Understanding Transistor Currents and Alpha
This question asks us to find the emitter current (\(I_E\)) of a transistor given its common-base current gain (\(\alpha\)) and base current (\(I_B\)). Transistors are fundamental semiconductor devices used for switching and amplification. The relationships between the different currents in a transistor (emitter, base, and collector) are governed by its characteristics, represented by parameters like \(\alpha\) and \(\beta\).
Transistor Current Relationships
In any bipolar junction transistor (BJT), the total emitter current is the sum of the base current and the collector current. This can be written as:
\[I_E = I_B + I_C\]The common-base current gain, \(\alpha\), is defined as the ratio of the collector current to the emitter current when the base is common to both the input and output circuits. Ideally, for DC current, it is given by:
\[\alpha = \frac{I_C}{I_E}\]This implies that \(I_C = \alpha I_E\).
Calculating Emitter Current from Base Current and Alpha
We are given the base current (\(I_B\)) and the common-base current gain (\(\alpha\)). We need to find the emitter current (\(I_E\)). We can use the two fundamental relationships mentioned above to derive a formula for \(I_E\) in terms of \(I_B\) and \(\alpha\).
Start with the current sum equation:
\[I_E = I_B + I_C\]Substitute the expression for \(I_C\) from the definition of \(\alpha\) (\(I_C = \alpha I_E\)):
\[I_E = I_B + \alpha I_E\]Now, rearrange the equation to isolate \(I_E\):
Subtract \(\alpha I_E\) from both sides:
\[I_E - \alpha I_E = I_B\]Factor out \(I_E\) from the left side:
\[I_E (1 - \alpha) = I_B\]Divide both sides by \((1 - \alpha)\) to solve for \(I_E\):
\[I_E = \frac{I_B}{1 - \alpha}\]This formula allows us to calculate the emitter current if we know the base current and the common-base current gain \(\alpha\).
Step-by-Step Calculation
Given values:
- Common-base current gain, \(\alpha = 0.995\)
- Base current, \(I_B = 200 \, \mu A\)
First, convert the base current from microamperes (\(\mu A\)) to amperes (A):
\[I_B = 200 \, \mu A = 200 \times 10^{-6} \, A\]Now, use the derived formula \(I_E = \frac{I_B}{1 - \alpha}\):
Calculate the denominator \(1 - \alpha\):
\[1 - \alpha = 1 - 0.995 = 0.005\]Substitute the values of \(I_B\) and \((1 - \alpha)\) into the formula for \(I_E\):
\[I_E = \frac{200 \times 10^{-6} \, A}{0.005}\] \[I_E = \frac{200}{0.005} \times 10^{-6} \, A\]To simplify the division, multiply the numerator and denominator by 1000:
\[I_E = \frac{200 \times 1000}{0.005 \times 1000} \times 10^{-6} \, A\] \[I_E = \frac{200000}{5} \times 10^{-6} \, A\] \[I_E = 40000 \times 10^{-6} \, A\]The result is in amperes. Convert it to milliamperes (mA) by multiplying by \(10^3\) (since \(1 \, mA = 10^{-3} \, A\)):
\[I_E = 40000 \times 10^{-6} \times 10^3 \, mA\] \[I_E = 40000 \times 10^{-3} \, mA\] \[I_E = 40 \, mA\]Thus, the value of the emitter current is 40 mA.
Summary of Results
| Parameter | Symbol | Value | Units |
|---|---|---|---|
| Common-base current gain | \(\alpha\) | 0.995 | (unitless) |
| Base current | \(I_B\) | 200 | \(\mu A\) |
| Base current (converted) | \(I_B\) | \(200 \times 10^{-6}\) | \(A\) |
| Emitter current | \(I_E\) | 40 | \(mA\) |
Comparing with Options
The calculated emitter current is 40 mA. Let's compare this with the given options:
- Option 1: 30 mA
- Option 2: 35 mA
- Option 3: 40 mA
- Option 4: 45 mA
The calculated value matches Option 3.
Revision Table: Transistor Currents and Parameters
| Parameter | Definition | Typical Value (BJT) | Relationship |
|---|---|---|---|
| Emitter Current | Total current entering/leaving the emitter terminal | Highest current (\(mA\) range) | \(I_E = I_B + I_C\) |
| Base Current | Current entering/leaving the base terminal | Lowest current (\(\mu A\) range) | \(I_B = I_E - I_C\) |
| Collector Current | Current entering/leaving the collector terminal | Similar to \(I_E\) (\(mA\) range) | \(I_C = I_E - I_B\) |
| Common-Base Current Gain | Ratio of \(I_C\) to \(I_E\) (\(\alpha\)) | 0.95 to 0.998 (close to 1) | \(\alpha = I_C / I_E\) |
| Common-Emitter Current Gain | Ratio of \(I_C\) to \(I_B\) (\(\beta\)) | 50 to 500 | \(\beta = I_C / I_B\) |
Additional Information: Alpha and Beta Relationship
The common-base current gain (\(\alpha\)) and the common-emitter current gain (\(\beta\)) are related. Since \(I_C = \alpha I_E\) and \(I_E = I_B + I_C\), we can substitute the second equation into the first:
\[I_C = \alpha (I_B + I_C)\] \[I_C = \alpha I_B + \alpha I_C\]Rearrange to find \(I_C\):
\[I_C - \alpha I_C = \alpha I_B\] \[I_C (1 - \alpha) = \alpha I_B\] \[I_C = \frac{\alpha}{1 - \alpha} I_B\]By definition, \(\beta = I_C / I_B\). Comparing this to the equation above, we get the relationship:
\[\beta = \frac{\alpha}{1 - \alpha}\]Conversely, we can express \(\alpha\) 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 very useful when analyzing transistor circuits in different configurations.
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