Question

The rise time of a BJT is 3.5 micro seconds. Find its transition frequency.

• A. 1 kHz
• B. 10 kHz
• C. 35 kHz
• D. 100 kHz

Hint:

The empirical relationship between 10%-90% rise time (\(t_r\)) and the upper 3dB frequency (\(f_H\)) or bandwidth (BW) of a single-pole system is \(t_r \cdot f_H \approx 0.35\).
The transition frequency \(f_T\) represents the maximum useful frequency of a transistor in terms of current gain. It's a key figure of merit for high-frequency applications.

Answer 1

The Correct Option is D

The relationship between the rise time (\(t_r\)) of a device (like a BJT or an amplifier) and its upper 3dB cutoff frequency (\(f_H\)) or bandwidth (BW) is approximately given by: \[ t_r \approx \frac{0.35}{f_H} \] 

The transition frequency (\(f_T\)) of a BJT is related to its high-frequency performance and is often considered as a measure of its bandwidth capability. In many contexts, especially for a single-pole response approximation, \(f_H\) can be related to \(f_T\). However, a more direct relation for transition frequency involves \(f_T = \text{gain} \times \text{bandwidth}\).

If we assume that the rise time is primarily limited by the dominant high-frequency pole, then \(f_H\) can be considered a measure of the bandwidth. The transition frequency \(f_T\) is the frequency at which the short-circuit common-emitter current gain (\(\beta(j\omega)\) or \(h_{fe}(j\omega)\)) drops to unity (0 dB).

The relation \(t_r \cdot BW \approx 0.35\) is commonly used. If we consider \(f_T\) to be the relevant bandwidth here: \[ f_T \approx \frac{0.35}{t_r} \]

Given \(t_r = 3.5\) micro seconds = \(3.5 \times 10^{-6}\) seconds.

\[ f_T \approx \frac{0.35}{3.5 \times 10^{-6} \text{ s}} = \frac{0.35}{3.5} \times 10^6 \text{ Hz} = 0.1 \times 10^6 \text{ Hz} = 100 \times 10^3 \text{ Hz} = 100 \text{ kHz} \]

This matches option (d).

Final Answer: \[ \boxed{100 \text{ kHz}} \]