answer:

Before the thermal excitation, the electron was located at the top of the valence band.

After the thermal excitation, the electron was present in the acceptor level.

This implies that the energy gap between the valence band and the acceptor level is equal to the energy absorbed by the electron in thermal excitation.

Thus,

2*kT* = Energy gap between the acceptor level and the valence band

$\Rightarrow E=2\times 1.38\times {10}^{-23}\times 300\phantom{\rule{0ex}{0ex}}\Rightarrow E=(2\times 1.38\times 3)\times {10}^{-21}\mathrm{J}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\left(\frac{6\times 1.38}{1.6}\right)\times \frac{{10}^{-21}}{{10}^{-19}}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\left(\frac{6\times 1.38}{1.6}\right)\times {10}^{-2}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow E=5.175\times {10}^{-2}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\Rightarrow E=51.75\mathrm{meV}\phantom{\rule{0ex}{0ex}}$

#### Page No 419:

#### Answer:

Given:

Band gap = 3.2 eV

As the electron in the conduction band combines with the hole in the valence band, the minimum energy band gap (because maximum energy is released) through which the electron has to jump will be equal to the band gap of the material.

This implies that the maximum energy released in this process will be equal to the band gap of the material.

$\mathrm{Here},\phantom{\rule{0ex}{0ex}}E\mathit{}=3.2\mathrm{eV}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}\Rightarrow 3.2\mathrm{eV}=\frac{1242\mathrm{eV}-\mathrm{nm}}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =388.1\mathrm{nm}$

#### Page No 419:

#### Answer:

Given:

Wavelength, $\lambda =820\mathrm{nm}$

The minimum energy released in the recombination of a conduction band electron with a valence band hole is equal to the band gap of the material.

Band gap, $E\mathit{=}\frac{hc}{\lambda}$

$\Rightarrow E=\frac{1240}{820}\frac{\mathrm{eV}-\mathrm{nm}}{\mathrm{nm}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=1.5\mathrm{eV}$

#### Page No 419:

#### Answer:

Given:

Band gap of germanium, *E* = 0.65 eV

Wavelength of the incident radiation, λ = ?

For the electron‒hole pair creation, the energy of the incident radiation should be at least equal to the band gap of the material.

So,

$E\mathit{=}\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\lambda \mathit{=}\frac{hc}{E}\mathit{=}\frac{1242\mathrm{eV}-\mathrm{nm}}{0.65\mathrm{eV}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\lambda =1910.7\times {10}^{-9}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\lambda \approx 1.9\times {10}^{-6}\mathrm{m}$

#### Page No 419:

#### Answer:

Conductivity of any material increases when the number of free charge carriers in the material increases. When a photo diode is exposed to light, additional electron hole pairs are created in the diode; thus, its conductivity increases. So to change the conductivity of a photo diode, the minimum energy of the incident radiation should be equal to the band gap of the material.

In other words,

Band gap = Energy of the incident radiation

$\Rightarrow E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1242\mathrm{eV}-\mathrm{nm}}{620\mathrm{nm}}=2.0\mathrm{eV}$

#### Page No 419:

#### Answer:

Given:

Number of electrons in the conduction band, *n** *= *e*^{−ΔE}^{/2kT}

Band gap of diamond, ΔE_{1} = 6 eV

Band gap of silicon, ΔE_{2} = 1.1 eV

Now,

${n}_{\mathrm{diamond}}={\mathrm{e}}^{\frac{-{\mathrm{\Delta E}}_{1}}{2\mathrm{k}T}}\phantom{\rule{0ex}{0ex}}{n}_{\mathrm{silicon}}={e}^{\frac{-{\mathrm{\Delta E}}_{2}}{2\mathrm{k}T}}\phantom{\rule{0ex}{0ex}}\therefore \frac{{n}_{\mathrm{diamond}}}{{n}_{\mathrm{silicon}}}={e}^{\frac{-1}{2kT}(\u2206{E}_{1}-\u2206{E}_{2})}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{n}_{\mathrm{diamond}}}{{n}_{\mathrm{silicon}}}={\mathrm{e}}^{\frac{-1}{2\times 8.62\times {10}^{-5}\times 300}(6.0-1.1)}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{n}_{\mathrm{diamond}}}{{n}_{\mathrm{silicon}}}=7.15\times {10}^{-42}$

Because of more band gap, the conduction electrons per cubic metre in diamond are very less as compared to those in silicon, or we can simply say that they are almost zero.

