Answer

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

Thinking Process

Go through the working of p-njunction.

Answer

(b) 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.

Thinking Process

Check the polarity of the diodes.

Answer

(b) In the given circuit $p$-side of $p$ - $n$ function $D_1$ is connected to lower voltage and $n$-side of $D_1$ to higher voltage.

Thus $D$ is reverse biased.

The $p$-side of $p$ - $n$ junction $D_2$ is at higher potential and $n$-side of $D_2$ is at lower potential. Therefore $D_2$ is forward biased.

Hence, current flows through the junction $B$ to $A$.

Thinking Process

$p$-n junction conducts during positive half cycle only.

Answer

(d) As $p$ - $n$ junction conducts during positive half cycle only, the diode connected here will work is positive half cycle. Potential difference across $C=$ peak voltage of the given $A C$ voltage $=V_0=V_{\text {rms }} \sqrt{2}=220 \sqrt{2} V$.

Answer

(b) The concept of hole describes the lack of an electron at a position where one could exist in an atom or atomic lattice. If an electron is excited into a higher state, it leaves a hole in its old state.

Thus, hole can be defined as a vacancy created when an electron leaves a covalent bond.

Thinking Process

For positive half cycle of input AC voltage, the p-n junction is forward biased and for negative half cycle of input $A C$ voltage the $p$ - $n$ junction is reversed biased.

Answer

(c) Due to forward biased during positive half cycle of input AC voltage, the resistance of $p-n$ junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in series of circuit. This result into zero output voltage across $p-n$ junction.

Due to reverse biase during negative half cycle of $A C$ voltage, the $p-n$ junction is reverse biased. The resistance of $p-n$ junction becomes high which will be more than resistance in series. That is why, there will be voltage across $p-n$ junction with negative cycle in output.

Answer

(b) Consider the fig. (b) given here, suppose the potential difference between $A$ and $B$. $r_1=5 k \Omega$ and $r_2=5 k \Omega$ are resistance in series connection.

$$ \begin{matrix} \text { Then, } \quad V-0.3 & .=[(r_1+r_2) 10^{3}] \times(0.2 \times 10^{-3})] & {[\because V=i r]} \\ & =[(5+5) 10^{3}] \times(0.2 \times 10^{-3}) & \\ & =10 \times 10^{3} \times 0.2 \times 10^{-3}=2 & \\ \Rightarrow \quad V & =2+0.3=2.3 V \end{matrix} $$

Answer

(c) Here,

$$ \begin{aligned} C & =A \cdot B \text { and } D=\bar{A} \cdot B \\ E & =C+D=(A \cdot B)+(\bar{A} \cdot B) \end{aligned} $$

Explanation The truth table of this arrangement of gates can be given by

$\newline$

Thinking Process

Electrons are negatively charged and its energy increases when electric fields is applied.

Answer

(a, c)

When electric field is applied across a semiconductor, the electrons in the conduction band get accelerated and acquire energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

Thinking Process

Draw the figure as given in the question.

Answer

$(a, c)$

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 crossover from emitter to collector.

Answer

$(b, c, d)$

From the given transfer characteristics of a base biased common emitter transistor, we note that

(i) when $V_{i}=0.4 V$, there is no collection current. The transistor circuit is in active state and is used as an amplifier.

(ii) when $V_{i}=1 V$ (This is in between $0.6 V$ to $2 V$ ), the transistor circuit is in active state and is used as an amplifier.

(iii) when $V_{i}=0.5 V$, there is no collector current. The transistor is in cut off state. The transistor circuit can be used as a switch turned off.

(iv) when $V_{i}=2.5 V$, the collector current becomes maximum and transistor is in saturation state and can used as switch turned on state.

Thinking Process

The collector current is the $95 %$ of electrons reaching the collector after emission.

Answer

$(b, c)$

Here,

Also

$\Rightarrow \quad I_{e}=\frac{10 \times 100}{95}=10.53 mA$

Also,

$$ \begin{aligned} & I_{C}=10 mA \\ & I_{c}=\frac{95}{100} I_{e} \\ & I_{e}=\frac{10 \times 100}{95}=10.53 mA \\ & I_{b}=I_{e}-I_{C}=10.53-10=0.53 mA \end{aligned} $$

Answer

$(a, b, d)$

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.

Answer

$(b, d)$

During regulation action of a Zener diode, the current through the $R_{s}$ changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

Thinking Process

Ripple factor ( $r$ ) of a full wave rectifier using capacitor filter is given by

$$ \begin{matrix} r & =\frac{1}{4 \sqrt{3} R_{L} C_{V}} \\ \text { i.e., } \quad r \propto \frac{1}{R_{L}} \Rightarrow r \propto \frac{1}{C}, r \propto \frac{1}{V} \end{matrix} $$

Answer

$(a, c, d)$

Ripple factor is inversely proportional to $R_{L}, C$ and $v$.

