I have seen a few circuits where a resistor is connected across the positive and negative (grounded) terminals of an ultrasonic receiver. It seems to "stabilize" the received signal. I have attached an example of one of these circuits, but I am using a simpler circuit with just a comparator.

What does this resistor do, and how does it work?


Image source

  • 1
    \$\begingroup\$ My guess is that the receiver will generate a small charge, and the resistor is there to remove the resulting bias voltage from the input of the amplifier (which has a high input impedance so the charge would otherwise just stay there). \$\endgroup\$ – 0x6d64 Apr 26 '15 at 6:32
  • 3
    \$\begingroup\$ It might be used to dampen the sharp resonance in pulse applications too. \$\endgroup\$ – Andy aka Apr 26 '15 at 8:47
  • 1
    \$\begingroup\$ As @0x6d64 says, the ultrasonic receiver acts as a capacitor and the resistor to ground helps empty the charge that accumulates in that capacitor. \$\endgroup\$ – m.Alin Apr 26 '15 at 10:43

It's a pull down resistor. It's job is to eliminate small noise, spurious energy that can capacitively develop. Kind of a filter. The value of the resistor matters, as well as the capactance specs for the receiver.

  • \$\begingroup\$ Do you have more information about what resistance value to choose? And how would you model this as a (I assume high-pass) filter using the receiver capacitance specs? (p.s. thanks for your answer) \$\endgroup\$ – Aralox May 1 '15 at 10:05
  • \$\begingroup\$ @Aralox: I would start with 100K for the resistor-- it needs to drain off residual capacitive current but not affect valid signal from the piezo/crystal. \$\endgroup\$ – GSLI May 1 '15 at 13:55
  • \$\begingroup\$ @Aralox- The value of the resistor is chosen to bleed off current, without upsetting the voltage level of anything else, so the resistor value can be high. It works in conjunction with the capacitor. \$\endgroup\$ – GSLI Jul 31 '19 at 22:49
  • \$\begingroup\$ Oh thank you for your comment, I forgot to mark this as the answer ages ago, sorry! \$\endgroup\$ – Aralox Aug 7 '19 at 9:03

The resistor does affect Q and hence affects resonant frequency slightly. leave the specified resister in while developing circuits.It could be enough to cause a matched transmitter/reciever pair to become mismatched costing sensitivity.


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.