Junction Transistor Questions in English

(A) In an $NPN$ transistor,the emitter is $N-$type,the base is $P-$type,and the collector is $N-$type. When used as an amplifier,the emitter-base junction is forward-biased and the collector-base junction is reverse-biased. Electrons,being the majority charge carriers in the $N-$type emitter,are injected into the base. Due to the thin base region,most of these electrons diffuse through the base and are collected by the collector. Thus,electrons move from the emitter to the base and then from the base to the collector.

(C) In a common emitter $(CE)$ transistor amplifier,the input signal is applied between the base and emitter,and the output is taken across the collector and 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 in the collector circuit,the output voltage decreases.
Conversely,when the input signal voltage decreases,the collector current decreases,leading to an increase in the output voltage.
This inverse relationship between the input and output signals results in a phase shift of $\pi$ radians,which is equivalent to $180^o$.

(A) An oscillator is a circuit that produces a continuous,repeated,alternating waveform without any input signal.
It functions as an amplifier with positive feedback.
In this configuration,a portion of the output signal is fed back to the input in phase with the original signal.
This positive feedback reinforces the input,allowing the circuit to sustain oscillations indefinitely,provided the Barkhausen criterion (loop gain $A\beta = 1$ and phase shift of $360^{\circ}$ or $0^{\circ}$) is satisfied.

(D) In a standard transistor configuration (such as the active region used for amplification),the emitter-base junction is forward-biased to allow charge carriers to flow from the emitter into the base.
Conversely,the collector-base junction is reverse-biased to collect the charge carriers that have crossed the base region.
Therefore,the correct configuration is forward-biased for the emitter-base junction and reverse-biased for the collector-base junction.

(D) Given that the collector current $i_c = 24 \, mA$.
Since $80\%$ of the electrons emitted from the emitter reach the collector,we have the relation: $i_c = 0.80 \times i_e$.
Substituting the value of $i_c$: $24 = 0.80 \times i_e$.
Solving for the emitter current: $i_e = \frac{24}{0.80} = 30 \, mA$.
Using the Kirchhoff's current law for a transistor: $i_e = i_b + i_c$.
Rearranging to find the base current: $i_b = i_e - i_c$.
Substituting the known values: $i_b = 30 \, mA - 24 \, mA = 6 \, mA$.
Thus,the base current is $6 \, mA$.

(B) For an $NPN$ transistor to conduct (operate in the active region),the emitter-base junction must be forward-biased and the collector-base junction must be reverse-biased.
In an $NPN$ transistor,the base is $P$-type and the emitter and collector are $N$-type.
$1$. To forward-bias the emitter-base junction,the $N$-type emitter must be at a lower potential than the $P$-type base (i.e.,emitter is negative with respect to the base).
$2$. To reverse-bias the collector-base junction,the $N$-type collector must be at a higher potential than the $P$-type base (i.e.,collector is positive with respect to the base).
Therefore,the transistor conducts when the collector is positive and the emitter is negative with respect to the base.

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

(B) In a transistor, the common emitter $(CE)$ configuration is the most widely used because it provides both high current gain $(\beta)$ and high voltage gain, resulting in a high power gain.
In the common base $(CB)$ configuration, the current gain $(\alpha)$ is < $1$.
In the common collector $(CC)$ configuration, the current gain is high, but the voltage gain is < $1$.
Therefore, the common emitter configuration is preferred for amplification purposes.

(B) The relationship between the common-base current gain $\alpha$ and the common-emitter current gain $\beta$ is given by the formula:
$\beta = \frac{\alpha}{1 - \alpha}$
Given $\alpha = 0.98$,we substitute the value into the formula:
$\beta = \frac{0.98}{1 - 0.98}$
$\beta = \frac{0.98}{0.02}$
$\beta = 49$
Thus,the value of $\beta$ is $49$.

(B) In a common base configuration,the current gain $\alpha$ is defined as the ratio of collector current $I_C$ to emitter current $I_E$.
Given: $\alpha = \frac{I_C}{I_E} = 0.96$.
The current gain in common emitter configuration,denoted by $\beta$,is related to $\alpha$ by the formula:
$\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the given value:
$\beta = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04}$.
$\beta = 24$.
Therefore,the maximum current gain in common emitter configuration is $24$.

(C) Given: Current gain $\alpha = 0.96$ and emitter current $I_e = 7.2 \, mA$.
In a common-base configuration,the current gain $\alpha$ is defined as $\alpha = \frac{I_c}{I_e}$.
Therefore,the collector current $I_c = \alpha \times I_e = 0.96 \times 7.2 \, mA = 6.912 \, mA$.
According to Kirchhoff's current law for a transistor,$I_e = I_c + I_b$.
Thus,the base current $I_b = I_e - I_c$.
$I_b = 7.2 \, mA - 6.912 \, mA = 0.288 \, mA$.
Rounding to two decimal places,$I_b \approx 0.29 \, mA$.