#### Page No 419:

#### Answer:

Here,

Conductivity at temperature ${T}_{1}$ = ${\sigma}_{1}$

Conductivity at temperature ${T}_{2}$ = ${\sigma}_{2}$

Given:

${T}_{1}$ = 4 K

${T}_{2}$ = 300 K

Variation in conductivity with respect to the temperature and band gap of the material is given by

$\sigma ={T}^{3/2}\mathrm{e}{-}^{\mathrm{\Delta E}/2kT}\phantom{\rule{0ex}{0ex}}\therefore \frac{{\sigma}_{2}}{{\sigma}_{1}}={\left(\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}\right)}^{3/2}\frac{{e}^{-{\mathrm{\Delta E}}_{2}/2k{T}_{2}}}{{e}^{-{\mathrm{\Delta E}}_{1}/2k{T}_{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\sigma}_{2}}{{\sigma}_{1}}={\left(\frac{300}{4}\right)}^{3/2}\frac{{\mathrm{e}}^{-0.67/(2\times 8.62\times {10}^{-5}\times 300)}}{{\mathrm{e}}^{-0.74/(2\times 8.62\times {10}^{-5}\times 4)}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\sigma}_{2}}{{\sigma}_{1}}={10}^{463}$^{}

#### Page No 419:

#### Answer:

Initially, the total number of charge carriers per cubic metre is given by

*n*_{i} = 2 × 7 × 10^{15}

*$\Rightarrow $**n*_{i} = 14 × 10^{15}

Finally, the total number of charge carriers per cubic metre is given by

*n*_{f}_{ }= 14 × 10^{17}/m^{3}

We know that the product of the concentrations of holes and conduction electrons remains almost the same.

Let *x* be the number of holes.

Thus,

$(7\times {10}^{15})\times (7\times {10}^{15})=x\times (14\times {10}^{17}-x)\phantom{\rule{0ex}{0ex}}\Rightarrow 14x\times {10}^{17}-{x}^{2}=49\times {10}^{30}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-14x\times {10}^{17}-49\times {10}^{30}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{14\times {10}^{17}\pm \sqrt{(14{)}^{2}\times {10}^{34}+4\times 49\times {10}^{30}}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{14\times {10}^{17}\pm \sqrt{(14{)}^{2}\times {10}^{34}+4\times 49\times {10}^{30}}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{28.0007}{2}\times {10}^{17}=14.00035\times {10}^{17}$

This is equal to the increased number of holes or the number of atoms of boron added.

Number of atoms of boron added = $(14.00035\times {10}^{17}-7\times {10}^{15})=1386.035\times {10}^{15}$

Now, 1386.035 × 10^{15} atoms are added per 5 × 10^{28} atoms of S*i* in 1 m^{3}.

Therefore, 1 atom of boron is added per $\frac{5\times {10}^{28}}{1386.035\times {10}^{15}}$ atoms of S*i* in 1 m^{3}.

Proportion of boron impurity is $3.607\times {10}^{-3}\times {10}^{13}=3.607\times {10}^{10}$.

#### Page No 419:

#### Answer:

We know that for an intrinsic semiconductor, the concentration of electrons is equal to the concentration of holes.

The product of electron hole pair concentration always remains constant.

i.e. Number of holes$\times $Number of conduction electrons = Constant

Initially, the number of conduction electron per cubic metre = 6 × 10^{19}

We know,

Number of holes per cubic metre = Number of electrons per cubic metre

∴ Number of holes per cubic metre = 6 ×10^{19}

After doping,

Number of conduction electrons per cubic metre = 2 × 10^{23}

Now,

Let the number of holes per cubic metre be *x*.

