GATE 2018 IN Q.41: solution and answer

Official answer: 17.5 to 17.9Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Assume the op-amp is ideal except for its slew-rate limit. The non-inverting amplifier has a closed-loop gain of 15, so $v_o=15v_i$.

2. Output peak: $V_m=15\times 0.3=4.5\text{ V}$. This is inside the $\pm 5\text{ V}$ supply, so the output does not clip. The only thing that limits the frequency is the slew rate.

3. For $v_o=V_m\sin \omega t$, we get $\frac{d{v}_o}{dt}=\omega V_m\cos \omega t$. Its largest magnitude occurs at the zero crossings: ${\frac{d{v}_o}{dt}|}_{\max }=\omega V_m=2\pi f{V}_m$.

4. For no slew-rate distortion we need $2\pi f{V}_m\le SR$. This gives $f_{\max }=\frac{SR}{2\pi V_m}$, the full-power bandwidth for this output amplitude.

5. $$f_{\max }=\frac{0.5\times 10^6}{2\pi \times 4.5}=\frac{0.5\times 10^6}{28.274}\approx 17684\text{ Hz}$$

6. $f_{\max }\approx 17.7\text{ kHz}$