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I am creating the following diode clipping circuit:

**enter image description here**

The input signal, Vin, offset is removed and the signal is amplified using an op-amp. Then there are two diodes designed to clip the signal to +/- 0.7V. The clipped signal then goes to another op-amp, R4, where a gain and an offset are applied (not showed).

The diodes I have chosen are MBD701 and have a forward voltage of 0.7V at 10mA. The op-amp used is an AD4177 and is capable of outputting 25mA. I was wondering if I can use R3 to control the current across each Diode? My understanding is if the current across R3 is equal to 15mA and the op-amp outputs the max current of -25mA since the impedance of the diode is low in the forward direction, then -10mA will then leave the R3 and output node. If R4 impedance is very large, I can assume that almost all of the -10mA current will go across the diode. Is this logic valid or are the OP amps capable of outputting greater current than on the datasheet? Also is there a way to calculate the impedance of a diode?

Thank you

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You can add a series resistor between the op-amp output and the diodes to limit the current. Something like 470 ohms will limit the current to less than 10mA. Take the feedback from the output rather than across the diodes or the op-amp will saturate when the diodes conduct.

The short circuit current of the op-amp ADA4177 is typically almost 50mA and can therefore be considerably more in specific cases.

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  • \$\begingroup\$ Thanks for clarifying a few things I was uncertain about. When adding the resistor in series how would I go about calculating the voltage drop across it? I am a student and haven't been able to find information on modeling the diodes for this scenario. Is modeling them as a voltage source and resistor valid for this case? Prior to the clipping circuit the voltage needs to be greater than 0.7V. Thanks \$\endgroup\$ – willindsey Apr 11 '19 at 12:26
  • \$\begingroup\$ Is this valid: I know the output at the op amp, I know the voltage at the diodes, and want a current equal to 10mA. So then is R = (Vopamp-Vdiode)/ currrent ? and the voltage drop across the resistor is just Vopamp-Vdiode. \$\endgroup\$ – willindsey Apr 11 '19 at 12:42
  • \$\begingroup\$ If the diodes were ideal they would have zero current at 0.6999V and any current you want at 0.70000V. So yes, your calculation is correct for the maximum voltage and current. In general the current will be very little when the output voltage is < 0.7V rising up to the maximum at the maximum op-amp output voltage. Real diodes are a bit "softer" than that so the clipping won't be so sharp but that's the idea. \$\endgroup\$ – Spehro Pefhany Apr 11 '19 at 12:47
  • \$\begingroup\$ Note too that the 470 or whatever resistor adds to R4 so you can either reduce R4 by the same amount to compensate or maybe the loading is light and you don't care (when the diodes are clipping the output resistance is R4, before clipping it's R4 + 470. The other circuit suggested by glen will work too (with an added R4) with some subtle differences (less dependency on loading but the clipping voltage will likely be less because you would pick higher feedback resistor values). \$\endgroup\$ – Spehro Pefhany Apr 11 '19 at 12:51
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R3 controls diode current to a very small extent. The opamp will go into current limiting and likely overheat, trying to supply large currents to the diodes when their forward voltage of 0.7V is exceeded.
You might try this arrangement. Diodes can even be LEDs - you may not need a following amplifier. Resistors R1, R2 control diode current: Driving diodes back-to-back

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