Junction Transistor Questions in English

(B) In a common-emitter configuration, the current gain $\beta$ is defined as $\beta = \frac{I_C}{I_B}$.
Given $\beta = 60$, we have $I_C = 60 I_B$.
We know that the emitter current $I_E$ is the sum of the base current $I_B$ and the collector current $I_C$:
$I_E = I_B + I_C$
Substituting $I_C = 60 I_B$ into the equation:
$I_E = I_B + 60 I_B = 61 I_B$
Given $I_E = 6.6 \,mA$, we can solve for $I_B$:
$6.6 \,mA = 61 I_B$
$I_B = \frac{6.6}{61} \,mA \approx 0.108 \,mA$.

(C) $1$. The emitter is heavily doped compared to the collector to provide a large number of charge carriers. This statement is correct.
$2$. Emitter and collector cannot be interchanged because they are designed differently in terms of doping concentration and physical size. This statement is incorrect.
$3$. The base region is very thin but is lightly doped to minimize recombination of charge carriers. This statement is incorrect.
$4$. In an $n-p-n$ transistor,the conventional current flows from the base to the emitter (as electrons flow from emitter to base). This statement is correct.
Therefore,statements $1$ and $4$ are correct.

(D) In a transistor,the three regions are the emitter,base,and collector.
$1$. The $Base$ is very thin and lightly doped to allow most charge carriers to pass through to the collector.
$2$. The $Collector$ is moderately doped and larger in size to dissipate heat.
$3$. The $Emitter$ is heavily doped to provide a large number of charge carriers.
Therefore,the order of increasing doping level is $Base < Collector < Emitter$.

(D) In a common emitter amplifier configuration,the input signal is applied to the base-emitter junction and the output is taken from the collector-emitter junction.
When the input voltage increases,the base current increases,which in turn increases the collector current.
Due to the voltage drop across the load resistor in the collector circuit,an increase in collector current leads to a decrease in the output voltage.
Because the output voltage decreases as the input voltage increases,they are in opposite phases.
Therefore,the phase difference between the input voltage and the output voltage in a common emitter amplifier is $180^{\circ}$.

(A) Given:
Change in collector current,$\Delta I_C = 8.9 \ mA$
Change in emitter current,$\Delta I_E = 9.0 \ mA$
We know that the emitter current is the sum of collector current and base current: $\Delta I_E = \Delta I_C + \Delta I_B$
Therefore,the change in base current is: $\Delta I_B = \Delta I_E - \Delta I_C = 9.0 \ mA - 8.9 \ mA = 0.1 \ mA$
The current amplification factor $\beta$ is defined as the ratio of change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{8.9 \ mA}{0.1 \ mA} = 89$

(A) Given: Power gain $(A_p)$ = $1800$, Voltage gain $(A_v)$ = $60$, Change in emitter current $(\Delta I_e)$ = $0.62 \text{ mA}$.
We know that Power gain $(A_p)$ = Current gain $(\beta)$ $\times$ Voltage gain $(A_v)$.
Therefore, $\beta = A_p / A_v = 1800 / 60 = 30$.
We also know that the current gain in common emitter configuration is $\beta = \Delta I_c / \Delta I_b$, and $\Delta I_e = \Delta I_b + \Delta I_c$.
Substituting $\Delta I_b = \Delta I_c / \beta$, we get $\Delta I_e = \Delta I_c / \beta + \Delta I_c = \Delta I_c (1 + 1/\beta) = \Delta I_c ((\beta + 1) / \beta)$.
Rearranging for $\Delta I_c$: $\Delta I_c = \Delta I_e \times (\beta / (\beta + 1))$.
Substituting the values: $\Delta I_c = 0.62 \times (30 / 31) = 0.62 \times (0.9677) \approx 0.60 \text{ mA}$.

