Common Base Configuration MCQ Quiz - Objective Question with Answer for Common Base Configuration - Download Free PDF

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.

Common base configuration:

Input Terminal Emitter – Base (EB)

Output Terminal Collector – Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

Various transistor configurations and their characteristics are shown in the table

Key PointsCurrent gain(α) of a Common base amplifier is unity (ideally).

The Input resistance is very low ( 30-150 Ω ).

The output resistance is very high ( ≈ 500K ).

The Voltage gain is high.

The Phase between input and output is the same.

Additional Information

Common base configuration:

Input Terminal Emitter – Base (EB)

Output Terminal Collector – Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

Various transistor configurations and their characteristics are shown in the table

The output resistance is boosted when a BJT is connected in a circuit in a common base configuration. 

Key PointsCurrent gain(α) of a Common base amplifier is unity (ideally).

The Input resistance is very low ( 30-150 Ω ).

The output resistance is very high ( ≈ 500K ).

The Voltage gain is high.

The Power gain is high.

The Phase between input and output is same.

Additional Information

Common base configuration:

Input Terminal Emitter – Base (EB)

Output Terminal Collector – Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

Various transistor configurations and their characteristics are shown in the table

Common base configuration:

Input Terminal Emitter – Base (EB)

Output Terminal Collector – Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

The various characteristics of different configuration are shown 

Concept:

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

IE = IB + IC

where IC = αIE

α = Current gain of the transistor

Calculation:

Given,

IC = 0.95 mA

IB = 0.05 mA

∴ IE = IB + IC = 0.05 + 0.95 = 1 mA

From the above concept,

IC = αIE

\(\alpha=\frac{I_C}{I_E}=\frac{0.95}{1}=0.95\)

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.

Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

\(\alpha = \frac{{{\rm{\Delta }}{I_C}}}{{{\rm{\Delta }}{I_E}}}\)

Where, IE = IC + IB

Calculation:

Given,

IB = 0.25 mA and IE = 1.4 mA

We know that,

IE = IC + IB

⇒ I_C = I_E - I_B = 1.4 - 0.25 = 1.15 mA

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

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

Calculation:

Given that,

α = .99

I_CBO = 1 μA

I_C = 1 mA

⇒ \(I_C= \frac{α}{1-α} I_B+\frac {1}{1-α}I_{CBO}\)

⇒ \(1 mA= \frac{.99}{1-.99} I_B+\frac {1}{1-.99} \times 1 μ A\)

⇒ I_B ≈ 9 μA

Concept:

For a BJT expression for collector current (I_c) is defined as-

I_C = β I_B + (1 + β) I_co

Where β = large signal current gain in CE mode of operation

I_C = collector current

I_co = collector current when emitter is open

I_B = Base current

α = large signal current gain in CB mode of operation

Relation between α and β:

\(\alpha = \frac{\beta }{{1 + \beta }}\;or\;\beta = \frac{\alpha }{{1 - \alpha }}\)

Calculation:

Given data: α = 0.98

I_co = 6 μA

I_B = 100 μA

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

I_c = β I_B + (1 + β) I_co

I_c = 49 × 100 + (49 + 1) 6 μA

I_c = 5200 μA

[ ∵ 1 μA = 10-6 A

1 mA = 10-3 A

1 mA = 103 μA ] 

I_c = 5.2 mA

The common base amplifier circuit output signal has 0 degrees phase shift with input

Additional Information

All the configurations are used in different circuits based on their applications:

Common-Base Configuration:

Input Terminal Emitter–Base (EB)

Output Terminal Collector–Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

Various transistor configurations and their characteristics are shown in the table

Current amplification factor: It is defined as the ratio of the output current to the input current. In common base configuration, the output current is emitter current IC, whereas the input current is base current IE.

Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

\(\alpha = \frac{{{\rm{\Delta }}{I_C}}}{{{\rm{\Delta }}{I_E}}}\)

Important Points:

β is the current amplification factor in the common-emitter configuration.

\(\beta = \frac{{{\rm{\Delta }}{I_C}}}{{{\rm{\Delta }}{I_B}}}\)

γ is the current amplification factor in the common collector configuration.

\(\gamma = \frac{{{\rm{\Delta }}{I_E}}}{{{\rm{\Delta }}{I_B}}}\)

Concept:

Here the output resistance is very high as compared to input resistance, since the input junction (base to emitter) of the transistor is forward biased while the output junction (base to collector) is reverse biased. Figure shows the conditions of the problem:

Calculation:

Input Current(I_E) = \(\frac{Signal}{R{in}}\) = \(\frac{500mV}{20ohm}\) = 25mA

Since \(α_{ac}\) is nearly 1, Output Current I_C = I_E = 25mA

Output Voltage (\(V_{out}\)) = I_CR_C = \(25mA×1Kohm\) = 25Volts

Voltage Amplification (A_v) = \(\frac{V{out}}{Signal}\) = \(\frac{25V}{500mV} = 50\)