Junction Transistor Questions in English

(C) Statement $I$ is incorrect because in a transistor,the three regions have different doping levels. The emitter is heavily doped,the base is lightly doped,and the collector is moderately doped.
Statement $II$ is correct because the collector is made the largest (thickest) to dissipate the heat generated during operation,and the base is made very thin to allow the majority of charge carriers from the emitter to pass through to the collector.
Therefore,Statement $I$ is incorrect and Statement $II$ is correct.

(C) Given collector current $I_C = 24\,mA$.
The current gain $\alpha$ is the ratio of collector current to emitter current,which is given as $80\% = 0.8$.
We know that $I_C = \alpha I_E$,so $I_E = \frac{I_C}{\alpha}$.
Substituting the values: $I_E = \frac{24\,mA}{0.8} = 30\,mA$.
Using Kirchhoff's current law for a transistor,$I_E = I_B + I_C$,so $I_B = I_E - I_C$.
$I_B = 30\,mA - 24\,mA = 6\,mA$.
In an $n-p-n$ transistor,the emitter current enters the transistor,while the collector and base currents leave the transistor. However,the question asks for the base current magnitude and direction relative to the base terminal. Since $I_E = I_B + I_C$,the base current $I_B$ flows into the base region to compensate for the recombination of charge carriers. Thus,the current is $6\,mA$ and entering the base.

(D) In the input circuit,applying Kirchhoff's Voltage Law $(KVL)$ to the base-emitter loop:
$V_{BB} - I_b R_b - V_{BE} = 0$
Assuming the transistor is in the active region and $V_{BE} \approx 0.7 \text{ V}$,we need $V_{BB}$ to find $I_b$. However,looking at the circuit,the base current is determined by the input bias. Given the standard approach for such problems where $V_{BB}$ is often assumed to be equal to $V_{CC}$ if not specified,or by calculating the required $I_b$ to reach saturation:
For saturation,$V_{CE} = 0 \text{ V}$.
Applying $KVL$ to the output loop:
$V_{CC} - I_c R_c - V_{CE} = 0$
$1 \text{ V} - I_c (1 \times 10^3 \text{ } \Omega) - 0 = 0$
$I_c = \frac{1}{1000} \text{ A} = 1 \text{ mA} = 1000 \text{ } \mu\text{A}$.
Since $\beta = \frac{I_c}{I_b}$,the base current required for saturation is:
$I_b = \frac{I_c}{\beta} = \frac{1000 \text{ } \mu\text{A}}{100} = 10 \text{ } \mu\text{A}$.

(A) The current gain $(\beta)$ in a common emitter $(CE)$ transistor is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.
Given:
$\Delta I_C = 16\,mA - 5\,mA = 11\,mA = 11 \times 10^{-3}\,A$
$\Delta I_B = 200\,\mu A - 100\,\mu A = 100\,\mu A = 100 \times 10^{-6}\,A$
Using the formula:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{11 \times 10^{-3}\,A}{100 \times 10^{-6}\,A} = \frac{11 \times 10^{-3}}{10^{-4}} = 11 \times 10^1 = 110$
Thus,the current gain of the transistor is $110$.

(C) The current gain $\beta$ is defined as the slope of the $I_C$ versus $I_B$ graph: $\beta = \frac{\Delta I_C}{\Delta I_B}$.
From the graph,taking two points $(100\,\mu A, 10\,mA)$ and $(200\,\mu A, 20\,mA)$:
$\beta = \frac{(20 - 10) \times 10^{-3} \, A}{(200 - 100) \times 10^{-6} \, A} = \frac{10 \times 10^{-3}}{100 \times 10^{-6}} = 100$.
The power gain $A_P$ is given by $A_P = \beta^2 \times \frac{R_C}{R_B}$.
Substituting the values: $A_P = (100)^2 \times \frac{1 \times 10^3 \, \Omega}{10 \times 10^3 \, \Omega} = 10000 \times 0.1 = 1000 = 10^3$.
Comparing this with $10^x$,we get $x = 3$.