As the product of electron hole pair concentration always remains constant,

$(6\times {10}^{19})(6\times {10}^{19})=(2\times {10}^{23})x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{6\times 6\times {10}^{19+19}}{2\times {10}^{23}}\phantom{\rule{0ex}{0ex}}\Rightarrow x=18\times {10}^{15}=1.8\times {10}^{16}$

#### Page No 419:

#### Answer:

Let the conductivity at temperature *T*_{1} be ${\sigma}_{1}$ and the conductivity at temperature *T* be ${\sigma}_{2}$.

Given:

${T}_{1}=300\mathrm{K}$

Band gap, *E* = 0.650 eV

Now,

According to the question,

$\sigma ={\sigma}_{0}\mathrm{e}{-}^{\frac{\Delta E}{2KT}}$

${\sigma}_{2}=2{\sigma}_{1}$

$\Rightarrow {\sigma}_{0}{\mathrm{e}}^{\frac{-\mathrm{\Delta E}}{2kT}}=2\times {\sigma}_{0}{\mathrm{e}}^{\frac{-\mathrm{\Delta}E}{2\times k\times {T}_{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\sigma}_{0}{\mathrm{e}}^{\frac{-\mathrm{\Delta E}}{2kT}}=2\times {\sigma}_{0}{\mathrm{e}}^{\frac{-\mathrm{\Delta}E}{2\times k\times 300}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{e}}^{\frac{-0.650}{2\times 8.62\times {10}^{-5}\times T}}=2\times {\mathrm{e}}^{\frac{-0.650}{2\times 8.62\times {10}^{-5}\times 300}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{e}}^{\frac{-0.650}{2\times 8.62\times {10}^{-5}\times T}}=6.96561\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

On taking natural natural log on both sides, we get

$\frac{-0.650}{2\times 8.62\times {10}^{-5}\times T}=-11.874525\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{T}=\frac{11.874525\times 2\times 8.62\times {10}^{-5}}{0.65}\phantom{\rule{0ex}{0ex}}\Rightarrow T=317.51178\approx 318\mathrm{K}$

#### Page No 419:

#### Answer:

Given:

Band gap = 1 eV

After doping,

Position of acceptor levels = 1 meV above the valence band

Net band gap after doping = (1 − 10^{−3}) eV = 0.999 eV

According to the question,

Any transition from one energy level to the other is almost forbidden if *kT* is less than 1/50 of the energy gap.

$\Rightarrow k{T}_{1}=\frac{0.999}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{1}=\frac{0.999}{50\times 8.62\times {10}^{-5}}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{1}=231.78\approx 232.8\mathrm{K}$

_{${T}_{1}$} is the temperature below which no transition is possible.

If *kT* is more than twice the gap, the upper levels have maximum population; that is, no more transitions are possible.

For the maximum limit,

$K{T}_{2}=2\times {10}^{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=\frac{2\times {10}^{-3}}{8.62\times {10}^{-5}}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=\frac{2}{8.62}\times {10}^{2}=23.2\mathrm{K}$

${T}_{2}$ is the temperature above which no transition is possible.

∴ Temperature range = 23.2−231.8

#### Page No 419:

#### Answer:

Let:

Depletion region width,* d* = 400 nm = 4 × 10^{−7} m

Electric field, *E* = 5 × 10^{5} Vm^{−1}

(a) Let the potential barrier be *V*.

The relation between the potential and the electric field is given by

*V = Ed
$\Rightarrow $V = E × d *= 5 ×

*d*

*$\Rightarrow $V*= 5 × 10

^{5}× 4 × 10

^{−7}= 0.2 V

(b) To find: Kinetic energy required

Energy of any electron accelerated through a potential of

*V*=

*eV*

Also, the minimum energy of the electron should be equal to the band gap of the material.

∴ Potential barrier ×

*e*= 0.2 eV (

*e*= Charge of the electron)

#### Page No 419:

#### Answer:

Potential barrier = 0.2 V

(a) The minimum kinetic energy of the hole should be equal to the band gap of the material.

Band gap = *eV*

KE = Potential difference × *e* = 0.2* *eV

(b) In forward biassing,

Kinetic energy = Effective potential of the barrier

∴ Kinetic energy = Potential under unbiased condition - Applied voltage

⇒ KE + *Ve *= 0.2* *eV

Here, *V* is the applied voltage.