(D) The emitter current $I_E$ is given by the rate of flow of charge: $I_E = \frac{q}{t} = \frac{n \cdot e}{t}$.
Given $n = 10^{10}$, $e = 1.6 \times 10^{-19} \text{ C}$, and $t = 0.4 \times 10^{-6} \text{ s}$.
$I_E = \frac{10^{10} \times 1.6 \times 10^{-19}}{0.4 \times 10^{-6}} = \frac{1.6 \times 10^{-9}}{0.4 \times 10^{-6}} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
Since $5 \%$ of the electrons are lost in the base, the collector current $I_C$ is $95 \%$ of the emitter current $I_E$.
$I_C = 0.95 \times I_E = 0.95 \times 4 \text{ mA} = 3.8 \text{ mA}$.

(A) transistor acts as a switch when it is operated between two states: the $OFF$ state and the $ON$ state.
In the $OFF$ state,the transistor is in the cutoff region,where the collector current is zero.
In the $ON$ state,the transistor is in the saturation region,where the collector current is maximum and the output voltage is very low.
The active region is used for amplification,not for switching.
Therefore,the transistor acts as a switch in the cutoff or saturation region.

(B) In a common emitter transistor configuration,the transfer characteristic curve shows three distinct regions:
$1$. Cutoff Region: When the input voltage $(V_{i})$ is low,the transistor is in the cutoff region,and the output voltage $(V_{o})$ is high (near $V_{CC}$). This corresponds to Region $I$.
$2$. Active Region: As $(V_{i})$ increases,the transistor enters the active region where the output voltage $(V_{o})$ decreases linearly with $(V_{i})$. This corresponds to Region $II$.
$3$. Saturation Region: When $(V_{i})$ is high,the transistor enters the saturation region,and the output voltage $(V_{o})$ becomes very low (near $0$). This corresponds to Region $III$.
Therefore,the active,saturation,and cutoff regions are represented by $II, III$ and $I$ respectively.

(D) In a common emitter configuration,the current amplification factor is given by $\beta = 80$.
The relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is $I_E = I_C + I_B$.
We also know that $\beta = \frac{I_C}{I_B}$,which implies $I_C = \beta I_B$.
Substituting $I_C$ in the first equation: $I_E = \beta I_B + I_B = I_B(\beta + 1)$.
Given $I_E = 2.43 \text{ mA} = 2430 \mu \text{A}$ and $\beta = 80$.
$I_B = \frac{I_E}{\beta + 1} = \frac{2430 \mu \text{A}}{80 + 1} = \frac{2430 \mu \text{A}}{81}$.
$I_B = 30 \mu \text{A}$.

(A) transistor operates as an amplifier in its active region.
In the active region,the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.
This configuration allows the transistor to provide current and voltage gain,making it suitable for amplification purposes.
In contrast,the cut-off region acts as an open switch ($OFF$ state),and the saturation region acts as a closed switch ($ON$ state).

(A) For a transistor in common emitter configuration, the voltage gain $A_V$ is given by the product of current gain $\beta$ and resistance gain $R_C/R_B$.
$A_V = \beta \times \frac{R_C}{R_B}$
Given: $A_V = 160$, $R_B = 1 \text{ k}\Omega$, $R_C = 4 \text{ k}\Omega$, and $\Delta I_B = 100 \mu A$.
We know that $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Substituting this into the voltage gain formula:
$A_V = \left(\frac{\Delta I_C}{\Delta I_B}\right) \times \frac{R_C}{R_B}$
$160 = \left(\frac{\Delta I_C}{100 \times 10^{-6} \text{ A}}\right) \times \left(\frac{4 \text{ k}\Omega}{1 \text{ k}\Omega}\right)$
$160 = \left(\frac{\Delta I_C}{100 \times 10^{-6}}\right) \times 4$
$40 = \frac{\Delta I_C}{100 \times 10^{-6}}$
$\Delta I_C = 40 \times 100 \times 10^{-6} \text{ A} = 4000 \times 10^{-6} \text{ A} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
Thus, the change in output current is $4 \text{ mA}$.