(B) For a transistor,the current gain $\alpha$ is defined as the ratio of collector current $(I_C)$ to emitter current $(I_E)$,i.e.,$\alpha = \frac{I_C}{I_E}$. Since $I_E = I_C + I_B$ and $I_B > 0$,it follows that $I_C < I_E$,which implies $\alpha < 1$.
The current gain $\beta$ is defined as the ratio of collector current $(I_C)$ to base current $(I_B)$,i.e.,$\beta = \frac{I_C}{I_B}$.
Since the base current $I_B$ is typically very small compared to the collector current $I_C$,the ratio $\beta = \frac{I_C}{I_B}$ is always greater than $1$.
Therefore,the correct relationship is $\beta > 1$ and $\alpha < 1$.

(C) Given that the ratio of collector current to emitter current is $\alpha = \frac{I_{C}}{I_{E}} = 0.95$.
For a common emitter transistor,the current gain $\beta$ is defined as the ratio of collector current to base current,given by the formula $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$ into the formula:
$\beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05}$.
$\beta = \frac{95}{5} = 19$.
Therefore,the current gain is $19$.

(C) The voltage gain $(A_v)$ of a transistor amplifier is given by the product of the current gain $(\beta)$ and the ratio of the load resistance $(R_L)$ to the input resistance $(R_{in})$.
Formula: $A_v = \beta \times \frac{R_L}{R_{in}}$
Given:
$AC$ current gain $(\beta)$ = $64$
Load resistance $(R_L)$ = $5400 \ \Omega$
Input resistance $(R_{in})$ = $540 \ \Omega$
Substituting the values:
$A_v = 64 \times \frac{5400}{540}$
$A_v = 64 \times 10$
$A_v = 640$
Therefore, the voltage gain is $640$.

(D) The power gain $(A_p)$ of a transistor is defined as the product of voltage gain $(A_v)$ and current gain $(A_i)$.
Mathematically,$A_p = A_v \times A_i$.
Therefore,the ratio of power gain to voltage gain is $\frac{A_p}{A_v} = A_i$.
In a common emitter configuration,the current gain is defined as $\beta = \frac{I_c}{I_b}$.
Thus,the ratio is equal to $\beta$.

(C) In a common emitter $(CE)$ amplifier configuration using an $n-p-n$ transistor:
$1$. The base-emitter junction is forward biased to allow current flow, and the collector-base junction is reverse biased.
$2$. The input signal is applied between the base and the emitter, while the output is taken across the collector and the emitter.
$3$. The input impedance is generally low, and the output impedance is high.
$4$. There is a phase difference of $180^{\circ}$ between the input and output signals.
Therefore, option $C$ is the correct statement.

(C) Given that $90 \%$ of the electrons emitted reach the collector,the current gain $\alpha$ (common base) is defined as the ratio of collector current to emitter current. Since $90 \%$ of the emitter current $(I_E)$ reaches the collector $(I_C)$,we have $I_C = 0.9 \ I_E$. Thus,$\alpha = \frac{I_C}{I_E} = 0.9$.
The current gain $\beta$ (common emitter) is related to $\alpha$ by the formula $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.9}{1 - 0.9} = \frac{0.9}{0.1} = 9$.
Therefore,$\alpha = 0.9$ and $\beta = 9$.

(C) Given: Collector current $I_C = 8 \ mA$.
Since $80 \%$ of the electrons from the emitter reach the collector,the current gain $\alpha$ is given by the ratio of collector current to emitter current,which is $0.8$.
Thus,$\alpha = \frac{I_C}{I_E} = 0.8$.
We can find the emitter current $I_E$ as $I_E = \frac{I_C}{\alpha} = \frac{8 \ mA}{0.8} = 10 \ mA$.
The base current $I_B$ is given by $I_E - I_C = 10 \ mA - 8 \ mA = 2 \ mA$.
The current gain $\beta$ is calculated as $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4.0$.
Therefore,$\alpha = 0.8$ and $\beta = 4.0$.

