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I have a circuit where I'm amplifying the output of a piezo transducer with an op-amp. By taking some readings, i've found that the piezo can output fairly high voltages when struck hard (~20V), and so I want to protect the op-amp against high currents.

To that end I thought about including a resister in series with the op-amp's non-inverting input, but I'm unsure on a few things. Namely, my questions are:

  1. What value to use for resistor (R1 below), and how to calculate it?
  2. How much current will be flowing through the op-amp due to the high voltage spike?
  3. What's the maximum current the op-amp can handle on the non-inverting input?

Below I've drawn up a simple schematic to represent my circuit - I've used a 1Ω resitor to represent the load, but in actuality the output signal will be fed into an op-amp comparator and then into a microcontroller subsequently:

schematic

simulate this circuit – Schematic created using CircuitLab

Thanks for your help!

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  • \$\begingroup\$ If your circuit can tolerate it, adding a capacitor across op-amp inputs (in conjunction with R1) will greatly limit the voltage peaks when you strike the piezo. However, it may also attenuate or distort the signal of interest, so you may not be able to do it. Adding a Schottky diode from the non-inverting input to the 5V rail will also likely protect the op-amp. A very small capacitor to GND might still be a good idea, even with the Schottky. Just to limit the voltage rise time so the Schottky has time to turn on. \$\endgroup\$ – mkeith Oct 14 at 23:28
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    \$\begingroup\$ Does what you've drawn really represent the circuit you're using? No feedback? No input biasing? \$\endgroup\$ – brhans Oct 15 at 0:40
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    \$\begingroup\$ ...and 1 Ohm load? A piezo transducer is open circuit to DC, so you need a bias resistor to set the voltage on the op amp input. The maximum resistance value will be determined by the op amp's bias current and acceptable voltage drop across the resistor. \$\endgroup\$ – Bruce Abbott Oct 15 at 3:32
  • \$\begingroup\$ @brhans no... perhaps I oversimplified. There is feedback, but I'm not sure what input biasing means? \$\endgroup\$ – jonny Oct 15 at 10:09
  • \$\begingroup\$ So show us your actual circuit!!! Your over-simplification results in people having to guess and draw potentially incorrect conclusions, as I think Tony may have done below, that you're using the LM358 as a comparator and not as an amplifier - leading to answers which may not help you as much as they could. \$\endgroup\$ – brhans Oct 15 at 13:00
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What value to use for resistor (R1 below), and how to calculate it?

Depends on input bias current, an opamp with very low bias current can have a R1 very high without affecting the measurement, several Mohms. Higher resistor also means higher noise.

How much current will be flowing through the op-amp due to the high voltage spike?

if the opamp has input protection diodes, then I=V/R1. If not, then current is almost zero and the whole voltage is applied to opamp input.

What's the maximum current the op-amp can handle on the non-inverting input?

It's in datasheet. Read absolute maximum ratings.

Use an opamp with ESD protection diodes; low bias current- usually a JFET or CMOS opamp comes into a play, then calculate a minimal resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT:

You won't find a max current for LM358, because it has no protection diode at the inputs.

I have added two diodes, which they should be low leakage diodes, for example BAS116, BAV199, BAS21, BAS45A (covered from light source). The output load resistor can't be 1 ohm, it is too low. Added feedack resistors to set the gain, else it will work as comparator. Added R2, C1 to act as low pass filter (with R1), this will prevent a voltage spike while diode is not conducting due to switching delay.

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  • \$\begingroup\$ Thanks for your quick response! I have a a couple of follow up questions: what would you consider a low bias current? The LM358 has an input bias current of -20uA at 1.4V on the output - is this low? How does the bias current change with respect to the output voltage? Additionally I couldn't find anything regarding current in the absolute maximum ratings part of the datasheet - only values in voltage. Do you have any inkling what value specifically I should be looking for? \$\endgroup\$ – jonny Oct 14 at 23:26
  • \$\begingroup\$ It has -20nA to -250nA not uA (nano, not micro). Still a lot of input bias current, but if you want to stay at low cost, you can use it. I shall edit my question with schematics, soon. \$\endgroup\$ – Marko Buršič Oct 15 at 6:37
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schematic

simulate this circuit – Schematic created using CircuitLab

If it generates more than 30V , consider this simple solution.

Op Amps make slow comparators without gain reduction. So be careful in your specs/choices. You probably want a comparator with a pull up R for open collector/drain types.

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  • \$\begingroup\$ Thanks for your answer! I'm confused as to how diodes in that configuration prevent voltage spikes, though. Would you be able to expand on that slightly? \$\endgroup\$ – jonny Oct 15 at 10:20
  • \$\begingroup\$ back to back diodes reduce differential voltage to 0.7V or so, long after the output has switched. Thus with a comparator output starts depends on offset Vin or positive feedback for small hysteresis and bangs are clipped to +/-0.7V. You can also use diodes to +5V,0 instead. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 15 at 10:42

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