(A) Given: Base current,$i_B = 10 \mu A$.
Emitter current,$i_E = 1 \text{ mA} = 1000 \mu A$.
According to the fundamental relation for a transistor,the emitter current is the sum of the base current and the collector current: $i_E = i_B + i_C$.
Rearranging the formula to find the collector current: $i_C = i_E - i_B$.
Substituting the values: $i_C = 1000 \mu A - 10 \mu A = 990 \mu A$.

(B) In the circuit symbol of a transistor,the arrow on the emitter terminal indicates the direction of conventional current flow.
For an $n-p-n$ transistor,the current flows from the base to the emitter,so the arrow on the emitter points outwards.
For a $p-n-p$ transistor,the current flows from the emitter to the base,so the arrow on the emitter points inwards.
In the given symbol,the arrow on the emitter is pointing outwards,which confirms that it is an $n-p-n$ transistor.

(C) In an $n-p-n$ transistor,the emitter is heavily doped to provide a large number of charge carriers. The base is very thin and lightly doped to allow most of the charge carriers from the emitter to pass through to the collector. The collector is moderately doped and is the largest in size to dissipate the heat generated during operation. Therefore,the statement that the collector is lightly doped is incorrect.

(B) In a $p-n-p$ transistor,the collector current $(I_C)$ is slightly less than the emitter current $(I_E)$.
This is because the emitter-base junction is forward-biased,causing the majority charge carriers (holes) from the emitter to move towards the base.
Since the base is very thin and lightly doped,only a small fraction of these holes recombine with the electrons in the base,resulting in a small base current $(I_B)$.
The majority of the holes cross the collector-base junction to reach the collector.
According to Kirchhoff's current law for a transistor,the relationship is $I_E = I_B + I_C$.
Since $I_B > 0$,it follows that $I_C < I_E$.

(D) In a $p-n-p$ transistor,the majority charge carriers are holes.
When the transistor is connected to an external circuit,the current is primarily carried by the movement of these holes through the semiconductor material.

(A) Given, the current gain $\alpha = 0.98$ (since $\alpha < 1$, it refers to the common base configuration).
For a common emitter configuration, the current gain $\beta$ is given by the formula:
$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
We know that the relationship between collector current change $\Delta I_C$ and base current change $\Delta I_B$ in a common emitter configuration is:
$\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given $\Delta I_B = 0.5 \,mA$, we can calculate $\Delta I_C$ as:
$\Delta I_C = \beta \times \Delta I_B = 49 \times 0.5 \,mA = 24.5 \,mA$.

(B) Given, base current in $n-p-n$ transistor, $I_B = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$.
Since $95 \%$ of emitted electrons reach the collector, the collector current $I_C$ is related to the emitter current $I_E$ as:
$I_C = 0.95 I_E \Rightarrow I_E = \frac{I_C}{0.95} \dots (i)$
We know that the relationship between emitter, collector, and base current is:
$I_E = I_C + I_B$
Substituting the value of $I_E$ from equation $(i)$:
$\frac{I_C}{0.95} = I_C + 2 \times 10^{-3}$
$\frac{I_C}{0.95} - I_C = 2 \times 10^{-3}$
$I_C \left( \frac{1}{0.95} - 1 \right) = 2 \times 10^{-3}$
$I_C \left( \frac{1 - 0.95}{0.95} \right) = 2 \times 10^{-3}$
$I_C \left( \frac{0.05}{0.95} \right) = 2 \times 10^{-3}$
$I_C \left( \frac{1}{19} \right) = 2 \times 10^{-3}$
$I_C = 38 \times 10^{-3} \text{ A} = 38 \text{ mA}$.