(B) The power gain $(A_p)$ of a transistor is given by the product of current gain $(\beta)$ and voltage gain $(A_v)$.
$A_p = \beta \times A_v$
Voltage gain $(A_v)$ is defined as the product of current gain $(\beta)$ and the ratio of load resistance $(R_L)$ to input impedance $(R_i)$.
$A_v = \beta \times \frac{R_L}{R_i}$
Given:
Current gain $(\beta)$ = $8$
Input impedance $(R_i)$ = $25 \, k\Omega$
Load resistance $(R_L)$ = $75 \, k\Omega$
First, calculate the voltage gain:
$A_v = 8 \times \frac{75 \, k\Omega}{25 \, k\Omega} = 8 \times 3 = 24$
Now, calculate the power gain:
$A_p = \beta \times A_v = 8 \times 24 = 192$
Note: The provided options in the original prompt appear to be incorrect based on the standard formula. The calculated power gain is $192$.

(D) In a common emitter $(CE)$ configuration of a transistor amplifier,the input signal is applied between the base and the emitter,while the output is taken across the collector and the emitter.
When the input signal voltage increases,the base current increases,which in turn increases the collector current.
Due to the voltage drop across the load resistor $R_L$ connected in the collector circuit $(V_{out} = V_{CC} - I_C R_L)$,an increase in collector current $I_C$ leads to a decrease in the output voltage $V_{out}$.
Since an increase in input voltage results in a decrease in output voltage,the output signal is inverted with respect to the input signal.
An inverted signal corresponds to a phase shift of $180^\circ$ or $\pi^c$ radians.

(C) Given,the ratio of collector current $(I_C)$ to emitter current $(I_E)$ is the current amplification factor $\alpha$.
So,$\alpha = \frac{I_C}{I_E} = 0.96$.
The relationship between current gain $\beta$ and $\alpha$ is given by the formula:
$\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$:
$\beta = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04}$.
$\beta = \frac{96}{4} = 24$.
Therefore,the current gain $\beta$ is $24$.

(C) Given that the current gain $\alpha = 0.8$.
First,we calculate the current gain $\beta$ for the common emitter configuration using the relation $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4$.
The relationship between collector current change $(\Delta I_C)$ and base current change $(\Delta I_B)$ is given by $\Delta I_C = \beta \times \Delta I_B$.
Given $\Delta I_B = 3 \mu A$,we have $\Delta I_C = 4 \times 3 \mu A = 12 \mu A$.
Therefore,the collector current changes by $12 \mu A$.

(B) Given: Current gain $(\beta)$ = $62$, Collector resistance $(R_C)$ = $5 \text{ k}\Omega = 5000 \text{ }\Omega$, Input resistance $(R_i)$ = $500 \text{ }\Omega$, Input voltage $(V_i)$ = $0.01 \text{ V}$.
First, calculate the voltage gain $(A_v)$ of the amplifier:
$A_v = \beta \times \frac{R_C}{R_i} = 62 \times \frac{5000}{500} = 62 \times 10 = 620$.
The output voltage $(V_o)$ is given by the product of voltage gain and input voltage:
$V_o = A_v \times V_i = 620 \times 0.01 \text{ V} = 6.2 \text{ V}$.
Therefore, the correct option is $B$.

(C) Given:
Input resistance,$R_i = 1.8 \text{ k}\Omega = 1800 \Omega$
Load resistance,$R_L = 9 \text{ k}\Omega = 9000 \Omega$
Current gain,$\beta = 70$
Input voltage,$V_i = 6 \text{ mV} = 6 \times 10^{-3} \text{ V}$
Step $1$: Calculate the input current $(I_b)$:
$I_b = \frac{V_i}{R_i} = \frac{6 \times 10^{-3}}{1800} = \frac{1}{3} \times 10^{-5} \text{ A}$
Step $2$: Calculate the output current $(I_c)$:
$I_c = \beta \times I_b = 70 \times \frac{1}{3} \times 10^{-5} = \frac{7}{3} \times 10^{-4} \text{ A}$
Step $3$: Calculate the output voltage $(V_o)$:
$V_o = I_c \times R_L = (\frac{7}{3} \times 10^{-4}) \times 9000 = 7 \times 3 \times 10^{-1} = 2.1 \text{ V}$
Alternatively,Voltage Gain $A_v = \beta \times \frac{R_L}{R_i} = 70 \times \frac{9000}{1800} = 70 \times 5 = 350$.
$V_o = A_v \times V_i = 350 \times 6 \text{ mV} = 2100 \text{ mV} = 2.1 \text{ V}$.

