I'm building a guitar tuner that can be used with a (piezo) microphone clip for acoustic, and from the pickup for electric. In my experiments they have comparable signal levels.

However, the piezo has a very small capacitance and large resistance. So the input circuit needs to have a high input impedance.

So far I've been using a good old non-inverting opamp with a 100k biasing resistor to ground. This seems to work okay, but then I found this collection of opamp circuits that includes a "Amplifier for Piezoelectric Transducer" that I don't really understand. What is going on here? How do I go about calculating the gain and impedance of this circuit? (and what is the 30pF cap doing?)

Amplifier for Piezoelectric Transducer

I've also found this other question about This guitar piezo opamp buffer lacks any sort of bass, why? which has the common problem of having too low input impedance. Their suggestion seems to be to just increase the biasing resistor of the classic circuit, but in my simulation this only goes so far before the input current of the opamp starts to create a noticeable offset.

  • 1
    \$\begingroup\$ Specifically, what op amp are you trying to use. Yes, it makes a difference. \$\endgroup\$ – Scott Seidman Jul 27 '17 at 16:29
  • \$\begingroup\$ Fair point. I've used LM358 and MCP6002 pretty much interchangeably and in LTspice I picked something at random because those were not available. My mouse seems to have landed on the LT1413. \$\endgroup\$ – Pepijn Jul 27 '17 at 16:45
  • 1
    \$\begingroup\$ You can immediately see its gain is 1. Then C1 couples its output back to the biassing resistor (R1+R2) to bootstrap the input impedance to a very high value. \$\endgroup\$ – Brian Drummond Jul 27 '17 at 17:44
  • 1
    \$\begingroup\$ You need an op amp with very high input impedance to deal well with low freqs with a piezoelement. the MPC6002 has this -- but the other amps you mention don't seem to spec it. the 358 has bipolar transistor inputs, so I suspect Rin is fairly low. \$\endgroup\$ – Scott Seidman Jul 27 '17 at 19:09

The LM108 is an old part, and the 30 pf capacitor is required for stability, especially at low gains, and as @Brian Drummond states this has a gain of 1. A more modern op amp will probably not require external compensation.

In answer to your question, the circuit's response is almost independent of the piezo characteristics in this circuit. This is because when the piezo voltage increases, the output tracks it and the voltage is pulled up to equal the piezo voltage by C1. Since the voltage is the same on both sides of R2, there is no current discharging the piezo device. The frequency response of this circuit is therefore determined not by the piezo characteristic, but by the RC time constant of C1 and R1. You may even want to make this time constant a little faster - right now it is 110 seconds (less than 0.01 Hz), way below any bass requirement. Also, be careful with layout because the impedance is so high that the circuit will pick up any hum.

| improve this answer | |
  • \$\begingroup\$ Can the gain be increased without ruining the circuit? It seems you'd have to be extremely careful to keep the positive feedback at unity. Any advice regarding layout is highly appreciated. The current revision is terrible in that regard. \$\endgroup\$ – Pepijn Jul 28 '17 at 7:15
  • \$\begingroup\$ I might stick with unity gain and add a second stage for amplification. As far as layout, keep all of the traces short and make sure the parts and grounds are close together, and that the ground of this part is not in any current path. Keep the line to the piezo short and I would use a twisted pair if you have to run it any distance. Also use decoupling caps from the supplies to ground near the op amp. \$\endgroup\$ – John Birckhead Jul 28 '17 at 11:56
  • \$\begingroup\$ Is there any reason in particular why 22MOhm was chosen, and why the other ones are half that? I came across something before about balancing resistor values to minimise offset, I think. Is the 22MOhm arbitrary, or related to the impedance of the opamp? \$\endgroup\$ – Pepijn Jul 29 '17 at 7:42
  • \$\begingroup\$ If you look in op amp specs there is a "bias current" number which is the amount of current flowing into or out of the positive and negative input. If you make the DC impedance at each input equal, the DC (IR) voltage associated with the bias current is also equal and so cancels. This reduces the DC offset at the output. Once again, this is less of a problem on newer op amps but it is good practice. \$\endgroup\$ – John Birckhead Jul 29 '17 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.