(C) Given: Load resistance $R_L = 3 k\Omega = 3000 \Omega$.
Input voltage $V_i = 30 mV = 30 \times 10^{-3} V$.
Change in base current $\Delta I_B = 30 \mu A = 30 \times 10^{-6} A = 3 \times 10^{-5} A$.
Change in collector current $\Delta I_C = 3 mA = 3 \times 10^{-3} A$.
First, calculate the current gain $\beta$:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3 \times 10^{-3}}{3 \times 10^{-5}} = 100$.
Next, calculate the input resistance $R_{in}$:
$R_{in} = \frac{V_i}{\Delta I_B} = \frac{30 \times 10^{-3}}{30 \times 10^{-6}} = 1000 \Omega$.
The power gain $A_P$ is given by the formula:
$A_P = \beta^2 \times \frac{R_L}{R_{in}}$.
Substituting the values:
$A_P = (100)^2 \times \frac{3000}{1000} = 10000 \times 3 = 30000$.
Thus, the power gain is $30000$.

(B) The voltage gain $(A_v)$ is given by:
$A_v = \frac{\Delta V_o}{\Delta V_i} = \frac{\Delta I_c \times R_L}{\Delta V_i} = \frac{4 \times 10^{-3} \times 10 \times 10^3}{40 \times 10^{-3}} = 1000$
The current gain $(A_i)$ is given by:
$A_i = \frac{\Delta I_c}{\Delta I_b} = \frac{4 \times 10^{-3}}{20 \times 10^{-6}} = 200$
The power gain $(P)$ is the product of voltage gain and current gain:
$P = A_v \times A_i = 1000 \times 200 = 2 \times 10^5$

(B) Given: Change in collector current $\Delta I_C = 6.8 \,mA$ and change in emitter current $\Delta I_E = 7 \,mA$.
Since $\Delta I_E = \Delta I_C + \Delta I_B$,the change in base current is $\Delta I_B = \Delta I_E - \Delta I_C = 7 \,mA - 6.8 \,mA = 0.2 \,mA$.
The current amplification factor $\beta$ for a common-emitter configuration is defined as $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Substituting the values: $\beta = \frac{6.8 \,mA}{0.2 \,mA} = 34$.
Thus,the current amplification factor is $34$.

(B) The output characteristics of a transistor are defined as the variation of the collector current $(I_C)$ with the collector-emitter voltage $(V_{CE})$ while keeping the base current $(I_B)$ constant.
This relationship is graphically represented by plotting $I_C$ on the y-axis and $V_{CE}$ on the x-axis for different fixed values of $I_B$.

(C) The current gain $\beta$ of a transistor in common-emitter configuration is defined as the ratio of the change in collector current to the change in base current.
$\beta = \frac{\Delta I_C}{\Delta I_B}$
Given:
$\Delta I_B = 140 \mu A - 45 \mu A = 95 \mu A = 95 \times 10^{-6} \text{ A}$
$\Delta I_C = 4.0 \text{ mA} - 0.2 \text{ mA} = 3.8 \text{ mA} = 3.8 \times 10^{-3} \text{ A}$
Substituting the values:
$\beta = \frac{3.8 \times 10^{-3}}{95 \times 10^{-6}}$
$\beta = \frac{3800 \times 10^{-6}}{95 \times 10^{-6}}$
$\beta = 40$

(C) The relationship between $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.
For $\alpha_1 = \frac{20}{21}$,$\beta_1 = \frac{20/21}{1 - 20/21} = \frac{20/21}{1/21} = 20$.
For $\alpha_2 = \frac{100}{101}$,$\beta_2 = \frac{100/101}{1 - 100/101} = \frac{100/101}{1/101} = 100$.
Therefore,the value of $\beta$ lies between $20$ and $100$.

(D) The current gain $\beta$ of a transistor in common-emitter configuration 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: $\beta = 80$ and $\Delta i_B = 250 \mu A = 250 \times 10^{-6} \text{ A}$.
Substituting the values into the formula:
$80 = \frac{\Delta i_C}{250 \times 10^{-6} \text{ A}}$
$\Delta i_C = 80 \times 250 \times 10^{-6} \text{ A}$
$\Delta i_C = 20,000 \times 10^{-6} \text{ A}$
$\Delta i_C = 20 \times 10^{-3} \text{ A} = 20 \text{ mA}$.
Therefore,the change in collector current is $20 \text{ mA}$.

