Question E6A08:
What is the term for the frequency at which the grounded-base current gain of a bipolar junction transistor has decreased to 0.7 of the gain obtainable at 1 kHz?  

Choices:  
A: Corner frequency  
B: Alpha rejection frequency  
C: Beta cutoff frequency  
D: Alpha cutoff frequency    

Correct Answer is  D: Alpha cutoff frequency  

To understand this question, remember:
- In a BJT, the grounded-base (common-base) current gain is called alpha, $\,\alpha = I_C/I_E\,$.
- As frequency increases, gain falls due to internal capacitances. The “cutoff” frequency is typically defined at the −3 dB point, i.e., where amplitude drops to about 0.707 of its low-frequency value.

Why D (Alpha cutoff frequency) is correct:
- “Grounded-base current gain” refers to $\,\alpha\,$.
- The alpha cutoff frequency, $f_\alpha$, is defined where the common-base current gain drops to $0.707$ of its low-frequency value (here referenced to 1 kHz):
  $$|\alpha(f_\alpha)| = 0.707 \, |\alpha(\text{1 kHz})|$$

Why the others are incorrect:
- A: Corner frequency  
  - This is a generic term for a −3 dB point in filters or responses, but the question asks specifically about the grounded-base current gain. The precise term is “alpha cutoff frequency,” not just “corner frequency.”
- B: Alpha rejection frequency  
  - Not a standard term in BJT frequency response.
- C: Beta cutoff frequency  
  - Beta ($\beta$) is the grounded-emitter (common-emitter) current gain, $\beta = I_C/I_B$. The question is about grounded-base (alpha), not beta.

Useful formulas and relations:
- $$\alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}$$
- Relationship between $\alpha$ and $\beta$:
  $$\beta = \frac{\alpha}{1-\alpha}, \quad \alpha = \frac{\beta}{\beta + 1}$$
- −3 dB amplitude ratio: $0.707$ of the reference amplitude.