Determine, to a first approximation, the frequency at which the small-signal current gain decreases to 1 / \sqrt{2} of its low-frequency value.
Consider the simplified hybrid-pi circuit shown in Figure 12.41. We are ignoring C_{\mu}, C_{s}, r_{\mu}, C_{j e}, r_{0}, and the series resistances. We must emphasize that this is a first-order calculation and that C_{\mu} normally cannot be neglected.
At very low frequency, we may neglect C_{\pi} so that
V_{b e}=I_{b} r_{\pi} \quad \text { and } \quad I_{c}=g_{m} V_{b e}=g_{m} r_{\pi} I_{b}
We can then write
h_{f e 0}=\frac{I_{c}}{I_{b}}=g_{m} r_{\pi}
where h_{f e 0} is the low-frequency, small-signal common-emitter current gain.
Taking into account C_{\pi}, we have
V_{b e}=I_{b}\left(\frac{r_{\pi}}{1+j \omega r_{\pi} C_{\pi}}\right)
Then
I_{c}=g_{m} V_{b e}=I_{b}\left(\frac{h_{f e 0}}{1+j \omega r_{\pi} C_{\pi}}\right)
or the small-signal current gain can be written as
A_{i}=\frac{I_{c}}{I_{b}}=\left(\frac{h_{f 0}}{1+j \omega r_{\pi} C_{\pi}}\right)
The magnitude of the current gain is then
\left|A_{i}\right|=\left|\frac{I_{c}}{I_{b}}\right|=\frac{h_{f e 0}}{\sqrt{1+\left(\omega r_{\pi} C_{\pi}\right)^{2}}}=\frac{h_{f e 0}}{\sqrt{1+\left(2 \pi f r_{\pi} C_{\pi}\right)^{2}}}
The magnitude of the current gain drops to 1 / \sqrt{2} of its low-frequency value at f=1 / 2 \pi r_{\pi} C_{\pi}.
If, for example, r_{\pi}=2.6 \mathrm{k} \Omega and C_{\pi}=4 \mathrm{pF}, then
f=15.3 \mathrm{MHz}
Comment
High-frequency transistors must have small-diffusion capacitances, implying the use of small devices.