Thus to reduce $r, R_{L}$ should be increased, input frequency $v$ should be increased and capacitance $C$ should be increased.

Answer

$(a, d)$

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.

Answer

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.

Thinking Process

The property of conduction level of any element depends on the energy gap between its conduction band and valence band.

Answer

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.

Answer

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.

Answer

As we know that the diode only works in forward biased, so the output is obtained only when positive input is given, so the output waveform is

Thinking Process

$\frac{\text { Output signal voltage }}{\text { Input Signal voltage }}=$ Total voltage amplification

Answer

Given,

$$ A v_{x}=10, A v_{y}=20, A v_{z}=30 $$

$$ \Delta V_{i}=1 mV=10^{-3} V $$

Now, $\quad \frac{\text { Output Signal Voltage }(\Delta V_0)}{\text { Input Singal Voltage }(\Delta V_{i})}=$ Total voltage amplification

$$ =A v_{x} \times A v_{y} \times A v_{z} $$

$\Rightarrow$

$$ \begin{aligned} \Delta V_0 & =A v_{x} \times A v_{y} \times A v_{z} \times \Delta V_{i} \\ & =10 \times 20 \times 30 \times 10^{-3}=6 V \end{aligned} $$

(i) If DC supply voltage is $10 V$, then output is $6 V$, since theoretical gain is equal to practical gain, i.e., output can never be greater than $6 V$.

(ii) If $DC$ supply voltage is $5 V$, i.e., $V_{c c}=5 V$. Then, output peak will not exceed $5 V$. Hence $V_0=5 V$.

Answer

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.

Answer

(i) The characteristic curve (a) is of Zener diode and curve (b) is of solar cell.

(ii) The point $P$ in fig. (a) represents Zener break down voltage.

(iii) In fig. (b), the point $Q$ represents zero voltage and negative current. It means light falling on solar cell with atleast minimum threshold frequency gives the current in opposite direction to that due to a battery connected to solar cell. But for the point $Q$, the battery is short circuited. Hence represents the short circuit current.

In fig. (b), the point $P$ represents some positive voltage on solar cell with zero current through solar cell.

It means, there is a battery connected to a solar cell which gives rise to the equal and opposite current to that in solar cell by virtue of light falling on it.

As current is zero for point $P$, hence we say $P$ represents open circuit voltage.

Answer

Given, wavelength of light $\lambda=6000 \AA=6000 \times 10^{-10} m$

Energy of the light photon

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

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 $D_2$. Then, diode $D_2$ will detect this radiation.

Answer

Consider the circuit in fig. (b) to find the change in reading As we know the formula for base current, $I_{B}=\frac{V_{BB}-V_{BE}}{R_{i}}$

As $R_{i}$ is increased, $I_{B}$ is decreased.

Now, the current in ammeter is collector current $I_{C}$.

$I_{C}=\beta I_{B}$ as $I_{B}$ decreased $I_{C}$ also decreased and the reading of voltmeter and ammeter also decreased.

Answer

As car enters in the gate, any one or both are opened.

The device is shown.

So, OR gate gives the desired output.

Answer

The NOT gate is a device which has only one input and one output i.e., $\bar{A}=Y$ 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 $R_{b}$ 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 $R_{b}$ and $R_{c}$ 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 $5 V$, the transistor is in saturation state.

Answer

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

Answer

The circuit resemble AND gate. The boolean expression of this circuit is, $V_0=A$.Bi.e., $V_0$ equals A AND B. The truth table of this gate is as given below

Answer

Given,

$$ \begin{aligned} \text { power } & =1 W \\ \text { Zener breakdown } V_{z} & =5 V \\ \text { Minimum voltage } V_{\min } & =3 V \\ \text { Maximum voltage } V_{\max } & =7 V \\ \text { Current } I_{Z_{\max }} & =\frac{P}{V_{Z}}=\frac{1}{5}=0.2 A \end{aligned} $$

The value of $R_{s}$ for safe operation $R_{s}=\frac{V_{\max }-V_{Z}}{I_{Z_{\max }}}=\frac{7-5}{0.2}=\frac{2}{0.2}=10 \Omega$

Answer

Given, forward biased resistance $=25 \Omega$

Reverse biased resistance $=\infty$

As the diode in branch $C D$ is in reverse biased which having resistance infinite,

$$ \text{So}\qquad I_3=0 $$

Resistance in branch $A B=25+125=150 \Omega$ say $R_1$

Resistance in branch $E F=25+125=150 \Omega$ say $R_2$

$A B$ is parallel to $E F$.

