I'm curious if the DC offset of a signal going into an op amp will affect the slew rate. Lets say you have an op amp configured to drive a 150 Ω load and is spec'ed to provided an output swing of ±3 V (say an AD829 or OP27). If I drive the op amp with a signal that has a DC offset of 2.7 V (close but not at the limit of its range) will this affect the slew rate? Here assume that the amplitude of the AC components are stay close or within the ±3 V.

I'm guessing the answer to this question will depend on whether the limiting factor is in the compensation or output driving stage of the op amp. I'm wondering if this is something that is known, that you shouldn't expect an op amp to provide its full slew rate unless its around the center point of its supply range or if it doesn't matter as long as you are within the output swing range.

I guess a good follow up question is whether it matters that you are starting with a DC offset that is positive or negative?

  • \$\begingroup\$ I imagine it depends on the amplifier, but I wouldn't expect a significant effect. I'm no expert, though, so I could be entirely wrong. \$\endgroup\$ – Hearth Apr 2 '19 at 3:58
  • \$\begingroup\$ The large signal response graphs in the datasheet should show the entire range. \$\endgroup\$ – CL. Apr 2 '19 at 7:01

In general, when driving to max swing the slew rate is current-limited into a rated [pF] load.
( e.g. scope probe and coupling capacitance to ground.)

As long as the offset is in the linear range, it is not affected by DC offset.


Slew rate is limited by two factors whichever is smaller;

The main reason for slew rate is current limit.

  1. \$dV/dt=I_{max}/C_{load}\$ by either + or - current limiting, if asymmetric .

However, we use rise time to estimate bandwidth, but if less than output driver current limit, then it is due to internal compensation capacitor and effective -3dB BW

  1. \$ \dfrac{ΔV}{Δt}= \dfrac{ΔV}{t_r}, ~~~t_r = \dfrac{0.35}{f_{-3dB}}\$

    • for Δt = 10% to 90% of ΔVmax.

Proof of "0.35":

An Op-Amp (OA) or oscilloscope can be modelled as a single-pole RC circuit.

There is a relationship between asymptote \$T=RC \$ , rise time, \$t_r\$ and bandwidth, \$BW=f_{−3dB}\$.

I should point out that for \$RC=T\$, only gives us the linear slope asymptote, but not the slew rate 10~90%.

\$T=~ 0 \text{ to } 64~\text{%} ~Vin = V_{in}(1-e^{-1})\$ for t=RC,

The RC step response is \$\dfrac{V_{out}}{ V_{in}} = (1 – e^{-\frac{t}{RC}})\$ .......... (1)

From the above equation, we find the V ratios ( =10%,90%)

10% = 0.105*T
90% = 2.30 *T

Thus (10% to 90%) \$t_r = ~2.2 RC\$ ........ (1a)
from (2.30-0.105)T = 2.195T and T=RC

For the RC model, \$RC=\dfrac{1}{2\pi \cdot BW }\$ .....(2)

Combining the equations (1a) & (2), we have:

10 to 90% \$t_R= \dfrac{0.35}{BW} = \dfrac{2.2 }{ 2π \cdot BW}\$


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