(D) In a transistor,the collector is designed to collect the majority of charge carriers coming from the emitter. This process generates significant heat,so the collector is made with the largest surface area to dissipate this heat effectively.
The base is the middle region,which is made very thin and lightly doped to allow charge carriers to pass from the emitter to the collector with minimal recombination.
The emitter is moderately sized and heavily doped to supply charge carriers.
Comparing the sizes: The collector $(l_3)$ is the largest,the base $(l_2)$ is the smallest,and the emitter $(l_1)$ is of intermediate size.
Therefore,the correct relation is ${l_3} > {l_1} > {l_2}$.

(D) Given that the collector current $i_C = 10 \,mA$.
Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation $i_C = 0.90 \times i_E$.
Substituting the value of $i_C$,we get $10 = 0.90 \times i_E$.
Therefore,$i_E = \frac{10}{0.9} \approx 11.11 \,mA$. Rounding to the nearest integer as per the options,$i_E = 11 \,mA$.
Using the Kirchhoff's current law for a transistor,$i_E = i_B + i_C$.
Substituting the values,$11 = i_B + 10$.
Thus,$i_B = 11 - 10 = 1 \,mA$.

(A) The current gain $\beta$ in a common emitter configuration is defined as the ratio of the change in collector current $\Delta I_C$ to the change in base current $\Delta I_B$.
Formula: $\beta = \frac{\Delta I_C}{\Delta I_B}$
Given: $\beta = 80$ and $\Delta I_B = 250 \ \mu A$.
To find the change in collector current $\Delta I_C$,we rearrange the formula:
$\Delta I_C = \beta \times \Delta I_B$
$\Delta I_C = 80 \times 250 \ \mu A$
Therefore,the change in collector current is $80 \times 250 \ \mu A$.

(B) In a bipolar junction transistor,the emitter is heavily doped to provide a large number of charge carriers. The collector is moderately doped to handle the power dissipation. The base is made very thin and is lightly (least) doped to ensure that most of the charge carriers injected from the emitter pass through to the collector without recombining in the base. Therefore,the base is the least doped region.

(B) In a transistor, the power gain is defined as the product of current gain $(\beta)$ and voltage gain $(A_v)$.
In a $Common \text{ } Emitter$ $(CE)$ configuration, both the current gain $(\beta)$ and the voltage gain are significantly greater than $1$.
Since the $CE$ configuration provides both high current gain and high voltage gain, it results in a high power gain.
Therefore, transistors provide good power amplification when they are used in the $Common \text{ } Emitter$ configuration.

(D) Given: Transfer ratio (current gain) $\beta = 50$,Input resistance $R_i = 1 \; k\Omega = 1000 \; \Omega$,and Peak input voltage $V_i = 0.01 \; V$.
First,calculate the peak input current $i_b$ using Ohm's law: $i_b = \frac{V_i}{R_i} = \frac{0.01 \; V}{1000 \; \Omega} = 10^{-5} \; A$.
The relationship between collector current $i_c$ and base current $i_b$ is given by $\beta = \frac{i_c}{i_b}$.
Therefore,the peak collector current is $i_c = \beta \times i_b = 50 \times 10^{-5} \; A = 500 \times 10^{-6} \; A = 500 \; \mu A$.

(B) The relationship between the common-emitter current gain $\beta$ and the common-base current gain $\alpha$ is given by the formula: $\alpha = \frac{\beta}{1 + \beta}$.
Given $\beta = 99$.
Substituting the value into the formula:
$\alpha = \frac{99}{1 + 99} = \frac{99}{100} = 0.99$.
Therefore,the value of $\alpha$ is $0.99$.

(D) In a common emitter $(CE)$ amplifier configuration,the transistor must be operated in the active region.
For the active region,the base-emitter junction must be forward-biased,and the collector-base junction must be reverse-biased.
Therefore,statement $(a)$ is correct.
Furthermore,in a $CE$ amplifier,the input signal is applied to the base-emitter circuit,which is in series with the $DC$ bias voltage applied to the base-emitter junction.
Therefore,statement $(c)$ is also correct.
Thus,both $(a)$ and $(c)$ are correct.

