After looking over this question and the answers received, it prompted me to wonder:

Given pulse signals generated by a uC, what kind of slew rate on an op-amp will you need to reproduce the waveform cleanly?

Example: Assume you have a project where you generate pulses 10uS in length on a uC pin that go into an op-amp in a buffer configuration. What is the minimum slew rate required to reproduce the waveform at the output of the buffer without introducing an integrating effect? What is a good rule of thumb to use?


2 Answers 2


You need to specify more information. This depends on

  • signal consumer requirements
    • what bandwidth is needed?
    • is error at a particular sampling point important?
    • is this being used by a digital or analog circuit?
    • is linearity important?
  • signal producer capabilities / signal properties
    • what is power supply voltage?
    • what is source slew rate?
    • what is source waveform timing?

If I just want to look on a scope, a 1us slew rate for a 10us pulse might be fine. If I'm sampling at particular instants in time, maybe the error during a particular sample might be important.

If your consumer is a digital circuit, you should be using a digital buffer, not an op-amp.

Also I challenge your problem statement:

Given pulse signals generated by a uC, what kind of slew rate on an op-amp will you need to reproduce the waveform cleanly?

Microcontroller outputs do not have "clean" voltage levels, so it really doesn't make sense to try to buffer their output waveform "cleanly".

If you need to maintain precise timing and particular reference voltage levels rather than the somewhat-indeterminate "Vol" and "Voh" output voltage levels of a microcontroller, which can drift, bounce, etc. as long as they stay within spec, then I would highly recommend using a fast analog SPDT multiplexer. (FSA4157 / NLAS4157 are really good choices) Hook the A/B inputs to stiff reference voltages of your choice, and hook the switching input to your microcontroller output. The multiplexer output is your analog output signal.

I can't think of any good reason whatsoever to spend money on a fast op-amp to buffer a cruddy microcontroller output signal.

  • \$\begingroup\$ Thanks for the response, wasn't trying to be confusing. I wrote the above question using as much as I understood about the problem, so after reading your response, it sounds like I really don't know about what I'm talking about and I have a lot more to learn. \$\endgroup\$
    – J. Polfer
    Dec 7, 2009 at 18:25
  • 1
    \$\begingroup\$ @sheepsimulator, I have run into a problem like this, as I only have a moment, look into schmitt trigger chips to buffer. The major thing to remember is a slew rate is often affected by the load on the device, if you want to drive 1A you will find the slew rate is going to be quite different than 1uA of current. \$\endgroup\$
    – Kortuk
    Dec 7, 2009 at 19:14
  • \$\begingroup\$ yeah, load capacitance (cables, MOSFET gates, etc.) really mattters. \$\endgroup\$
    – Jason S
    Dec 7, 2009 at 20:37
  • \$\begingroup\$ (Δv/Δt = I / C) \$\endgroup\$
    – Jason S
    Dec 7, 2009 at 20:38
  • \$\begingroup\$ @sheepsimulator, one more thing, if you are just trying to at a simple level make sure everything will work, compare your uC slew rate to your buffer slew rate. if the uC slew rate is slower, you will not notice a difference. \$\endgroup\$
    – Kortuk
    Dec 8, 2009 at 2:31

There's no exact answer to this; it all depends on what you think is acceptable. Take for instance the 10\$\mu\$s pulse. The opamp will produce less than perfect edges, so the 10\$\mu\$s high may become 9\$\mu\$s, or 8. What's it worth to you?

One thing worth noting is that opamps in general are pretty bad at reproducing steep edges in a signal. That's because (most) opamps have internal compensation to prevent them from oscillating. Slew rates may be as low as 0.5V/\$\mu\$s for the 741. This means that your 10\$\mu\$s square pulse at a 5V output level will be reduced to a triangle. More modern opamps may have slew rates of a few V/\$\mu\$s. At 5V/\$\mu\$s the 10\$\mu\$s pulse will be reduced to 9\$\mu\$s. It's up to you to decide whether that's acceptable. If it isn't there's a better solution.

Comparators are much like opamps, but don't need the compensation because their amplification is much lower. This makes them much faster than opamps.

enter image description here

The graph illustrates the speed of high-speed TLV3501, a whopping 2000V/\$\mu\$s, but even a standard LM339 comparator achieves more than 100V/\$\mu\$s. The 10\$\mu\$s pulse will then still be 9.95\$\mu\$s wide. (Rise and fall delays may be different, especially for low overdrive voltages, so that's something to take into account.)

Further reading
"Amplifiers as comparators?"


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