⇒ KE = 0.2 − 0.1 = 0.1* *eV

(c) In reverse biassing,

Kinetic energy = Effective potential of the barrier

∴ Kinetic energy = Potential under unbiased condition + Applied voltage

⇒ KE − *Ve *= 0.2* *eV

Here, *V* is the applied voltage.

⇒ KE = 0.2 + 0.1 = 0.3 eV

#### Page No 419:

#### Answer:

Given:

Potential barrier, *d* = 250 meV

Initially,

Kinetic energy of a hole = 330 meV

We know that the kinetic energy of a hole decreases when the junction is forward biassed (because of the energy loss in crossing the junction).

Also, the kinetic energy of a hole increases when the junction is reverse biassed (because reverse bias voltage pushes the hole on the *n*-side towards the junction) in the given the *p-n* junction diode.

(a) The kinetic energy of a hole decreases under forward bias.

∴ Final kinetic energy = (300 − 250) meV

= 50 meV

(b) The kinetic energy of a hole increases under reverse bias.

∴ Final kinetic energy = (300 + 250) meV

= 550 meV

#### Page No 419:

#### Answer:

Given:

Drift current (current under reverse bias), *i*_{1} = 25 µA

Forward bias voltage, *V* = 200 mV

Net current under forward bias, *i*_{2} = 75 µA

(a) When the *p‒n* junction is in unbiased condition, no net current flows across the junction.

i.e. Drift current = Diffusion current

∴ Diffusion current = 25 µA

(b) Under reverse bias, the built in the potential and applied voltage opposes the motion of the majority carriers across the junction.

Thus, the diffusion current becomes zero.

(c) Under forward bias, the voltage supports the motion of majority carriers across the junction.

Let the actual current be *x*.

So,

(*x *− Drift current) = Forward-biassed current

$\Rightarrow x-25\mathrm{\mu A}=75\mathrm{\mu A}\phantom{\rule{0ex}{0ex}}\Rightarrow x=(75+25)\mathrm{\mu A}\phantom{\rule{0ex}{0ex}}\Rightarrow x=100\mathrm{\mu A}$

#### Page No 420:

#### Answer:

Given:

Drift current,* i _{d} *= 20 µA = 20 × 10

^{−6}A

Both holes and electrons are moving and contributing to the current flow.

We know that current is the rate of the flow of charge.

Thus, we need to find the number of electrons crossing unit area per second.

Now,

*t*= 1 s

${i}_{d}=\frac{Q}{T}\phantom{\rule{0ex}{0ex}}\because T=1\mathrm{s}\phantom{\rule{0ex}{0ex}}\therefore {i}_{d}=Q=ne\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{{i}_{d}}{e}$

So, the total number of charge carriers crossing the depletion region is given by

$n=\frac{20\times {10}^{-6}}{2\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=6.25\times {10}^{13}$

Also, the number of electrons crossing the depletion region is given by

${n}_{e}=\frac{n}{2}=\frac{6.25\times {10}^{13}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {n}_{e}=3.1\times {10}^{13}$

#### Page No 420:

#### Answer:

(a) The current‒voltage relationship of a diode is given by

$i={i}_{0}\left({e}^{eV/kT-1}\right)$

For a large value of voltage, 1 can be neglected.

$i\approx {i}_{0}{e}^{eV/kT}$

Again, we need to find the voltage at which

${e}^{eV/k\mathrm{T}}=100$

$\Rightarrow \frac{eV}{kT}=\mathrm{ln}100\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{\mathrm{ln}100\times kT}{e}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{2.303\times \mathrm{log}100\times 8.62\times {10}^{-5}\times 300}{e}\phantom{\rule{0ex}{0ex}}\Rightarrow V=0.12\mathrm{V}$

(b) Given:

$i={i}_{0}\left({e}^{eV/kT-1}\right)$ ...(1)

We know that the dynamic resistance of a diode is the rate of change of voltage w.r.t. current.

i.e. $R=\mathit{}\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{i}}$

As the exponential factor dominates the factor of 1, we can neglect this factor.

