Bipolar Junction Transistors MCQ Quiz - Objective Question with Answer for Bipolar Junction Transistors - Download Free PDF

Concept:

The input impedance (r_π) of a BJT in the active region is given by: \( r_\pi = \frac{\beta V_T}{I_C} \)

Given:

Common emitter current gain, \( \beta = 100 \)

Thermal voltage, \( V_T = 25~mV = 25 \times 10^{-3}~V \)

Collector current, \( I_C = 2~mA = 2 \times 10^{-3}~A \)

Calculation:

\( r_\pi = \frac{100 \times 25 \times 10^{-3}}{2 \times 10^{-3}} \)
\( r_\pi = \frac{2.5}{2 \times 10^{-3}} = 1250~\Omega = 1.25~k\Omega \)

Answer:

Option 3) 1.25 kΩ

Explanation:

Key Formula:

• Collector Current (I_C) in common emitter configuration: I_C = β × I_B + I_CBO
• Where:
  • β = Current gain in common emitter configuration (related to α)
  • I_B = Base current
  • I_CBO = Collector-base reverse saturation current

Given Data:

• Common base current gain (α) = 0.99
• Collector-base reverse saturation current (I_CBO) = 1 μA
• Base current (I_B) = 50 μA

Step-by-Step Calculation:

1. Relation between α and β:

In a BJT, α and β are related as follows:

β = α / (1 - α)

Substitute the value of α:

β = 0.99 / (1 - 0.99)

β = 0.99 / 0.01 = 99

2. Collector Current (I_C):

Using the formula:

I_C = β × I_B + I_CBO

Substitute the values:

I_C = 99 × 50 μA + 1 μA

I_C = 4950 μA + 1 μA

I_C = 4951 μA

Convert μA to mA:

I_C = 4.951 mA

Explanation:

Voltage Divider Bias Circuit in CE Configuration

Problem Statement: For a voltage divider bias circuit in CE configuration using NPN BJT transistor, with given values of V_CC = 10 V, R_C = 2 kΩ, and R_E = 500 Ω, we are tasked to determine the approximate maximum value of the collector current (I_C) for the circuit, assuming a very high DC current gain (β).

Solution:

The maximum collector current (I_C) can be calculated by considering the saturation condition of the transistor, where the transistor acts as a closed switch. In this state, the voltage across the collector-emitter junction (V_CE) is approximately zero.

Step 1: Voltage Distribution in the Circuit

In the saturation condition, the voltage across the collector resistor (R_C) and emitter resistor (R_E) will be:

V_RC + V_RE = V_CC

Where:

• V_RC = Voltage drop across the collector resistor (R_C)
• V_RE = Voltage drop across the emitter resistor (R_E)

Step 2: Current Flow in the Circuit

The current flowing through R_C and R_E is the collector current (I_C), which is equal to the emitter current (I_E) in the saturation condition because the base current (I_B) is negligible for high β values.

Step 3: Maximum Collector Current Calculation

To calculate the maximum collector current (I_C), we use Ohm's Law:

I_C = I_E = V_CC ÷ (R_C + R_E)

Substituting the given values:

V_CC = 10 V, R_C = 2 kΩ = 2000 Ω, R_E = 500 Ω

I_C = 10 ÷ (2000 + 500)

I_C = 10 ÷ 2500

I_C = 0.004 A

I_C = 4 mA

Thus, the maximum collector current for the voltage divider bias circuit is approximately 4 mA.

Important Information

Let’s analyze other options:

Option 1: 5 mA

This value is incorrect because the calculation shows that the maximum collector current (I_C) is 4 mA, based on the circuit parameters. The resistance values and supply voltage do not support a collector current of 5 mA.

Option 3: 2.5 mA

This value is incorrect because it underestimates the current. Using the formula I_C = V_CC ÷ (R_C + R_E), the calculated value is 4 mA, not 2.5 mA.

Option 4: 20 mA

This value is incorrect because it overestimates the current. The resistance values and supply voltage do not support such a high collector current. The calculated maximum value is 4 mA.

Conclusion:

The correct option is Option 2, which gives the maximum collector current (I_C) as approximately 4 mA. This value aligns with the calculations based on the given circuit parameters, considering the saturation condition of the transistor and assuming a very high DC current gain (β).

Emitter Current in BJT (Bipolar Junction Transistor):

Definition: The emitter current (I_E) in a Bipolar Junction Transistor (BJT) is the total current flowing out of the emitter terminal. It is the sum of the base current (I_B) and the collector current (I_C).

Important Parameters:

• Collector Current (I_C): The current flowing through the collector terminal.
• Base Current (I_B): The current flowing through the base terminal.
• Emitter Current (I_E): The total current flowing out of the emitter terminal, given by:

Formula:

The emitter current can be expressed as:

I_E = I_C + I_B

Additionally, the common base short circuit current gain, denoted by α, relates the collector current to the emitter current:

α = I_C / I_E

Rearranging the formula, we can write:

I_E = I_C / α

Given Data:

• Collector Current, I_C = 1 mA
• Common Base Short Circuit Current Gain, α = 1 (unity)

Solution:

Using the formula:

I_E = I_C / α

Substitute the given values:

I_E = 1 mA / 1

I_E = 1 mA

Hence, the emitter current is 1 mA.