(A) In a transistor,the base region is intentionally made very thin and lightly doped compared to the emitter and collector regions.
This design ensures that the majority of charge carriers injected from the emitter into the base can diffuse through the base and reach the collector region before they can recombine.
Therefore,in a $PNP$ transistor,the width of the $N$-type base is much smaller than the width of the $P$-type emitter and collector regions.

(C) The voltage gain $(A_v)$ of a common emitter amplifier is given by the product of the current gain $(\beta)$ and the resistance gain.
First,calculate the current gain $\beta$ using the relation $\beta = \frac{\alpha}{1 - \alpha}$:
$\beta = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99$.
Next,calculate the resistance gain,which is the ratio of the load resistance $(R_L)$ to the input impedance $(R_i)$:
Resistance gain $= \frac{R_L}{R_i} = \frac{10 \ k\Omega}{1 \ k\Omega} = 10$.
Finally,calculate the voltage gain:
$A_v = \beta \times \text{Resistance gain} = 99 \times 10 = 990$.

(B) In the given circuit diagram,the emitter terminal of the $NPN$ transistor is connected to the ground,which is common to both the input circuit (between base and emitter) and the output circuit (between collector and emitter). Since the emitter is the common terminal between the input and output,this configuration is known as a common emitter amplifier circuit.

(B) In a transistor,the $Emitter$ is the section that is heavily doped. The purpose of heavy doping is to provide a large number of majority charge carriers,which are then injected into the $Base$ region. The $Base$ is lightly doped and thin,while the $Collector$ is moderately doped. Therefore,the correct option is $B$.

(C) Given,current amplification factor $\alpha = 0.8$.
For a common emitter configuration,the current gain $\beta$ is related to $\alpha$ by the formula $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$:
$\beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4$.
We know that the current gain $\beta$ is defined as the ratio of the change in collector current $\Delta I_c$ to the change in base current $\Delta I_b$:
$\beta = \frac{\Delta I_c}{\Delta I_b}$.
Given $\Delta I_b = 6 \, mA$,we can calculate $\Delta I_c$ as:
$\Delta I_c = \beta \times \Delta I_b = 4 \times 6 \, mA = 24 \, mA$.
Therefore,the change in collector current is $24 \, mA$.

(A) In a transistor,the relationship between currents is given by $\Delta i_e = \Delta i_c + \Delta i_b$.
Given $\alpha = \frac{\Delta i_c}{\Delta i_e} = 0.98$ and $\Delta i_e = 2\, mA$.
First,calculate the change in collector current: $\Delta i_c = \alpha \times \Delta i_e = 0.98 \times 2 = 1.96\, mA$.
Now,calculate the change in base current: $\Delta i_b = \Delta i_e - \Delta i_c$.
$\Delta i_b = 2\, mA - 1.96\, mA = 0.04\, mA$.

(B) In an $NPN$ transistor,the emitter current $(I_E)$ is the sum of the base current $(I_B)$ and the collector current $(I_C)$,expressed as $I_E = I_B + I_C$.
Since a small fraction of electrons injected from the emitter recombine with holes in the base region,the base current $(I_B)$ is non-zero.
Therefore,the collector current is given by $I_C = I_E - I_B$.
Because $I_B > 0$,it follows that $I_C < I_E$.

(B) From the given circuit diagram,the base-emitter loop consists of the $9 \text{ V}$ battery and the base resistor $R_b$.
Applying Kirchhoff's voltage law to the base-emitter loop (assuming the base-emitter voltage $V_{BE} \approx 0 \text{ V}$ for an ideal transistor or that the $9 \text{ V}$ is applied directly across $R_b$):
$V_b = i_b \times R_b$
Given $V_b = 9 \text{ V}$ and $i_b = 35 \mu A = 35 \times 10^{-6} \text{ A}$.
$R_b = \frac{V_b}{i_b} = \frac{9}{35 \times 10^{-6}} \Omega$
$R_b = \frac{9}{35} \times 10^6 \Omega \approx 0.25714 \times 10^6 \Omega$
$R_b \approx 257.14 \times 10^3 \Omega = 257.14 \text{ k}\Omega$.
Rounding to the nearest integer,we get $257 \text{ k}\Omega$.

(D) For a transistor,the relationship between the changes in currents is given by the equation: $\Delta I_e = \Delta I_c + \Delta I_b$.
Given: $\Delta I_e = 8.0\, mA$ and $\Delta I_c = 7.8\, mA$.
Substituting these values into the equation: $8.0\, mA = 7.8\, mA + \Delta I_b$.
Solving for the change in base current: $\Delta I_b = 8.0\, mA - 7.8\, mA = 0.2\, mA$.
Converting the result to microamperes $(\mu A)$: $0.2\, mA = 0.2 \times 10^3\, \mu A = 200\, \mu A$.