Now, on differentiating eq. (1) w.r.t. *V*, we get

$\frac{di}{dV}={i}_{0}\frac{e}{kT}{\mathrm{e}}^{eV/kT}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{R}=\frac{e{i}_{0}}{kT}{\mathrm{e}}^{eV/kT}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{kT}{e{i}_{0}}{\mathrm{e}}^{-e\mathrm{V}/k\mathrm{T}}...\left(2\right)$

(c) Given*:
R* = 2 Ω

On substituting this value in eq. (2), we get

$2=\frac{8.62\times {10}^{-5}\times 300}{e\times 10\times {10}^{-6}}{\mathrm{e}}^{-eV/8.62\times {10}^{-5}\times 300}\phantom{\rule{0ex}{0ex}}\Rightarrow V=0.25\mathrm{V}$

#### Page No 420:

#### Answer:

(a) Given:

Drift current, *i*_{0} = 20 × 10^{−6} A

Temperature, *T* = 300 K

Applied voltage,* V *= 300 mV

The variation in the current with respect to the voltage is given by

$i={i}_{0}\left({e}^{\frac{eV}{KT}}-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow i=20\times {10}^{-6}\left({e}^{\frac{0.3}{8.62\times 300\times {10}^{-5}}}-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow i=20\times {10}^{-5}\left({e}^{\frac{100}{8.61}}-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow i=2.18\mathrm{A}\approx 2\mathrm{A}$

(b) We need to find the voltage at which the current doubles so that the new value of the current becomes 4 A.

$\Rightarrow 4=20\times {10}^{-6}\left({e}^{\frac{eV}{8.62\times 3\times {10}^{-3}}}-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{e}}^{\frac{V\times {10}^{3}}{8.62\times 3}}-1=\frac{4\times {10}^{6}}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{e}}^{\frac{V\times {10}^{3}}{8.62\times 3}}=200001\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{V\times {10}^{3}}{8.62\times 3}=12.2060\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{12.206\times 8.63\times 3}{{10}^{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=318\mathrm{mV}$

#### Page No 420:

#### Answer:

(a) From the circuit diagram, it can be said that the diode is reverse biassed, with applied voltage of 5.0 V.

Under reverse bias condition,

Current in the circuit = Drift current

So, the current in the circuit is 20 µA.

(b) Voltage across the diode will be equal to the voltage of the battery minus the voltage drop across the 20 ohm resistor.

$\Rightarrow V=5-iR\phantom{\rule{0ex}{0ex}}\Rightarrow V=5-(20\times 20\times {10}^{-6})\phantom{\rule{0ex}{0ex}}\Rightarrow V=5-(4\times {10}^{-4})\phantom{\rule{0ex}{0ex}}\Rightarrow V={10}^{-4}(50000-4)\phantom{\rule{0ex}{0ex}}\Rightarrow V=49996\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\Rightarrow V=4.9996\mathrm{V}\cong 5\mathrm{V}$

#### Page No 420:

#### Answer:

According to the Wheatstone bridge principle, if the bridge is balanced, then the current flow across the central resistor is zero. So, to simplify the circuit, we can remove the central resistor.

The given circuit is also balanced, so there is no current through the diode.

Hence, the net resistance of the circuit is given by

${R}_{\mathrm{P}}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}\phantom{\rule{0ex}{0ex}}\mathrm{For}{R}_{1}={R}_{2}=R,\phantom{\rule{0ex}{0ex}}{R}_{\mathrm{P}}=\frac{R}{2}=\frac{40}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{\mathrm{P}}=20\mathrm{\Omega}$

#### Page No 420:

#### Answer:

We know that under forward bias, an ideal diode acts as a short circuit; and under reverse bias, an ideal diode acts as an open circuit.

(a) Since both the diodes are forward biassed, they can be replaced by short circuits.

Net resistance = 2 Ω

$i=\frac{2\mathrm{V}}{2\mathrm{\Omega}}=1\mathrm{A}$

(b) One diode is forward biassed and the other is reverse biassed; thus, the resistance of one becomes ∞. The circuit becomes open at the position of the second diode, so no current flows through this arm.

$\therefore i=\frac{2}{2+\infty}=0\mathrm{A}$

(c) Both of them are forward biassed. Thus, the resistance is zero.

We can replace both the resistors by short circuits.