(C) In a Common-Emitter $(C-E)$ configuration, the transistor acts as an amplifier for both current and voltage.
$1$. Current gain $(\beta = I_C / I_B)$ is typically much greater than $1$.
$2$. Voltage gain $(A_V = \beta \times (R_L / R_{in}))$ is also significantly greater than $1$ because the output resistance $(R_L)$ is much higher than the input resistance $(R_{in})$.
$3$. Since both current and voltage are amplified, the power gain $(A_P = A_V \times \beta)$ is also significantly greater than $1$.
Therefore, the $C-E$ configuration provides both current and voltage amplification.

(C) $1$. The emitter is heavily doped compared to the collector to provide a large number of charge carriers. This statement is correct.
$2$. Emitter and collector cannot be interchanged because they have different doping levels and physical dimensions. This statement is incorrect.
$3$. The base region is very thin and lightly doped to allow most charge carriers from the emitter to pass through to the collector. This statement is incorrect.
$4$. In an $n-p-n$ transistor,the conventional current flows from the collector to the base and from the base to the emitter (or simply,the emitter current $I_E$ flows out of the emitter). The conventional current direction is from base to emitter in the $n-p-n$ transistor structure. This statement is correct.
Therefore,statements $1$ and $4$ are correct.

(A) In an $n-p-n$ transistor,the emitter is $n$-type,the base is $p$-type,and the collector is $n$-type.
When used as an amplifier,the base-emitter junction is forward-biased and the base-collector junction is reverse-biased.
Due to the forward bias,electrons are injected from the $n$-type emitter into the $p$-type base.
Because the base is very thin and lightly doped,most of these electrons diffuse through the base into the collector region.
The collector is kept at a positive potential relative to the base,which attracts these electrons from the base to the collector.
Therefore,electrons move from the base to the collector.

(D) Given:
$Current \ gain \ (\beta) = 50$
$Load \ resistance \ (R_L) = 4 \ k\Omega = 4000 \ \Omega$
$Input \ resistance \ (R_i) = 500 \ \Omega$
The voltage gain $(A_v)$ of a common emitter amplifier is given by the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Substituting the values:
$A_v = 50 \times \frac{4000}{500}$
$A_v = 50 \times 8$
$A_v = 400$
Therefore, the voltage gain of the amplifier is $400$.

(A) Given:
Increase in base current,$\Delta I_b = 50 \mu A = 50 \times 10^{-6} \ A$.
Increase in collector current,$\Delta I_c = 1 \ mA = 1 \times 10^{-3} \ A$.
The collector voltage is kept constant,which is the condition for calculating the common-emitter current gain $\beta$.
The current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_c}{\Delta I_b}$
Substituting the values:
$\beta = \frac{1 \times 10^{-3} \ A}{50 \times 10^{-6} \ A}$
$\beta = \frac{1000}{50} = 20$.
Therefore,the current gain of the transistor is $20$.

(A) Given:
Change in collector current,$\Delta I_c = 8.2 \,mA$
Change in emitter current,$\Delta I_e = 8.3 \,mA$
We know that the emitter current is the sum of base current and collector current: $\Delta I_e = \Delta I_b + \Delta I_c$
Therefore,the change in base current is: $\Delta I_b = \Delta I_e - \Delta I_c = 8.3 \,mA - 8.2 \,mA = 0.1 \,mA$
The forward current ratio (current gain $\beta$) is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{8.2 \,mA}{0.1 \,mA} = 82$

(A) Given:
$\beta = 48$
$I_{B} = 200 \mu A = 200 \times 10^{-6} \ A$
$R_{C} = 500 \ \Omega$
$V_{CC} = 5 \ V$
Step $1$: Calculate the collector current $I_{C}$.
$I_{C} = \beta I_{B} = 48 \times 200 \times 10^{-6} \ A = 9600 \times 10^{-6} \ A = 9.6 \times 10^{-3} \ A = 9.6 \ mA$.
Step $2$: Apply Kirchhoff's voltage law to the output loop.
$V_{CC} = I_{C} R_{C} + V_{CE}$
$V_{Y} = V_{CE} = V_{CC} - I_{C} R_{C}$
$V_{Y} = 5 \ V - (9.6 \times 10^{-3} \ A) \times (500 \ \Omega)$
$V_{Y} = 5 \ V - 4.8 \ V = 0.2 \ V$.
Thus,the voltage at terminal $Y$ is $0.2 \ V$.