(B) In a common-emitter transistor configuration,the current gain $\beta$ is defined as the ratio of the collector current $(I_c)$ to the base current $(I_b)$.
Mathematically,$\beta = \frac{I_c}{I_b}$.
Therefore,the correct option is $B$.

(B) unipolar transistor is a device in which the current conduction is due to only one type of charge carrier (either electrons or holes).
$FET$ (Field Effect Transistor) is a unipolar device because the current in it is carried by only one type of majority charge carrier.
In contrast,$PNP$ and $NPN$ transistors are bipolar junction transistors $(BJT)$ because their current conduction involves both electrons and holes.

(A) In a common emitter configuration,the alternating current gain $\beta$ (also known as the forward current gain) 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,this is expressed as: $\beta = \left( \frac{\Delta I_C}{\Delta I_B} \right)_{V_{CE}}$.
Since it is a ratio of two currents,$\beta$ is a dimensionless quantity. $A$ small change in the base current results in a significantly larger change in the collector current,which is the fundamental principle behind transistor amplification.

(B) In a transistor,the emitter current $i_e$ is the sum of the base current $i_b$ and the collector current $i_c$:
$i_e = i_b + i_c$
Dividing both sides by $i_c$:
$\frac{i_e}{i_c} = \frac{i_b}{i_c} + 1$
We know that $\alpha = \frac{i_c}{i_e}$ (so $\frac{i_e}{i_c} = \frac{1}{\alpha}$) and $\beta = \frac{i_c}{i_b}$ (so $\frac{i_b}{i_c} = \frac{1}{\beta}$).
Substituting these values:
$\frac{1}{\alpha} = \frac{1}{\beta} + 1$
$\frac{1}{\alpha} = \frac{1 + \beta}{\beta}$
Therefore,$\alpha = \frac{\beta}{1 + \beta}$.

(B) In an $NPN$ transistor,the emitter is heavily doped with electrons. When the emitter-base junction is forward-biased,the potential barrier is lowered. This allows the majority charge carriers (electrons) in the emitter to be injected into the base region. Therefore,electrons move from the emitter to the base.

(A) The output resistance $(R_{out})$ in $CB$ mode is defined as the ratio of the change in collector voltage $(\Delta V_C)$ to the change in collector current $(\Delta I_C)$ while keeping the emitter current constant.
Given: $\Delta V_C = 0.5\,V$ and $\Delta I_C = 0.05\,mA = 0.05 \times 10^{-3}\,A$.
Using the formula: $R_{out} = \frac{\Delta V_C}{\Delta I_C}$.
Substituting the values: $R_{out} = \frac{0.5}{0.05 \times 10^{-3}} = \frac{0.5}{5 \times 10^{-5}} = 0.1 \times 10^5 = 10^4\,\Omega$.
Since $10^4\,\Omega = 10 \times 10^3\,\Omega = 10\,k\Omega$.
Therefore, the output resistance is $10\,k\Omega$.

(A) In a transistor,the emitter current $I_e$ is the sum of collector current $I_c$ and base current $I_b$,so $\Delta I_e = \Delta I_c + \Delta I_b$.
Given: $\Delta I_c = 8.2 \, mA$ and $\Delta I_e = 8.3 \, mA$.
First,calculate the change in base current: $\Delta I_b = \Delta I_e - \Delta I_c = 8.3 \, mA - 8.2 \, mA = 0.1 \, mA$.
The forward current gain $h_{fe}$ (also denoted as $\beta$) is defined as the ratio of change in collector current to the change in base current at a constant collector-emitter voltage: $h_{fe} = \frac{\Delta I_c}{\Delta I_b}$.
Substituting the values: $h_{fe} = \frac{8.2 \, mA}{0.1 \, mA} = 82$.

(B) Given: Current gain $\beta = 100$,Change in collector current $\Delta I_c = 1\, mA$.
We know that the current gain in common-emitter configuration is defined as $\beta = \frac{\Delta I_c}{\Delta I_b}$.
Therefore,the change in base current is $\Delta I_b = \frac{\Delta I_c}{\beta} = \frac{1\, mA}{100} = 0.01\, mA$.
The relationship between emitter,collector,and base currents is $\Delta I_e = \Delta I_c + \Delta I_b$.
Substituting the values,we get $\Delta I_e = 1\, mA + 0.01\, mA = 1.01\, mA$.
Thus,the change in emitter current is $1.01\, mA$.

