Question 1 of 5: Slew Rate and Rise Time of a Unity-Gain Follower

Question 1: Slew Rate and Rise Time of a Unity-Gain Follower (20 marks)

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Given. Slew rate $SR=1\ \text{V/}\mu\text{s}=10^{6}\ \text{V/s}$; unity-gain bandwidth $f_t=1\ \text{MHz}$; unity-gain follower ($v_{OUT}/v_{IN}=1/(1+s/\omega_t)$, single pole at $\omega_t=2\pi f_t$); rails $\pm 10\ \text{V}$.

Find. (a) the largest step that keeps the output a pure exponential (not slew-limited); (b) its $10\%\!-\!90\%$ rise time; (c) the rise time when the step is ten times larger.

Approach. For a small step the follower’s output is the single-pole response $v_{OUT}(t)=V(1-e^{-\omega_t t})$, whose steepest slope is $V\omega_t$ at $t=0$; the response stays exponential only while that slope does not exceed the slew rate. A large step forces the output to ramp at the constant slew rate instead.

1. Pole frequency. $\omega_t=2\pi f_t=2\pi(10^{6})=6.283\times10^{6}\ \text{rad/s}$, so the small-signal time constant is $\tau=1/\omega_t=0.159\ \mu\text{s}$.
2. (a) Largest exponential step. The initial slope of $V(1-e^{-\omega_t t})$ is $\left.\dfrac{dv_{OUT}}{dt}\right|_{0}=V\,\omega_t$. Setting this equal to the slew rate gives the boundary between linear and slew-limited operation: $$V_{\max}=\frac{SR}{\omega_t}=\frac{10^{6}}{6.283\times10^{6}}\ \Rightarrow\ \boxed{V_{\max}\approx 0.159\ \text{V}=159\ \text{mV}}$$
3. (b) Rise time of the exponential. A single-pole $10\%\!-\!90\%$ rise time is $t_r=\dfrac{\ln 9}{\omega_t}=\dfrac{2.2}{\omega_t}$, independent of amplitude: $$t_r=\frac{2.2}{6.283\times10^{6}}=\boxed{0.35\ \mu\text{s}}$$ For the $159\ \text{mV}$ step this is the actual rise time because the response is still a clean exponential.
4. (c) Ten-times-larger step. The step is $10\times0.159=1.59\ \text{V}$. Its would-be initial slope $V\omega_t=10\,SR$ far exceeds $SR$, so the amplifier slews: the output is a straight ramp of slope $SR$. The $10\%\!-\!90\%$ excursion covers $0.8V$, hence $$t_r=\frac{0.8\,V}{SR}=\frac{0.8(1.59)}{10^{6}}=\boxed{1.27\ \mu\text{s}}$$ about $3.6\times$ longer than the small-signal value — the signature of slewing.