Discussion
A transistor has \( \beta = 100 \) and \( f_{T} = 300 \ \mathrm{MHz} \). Its beta cut-off frequency is
Answer: D
Dividing the transition frequency by the low-frequency gain:
\[ \begin{aligned} f_{\beta} &= \frac{300 \ \mathrm{MHz}}{100} \\ &= 3 \ \mathrm{MHz} \end{aligned} \]
giving 3 MHz, beyond which the current gain starts to fall. The other options misplace a decade. The reciprocal relation between gain and bandwidth is what the transition frequency encodes, and it means a stage run at lower gain automatically has more bandwidth available.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development