So, $\quad$ resultant resistance $\frac{1}{R^{\prime}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}$

$$ \Rightarrow \qquad R^{\prime}=75 \Omega $$

Total resistance $R=R^{\prime}+25=75+25=100 \Omega$

$$ \text{Current}\hspace{1mm}I_1=\frac{V}{R}=\frac{5}{100}=0.05 A$$

$$ \begin{aligned} & I_1=I_4+I_2+I_3 \qquad \text { (Here } I_3=0 \text { ) } \\ So,\qquad & I_1=I_4+I_2 \end{aligned} $$

Here,the resistances $R_1$ and $R_2$ is same.

$$ i.e.,\qquad I_4=I_2 $$

$$ \begin{matrix} \therefore & I_1=2 I_2 \\ \Rightarrow \qquad & I_2=\frac{I_1}{2}=\frac{0.05}{2}=0.025 A \end{matrix} $$

$$ \text{and}\qquad I_4=0.025 A $$

Thus, $\quad I_1=0.05 A, I_2=0.025 A, I_3=0$ and $I_4=0.025 A$

Answer

Given,

$$ \begin{aligned} \text { voltage across } R_{B} & =10 V \\ \text { Resistance } R_{B} & =400 k \Omega \\ V_{BE} & =0, V_{CE}=0 R_{C}=3 k \Omega \\ I_{B} & =\frac{\text { Voltage across } R_{B}}{R_{B}} \\ & =\frac{10}{400 \times 10^{3}}=25 \times 10^{-6} A=25 \mu A \end{aligned} $$

Voltage across $R_{C}=10 V$

$$ \begin{aligned} I_{C} & =\frac{\text { Voltage across } R_{C}}{R_{C}}=\frac{10}{3 \times 10^{3}} \\ & =3.33 \times 10^{-3} A=3.33 mA \\ \beta & =\frac{I_{C}}{I_{B}}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}} \\ & =1.33 \times 10^{2}=133 \end{aligned} $$

Semiconductor Electronics : Material, Devices and Simple Circuit 241

Answer

First draw the truth table of $C_1$ and $C_2$.

Answer

Given,

$$ \begin{aligned} V_{B E} & =0.7 V, V_{C C}=V_{B B}=16 V \\ V_{C E} & =8 V \qquad \text{(from graph)} \\ I_{C} & =4 mA=4 \times 10^{-3} A \\ I_{B} & =30 \mu A=30 \times 10^{-6} A \end{aligned} $$

For the output characteristic at $\theta$

$$ \begin{aligned} & V_{C C}=I_{C} R_{C}+V_{C E} \\ & R_{C}=\frac{V_{C C}-V_{C E}}{I_{C}}=\frac{16-8}{4 \times 10^{-3}}=\frac{8 \times 1000}{4}=2 k \Omega \end{aligned} $$

Using the relation,

$$ V_{BB}=I_{B} R_{B}+V_{BE} $$

$$ \begin{aligned} R_{B} & =\frac{V_{BB}-V_{BE}}{I_{B}}=\frac{16-0.7}{30 \times 10^{-6}} \\ & =510 \times 10^{3} \Omega=510 k \Omega \\ \beta & =\frac{I_{C}}{I_{B}}=\frac{4 \times 10^{-3}}{30 \times 10^{-6}}=133 \end{aligned} $$

Voltage gain $=\beta \frac{R_{C}}{R_{B}}=\frac{133 \times 2 \times 10^{3}}{510 \times 10^{3}}=0.52$

Power gain $=\beta \times$ Voltage gain $=133 \times 0.52=69$

Thinking Process

An ideal diode is a diode that acts like a perfect conductor when voltage is applied forward biased and like a perfect insulator when voltage is applied reverse biased.

Answer

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 voltage beyond $+5 \vee$ will appear across $R$.

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 wave form is then passed on to the output terminals.