Correct Option Analysis:

The correct option is:

Option 3: 1 mA

This option is correct because the emitter current is calculated to be 1 mA based on the given collector current and the common base short circuit current gain (α = 1). The result directly matches the provided data and the relationship between these parameters in the BJT.

Additional Information

To further understand the analysis, let’s evaluate why the other options are incorrect:

Option 1: 1.5 mA

This option suggests that the emitter current is 1.5 mA. However, the given parameters (I_C = 1 mA and α = 1) clearly show that the emitter current is 1 mA. A value of 1.5 mA would imply a different value of α, which is not consistent with the problem statement.

Option 2: 2 mA

This option proposes an emitter current of 2 mA. This value is incorrect because, with α = 1, the emitter current must equal the collector current. An emitter current of 2 mA would require a collector current greater than 1 mA, which contradicts the given data.

Option 4: 0.5 mA

This option suggests an emitter current of 0.5 mA. This is incorrect because, with α = 1, the emitter current cannot be less than the collector current. A value of 0.5 mA for the emitter current would imply either an incorrect value of α or collector current, which is not the case here.

Conclusion:

Understanding the relationship between emitter current, collector current, and the common base current gain (α) is essential in analyzing BJTs. In this case, the emitter current is directly equal to the collector current due to the unity gain (α = 1). This simplicity highlights the importance of correctly applying the fundamental formulas of BJT operation.

Explanation:

Junction Capacitance in a p-n Diode

Definition: Junction capacitance, also known as depletion capacitance, is a property of a p-n junction that arises due to the separation of charges in the depletion region. It is a significant parameter in understanding the electrical behavior of diodes, especially under varying voltage conditions.

Correct Option: The correct answer is:

Option 3: Change in depletion region width with applied voltage.

Explanation:

In a p-n diode, the depletion region is formed at the junction of the p-type and n-type materials. This region is devoid of free charge carriers (electrons and holes) because they combine near the junction, leaving behind immobile ions. This creates an electric field across the depletion region. The capacitance associated with this region is termed as junction capacitance or depletion capacitance.

The junction capacitance is defined as:

C = ε × A / W

Where:

• C: Junction capacitance
• ε: Permittivity of the semiconductor material
• A: Cross-sectional area of the junction
• W: Width of the depletion region

The width of the depletion region (W) is dependent on the applied voltage across the diode. When a reverse bias voltage is applied, the depletion region widens, increasing W and thereby decreasing the capacitance (C). Conversely, when a forward bias voltage is applied, the depletion region narrows, reducing W and increasing the capacitance (C).

Thus, the primary cause of junction capacitance in a p-n diode is the change in the width of the depletion region (W) with the applied voltage. This behavior is crucial in applications like varactor diodes, where the junction capacitance is intentionally varied by changing the applied voltage.

Important Information:

Let us analyze the other options to understand why they are incorrect:

Option 1: Variation in minority carrier concentration with applied voltage

This option is incorrect because the minority carrier concentration does not directly contribute to the junction capacitance. Junction capacitance is primarily related to the immobile charges in the depletion region and the change in its width, not to the behavior of minority carriers.

Option 2: Thermal generation of carriers in the depletion region

This option is incorrect because thermal generation of carriers primarily affects the leakage current in a diode under reverse bias. While it influences the diode's overall behavior, it is not the cause of junction capacitance.

Option 4: Variation in majority carrier concentration with applied voltage

This option is also incorrect because majority carriers are not directly involved in the formation of the depletion region or its associated capacitance. The majority carriers are swept away from the junction, creating the depletion region, but their concentration does not vary significantly with the applied voltage in a way that influences capacitance.

Conclusion:

The correct answer is Option 3 because the junction capacitance arises from the change in the width of the depletion region with the applied voltage. This is a fundamental property of the p-n junction and plays a significant role in the behavior of diodes, especially in high-frequency and voltage-variable applications. Understanding this concept is essential for analyzing and designing circuits involving diodes, such as rectifiers, varactors, and RF circuits.

Concept:

For a transistor, the base current, the emitter current, and the collector current are related as:

IE = IB + IC

where IC = β IB

β = Current gain of the transistor

Typical base-emitter voltages, VBE for both NPN and PNP transistors are as follows:

• If the transistor is made up of a silicon material, the base-emitter voltage VBE will be 0.7 V.
• If the transistor is made up of a germanium material, the base-emitter voltage VBE will be 0.3 V.
   