(B) We know the relationship between the current gain parameters $\alpha_{dc}$ and $\beta_{dc}$ of a transistor is given by $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$.
Rearranging this,we get $\frac{1}{\beta_{dc}} = \frac{1 - \alpha_{dc}}{\alpha_{dc}} = \frac{1}{\alpha_{dc}} - 1$.
This implies $\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = 1$.
Taking the common denominator,we get $\frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \times \beta_{dc}} = 1$.

(A) The emitter current $I_e$ is given by $I_e = \frac{n_e \times e}{t}$.
Since $2 \%$ of electrons recombine in the base,the number of electrons reaching the collector is $98 \%$ of the emitter electrons.
Thus,the collector current $I_c = 0.98 \ I_e$.
The current gain $\alpha$ is defined as $\alpha = \frac{I_c}{I_e} = \frac{0.98 \ I_e}{I_e} = 0.98$.
The current gain $\beta$ is defined as $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
Therefore,the values are $\alpha = 0.98$ and $\beta = 49$.

(B) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the product of the current gain $(\beta)$ and the ratio of the load resistance $(R_L)$ to the input resistance $(R_{in})$.
$A_v = \beta \times \frac{R_L}{R_{in}}$
Given:
$\beta = 78$
$R_L = 6.5 \text{ k}\Omega = 6.5 \times 10^3 \Omega$
$R_{in} = 1.3 \text{ k}\Omega = 1.3 \times 10^3 \Omega$
Substituting the values:
$A_v = 78 \times \frac{6.5 \times 10^3}{1.3 \times 10^3}$
$A_v = 78 \times 5$
$A_v = 390$

(B) An $n-p-n$ transistor consists of a $p$-type base sandwiched between two $n$-type regions (emitter and collector).
This structure creates two $p-n$ junctions: the emitter-base junction and the collector-base junction.
In an $n-p-n$ transistor,the base is $p$-type and both the emitter and collector are $n$-type.
When represented as two diodes,the $p$-type base acts as the common terminal for the anodes of both diodes.
Therefore,the two diodes are connected such that their anodes are joined at the base terminal $(B)$,while their cathodes point towards the emitter $(E)$ and collector $(C)$ terminals respectively.
This configuration is correctly represented by two diodes with their anodes connected to the base terminal.

(B) Given: Collector voltage drop $V_C = 0.6 \text{ V}$,Resistance $R_C = 600 \Omega$,Current gain $\beta = 20$.
First,calculate the collector current $I_C$ using Ohm's law: $I_C = \frac{V_C}{R_C} = \frac{0.6}{600} = 0.001 \text{ A} = 1 \text{ mA}$.
Using the formula for current gain in common emitter configuration: $\beta = \frac{I_C}{I_B}$.
Substituting the values: $20 = \frac{1 \text{ mA}}{I_B}$.
Therefore,$I_B = \frac{1}{20} \text{ mA} = 0.05 \text{ mA}$.

(D) Given: Current amplification factor $\beta = 50$,Input resistance $R_i = 1 \text{ k}\Omega = 10^3 \Omega$,Input voltage $V_i = 0.01 \text{ V}$.
Using Ohm's law for the input circuit,the base current $I_B$ is given by:
$I_B = \frac{V_i}{R_i} = \frac{0.01 \text{ V}}{10^3 \Omega} = 10^{-5} \text{ A}$.
The relationship between collector current $I_C$ and base current $I_B$ is $I_C = \beta \times I_B$.
Substituting the values:
$I_C = 50 \times 10^{-5} \text{ A} = 500 \times 10^{-6} \text{ A} = 500 \mu\text{A}$.

