I'm currently studying the concept of slew rate and everything I read or watch I don't fully understand. I was hoping that someone could explain it like you were teaching a child. Another question I had is for op amp design: Do we generally want a high or low slew rate? and why? (I know this is a basic concept but for some reason I cannot wrap my head around it).


2 Answers 2


OK, I'll give it a try. Inside an op-amp there's a compensation capacitor, typically on the stage after the differential pair.

This capacitor is charged and discharged by a current source. This current source has a finite amount of current available to charge and discharge the cap. I.e. it saturates at some maximum current value.

When that saturation happens due to a large fast-changing signal at the input of the amplifier, the amplifier no longer behaves linearly but will "slew" at a rate defined by the compensation cap and the max available charge/discharge current. dv/dt = I/C.

This means that for example when a large sine wave causes slew rate limitation or distortion, the sine starts to resemble a triangle due to the constant current charging the cap.

Typically, a higher slew rate is better, but if your signals are such that you don't get into slew rate distortion then a less expensive amp might be the better choice.

Here is a good discussion of slew rate.


Slew rate is one of those things that exists because we don't live in an ideal world. Your early education with electronics could ignore slew rate but now that you're starting to get a little deeper some of the nastiness of "the real world" begins to creep in. Slew rate is one of those things.

If we think of a simple op-amp configuration: we can think of either an open-loop op-amp circuit (which is basically a comparator), or even a simple voltage follower. In the open-loop configuration, the output will move to one of the rails: the positive rail if the + input is more positive than the - input, and the negative rail for the opposite condition. In the voltage follower case, the output will follow the input. If the input conditions change suddenly, how long does it take for the op-amp to change its output to the new level? In ideal world, it's instantaneous. However the op-amp is an electronic device and has limitations; one of these limitations is how quickly it can move its output. The rate of how fast the output can change is called slew rate and is usually measured in V/us some similar scaling thereof. I've personally seen V/us and kV/s, depending on the age of the technology.

The datasheet of an op-amp will usually list the slew rate under its AC or switching characteristics. I quickly looked up an ancient op-amp, the uA741; it's got a unity-gain slew rate of 0.5V/us. This means if (in the voltage follower configuration example) the input moves from 1V to 3V instantaneously, the output will take 4us to reflect that change. If you were to view this on an oscilloscope, the input would look like a single "step" from 1V to 3V, and the output would ramp linearly from 1V to 3V over the course of 4us. If we do the same thing with an OPA855 which has a gain-bandwidth product of 8GHz, its slew rate is 2750V/us; the same input change in the same configuration would show an output that instantaneously changed with the input (up until your switching speed approached 100ps, that is).

What determines the slew rate of an op-amp (or any output driver circuit)? Mainly it's a product of the amount of current the output stage can sink or source into the load, but the internal construction of the driver stage also affects this.

What's better, a fast or slow slew rate? That depends entirely on the application. There are many applications where the slew rate of a driver is intentionally slowed down, particularly in digital systems. This is because the faster you can move from one state to another, the more harmonic content your output waveform will have, and the easier it is for your to radiate, potentially causing EMI or RFI. There are other elements at play here, but if you slow down the slew rate (generally by limiting the instantaneous current available), you will reduce the harmonic content of your output. Also, capacitive loads can cause an op-amp to oscillate, which is usually undesirable.

If you are dealing with high frequency signals, you will need to faithfully reproduce those signals (think of any amplifier application -- you want the output to be louder than the input, but generally not otherwise changed). In order for you to be able to reproduce a signal with a lot of fast changes in it, you need to be able to make your output move quickly, and that requires current. If the thing you're driving has a lot of capacitance (think long wires) then you'll need even more current or you won't be able to charge and discharge the distributed capacitance in those cables quickly enough to faithfully reproduce the input signal.


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