Application:

From the given figure, Apply KVL

10 - I_B × R_B - V_BE = 0

Let us assume V_BE = 0.7 V

10 - I_B (1 × 10⁶) - 0.7 = 0

I_B = 9.3 μA

We know that,

I_C = β I_B

Where,

I_C  & I_B = collector current and base current

Therefore,

I_C = 100 × 9.3 μA

= 930 μA

= 0.93 mA

≈ 1 mA

BJT Amplifier:

• Transistors biasing is done to keep stable DC operating conditions needed for its functioning as an amplifier.
• A properly biased transistor must have it's Q-point (DC operating parameters like IC and VCE) at the center of saturation mode and cut-off mode i.e. active mode.
• In the active mode of transistor operation, the emitter-base junction is forward biased and the collector-base junction is reverse biased.​
• In the cut-off mode of transistor operation, the emitter-base junction is reverse biased and the collector-base junction is reverse biased.​

​

Different modes of BJT operations are:

Early Effect:

• A large collector base reverse bias is the reason behind the early effect manifested by BJTs.
• As reverse biasing of the collector to base junction increases, the depletion region penetrates more into the base, as the base is lightly doped.
• This reduces the effective base width and hence the concentration gradient in the base increases.
• This reduction in the effective base width causes less recombination of carriers in the base region which results in an increase in collector current. This is known as the Early effect.
• The decrease in base width causes ß to increase and hence collector current increases with collector voltage rather than staying constant.
• The slope introduced by the Early effect is almost linear with IC and the common-emitter characteristics extrapolate to an intersection with the voltage axis VA, called the Early voltage.

This is explained with the help of the following VCE (Reverse voltage) vs IC (Collector current) curve:

Concept:

\(\beta = \frac{\alpha }{{1 - \alpha }}\)

Where β = common-emitter current gain

α = Common base current gain

Calculation:

Common base current gain = α = 0.98

\(\beta = \frac{{0.98}}{{1 - 0.98}} = 49\)

Note: \(\alpha = \frac{{{I_C}}}{{{I_E}}}\) & \(\beta = \frac{{{I_C}}}{{{I_B}}}\)

Where I_C = Collector current

I_E = Emitter current

Concept:

In a common base connection,

I_E = I_B + I_C

And I_C = α I_E + I_CBO

Where α is the current amplification factor

I_E = I_B + α I_E + I_CBO

⇒ I_E (1 – α) = I_B + I_CBO

\( \Rightarrow {I_E} = \frac{1}{{\left( {1 - \alpha } \right)}}\left( {{I_B} + {I_{CBO}}} \right)\)

Calculation:

Given that, I_B = 0.02 mA

Current amplification factor (α) = 0.9

I_CBO = 30 μA = 0.03 mA

The emitter current is,

\({I_E} = \frac{1}{{1 - 0.9}}\left( {0.02 + 0.03} \right) = 0.5\;mA\)

Transistor

A transistor is a three-layer, three-terminal device.

The three layers can be (n-p-n or p-n-p) and the three terminals are the collector, base, and emitter.

Transistor when working in the saturation region acts as a closed switch (ON Button) and in the cut-off region acts as an open switch (OFF Button).

Working modes of transistor

Common Emitter(CE) Configuration:

In CE configuration input is connected between base and emitter while the output is taken between collector and emitter.

\(β = \frac{α }{{1 - α }} = \frac{{{I_C}}}{{{I_B}}} = \frac{{{I_C}}}{{{I_E} - {I_C}}}\)

I_E = I_B + I_C  

I_C = β I_B + I_CEO 

IC = α I_E + ICBO

I_C = α (I_C + I_B) + ICBO

IC (1 - α ) = α I_B + ICBO

\(\Large{I_C=\frac{\alpha I_B}{1-\alpha}+\frac{I_{CBO}}{1-\alpha}}\)

In CE configuration, when I_B = 0 then I_C = I_CEO

\(\Large{I_{CEO} = \frac{I_{CBO}}{1 - α}} \)

Where, α = Current gain

β = Current Amplification Factor

IE, IB, IC = Emitter, Base and Collector current respectively

ICEO = Collector emitter cutoff current

ICBO = Collector base cutoff current

Calculation: 

Given: α = 0.98, ICBO = 5 μA, IB = 95 μA

\(β = \frac{0.98 }{(1 - 0.98)}= \frac{0.98 }{0.02 }= 49\)

\(I_{CEO} = \frac{5 × 10^{-6}}{(1 - 0.98)} = 250\ \mu A\)

I_C = 49 x (95 × 10^-6) + 250 × 10^-6 = 4905 × 10^-6 A

\(I_E=\frac{(4905 - 5) × 10^{-6}}{0.98}=\frac{4900 × 10^{-6}}{0.98}=5\ \ mA\)

Concept:

• Transistor: A transistor has two PN junctions i.e., it is like two diodes. The junction between base and emitter may be called emitter diode. The junction between base and collector may be called collector diode.

• The transistor can act in one of the three states:
• CUT-OFF: EMITTER DIODE AND COLLECTOR DIODE ARE OFF.
• ACTIVE: EMITTER DIODE IS ON AND COLLECTOR DIODE IS OFF.
• SATURATED: EMITTER DIODE AND COLLECTOR DIODE ARE ON.

Note: Please understand that the question is asking about the regions of a biased transistor. Biasing is a set of DC voltages that we apply at the Base-emitter 'or' emitter-collector terminal. There are three possibilities based on this voltage, i.e. Active Region, Cut-off region, and Saturation region.