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I am trying to design an op-amp integrator for a Rogowski Coil based current sensor. The op amp I use is THS4631D from Texas Instruments. It has a slew rate of 1000 V/us according to the datasheet. However when I am using it as an integrator the output voltage slews at a much lesser dv/dt and distorts the current waveform. I also see that the current measured at the output terminal hits the current limit of 95 mA. Does the integration capacitor affect the slew rate of an op-amp in any way? If yes, what might be a solution?

Spice ModelOnly SlewAll waveforms

I am also adding the response with 1 nF capacitor and 200 ohm resistance. I do not hit the current limit as rightly pointed out by @Andy aka. But the waveform is still distorted at the top edge where the transition appears sharper.

proper slew

A response with 0.1 nF capacitor and 2k resistor makes the response even sharper. How does making the integration capacitor smaller affect this transition?

smaller_cap

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2 Answers 2

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Because of the nature of op-amps and their use of a virtual earth, inevitably, when you place a capacitor from output to inverting input, the capacitor itself becomes a load on the op-amp output. This can seriously affect the performance of the op-amp and particular so at higher frequencies (because the capacitive impedance drops proportionately with frequency rising).

In other words, you have to be very careful to avoid eroding the naturally good slew rate of the unloaded op-amp. You have used a 10 nF integrating capacitor and, I'm not surprised that you have hit the op-amp output current limit. Maybe try reducing that capacitor to 1 nF and increasing R2 from 20 Ω to 200 Ω: -

enter image description here

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  • \$\begingroup\$ I made the capacitor smaller and it does not hit the current limit, but the waveform now has really fast transitions. How does a smaller capacitor cause this as the integration time constant has not yet changed? I have modified the question and added the waveforms to reflect the same. \$\endgroup\$ Commented Feb 8, 2022 at 14:00
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    \$\begingroup\$ @SambeetPanigrahi maybe you are now seeing the true integrated waveform without excessive slew rate limiting due to the capacitive loading on the output being reduced by a factor of ten. \$\endgroup\$
    – Andy aka
    Commented Feb 8, 2022 at 14:08
  • \$\begingroup\$ The idea was that for higher frequencies the integration happens with the R3 and C3, whereas the C2 effectively is effectively shorted but for lower frequencies C3 is open and integration happens via R3 and C2. All this was done to create a non-inverting integrator. So are you saying the sharper transitions happen due to non-ideal integration by the R3 and C3? \$\endgroup\$ Commented Feb 8, 2022 at 14:11
  • \$\begingroup\$ I don't understand what you mean regards your explanation of the capacitors and resistors. A non-inverting integrator has a specific circuit configuration as per this for example and your explanation didn't appear to fit that. \$\endgroup\$
    – Andy aka
    Commented Feb 8, 2022 at 14:23
  • \$\begingroup\$ I have been using the same circuit as per the example you mentioned. The question is, the opamp is not hitting the slew rate after I used a smaller capacitor but the waveform with 1nF and 200ohm or 0.1nF and 2kohm (see edits in the question) still looks like the top is clipped. Why does that happen even if I am not hitting the current limit now? \$\endgroup\$ Commented Feb 9, 2022 at 5:28
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Does the integration capacitor affect the slew rate of an op-amp in any way

No - the slew rate is caused by an internal effect within the opamp. And - of course, it remains valid also for integrating circuits. It remains valid for ALL applications.

However, as mentioned already earlier, it is another question if the opamp can drive a pretty large capacitor at its output. But this has nothing to do with the parameter we call "large signal slew rate".

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  • \$\begingroup\$ However, if we drive a large capacitor, there is less current available for charging the compensation capacitor, and as we hit the output current limit we can still have a constant current passing through it but at a lower value than that of an unloaded op-amp. Hence though not directly related, driving heavy loads can lead to opamp slewing less than their datasheet value. Is this explanation correct? \$\endgroup\$ Commented Feb 9, 2022 at 5:37
  • \$\begingroup\$ Thank you for this comment - but I am afraid, the explanation is not correct. The compenation cap is charged by the input voltage which is responsible for overdriving the 1st stage - until the feedback comes into play and brings the device back to the linear operation. \$\endgroup\$
    – LvW
    Commented Feb 9, 2022 at 9:00
  • \$\begingroup\$ Why then do I see a constant dv/dt when the output current hits the limit as shown by waveforms constant slew output and current limit. But when I decrease the feedback cap, and hence the capacitive loading on the op-amp, I could get more dv/dt in the output without a lot of slew. \$\endgroup\$ Commented Feb 9, 2022 at 10:28
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    \$\begingroup\$ Yes - as mentioned earlier, there are two effects overlapping each other: The "classical" slew rate (responsible is the 1st stage) as well as the maximum output current capabilities of the last opamp stage. \$\endgroup\$
    – LvW
    Commented Feb 9, 2022 at 15:43

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