Slew rate and output swing

A small step and a large step

Put a 10 mV step and a 400 mV step into two copies of the buffer, and show each output in percent of its own step. A linear circuit gives two identical curves, because doubling the input doubles every point of the output.

Here is why. A step of 400 mV appears almost entirely between the two gates at first, because the output has not moved yet. The pair of E5.1 steers all of the tail current I_SS to one side once the difference passes a few hundred millivolts. It can never deliver more than I_SS. That current charges C_c, so the output ramps at a fixed rate: SR = I_SS/C_c.

A 10 mV step stays inside the pair's linear range. The current that flows is proportional to the error, so the response stays the scaled small-signal one, however fast it is.

Where the current goes on each edge

The lab measures the slope at the middle of each edge (v_out = 1.05 V) and the currents at that instant.

The falling edge is slower than even the first falling formula, which treats C_c as a second load because it assumes o1 stands still. But M4 drives o1 upward at 5.73 V/µs while the output falls at 11.13 V/µs. The voltage across C_c changes by both, so it carries 1 pF × 16.86 V/µs = 16.86 µA into the output. C_L needs 2 pF × 11.13 V/µs = 22.25 µA. Together that is 39.11 µA of M7's 41.13 µA; M6 still supplies 0.77 µA and the rest charges the transistors' own capacitance.

Most of M4's I_SS flows on through C_c, so the better hand rule leaves M7 only I₇ − I_SS for C_L: 11.0 V/µs, within 1 % of the run.

Output swing and full-power bandwidth

The output swing is the range the output covers while the op-amp is still an amplifier. The hand rule: M7 needs V_DS above its saturation voltage, and M6 needs V_SD above its own. That gives V_Dsat7 to VDD − V_SDsat6, 0.134 V to 1.614 V here.

A buffer cannot show it, because its input moves with its output. The figure sweeps a buffer and an inverting amplifier of gain −1 (100 kΩ in, 100 kΩ feedback, + input at 0.9 V), whose inputs stay near 0.9 V.

So the brief's 1.4 V buffer stops at 1.370 V because of the input range, not the output stage. But tracking within 10 mV flatters the output too: a loop with 80 dB to spare hides a gain that has fallen a long way. Sweep the op-amp open loop instead. Its gain peaks at 11 022 and falls to half at 0.293 V and 1.489 V of output. That is well inside the hand rule's 0.134 V to 1.614 V. r_o drops long before a device leaves saturation (E2.4).

Slew rate also limits large sines. A sine of amplitude A at frequency f has a steepest slope of 2πfA. The full-power bandwidth is the frequency where that equals the slew rate: f = SR/(2πA). For A = 0.2 V and the falling 11.13 V/µs it is 8.85 MHz.

Try it

Run it and open Plots: v(out) ramps up, ramps down more slowly and dips before it settles. sr_rise and sr_fall are the slopes in V/µs where the output crosses 1.05 V; i5_rise, i6_fall and i7_fall are the currents at those instants.

Now set cc=2p and predict sr_rise before you run: the tail current is the same, so the rising slope should halve. Then put cc back and set w7=32. M7 sinks twice the current: watch the falling slope and the dip.

Check yourself

* the first stage while slewing: the whole tail current, 18 uA, into C_c = 1 pF
I1 0 c 18u
C1 c 0 1p
.control
tran 1n 100n uic
* the slope in V/us
let sr = deriv(v(c))/1e6
meas tran sr FIND sr AT=50n
.endc
.end