Question 12.12: Increasing the Gain-Bandwidth Product by Means of Amplifiers...

Known Quantities: Gain-bandwidth product and gain of each amplifier.

Find: \omega_{3\mathrm{~dB}} of cascade amplifier.

Schematics, Diagrams, Circuits, and Given Data: A_{0} \times \omega_{0}=K=4  \pi \times 10^{6} for each amplifier. R_{F} / R_{S}=100 for each amplifier.

Assumptions: Assume gain-bandwidth product-limited (otherwise ideal) op-amps.

Analysis: Let A_{1} and \omega_{1} denote the gain and the 3-dB bandwidth of the first amplifier, respectively, and A_{2} and \omega_{2} those of the second amplifier.

The 3-dB bandwidth of the first amplifier is:

\omega_{1}=\frac{K}{A_{1}}=\frac{4 \pi \times 10^{6}}{10^{2}}=4 \pi \times 10^{4} \quad \frac{\mathrm{rad}}{\mathrm{s}}

The second amplifier will also have:

\omega_{2}=\frac{K}{A_{2}}=\frac{4 \pi \times 10^{6}}{10^{2}}=4 \pi \times 10^{4} \quad \frac{\mathrm{rad}}{\mathrm{s}}

Thus, the approximate bandwidth of the cascade amplifier is 4 \pi \times 10^{4} and the gain of the cascade amplifier is A_{1} \times A_{2}=100 \times 100=10^{4}.

Had we attempted to achieve the same gain with a single-stage amplifier having the same K, we would have achieved a bandwidth of only:

\omega_{3}=\frac{K}{A_{3}}=\frac{4 \pi \times 10^{6}}{10^{4}}=4 \pi \times 10^{2} \quad \frac{\mathrm{rad}}{s}

Comments: In practice, the actual 3-dB bandwidth of the cascade amplifier is not quite as large as that of each of the two stages, because the gain of each amplifier starts decreasing at frequencies somewhat lower than the nominal cutoff frequency. The calculation of the actual 3-dB bandwidth of the cascade amplifier is illustrated in Check Your Understanding Exercise 12.17.