(C) Common Emitter Configuration:
In this configuration,the emitter terminal is common to both the input and output sides of the circuit.
The input signal is applied across the base-emitter junction.
The output is taken across the collector-emitter junction.
Biasing:
For proper operation,the base-emitter junction should be forward-biased,not reverse-biased as mentioned in Option $B$. Therefore,Option $B$ is incorrect.
Phase Relationship:
The output voltage is inverted with respect to the input voltage,meaning they are $180^{\circ}$ out of phase. Hence,Option $D$ is incorrect.
Impedance Characteristics:
The input impedance is generally low to moderate,not high,and the output impedance is generally moderate to high. Thus,Option $A$ is incorrect.
In summary,the common emitter configuration is widely used due to its ability to amplify the input signal with phase inversion,making Option $C$ the accurate description.

(D) The current gain $(\beta)$ of a transistor is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.
$\beta = \frac{\Delta I_C}{\Delta I_B}$
Given that $\beta = 50$ and the change in collector current $\Delta I_C = 350 \mu A$.
Rearranging the formula to solve for the change in base current $(\Delta I_B)$:
$\Delta I_B = \frac{\Delta I_C}{\beta}$
Substituting the given values:
$\Delta I_B = \frac{350 \mu A}{50} = 7 \mu A$.
Thus,the base current should be changed by $(\frac{350}{50}) \mu A$.

(A) Given input signal: $V_i = 2 \cos \left(12 t + \frac{\pi}{3}\right)$.
Voltage gain $A_v = 126$.
In a Common Emitter $(C.E.)$ amplifier,the output signal is phase-shifted by $\pi$ radians $(180^\circ)$ relative to the input signal.
The output voltage $V_o$ is given by $V_o = A_v \times V_i$ with a phase shift of $\pi$.
$V_o = 126 \times 2 \cos \left(12 t + \frac{\pi}{3} + \pi\right)$.
$V_o = 252 \cos \left(12 t + \frac{4 \pi}{3}\right)$.

(D) The power gain of an amplifier is defined as the product of voltage gain and current gain.
$\text{Power Gain} = \text{Voltage Gain} \times \text{Current Gain}$
Therefore,the ratio of power gain to voltage gain is equal to the current gain.
In a common emitter configuration,the current gain is denoted by $\beta$.
Thus,$\frac{\text{Power Gain}}{\text{Voltage Gain}} = \beta$.

(C) In a transistor,the emitter-base junction is always forward-biased to allow charge carriers to move into the base.
In an $n-p-n$ transistor,the emitter is $n$-type,so it injects electrons (majority carriers) into the base.
In a $p-n-p$ transistor,the emitter is $p$-type,so it injects holes (majority carriers) into the base.
Therefore,the primary difference in operation is the type of charge carrier injected by the emitter into the base region: $p-n-p$ injects holes,while $n-p-n$ injects electrons.

(D) We know the relationship between current gains $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.
Alternatively,we can use the fundamental current relation $I_E = I_C + I_B$.
Dividing by $I_C$,we get $\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$.
This implies $\frac{1}{\alpha} = 1 + \frac{1}{\beta}$.
Rearranging the terms,we get $\frac{1}{\alpha} - \frac{1}{\beta} = 1$.
Substituting the given values: $\frac{1}{0.98} - \frac{1}{49} \approx 1.0204 - 0.0204 = 1$.

(D) It is given that $80 \%$ of the emitted electrons reach the collector, which means the collector current $I_c$ is $80 \%$ of the emitter current $I_e$.
$I_c = 0.80 \times I_e$
Given $I_c = 28 \,mA$, we have $28 = 0.80 \times I_e$.
$I_e = \frac{28}{0.80} = 35 \,mA$.
Using the relation for transistor currents, $I_e = I_c + I_b$, we can find the base current $I_b$.
$I_b = I_e - I_c = 35 \,mA - 28 \,mA = 7 \,mA$.