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

Answer

When As is implanted in Si crystal, $n$ - type wafer is created. The number of majority carriers electrons due to doping of As is

$$ \begin{aligned} n_{e} & =N_{D}=\frac{1}{10^{6}} \times 5 \times 10^{28} \\ & =5 \times 10^{22} / m^{3} \end{aligned} $$

Number of minority carriers (holes) in $n$-type wafer is

$$ \begin{aligned} n_{h} & =\frac{n_i^{2}}{n_{e}}=\frac{(1.5 \times 10^{16})^{2}}{5 \times 10^{22}} \\ & =0.45 \times 10^{10} / m^{3} \end{aligned} $$

When $B$ is implanted in Si crystal, $p$-type wafer is created with number of holes,

$$ n_{h}=N_{A}=\frac{200}{10^{6}} \times(5 \times 10^{28})=1 \times 10^{25} / m^{3} $$

Minority carriers (electrons) created in $p$ - type wafer is

$$ \begin{aligned} n_{e} & =\frac{n_i^{2}}{n_{h}}=\frac{(1.5 \times 10^{16})^{2}}{1 \times 10^{25}} \\ & =2.25 \times 10^{27} / m^{3} \end{aligned} $$

When $p-n$ junction is reverse biased, the minority carrier holes of $n$-region wafer $(n_{h}=0.45 \times 10^{10} / m^{3})$ would contribute more to the reverse saturation current than minority carrier electrons $(n_{e}=2.25 \times 10^{7} / m^{3})$ of pregion wafer.

Answer

Given, the logic relation for the given truth table is

$$ \begin{aligned} Y & =\bar{A} \cdot B+A \cdot \bar{B}=Y_1+Y_2 \\ Y_1 & =A \cdot B \text { and } Y_2=A \cdot \bar{B} \end{aligned} $$

$Y_1$ can be obtained as output of AND gate I for which one Input is of $A$ through NOT gate and another input is of $B$. $Y_2$ 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 $Y_2$ can be obtained as output from OR gate, where, $Y_1$ and $Y_2$ are input of $O R$ gate. Thus, the given table can be obtained from the logic circuit given below

Answer

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

(b) In V-I 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 $=(1 / 1000) \Omega$.

It means $A$ is connected to $n$-side of $p-n$ junction $I$ and $B$ is connected to $p$-side of $p-n$ junction $I$ and resistance $R$ is in series of $p-n$ junction $I$ between $A$ and $B$.

(c) In $V$ - $I$ 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 $p-n I, p-n |$ and resistance $R$ between $A, B$ and $C$ will be as shown in the figure

Answer

Consider the fig. (b) given here to solve this problem

$$ \begin{aligned} I_{C} & \approx I_{E} \qquad \text{[As base current is very small.]} \\ R_{C} & =7.8 k \Omega \\ \text { From the figure, } I_{C}(R_{C}+R_{E})+V_{CE} & =12 \\ (R_{E}+R_{C}) \times 1 \times 10^{-3}+3 & =12 \\ R_{E}+R_{C} & =9 \times 10^{3}=9 k \Omega \\ R_{E} & =9-7.8=1.2 k \Omega \\ V_{E} & =I_{E} \times R_{E} \\ & =1 \times 10^{-3} \times 1.2 \times 10^{3}=1.2 V \\ \text { Voltage } V_{B} & =V_{E}+V_{B E}=1.2+0.5=1.7 V \\ \text { Current } I & =\frac{V_{B}}{20 \times 10^{3}}=\frac{1.7}{20 \times 10^{3}}=0.085 mA \\ \text { Resistance } R_{B} & =\frac{12-1.7}{\frac{I_C}{\beta}+0.085}=\frac{10.3}{0.01+0.085} \qquad \text{[Given,$\beta$ =100]} \\ & =108k\Omega \end{aligned} $$

Answer

Consider the fig. (b) to solve this question,

$$ \begin{aligned} I_E & =I_C+I_B \text { and } I_C=\beta I_B \qquad \text{…(i)} \\ I_C R_C+V_{C E}+I_E R_E & =V_{C C} \qquad \text{…(ii)} \\ R I_B+V_{B E}+I_E R_E & =V_{C C} \qquad \text{…(iii)} \\ \because \qquad I_E & \approx I_C=\beta I_B \end{aligned} $$

From Eq.(iii),

$$ \begin{aligned} (R+\beta R_{E}) I_{B} & =V_{CC}-V_{BE} \\ \Rightarrow \qquad I_{B} & =\frac{V_{CC}-V_{BE}}{R+\beta \cdot R_{E}} \\ & =\frac{12-0.5}{80+1.2 \times 100}=\frac{11.5}{200} mA \end{aligned} $$

From Eq.(ii),

$$ (R_{C}+R_{E})=\frac{V_{C E}-V_{B E}}{I_{C}}=\frac{V_{C C}-V_{C E}}{\beta I_{B}} \qquad(\because I_{C}=\beta I_{B}) $$

$$ \begin{aligned} (R_{C}+R_{E}) & =\frac{2}{11.5}(12-3) k \Omega=1.56 k \Omega \\ R_{C}+R_{E} & =1.56 \\ R_{C} & =1.56-1=0.56 k \Omega \end{aligned} $$