(C) Given: Current gain $\beta = 45$,Collector resistance $R_C = 1 \ k\Omega = 1000 \ \Omega$,Potential drop across collector resistance $V_C = 5 \ V$.
We know that the collector current $I_C$ is given by $I_C = \frac{V_C}{R_C}$.
Substituting the values: $I_C = \frac{5 \ V}{1000 \ \Omega} = 5 \times 10^{-3} \ A = 5 \ mA$.
The relationship between collector current $I_C$ and base current $I_B$ is given by $\beta = \frac{I_C}{I_B}$.
Therefore,$I_B = \frac{I_C}{\beta}$.
Substituting the values: $I_B = \frac{5 \times 10^{-3} \ A}{45} = \frac{1}{9} \times 10^{-3} \ A \approx 0.111 \times 10^{-3} \ A$.
Converting to microamperes: $I_B \approx 111 \ \mu A$.

(B) In a common emitter configuration of a transistor,the base current $I_{B}$ is very small and is typically in the order of microamperes $(\mu A)$.
Since the collector current $I_{C} = \beta I_{B}$ and the current gain $\beta$ is typically large (around $100$),the collector current $I_{C}$ is in the order of milliamperes $(mA)$.
The collector-emitter voltage $V_{CE}$ is typically maintained in the order of volts $(V)$ to keep the transistor in the active region.
Therefore,the correct order of magnitude is $I_{B}$ in $\mu A$,$I_{C}$ in $mA$,and $V_{CE}$ in volts.

(A) In an $n-p-n$ or $p-n-p$ transistor,the emitter is heavily doped to provide a large number of majority charge carriers. The base is very thin and lightly doped to minimize recombination. The collector is moderately doped and is physically larger than the emitter to handle the power dissipation. Therefore,the emitter has a higher degree of doping compared to the collector.

(D) Given:
$\beta = 50$
Input resistance $R_i = 1 \text{ k}\Omega = 10^3 \Omega$
Peak input voltage $V_i = 0.01 \text{ V}$
Step $1$: Calculate the peak value of base current $(\Delta I_B)$:
$\Delta I_B = \frac{V_i}{R_i} = \frac{0.01 \text{ V}}{10^3 \Omega} = 10^{-5} \text{ A}$
Step $2$: Calculate the peak value of collector current $(\Delta I_C)$:
Using the relation $\beta = \frac{\Delta I_C}{\Delta I_B}$,we get:
$\Delta I_C = \beta \times \Delta I_B$
$\Delta I_C = 50 \times 10^{-5} \text{ A}$
$\Delta I_C = 5 \times 10^{-4} \text{ A} = 500 \times 10^{-6} \text{ A} = 500 \mu\text{A}$

(D) The voltage gain $A_v$ of a common emitter amplifier is defined as the ratio of the change in output voltage to the change in input voltage.
$A_v = \frac{\Delta V_{out}}{\Delta V_{in}} = \frac{\Delta I_C \times R_L}{\Delta V_{BE}}$
Given:
Load resistance $R_L = 6 k\Omega = 6000 \Omega$
Input signal voltage $\Delta V_{BE} = 15 mV = 15 \times 10^{-3} V$
Change in collector current $\Delta I_C = 1.8 mA = 1.8 \times 10^{-3} A$
Substituting the values into the formula:
$A_v = \frac{1.8 \times 10^{-3} A \times 6000 \Omega}{15 \times 10^{-3} V}$
$A_v = \frac{1.8 \times 6000}{15} = \frac{10800}{15} = 720$
Thus, the voltage gain of the amplifier is $720$.