Net resistance = 2 Ω

$\therefore i=\frac{2}{2}=1\mathrm{A}$

(d) One of them is forward biassed and the other is reverse biassed.

Thus, the current passes through the forward-biassed diode and the other acts as open circuit.

$\therefore i=\frac{2}{2}=1\mathrm{A}$

#### Page No 420:

#### Answer:

From the circuit diagram, we can see that one diode (the upper one) is reverse biassed.

We know that when a diode is forward biassed, it has zero resistance ideally. So, it can be replaced by a short circuit. When a diode is reverse biassed, it has infinite resistance ideally. So, it can be replaced by an open circuit.

Thus,

In the given circuit, one diode is reverse biassed. So, it can be replaced by an open circuit. Hence, the current in this branch will be zero.

So, the current through A_{1} is zero.

For A_{2},

Current = $\frac{2}{10}$ = 0.2 A

#### Page No 420:

#### Answer:

We know that when a diode is forward biassed, it has zero resistance ideally. So, it can be replaced by a short circuit. When a diode is reverse biassed, it has infinite resistance ideally. So, it can be replaced by an open circuit.

(a) In the given circuit diagram, both diodes are forward biassed. So, the resistance of both of them is zero. Thus, the diode resistance is zero.

$i=\frac{5}{{\displaystyle \frac{10\times 10}{10+10}}}=\frac{5}{5}=1\mathrm{A}$

(b) One diode is forward biassed and the other is reverse biassed. The reverse-biassed diode is replaced by an open circuit, so no current flows through this branch.

The current passes through the forward-biassed diode only.

$i=\frac{V}{{R}_{\mathrm{net}}}=\frac{5}{10}=0.5\mathrm{A}$

#### Page No 420:

#### Answer:

(a) When *R* = 12 Ω:

The 4 V battery is forward biassing the diode and the 6 V battery is reverse biassing the diode, so the diode is effectively reverse biassed. It acts like an open circuit so that no current flows through this branch. Hence, to simplify the circuit, this branch can be removed.

The current through *R* on applying the KVL in the circuit is given by

$i=\frac{10}{24}=0.4166=0.42\mathrm{A}$

(b) Similarly, for *R* = 48 Ω,

$i=\frac{10}{48+12}=\frac{10}{60}=0.16\mathrm{A}$

#### Page No 420:

#### Answer:

(a) If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the diode will get forward biassed by the applied voltage. So, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forward-biassed diode.

(b) If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the upper diode will get forward biassed and the lower diode will get reverse biassed by the applied voltage. So, this lower branch can be replaced by an open circuit; hence, the current flow through this branch will be zero. The current flows only through the upper diode, so the circuit on simplification will become identical to the circuit in part (a). Hence, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forward-biassed diode.

#### Page No 420:

#### Answer:

Let the potentials at A and B be *V*_{A} and *V*_{B}, respectively.

(i) When V_{A} > V_{B}, that is, a battery is connected between points A and B, with its positive terminal connected to point A and its negative terminal connected to point B:

As the potential on the *p*-side of the diode is greater than the potential on the *n*-side of the diode, the diode is forward biased; thus, it can be replaced by a short circuit.

As the two resistances are connected in parallel, the effective resistance becomes

Equivalent resistance$=\frac{10}{2}=5\mathrm{\Omega}$

(ii) When V_{A} < V_{B}:

The diode is reverse biased, so it is replaced by an open circuit. Thus, no current flows through this branch.

∴ Equivalent resistance = 10 Ω

#### Page No 420:

#### Answer:

Given:

Change in the base current, $\delta {I}_{\mathrm{b}}=(80-30)\mathrm{\mu A}$

Change in the collector current, $\delta {I}_{c}=(3.5-1)\mathrm{mA}$

Base current gain = Rate of change of collector current with respect to base current

Thus,

$\beta =\left(\frac{\delta {l}_{\mathrm{c}}}{\delta {l}_{\mathrm{b}}}\right)\mathrm{at}\mathrm{constant}{V}_{\mathrm{cc}}\phantom{\rule{0ex}{0ex}}\Rightarrow \beta =\frac{2.5\times {10}^{-3}}{50\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow \beta =\frac{250}{50}=50$