(B) The correct option is $B$.
Concept: The current gain $\beta$ is given by $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given: $\Delta I_C = 1.5 \ mA = 1.5 \times 10^{-3} \ A$ and $\Delta I_B = 20 \ \mu A = 20 \times 10^{-6} \ A$.
Calculating $\beta$: $\beta = \frac{1.5 \times 10^{-3}}{20 \times 10^{-6}} = \frac{1500}{20} = 75$.
The voltage gain $A_v$ is given by $A_v = \beta \times \frac{R_L}{R_i}$.
Given: Load resistance $R_L = 2 \ k\Omega = 2000 \ \Omega$ and input resistance $R_i = 150 \ \Omega$.
Substituting the values: $A_v = 75 \times \frac{2000}{150} = 75 \times \frac{40}{3} = 25 \times 40 = 1000$.

(A) In a common emitter $(CE)$ configuration of a transistor amplifier,the input signal is applied to the base-emitter junction and the output is taken from the collector-emitter junction.
When the input signal voltage increases,the base current increases,which leads to an increase in the collector current.
Due to the voltage drop across the load resistor $R_C$ connected in the collector circuit,the collector voltage decreases.
Thus,an increase in input voltage results in a decrease in output voltage,and vice versa.
This inverse relationship corresponds to a phase shift of $180^{\circ}$ or $\pi$ radians between the input and output signals.

(D) In an $n-p-n$ transistor,the emitter is $n$-type,the base is $p$-type,and the collector is $n$-type.
$1$. Electrons are the majority charge carriers in the emitter and they move from the emitter to the base.
$2$. In the base,electrons recombine with holes,but since the base is very thin and lightly doped,most electrons cross into the collector region.
$3$. The collector is reverse-biased,which attracts the electrons coming from the base.
$4$. Holes are the majority charge carriers in the base,but they do not move from the base to the collector in significant numbers to constitute the main current; rather,the collector current is primarily due to the flow of electrons from the emitter through the base to the collector.
$5$. Statement $D$ is incorrect because electrons move from the emitter to the collector,not from the collector to the base.

(C) The output characteristics of a transistor in common emitter $(CE)$ configuration represent the relationship between the output current $(I_C)$ and the output voltage $(V_{CE})$ while keeping the input current $(I_B)$ constant.
In this configuration,the collector current $I_C$ is plotted on the $y$-axis and the collector-emitter voltage $V_{CE}$ is plotted on the $x$-axis for different fixed values of base current $I_B$.
Therefore,the correct graph is obtained by plotting $I_C$ against $V_{CE}$ at constant $I_B$.

(B) The output resistance $(R_o)$ of a transistor in common emitter configuration is defined as the ratio of the change in collector-emitter voltage $(\Delta V_{CE})$ to the change in collector current $(\Delta I_C)$ at a constant base current.
$R_o = \frac{\Delta V_{CE}}{\Delta I_C}$
Given:
$\Delta V_{CE} = 0.4 \, V$
$\Delta I_C = 0.04 \, mA = 0.04 \times 10^{-3} \, A$
Substituting the values:
$R_o = \frac{0.4}{0.04 \times 10^{-3}} = \frac{0.4}{4 \times 10^{-5}} = 0.1 \times 10^5 \, \Omega = 10,000 \, \Omega = 10 \, k\Omega$

(D) We know that for a transistor,the current amplification factors are defined as $\alpha_{DC} = \frac{I_C}{I_E}$ and $\beta_{DC} = \frac{I_C}{I_B}$.
Substituting these into the expression:
$\frac{1}{\alpha_{DC}} - \frac{1}{\beta_{DC}} = \frac{I_E}{I_C} - \frac{I_B}{I_C}$
Since $I_E = I_C + I_B$,we have $I_E - I_B = I_C$.
Therefore,$\frac{I_E - I_B}{I_C} = \frac{I_C}{I_C} = 1$.

