I have piezo sensor which outputs around 1mV of signal. I am using an already build analog front end to measure the signal, with Vcc = 5V and Vss = 0V. To measure the signal, the front end has internal PGA and DAC and needs the input signals to be within range of 2.3V to 2.7V. I don't want to use any op-amps because I want to minimize the components and no additional PCBs.

I have thought something like this:

enter image description here

If R1 = R2, will the signal be centered to 2.5V? Are there any problems with this circuit? Or any other solutions?

  • \$\begingroup\$ what is your Analog Front End (AFE) \$\endgroup\$ Commented Jun 19, 2015 at 9:44

1 Answer 1


The piezo element is capacitive in nature having a very high DC resistance. The effect of R1 and R2 are therefore insignificant. Use two potential dividers connected to either side of the piezo element and that should work.

Use high value resistors (> 1M ohm) so that the signal is not affected. However, you might find that high value resistors create an offset problem with the monitoring circuit you are using; bias currents from the circuit pins may not be equal and you end up with a small DC offset. This may be acceptable but, as I don't know the application (or the device) I can't tell.

  • \$\begingroup\$ Thanks. What if I use only one potential divider and connect one of the terminal of the sensor to the divider. And I measure the signal difference between that mid-point and the other end of the sensor. (schematics.com/editor/piezo-19870) Will it work? \$\endgroup\$
    – zud
    Commented Jun 19, 2015 at 9:32
  • \$\begingroup\$ The other end will not be biased - the piezo is capacitive and does not pass thru it any appreciable midpoint bias voltage. You need two dividers or 1 divider and a resistor across the piezo. \$\endgroup\$
    – Andy aka
    Commented Jun 19, 2015 at 10:36

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