Understanding Transistor Currents and Gain

In a bipolar junction transistor (BJT), the currents flowing into and out of the three terminals — emitter (E), base (B), and collector (C) — are related. The emitter current ($I_E$) is the largest and is the sum of the base current ($I_B$) and the collector current ($I_C$). This fundamental relationship is given by:

$$I_E = I_B + I_C$$

Transistors are often characterized by their current gain, which describes how much the current is amplified. There are two primary current gains:

• Common-Emitter Current Gain (\(\beta\)): This relates the collector current to the base current (\(\beta = I_C / I_B\)).
• Common-Base Current Gain (\(\alpha\)): This relates the collector current to the emitter current (\(\alpha = I_C / I_E\)).

The question provides the common-base DC current gain (\(\alpha_{DC}\)) and the emitter current (\(I_E\)). We need to find the base current (\(I_B\)).

Calculating Collector Current (\(I_C\))

We know the definition of the common-base DC current gain:

$$\alpha_{DC} = \frac{I_C}{I_E}$$

We are given \(\alpha_{DC} = 0.967\) and \(I_E = 10 \text{ mA}\). We can use this relationship to find the collector current (\(I_C\)):

$$I_C = \alpha_{DC} \times I_E$$ $$I_C = 0.967 \times 10 \text{ mA}$$ $$I_C = 9.67 \text{ mA}$$

Finding the Base Current (\(I_B\))

Now that we have the emitter current (\(I_E\)) and the collector current (\(I_C\)), we can use the fundamental transistor current relationship to find the base current (\(I_B\)):

$$I_E = I_B + I_C$$

Rearranging the formula to solve for \(I_B\):

$$I_B = I_E - I_C$$

Substitute the given value of \(I_E\) and the calculated value of \(I_C\):

$$I_B = 10 \text{ mA} - 9.67 \text{ mA}$$ $$I_B = 0.33 \text{ mA}$$

Alternative Method: Using \(\alpha_{DC}\) and \(I_E\) Directly

We can also derive a formula for \(I_B\) directly in terms of \(I_E\) and \(\alpha_{DC}\). Starting with the two main equations:

$$I_E = I_B + I_C \quad \text{(Equation 1)}$$ $$\alpha_{DC} = \frac{I_C}{I_E} \Rightarrow I_C = \alpha_{DC} I_E \quad \text{(Equation 2)}$$

Substitute Equation 2 into Equation 1:

$$I_E = I_B + \alpha_{DC} I_E$$

Rearrange to solve for \(I_B\):

$$I_B = I_E - \alpha_{DC} I_E$$ $$I_B = I_E (1 - \alpha_{DC})$$

Now, substitute the given values:

$$I_B = 10 \text{ mA} (1 - 0.967)$$ $$I_B = 10 \text{ mA} (0.033)$$ $$I_B = 0.33 \text{ mA}$$

Both methods yield the same result for the base current.

Summary of Calculation Steps

To find the base current given the emitter current and common-base current gain:

1. Recall the relationship between emitter, base, and collector currents: \(I_E = I_B + I_C\).
2. Recall the definition of common-base current gain: \(\alpha_{DC} = I_C / I_E\).
3. Option 1: Calculate \(I_C\) using \(I_C = \alpha_{DC} \times I_E\), then calculate \(I_B\) using \(I_B = I_E - I_C\).
4. Option 2: Use the derived formula \(I_B = I_E (1 - \alpha_{DC})\) directly.
5. Substitute the given values and calculate the result.

Result Verification

Given \(I_E = 10 \text{ mA}\) and calculated \(I_B = 0.33 \text{ mA}\) and \(I_C = 9.67 \text{ mA}\):

• Check current relationship: \(I_B + I_C = 0.33 \text{ mA} + 9.67 \text{ mA} = 10.00 \text{ mA}\). This matches \(I_E\).
• Check common-base gain: \(\alpha_{DC} = I_C / I_E = 9.67 \text{ mA} / 10 \text{ mA} = 0.967\). This matches the given \(\alpha_{DC}\).

The calculated base current of 0.33 mA is consistent with the given parameters.

Revision Table: Transistor Current Gains

Additional Information: Relationship Between \(\alpha\) and \(\beta\)

The common-base current gain (\(\alpha\)) and common-emitter current gain (\(\beta\)) are related. We can derive this relationship using the fundamental current equation \(I_E = I_B + I_C\) and the definitions of \(\alpha\) and \(\beta\).

We know \(I_C = \alpha I_E\) and \(I_C = \beta I_B\).

From \(I_E = I_B + I_C\), divide all terms by \(I_C\):

$$\frac{I_E}{I_C} = \frac{I_B}{I_C} + \frac{I_C}{I_C}$$

Since \(\alpha = I_C / I_E\), \(I_E / I_C = 1 / \alpha\). Since \(\beta = I_C / I_B\), \(I_B / I_C = 1 / \beta\).

Substitute these into the equation:

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

This can be rearranged to find \(\beta\) in terms of \(\alpha\):

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

Or, to find \(\alpha\) in terms of \(\beta\):

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

These relationships are useful for converting between the two current gain parameters. For the value given in the question, \(\alpha_{DC} = 0.967\), the corresponding \(\beta_{DC}\) would be:

$$\beta_{DC} = \frac{0.967}{1 - 0.967} = \frac{0.967}{0.033} \approx 29.3$$

This shows that a small change in \(\alpha\) (close to 1) corresponds to a large change in \(\beta\).