In the common-base configuration, the collector current is given by:

Question

Prepp Team · Accepted Answer

Understanding Collector Current in Common-Base Configuration

The common-base configuration is one of the three basic transistor amplifier configurations. In this configuration, the base terminal is common to both the input and output circuits. The input signal is applied between the emitter and the base, and the output is taken between the collector and the base.

Transistor Current Relationships

For any bipolar junction transistor (BJT), the total emitter current ($I_E$) is the sum of the collector current ($I_C$) and the base current ($I_B$). This is represented by the equation:

$\label{eq:1} I_E = I_C + I_B$

In the common-base configuration, the collector current ($I_C$) is primarily controlled by the emitter current ($I_E$). The relationship is given by:

$\label{eq:2} I_C = \alpha I_E + I_{CBO}$

Where:

$\alpha$ is the common-base forward current gain (typically close to but less than 1). It is defined as the ratio of collector current to emitter current ($\Delta I_C / \Delta I_E$) at constant collector-base voltage ($V_{CB}$).
$I_{CBO}$ is the collector-base leakage current with the emitter open. This represents a small reverse saturation current.

Deriving Collector Current $I_C$ in terms of Base Current $I_B$

We need to express the collector current $I_C$ in terms of the base current $I_B$ for the common-base configuration. We can use the two fundamental equations mentioned above.

Start with the equation relating $I_C$, $I_E$, and $I_{CBO}$:

$I_C = \alpha I_E + I_{CBO}$

From the first equation, we know that $I_E = I_C + I_B$. Substitute this expression for $I_E$ into the equation above:

$I_C = \alpha (I_C + I_B) + I_{CBO}$

Now, distribute $\alpha$ on the right side:

$I_C = \alpha I_C + \alpha I_B + I_{CBO}$

We want to isolate $I_C$. Move the term $\alpha I_C$ to the left side of the equation:

$I_C - \alpha I_C = \alpha I_B + I_{CBO}$

Factor out $I_C$ from the terms on the left side:

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

Finally, divide both sides by $(1 - \alpha)$ to solve for $I_C$:

$I_C = \frac{\alpha}{1 - \alpha} I_B + \frac{1}{1 - \alpha} I_{CBO}$

This formula gives the collector current $I_C$ in the common-base configuration as a function of the base current $I_B$, the common-base current gain $\alpha$, and the collector-base leakage current $I_{CBO}$.

Comparison with Options

Comparing the derived formula with the given options:

${I_C} = \frac{\alpha }{{1 + \alpha }}{I_B} + \frac{1}{{1 - \alpha }}{I_{CBO}}$
${I_C} = \frac{\alpha }{{1 - \alpha }}{I_B} + \frac{1}{{1 + \alpha }}{I_{CBO}}$
${I_C} = \frac{\alpha }{{1 + \alpha }}{I_B} + \frac{1}{{1 + \alpha }}{I_{CBO}}$
${I_C} = \frac{\alpha }{{1 - \alpha }}{I_B} + \frac{1}{{1 - \alpha }}{I_{CBO}}$

The derived formula matches option 4.

Revision Table: Transistor Parameters

Parameter	Symbol	Description	Typical Value (for small signal)
Common-Base Current Gain	$\alpha$	Ratio of change in $I_C$ to change in $I_E$ at constant $V_{CB}$.	0.95 to 0.999
Common-Emitter Current Gain	$\beta$	Ratio of change in $I_C$ to change in $I_B$ at constant $V_{CE}$.	50 to 300+
Collector-Base Leakage Current (Emitter Open)	$I_{CBO}$	Reverse saturation current between collector and base with emitter open.	nA to µA range
Collector-Emitter Leakage Current (Base Open)	$I_{CEO}$	Reverse saturation current between collector and emitter with base open.	$\approx (1+\beta)I_{CBO}$

Additional Information: Relating Transistor Parameters ($\alpha$ and $\beta$)

The current gains in common-base ($\alpha$) and common-emitter ($\beta$) configurations are related. The common-emitter current gain $\beta$ is defined as $I_C / I_B$. We can find the relationship between $\alpha$ and $\beta$ using the fundamental current equations.

We know:

$I_E = I_C + I_B$
$I_C \approx \alpha I_E$ (ignoring $I_{CBO}$ for simplicity in relating gains)

Substitute $I_E = I_C + I_B$ into the second equation:

$I_C = \alpha (I_C + I_B)$

$I_C = \alpha I_C + \alpha I_B$

$I_C - \alpha I_C = \alpha I_B$

$I_C (1 - \alpha) = \alpha I_B$

Divide by $I_B$:

$\frac{I_C}{I_B} (1 - \alpha) = \alpha$

Since $\beta = I_C / I_B$:

$\beta (1 - \alpha) = \alpha$

$\beta - \beta \alpha = \alpha$

$\beta = \alpha + \beta \alpha$

$\beta = \alpha (1 + \beta)$

Solving for $\alpha$:

$\alpha = \frac{\beta}{1 + \beta}$

Solving for $\beta$:

$\beta = \frac{\alpha}{1 - \alpha}$

Using this relationship $\beta = \frac{\alpha}{1 - \alpha}$, the formula for $I_C$ derived earlier:

$I_C = \frac{\alpha}{1 - \alpha} I_B + \frac{1}{1 - \alpha} I_{CBO}$

can also be written in terms of $\beta$ and $I_{CEO}$ (where $I_{CEO} \approx (1+\beta)I_{CBO}$ or $I_{CBO} \approx \frac{I_{CEO}}{1+\beta}$). However, the question specifically asks for the common-base case with terms $I_B$, $\alpha$, and $I_{CBO}$.

Prepp Team · Answer

${I_C} = \frac{\alpha }{{1 + \alpha }}{I_B} + \frac{1}{{1 - \alpha }}{I_{CBO}}$

Prepp Team · Answer

${I_C} = \frac{\alpha }{{1 - \alpha }}{I_B} + \frac{1}{{1 + \alpha }}{I_{CBO}}$

Prepp Team · Answer

${I_C} = \frac{\alpha }{{1 + \alpha }}{I_B} + \frac{1}{{1 + \alpha }}{I_{CBO}}$

Understanding Collector Current in Common-Base Configuration

Transistor Current Relationships

Deriving Collector Current \(I_C\) in terms of Base Current \(I_B\)

Comparison with Options

Revision Table: Transistor Parameters

Additional Information: Relating Transistor Parameters (\(\alpha\) and \(\beta\))

Important Questions from Bipolar Junction Transistors

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Parameter	Symbol	Description	Typical Value (for small signal)
Common-Base Current Gain	\(\alpha\)	Ratio of change in \(I_C\) to change in \(I_E\) at constant \(V_{CB}\).	0.95 to 0.999
Common-Emitter Current Gain	\(\beta\)	Ratio of change in \(I_C\) to change in \(I_B\) at constant \(V_{CE}\).	50 to 300+
Collector-Base Leakage Current (Emitter Open)	\(I_{CBO}\)	Reverse saturation current between collector and base with emitter open.	nA to µA range
Collector-Emitter Leakage Current (Base Open)	\(I_{CEO}\)	Reverse saturation current between collector and emitter with base open.	\(\approx (1+\beta)I_{CBO}\)