(B) The total charge $q$ entering the emitter is given by $q = n \times e = 200 \times 1.6 \times 10^{-19} \ C = 3.2 \times 10^{-17} \ C$.
The emitter current $I_e$ is calculated as $I_e = \frac{q}{t} = \frac{3.2 \times 10^{-17} \ C}{10^{-8} \ s} = 3.2 \times 10^{-9} \ A$.
Given that $1 \%$ of electrons are lost in the base,the base current $I_b = 0.01 \times I_e$.
The collector current $I_c$ is the remaining current,so $I_c = I_e - I_b = 0.99 \times I_e$.
The current amplification factor $\beta$ is defined as $\beta = \frac{I_c}{I_b} = \frac{0.99 \times I_e}{0.01 \times I_e} = 99$.

(B) The power gain of an amplifier is given by the product of voltage gain and current gain.
$\text{Power Gain} = \text{Voltage Gain} \times \text{Current Gain}$
Given:
Voltage Gain $(A_v)$ = $40$
Input Impedance $(Z_i)$ = $100 \ \Omega$
Output Impedance $(Z_o)$ = $400 \ \Omega$
First, we calculate the current gain $(\beta)$:
$\text{Current Gain} = \frac{\text{Output Current}}{\text{Input Current}} = \frac{V_o / Z_o}{V_i / Z_i} = \left( \frac{V_o}{V_i} \right) \times \left( \frac{Z_i}{Z_o} \right)$
$\text{Current Gain} = A_v \times \left( \frac{Z_i}{Z_o} \right) = 40 \times \left( \frac{100}{400} \right) = 40 \times 0.25 = 10$
Now, calculate the power gain:
$\text{Power Gain} = 40 \times 10 = 400$

(D) The current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_b} = \frac{5 \text{ mA}}{0.2 \text{ mA}} = 25$
The voltage gain $A_v$ for a common emitter amplifier is given by the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Given that $A_v = 75$,$R_i = 2 \text{ k}\Omega$,and $\beta = 25$,we can solve for the load resistance $R_L$:
$75 = 25 \times \frac{R_L}{2 \text{ k}\Omega}$
Rearranging the equation to solve for $R_L$:
$R_L = \frac{75 \times 2 \text{ k}\Omega}{25}$
$R_L = 3 \times 2 \text{ k}\Omega = 6 \text{ k}\Omega$
Therefore,the load resistance is $6 \text{ k}\Omega$.

(A) The relationship between emitter current, collector current, and base current is given by the equation: $\Delta I_{E} = \Delta I_{C} + \Delta I_{B}$.
Given values are $\Delta I_{E} = 8.0 \,mA$ and $\Delta I_{C} = 7.8 \,mA$.
Substituting these values into the equation: $8.0 \,mA = 7.8 \,mA + \Delta I_{B}$.
Therefore, $\Delta I_{B} = 8.0 \,mA - 7.8 \,mA = 0.2 \,mA$.
Converting to microamperes: $0.2 \,mA = 0.2 \times 1000 \mu A = 200 \mu A$.

(B) The relationship between the current gain $\beta_{dc}$ and the current ratio $\alpha_{dc}$ is given by the formula: $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$.
Given $\alpha_{dc} = \frac{69}{70}$.
Substituting the value: $\beta_{dc} = \frac{69/70}{1 - 69/70} = \frac{69/70}{1/70} = 69$.
Therefore,the current gain $\beta_{dc}$ is $69$.

(D) In a common-emitter transistor amplifier configuration,the current gain (denoted by $\beta$) is defined as the ratio of the change in collector current $(\Delta I_{C})$ to the change in base current $(\Delta I_{B})$ while keeping the collector-emitter voltage $(V_{CE})$ constant.
Mathematically,$\beta = \frac{\Delta I_{C}}{\Delta I_{B}}$.
Therefore,the correct option is $D$.