(A) In a common base $(CB)$ amplifier configuration,the input signal is applied between the emitter and the base,and the output is taken between the collector and the base.
Since the base is common to both the input and output circuits,the input voltage and output voltage signals are in the same phase.
Therefore,the phase difference between the input signal voltage and the output voltage is $0$.

(B) Given that the collector current $I_C = 10\, mA$.
Since $90\%$ of the electrons emitted reach the collector,the collector current is $90\%$ of the emitter current $(I_E)$.
Therefore,$I_C = 0.90 \times I_E$.
Substituting the value of $I_C$: $10\, mA = 0.90 \times I_E$.
Solving for $I_E$: $I_E = \frac{10}{0.90} = 11.11\, mA$.
Using the relation $I_E = I_B + I_C$,we can find the base current $I_B$:
$I_B = I_E - I_C = 11.11\, mA - 10\, mA = 1.11\, mA$.
Comparing this with the given options,the correct statement is that the emitter current is $11.1\, mA$.

(D) For a transistor in $CE$ configuration,the voltage gain $(A_v)$ is given by $A_v = \beta \times \frac{R_L}{R_i}$,where $\beta$ is the current gain,$R_L$ is the load resistance,and $R_i$ is the input resistance.
The power gain $(A_p)$ is given by $A_p = \beta^2 \times \frac{R_L}{R_i}$.
Therefore,the ratio of power gain to voltage gain is:
$\frac{A_p}{A_v} = \frac{\beta^2 \times (R_L / R_i)}{\beta \times (R_L / R_i)} = \beta$.
Thus,the correct option is $D$.

(B) We know that the emitter current is the sum of the collector current and the base current:
$I_{E} = I_{C} + I_{B}$
Dividing both sides by $I_{C}$:
$\frac{I_{E}}{I_{C}} = 1 + \frac{I_{B}}{I_{C}}$
Since $\alpha = \frac{I_{C}}{I_{E}}$,it follows that $\frac{1}{\alpha} = \frac{I_{E}}{I_{C}}$.
Since $\beta = \frac{I_{C}}{I_{B}}$,it follows that $\frac{1}{\beta} = \frac{I_{B}}{I_{C}}$.
Substituting these into the equation:
$\frac{1}{\alpha} = 1 + \frac{1}{\beta}$
Rearranging to solve for $\frac{1}{\beta}$:
$\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1 - \alpha}{\alpha}$
Therefore,$\beta = \frac{\alpha}{1 - \alpha}$.

(D) In a common emitter configuration,the current gain $A_i$ is given by the formula:
$A_i = \frac{-h_{fe}}{1 + h_{oe} R_L}$
Given:
$h_{fe} = 50$
$h_{oe} = 25 \ \mu A/V = 25 \times 10^{-6} \ S$
$R_L = 1 \ k\Omega = 10^3 \ \Omega$
Substituting the values:
$A_i = \frac{-50}{1 + (25 \times 10^{-6}) \times 10^3}$
$A_i = \frac{-50}{1 + 0.025}$
$A_i = \frac{-50}{1.025}$
$A_i \approx -48.78$

(C) Given: Input voltage $V_{in} = 1 \, mV = 10^{-3} \, V$, current gain $\beta = 100$, input resistance $R_{in} = 1 \, k\Omega = 10^3 \, \Omega$, and load resistance $R_L = 10 \, k\Omega = 10^4 \, \Omega$.
Voltage gain $A_v = \beta \times \frac{R_L}{R_{in}}$.
$A_v = 100 \times \frac{10 \, k\Omega}{1 \, k\Omega} = 100 \times 10 = 1000$.
Output voltage $V_{out} = A_v \times V_{in}$.
$V_{out} = 1000 \times 10^{-3} \, V = 1 \, V$.

(C) $1$. Apply Kirchhoff's Voltage Law $(KVL)$ to the base-emitter loop:
$V_{BB} - I_B R_B - V_{BE} = 0$
$5 \, V - I_B (8.6 \, k\Omega) - 0.7 \, V = 0$
$I_B = \frac{5 - 0.7}{8.6 \times 10^3} = \frac{4.3}{8.6 \times 10^3} = 0.5 \times 10^{-3} \, A = 0.5 \, mA$.
$2$. Calculate the collector current $(I_C)$:
$I_C = \beta I_B = 100 \times 0.5 \, mA = 50 \, mA = 0.05 \, A$.
$3$. Apply $KVL$ to the collector-emitter loop:
$V_{CC} - I_C R_L - V_{CE} = 0$
$18 \, V - (0.05 \, A)(100 \, \Omega) - V_{CE} = 0$
$18 \, V - 5 \, V - V_{CE} = 0$
$V_{CE} = 13 \, V$.