Question

For a transistor, $\frac{1}{\alpha_{DC}} - \frac{1}{\beta_{DC}}$ is equal to [$\alpha_{DC}$ and $\beta_{DC}$ are current amplification factors]

• A. three
• B. two
• C. zero
• D. one

Hint:

Always remember the basic current relation for a transistor: the emitter current is the sum of collector and base currents ($I_E = I_C + I_B$). This relationship is the key to deriving most transistor gain formulas.

Answer 1

The Correct Option is D

Step 1: What is being asked.
For a transistor we must work out the value of $\dfrac{1}{\alpha_{DC}} - \dfrac{1}{\beta_{DC}}$, where these are the two current gains.

Step 2: Recall the definitions.
$\alpha = \dfrac{I_C}{I_E}$ (collector over emitter) and $\beta = \dfrac{I_C}{I_B}$ (collector over base).

Step 3: Flip each one over.
So $\dfrac{1}{\alpha} = \dfrac{I_E}{I_C}$ and $\dfrac{1}{\beta} = \dfrac{I_B}{I_C}$.

Step 4: Subtract them.
$\dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{I_E}{I_C} - \dfrac{I_B}{I_C} = \dfrac{I_E - I_B}{I_C}$.

Step 5: Use the current law of a transistor.
The emitter current splits into base and collector, so $I_E = I_B + I_C$, which means $I_E - I_B = I_C$.

Step 6: Finish it.
The top becomes $I_C$, so the whole thing is $\dfrac{I_C}{I_C} = 1$. The answer is one, option (D).
\[ \boxed{\frac{1}{\alpha_{DC}} - \frac{1}{\beta_{DC}} = 1} \]