(D) Given: $\beta = 20$, $I_{E} = 7 \,mA$.
We know that the current gain in common emitter mode is defined as $\beta = \frac{I_{C}}{I_{B}}$.
Also, the relationship between emitter current, collector current, and base current is $I_{E} = I_{C} + I_{B}$, which implies $I_{B} = I_{E} - I_{C}$.
Substituting $I_{B}$ in the gain formula: $\beta = \frac{I_{C}}{I_{E} - I_{C}}$.
Rearranging the equation: $20 = \frac{I_{C}}{7 - I_{C}}$.
$20(7 - I_{C}) = I_{C}$.
$140 - 20I_{C} = I_{C}$.
$140 = 21I_{C}$.
$I_{C} = \frac{140}{21} = \frac{20}{3} \,mA$.

(B) Given: Input resistance $R_{i} = 1000 \Omega$,Input signal voltage $v_{i} = 5 mV = 5 \times 10^{-3} V$,and current gain $\beta = 60$.
First,calculate the input current $I_{b}$ using Ohm's law: $I_{b} = \frac{v_{i}}{R_{i}} = \frac{5 \times 10^{-3} V}{1000 \Omega} = 5 \times 10^{-6} A$.
The output current in a common emitter amplifier is the collector current $I_{c}$.
Using the relation $I_{c} = \beta \times I_{b}$,we get: $I_{c} = 60 \times 5 \times 10^{-6} A = 300 \times 10^{-6} A$.
Therefore,the peak value of the output current is $3 \times 10^{-4} A$.

(B) The voltage gain of an amplifier with feedback is given by $A_f = \frac{A}{1 - A\beta}$,where $A$ is the open-loop voltage gain and $\beta$ is the feedback factor.
For an oscillator to produce sustained oscillations,the Barkhausen criterion must be satisfied.
The Barkhausen criterion states that the loop gain must be equal to unity,which is expressed as $A\beta = 1$.

(C) Given: $\alpha = 0.8$ and $\Delta I_B = 6 \text{ mA}$.
First, we calculate the current gain $\beta$ for the common-emitter configuration using the relation $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4$.
The relationship between the change in collector current $\Delta I_C$ and the change in base current $\Delta I_B$ is given by $\Delta I_C = \beta \times \Delta I_B$.
Substituting the values: $\Delta I_C = 4 \times 6 \text{ mA} = 24 \text{ mA}$.
Therefore, the change in collector current is $24 \text{ mA}$.

(D) In a transistor,the emitter current is the sum of the base current and the collector current: $I_{e} = I_{b} + I_{c}$.
Since $I_{b}$ and $I_{c}$ are both positive currents,it follows that $I_{c} < I_{e}$ and $I_{b} < I_{e}$.
The current gain in common emitter configuration is defined as $\beta = \frac{I_{c}}{I_{b}}$.
For a transistor,$I_{c}$ is typically much larger than $I_{b}$ (since the base region is very thin and lightly doped),which makes $\beta > 1$.
Therefore,the condition $I_{c} > I_{b}$ is the fundamental reason why the current gain $\beta$ is greater than $1$.

(D) Given: The ratio of collector current $(i_c)$ to emitter current $(i_e)$ is $\alpha = \frac{i_c}{i_e} = 0.98$.
The collector current is $i_c = 3 \text{ mA}$.
We know that the emitter current is the sum of collector current and base current: $i_e = i_c + i_b$.
Substituting this into the ratio: $\frac{i_c}{i_c + i_b} = 0.98$.
Rearranging the equation: $\frac{i_c + i_b}{i_c} = \frac{1}{0.98} = \frac{100}{98}$.
$1 + \frac{i_b}{i_c} = \frac{100}{98} \implies \frac{i_b}{i_c} = \frac{100}{98} - 1 = \frac{2}{98} = \frac{1}{49}$.
Therefore,$i_b = \frac{i_c}{49} = \frac{3 \text{ mA}}{49} = \frac{3000 \mu\text{A}}{49} \approx 61.22 \mu\text{A}$.
Rounding to the nearest given option,the base current is approximately $60 \mu\text{A}$.