∴ Current gain = 50

#### Page No 420:

#### Answer:

Given:

Base current gain, $\beta =50$

Change in base current, $\delta {I}_{\mathrm{b}}=50\mathrm{\mu A}$

Load resistance,${R}_{\mathrm{L}}$ = 2 kΩ

Input resistance, ${R}_{\mathrm{i}}$ = 0.50 kΩ

(a) The change in output voltage is given by

${V}_{0}={I}_{\mathrm{c}}\times {R}_{\mathrm{L}}\phantom{\rule{0ex}{0ex}}\because {I}_{\mathrm{c}}=\beta \times {I}_{\mathrm{b}}\phantom{\rule{0ex}{0ex}}\therefore {V}_{0}=\beta \times {I}_{\mathrm{b}}\times {R}_{\mathrm{L}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}=50\times 50\mathrm{\mu A}\times 2\mathrm{k\Omega}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{0}=5\mathrm{V}$

(b) The change in input voltage is given by

$\delta {V}_{\mathrm{i}}=\delta {l}_{\mathrm{b}}\times {R}_{\mathrm{i}}\phantom{\rule{0ex}{0ex}}\Rightarrow \delta {V}_{\mathrm{i}}=50\times {10}^{-6}\times 5\times {10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \delta {V}_{\mathrm{i}}=25\times {10}^{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow \delta {V}_{\mathrm{i}}=25\mathrm{mV}\phantom{\rule{0ex}{0ex}}$

(c) Power gain is given by

${\mathrm{\beta}}^{2}\times \frac{{R}_{\mathrm{L}}}{{R}_{\mathrm{i}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2500\times \frac{2}{0.5}\phantom{\rule{0ex}{0ex}}\Rightarrow 2500\times \frac{20}{5}={10}^{4}$

#### Page No 420:

#### Answer:

Given:

Output $X=A\overline{)BC}+B\overline{)CA}+C\overline{)AB}$

(a) *A* = 1, *B* = 0,* C* = 1

$X=1.(\overline{0.1)}+0.(\overline{1.1)}+1.(\overline{1.0)}\phantom{\rule{0ex}{0ex}}=1.\overline{0}+0.\overline{1}+1.\overline{0}\phantom{\rule{0ex}{0ex}}=1.1+0.0+1.1\phantom{\rule{0ex}{0ex}}=1+0+1\phantom{\rule{0ex}{0ex}}=1+1\phantom{\rule{0ex}{0ex}}=1$

(b) A = B = C = 1

$X=1.(\overline{1.1)}+1.(\overline{1.1)}+1.(\overline{1.1)}\phantom{\rule{0ex}{0ex}}=1.\overline{1}+1.\overline{1}+1.\overline{1}\phantom{\rule{0ex}{0ex}}=1.0+1.0+1.0\phantom{\rule{0ex}{0ex}}=0+0+0\phantom{\rule{0ex}{0ex}}=0$

(c) A = B = C = 0

$X=0.(\overline{0.0)}+0(\overline{0.0)}+0.(\overline{0.0)}\phantom{\rule{0ex}{0ex}}=0.\overline{0}+0.\overline{0}+0.\overline{0}\phantom{\rule{0ex}{0ex}}=0.1+0.1+0.1\phantom{\rule{0ex}{0ex}}=0+0+0\phantom{\rule{0ex}{0ex}}=0$

#### Page No 421:

#### Answer:

*$X=A\overline{BC}+B\overline{CA}\phantom{\rule{0ex}{0ex}}=A(\overline{B}+\overline{C})+B(\overline{C}+\overline{A})\phantom{\rule{0ex}{0ex}}=A\overline{B}+A\overline{C}+B\overline{C}+B\overline{A}\phantom{\rule{0ex}{0ex}}=A\overline{B}+\overline{C}(A+B)+B\overline{A}$*

#### Page No 421:

#### Answer:

Given:

$X=AB+\overline{)AB}$

Let:

*AB* =* y*

∴ $X=y+\overline{)y}$

For *y *= 0,

*X* = 0 + 1

= 1

For *y* = 1,

*X* = 1 + 0

= 1

Thus, *X* = 1 for all values of* y = AB.*

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