Question

For a transistor amplifier, the frequency response is such that the mid band voltage gain is 200. The cutoff frequencies are 20 Hz and 20 kHz. What is the ratio (rounded off to two decimal places) of the voltage gain at 10 Hz to that at 100 kHz?

Hint:

For frequency-dependent voltage gain, use the relation with cutoff frequencies to calculate the gain at any given frequency.

Answer 1

The voltage gain at any frequency is related to the midband gain by the following expression: \[ G(f) = \frac{G_{\text{mid}}}{\sqrt{1 + (\frac{f}{f_c})^2}}, \] where $f_c$ is the cutoff frequency.
For the given amplifier:
- Midband gain $G_{\text{mid}} = 200$
- $f_1 = 10~\text{Hz}$ and $f_2 = 100~\text{kHz}$
- The lower cutoff frequency is $f_c = 20~\text{Hz}$.
Thus, the voltage gain at $f_1$ is: \[ G(f_1) = \frac{200}{\sqrt{1 + (\frac{10}{20})^2}} = \frac{200}{\sqrt{1 + 0.25}} = \frac{200}{\sqrt{1.25}} = 200 / 1.118 = 178.88. \] For $f_2$ (at $100~\text{kHz}$), \[ G(f_2) = \frac{200}{\sqrt{1 + (\frac{100000}{20000})^2}} = \frac{200}{\sqrt{1 + 25}} = \frac{200}{\sqrt{26}} = 200 / 5.099 = 39.23. \] Thus, the ratio of the gain at $10~\text{Hz}$ to that at $100~\text{kHz}$ is: \[ \text{Ratio} = \frac{G(f_1)}{G(f_2)} = \frac{178.88}{39.23} \approx 4.56 \approx 45. \]