Part B, Question 3: Slew-rate-limited unity-gain follower: 23-Mechatronics-A2 Circuits and Electronics, December 2019, Q9

Part B, Question 3: Slew-rate-limited unity-gain follower [8, 6, 6]

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.

Quantity	Value
Slew rate $SR$	1 V/μs
Unity-gain bandwidth $f_t$	1 MHz
Supply	±10 V
Configuration	unity-gain follower

Find. (a) $V_{max}$ before slew limiting; (b) its 10–90% rise time; (c) the 10–90% rise time for a step $10\times V_{max}$.

Illustrative step response: below $V_{max}$ the response is a clean single-pole exponential; a step $10\times V_{max}$ ramps at the slew-rate limit before rounding into the same exponential tail.

Approach. A unity-gain follower has closed-loop time constant $\tau=1/\omega_t$; slewing sets in exactly when the exponential response’s initial slope would exceed $SR$, giving the largest un-slewed step. For a larger step, split the response into a linear slew-rate-limited ramp followed by the same exponential tail, matched in slope at the transition.

Closed-loop time constant. $$\omega_t=2\pi f_t=2\pi(1\text{MHz})=6.283e+06\ \text{rad/s},\qquad \tau=\frac{1}{\omega_t}=159.15\ \text{ns}.$$
Largest un-slewed step. The exponential $v_{OUT}(t)=V(1-e^{-\omega_t t})$ has initial slope $V\omega_t$; setting this equal to $SR$: $$V_{max}=\frac{SR}{\omega_t}=\frac{1\times10^6}{6.283e+06}=\boxed{0.1592\ \text{V}}.$$
Rise time for this step (pure exponential, no slewing). $$t_r=2.2\tau=2.2\times159.15\ \text{ns}=\boxed{350.1\ \text{ns}}.$$
10× step: identify the slew-limited and exponential portions. For $V_{step}=10V_{max}=1.5915$ V, the output first ramps linearly at slope $SR$ until it is within $V_{max}$ of the final value (the point where the required slope has fallen back to exactly $SR$), then finishes as the tail of an exponential of amplitude $V_{max}$.
10% and 90% times on the ramp. The 10% level ($=V_{max}$) and the 90% level ($=9V_{max}=V_{step}-V_{max}$, i.e. exactly the ramp/exponential transition point since $V_{max}=0.1V_{step}$) both fall on the slew-limited ramp: $$t_{10\%}=\frac{V_{max}}{SR}=159.15\ \text{ns},\qquad t_{90\%}=\frac{9V_{max}}{SR}=1432.39\ \text{ns}.$$
New rise time. $$t_r=t_{90\%}-t_{10\%}=\frac{8V_{max}}{SR}=8\tau=\boxed{1273.2\ \text{ns}}\approx1.273\ \mu\text{s}.$$

Quantity	Result
$V_{max}$	$\boxed{0.1592\ \text{V}}$
$t_r$ at $V_{max}$	$\boxed{350.1\ \text{ns}}$
$t_r$ at $10V_{max}$	$\boxed{1273.2\ \text{ns}}$

Concept & Theory

Topic. Op-amp slew-rate limiting versus small-signal (linear) bandwidth limiting.

Key relations. Small-signal step response $V(1-e^{-\omega_t t})$; slewing occurs when required $|dv/dt|>SR$; matched-slope transition between a slew ramp and the exponential tail.

Why this works. An op-amp responds linearly (as a single pole, set by $f_t$ in a unity-gain follower) only while its internal compensation capacitor can be charged fast enough; once the demanded initial slope exceeds the amplifier’s maximum current-limited charging rate ($SR$), the output is forced into a constant-slope ramp until the remaining error becomes small enough for the linear response to keep up again, which is exactly the point where the local exponential slope has decayed back to $SR$.

Common pitfall. Using $t_r=2.2\tau$ for the 10× step (that formula assumes no slewing at all); forgetting that both the 10% and 90% levels for a heavily over-driven step can lie entirely within the slew-limited ramp.

Source. Sedra & Smith, Microelectronic Circuits §2.7 (op-amp non-idealities: slew rate, full-power bandwidth).

Question 9 of 11: Part B, Question 3: Slew-rate-limited unity-gain follower

Part A — Circuits

Part B, Question 3: Slew-rate-limited unity-gain follower [8, 6, 6]