Discussion
A Gunn diode of thickness 7 μm has a drift velocity of 1 × \(10^{5}\) m/s, find the transit time of an electron :
Answer: A
The transit time is 7 × 10-11 s.
The transit time is the thickness of the active region divided by the drift velocity:
\[ \begin{aligned} t &= \frac{L}{v_{d}} = \frac{7 \times 10^{-6}\ \text{m}}{1 \times 10^{5}\ \text{m/s}} \\ &= 7 \times 10^{-11}\ \text{s} \end{aligned} \]
That is 70 ps, corresponding to an oscillation frequency of about \(1/t \approx 14\) GHz, typical of a Gunn oscillator. The value \(7 \times 10^{-6}\) s forgets to divide by the velocity (it is the thickness in metres), and the other options are many orders of magnitude off.
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