Question 3 of 5: Non-Inverting Amplifier — Bandwidth and Slew Rate

Question 3: Non-Inverting Amplifier — Bandwidth and Slew Rate (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Non-inverting amplifier, $R_1=1\text{ k}\Omega$, $R_2=100\text{ k}\Omega$. Op-amp open-loop response (from the Bode plot): DC gain $100\text{ dB}=10^5$, a single pole at $10\text{ Hz}$, and the $-20\text{ dB/dec}$ roll-off crossing $0\text{ dB}$ at $10^6\text{ Hz}$ (unity-gain frequency $f_t=1\text{ MHz}$). Slew rate $SR=0.5\text{ V/}\mu\text{s}$.

Find. (a) closed-loop $f_{3dB}$ and the response sketch; (b) output amplitude and shape at 10 kHz, 20 mV$_{pp}$; (c) output amplitude and shape at 100 kHz, 5 V$_{pp}$.

Approach. Use the gain–bandwidth constancy of a single-pole op-amp to get the closed-loop corner, then test each input against both the small-signal bandwidth and the large-signal slew-rate limit.

1. Closed-loop gain and bandwidth (part a). The non-inverting gain is $$A_{CL}=1+\frac{R_2}{R_1}=1+\frac{100\text{ k}\Omega}{1\text{ k}\Omega}=101\quad(40.1\text{ dB}).$$ For a single-pole (internally compensated) op-amp the gain–bandwidth product is constant and equals $f_t$, so $$f_{3dB}=\frac{f_t}{A_{CL}}=\frac{1\text{ MHz}}{101}=\boxed{9.9\text{ kHz}.}$$ The closed-loop response is flat at 40.1 dB out to 9.9 kHz, then rolls off at $-20\text{ dB/dec}$, rejoining the open-loop curve and reaching 0 dB at $f_t=1\text{ MHz}$.

2. Part (b): 10 kHz, 20 mV$_{pp}$. First the slew check: an undistorted output would have amplitude $\le A_{CL}(20\text{ mV})=2.02\text{ V}_{pp}$, i.e. $1.01\text{ V}$ peak, whose maximum slope is $2\pi f\,\hat V=2\pi(10\text{ kHz})(1.01\text{ V})=0.063\text{ V/}\mu\text{s}\ll SR$. So the output is not slew-limited — it stays a clean sinusoid. However, 10 kHz sits essentially at the closed-loop corner ($f_{3dB}=9.9\text{ kHz}$), so the gain is down by the frequency-response factor: $$\lvert A\rvert=\frac{A_{CL}}{\sqrt{1+(f/f_{3dB})^2}}=\frac{101}{\sqrt{1+(10/9.9)^2}}=71.1.$$ $$V_{o,pp}=71.1\times 20\text{ mV}=\boxed{1.42\ \text{V}_{pp}}$$ (about $-3\text{ dB}$, i.e. $0.71\times$, of the 2.02 V$_{pp}$ mid-band value), a 10 kHz sinusoid lagging the input by roughly $45^\circ$.

3. Part (c): 100 kHz, 5 V$_{pp}$. Now the large-signal (slew) limit dominates. The maximum undistorted sinusoidal amplitude at 100 kHz is $$\hat V_{max}=\frac{SR}{2\pi f}=\frac{0.5\text{ V/}\mu\text{s}}{2\pi(100\text{ kHz})}=0.80\text{ V}\;(1.59\text{ V}_{pp}).$$ The amplifier is being asked for far more than this (even the small-signal gain $f_t/f=10$ would demand $50\text{ V}_{pp}$), so the output cannot follow the sinusoid and instead ramps at $\pm SR$ the whole time — it becomes a triangle wave. Over each half period $T/2=5\ \mu\text{s}$ the output swings $$V_{o,pp}=SR\cdot\frac{T}{2}=(0.5\text{ V/}\mu\text{s})(5\ \mu\text{s})=\boxed{2.5\ \text{V}_{pp}}$$ i.e. a symmetric $\pm 1.25\text{ V}$ triangle at 100 kHz (well within the ±15 V rails, so no clipping — the shape is set purely by slewing).