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so my question is pretty straightforward: what are the ways in which this circuit is likely to fail, and how can I protect against that?

I play music (mandolin and viola) and my instruments have piezoelectric pickups, which sound kind of crap connected directly to a mixer.

I've built this buffer circuit (see image) and it works - my instruments sound significantly better. I intend to build it into a module for my instruments for use while performing live.

It is my impression that, because this circuit is so basic there shouldn't be much that can go wrong. Feel free to disabuse me of this notion - but please explain why as I wish to get a clear understanding of what I'm missing.

The circuit I've come up with, which works

EDIT 1: I also attempted to measure the current draw, but it's so low (well under 10mA) that my cheap multimeter couldn't measure it, and the 5V USB power bank I was using to power the circuit kept powering down. For this reason, I presume that the voltage levels will generally remain stable during use ... is that a safe assumption?

EDIT 2: Updated circuit design based on suggestions from @richardthespacecat and @analogsystemsrf

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Or buffering the reference with say another opamp. \$\endgroup\$ – Richard the Spacecat May 1 '18 at 3:28
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    \$\begingroup\$ Try @RichardtheSpacecat's suggestion, then; buffer the reference with another op amp. (many op amps come in dual packages, two amplifiers in a single package, so you wouldn't be adding to the size.) Explaining why exactly it's not a good idea to make a ground reference with a voltage divider would take more awakeness than I have at the moment, though. \$\endgroup\$ – Hearth May 1 '18 at 3:50
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    \$\begingroup\$ Or place a large capacitor on the 4.5 volt node, so RFI will not cause high-frequency common-mode problems. \$\endgroup\$ – analogsystemsrf May 1 '18 at 4:00
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    \$\begingroup\$ Basically yes! Just follow the midrail divider, and use the follower's output as your reference. @Felthry has already suggested using a dual opamp, which I second. It may also help adding some capacitor(s) from the divider's output to ground (and perhaps to the 9V as well). \$\endgroup\$ – Richard the Spacecat May 1 '18 at 4:10
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    \$\begingroup\$ C1 is unnecessary, but won't cause any problems. Other than that, it looks like that'll work fine. \$\endgroup\$ – Hearth May 1 '18 at 12:43
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's Figure 2 has a few odities.

I have a few concerns about this circuit.

  • Using the op-amp output as ground is unusual. Normally the 0 V line would be considered ground. This simplifies power supply, debug and, in guitar effects units, for example, allows the battery to be connected into circuit by insertion of the jack plug.
  • The piezo pickup will have an extremely high input impedance. That means there is nowhere for IC1a's bias current to go. That means that the bias current will drive the + input to one supply rail or the other.

schematic

simulate this circuit

Figure 2. A more traditional approach.

The normal way to do this is to use the mid-supply as a bias for the operating point of the amplifiers. Since you have chosen a non-inverting arrangement then you need to apply the bias as shown via R3.

C2 and C3 decouple the input and output and prevent the DC being affected by external connections.

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  • \$\begingroup\$ A few technical questions. 1) why is it frowned upon to use the voltage divider as a reference ground (buffered or not)? I always thought voltages were relative ... and my original circuit works, as built in the first image. 2) Could you tell me what is a good value/type of cap to use for C2 and C3 that will have minimal effect on sound quality? 3) it was my understanding that coupling capacitors are only necessary for amplifiers, or for multiple stages of circuitry. Is there significant danger in omitting the caps for this single stage low power build? \$\endgroup\$ – Boloar May 2 '18 at 2:03
  • \$\begingroup\$ 4) In my Figure 2 design, I thought that op amps can both source and sink current? So I'd have assumed that bias current could equalize between the outputs of IC1a and IC1b? To be clear, I'm not questioning your design at all, just making sure I understand what I'm doing wrong. \$\endgroup\$ – Boloar May 2 '18 at 2:21

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