Junction Transistor Questions in English

(B) Given:
Common emitter current gain, $\beta = 20$
Emitter current, $I_E = 7 \text{ mA}$
We know the relationship between emitter current and collector current in common emitter mode is $I_E = I_C + I_B$.
Also, $\beta = I_C / I_B$, which implies $I_B = I_C / \beta$.
Substituting $I_B$ in the first equation:
$I_E = I_C + (I_C / \beta) = I_C(1 + 1/\beta) = I_C((\beta + 1) / \beta)$.
Rearranging for $I_C$:
$I_C = I_E \times (\beta / (\beta + 1))$.
Substituting the values:
$I_C = 7 \times (20 / (20 + 1)) = 7 \times (20 / 21)$.
$I_C = 7 \times (20 / 21) = 20 / 3 \text{ mA}$.

(A) In a common-emitter transistor amplifier configuration, the input current is the base current $(I_B)$ and the output current is the collector current $(I_C)$.
The current gain, denoted by $\beta$, is defined as the ratio of the change in collector current to the change in base current, keeping the collector-emitter voltage constant.
Mathematically, $\beta = \Delta I_C / \Delta I_B$.

(A) Given:
$d.c. \text{ current gain } (\beta) = 20$
$Emitter current (I_E) = 7 \text{ mA}$
We know the relationship between current gain $(\beta)$ and collector current $(I_C)$ is given by $I_C = \beta I_B$, where $I_B$ is the base current.
Also, $I_E = I_B + I_C$.
Substituting $I_B = I_C / \beta$ into the equation:
$I_E = I_C / \beta + I_C$
$I_E = I_C (1/\beta + 1)$
$I_E = I_C (1 + \beta) / \beta$
Rearranging to solve for $I_C$:
$I_C = I_E \times \beta / (1 + \beta)$
$I_C = 7 \times 20 / (1 + 20)$
$I_C = 140 / 21$
Dividing both numerator and denominator by $7$:
$I_C = 20 / 3 \text{ mA}$.

(A) Given: $\alpha = 0.9$ and change in base current $\Delta I_B = 0.4 \text{ mA}$.
First, we calculate the current gain $\beta$ for the $CE$ configuration using the relation: $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$: $\beta = \frac{0.9}{1 - 0.9} = \frac{0.9}{0.1} = 9$.
The relationship between collector current change $\Delta I_C$ and base current change $\Delta I_B$ is given by: $\Delta I_C = \beta \times \Delta I_B$.
Substituting the values: $\Delta I_C = 9 \times 0.4 \text{ mA} = 3.6 \text{ mA}$.
Therefore, the change in collector current is $3.6 \text{ mA}$.

(D) The current gain $\beta$ is given by the ratio of the change in collector current $\Delta I_C$ to the change in base current $\Delta I_B$.
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \text{ mA}}{0.2 \text{ mA}} = 25$.
The voltage gain $A_V$ is given by the formula $A_V = \beta \times \frac{R_L}{R_{in}}$, where $R_L$ is the load resistance and $R_{in}$ is the input resistance.
Given $A_V = 75$, $\beta = 25$, and $R_{in} = 2 \text{ k}\Omega$.
Substituting the values: $75 = 25 \times \frac{R_L}{2 \text{ k}\Omega}$.
$3 = \frac{R_L}{2 \text{ k}\Omega}$.
$R_L = 3 \times 2 \text{ k}\Omega = 6 \text{ k}\Omega$.

(B) The current gain $\beta$ is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \text{ mA}}{0.2 \text{ mA}} = 25$.
The voltage gain $A_V$ is given by the formula $A_V = \beta \times \frac{R_L}{R_{in}}$, where $R_L$ is the load resistance and $R_{in}$ is the input resistance.
Given $A_V = 75$, $\beta = 25$, and $R_{in} = 2 \text{ k}\Omega$.
Substituting the values: $75 = 25 \times \frac{R_L}{2 \text{ k}\Omega}$.
$3 = \frac{R_L}{2 \text{ k}\Omega}$.
$R_L = 3 \times 2 \text{ k}\Omega = 6 \text{ k}\Omega$.
Therefore, the load resistance is $6 \text{ k}\Omega$.

(A) In a common emitter transistor amplifier:
$1$. The voltage gain $A_V$ is defined as the ratio of the output voltage to the input voltage.
$A_V = \frac{V_{out}}{V_{in}} = \frac{I_c R_L}{I_b r_i} = \beta \frac{R_L}{r_i}$, where $\beta = \frac{I_c}{I_b}$ is the a.c. current gain.
$2$. The power gain $A_P$ is defined as the product of the current gain and the voltage gain.
$A_P = \beta \times A_V = \beta \times (\beta \frac{R_L}{r_i}) = \beta^2 \frac{R_L}{r_i}$.
Therefore, the voltage gain is $\beta (R_L/r_i)$ and the power gain is $\beta^2 (R_L/r_i)$.

(A) In a transistor, the emitter current $(I_E)$ is the sum of the collector current $(I_C)$ and the base current $(I_B)$, i.e.,$I_E = I_C + I_B$.
Given that $80\%$ of the electrons emitted reach the collector, we have $I_C = 0.80 \times I_E$.
Given $I_C = 28 \text{ mA}$, we can find $I_E$ as:
$I_E = \frac{I_C}{0.80} = \frac{28}{0.80} = 35 \text{ mA}$.
Now, using the relation $I_B = I_E - I_C$, we get:
$I_B = 35 \text{ mA} - 28 \text{ mA} = 7 \text{ mA}$.
Therefore, the base current is $7 \text{ mA}$.

(B) In a bipolar junction transistor $(BJT)$, the three regions are doped differently to optimize their functions:
$1$. The $Emitter$ is heavily doped to provide a large number of charge carriers.
$2$. The $Base$ is very thin and lightly doped to allow most of the charge carriers from the emitter to pass through to the collector.
$3$. The $Collector$ is moderately doped compared to the emitter and base.
Therefore, the part that is heavily doped is the $Emitter$, and the part that is lightly doped is the $Base$.

(C) The voltage gain $(A_v)$ of a common emitter transistor amplifier is given by the formula:
$A_v = \beta \times \frac{R_L}{R_i}$
Where:
$\beta$ ($AC$ current gain) = $65$
$R_L$ (Load resistance) = $5400 \text{ }\Omega$
$R_i$ (Input resistance) = $450 \text{ }\Omega$
Substituting the values into the formula:
$A_v = 65 \times \frac{5400}{450}$
$A_v = 65 \times 12$
$A_v = 780$
Therefore, the voltage gain is $780$.