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I was fiddling around in LTSpice for a project I'm working on.

This circuit is a basic preamplifier and a low-pass filter. But this question refers solely to the amplifier stage. The opamp is the "universal opamp2" model.

preamp and filter

However, the resulting waveform has a 4.5V DC bias. Since I intend to run this from a single-supply source I was planning on biasing the input signal at 4.5V so the opamp can swing the signal to 0 and 9V. But I hadn't added the bias yet.

Without C3 there is no bias and the output waveform is as expected. If I add a, say, 1M resistor from the noninverting input to ground, creating a high-pass filter, the bias is gone.

The C3 is supposed to remove any DC component from the input signal.

I figured this was just SPICE being stupid, but is this something that would be expected in practice?

output waveform

Nodes:

  • V[n004]: Input signal (green)
  • V[n005]: Signal at non-inverting input (cyan)
  • V[n006]: Signal at opamp output (red)
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  • \$\begingroup\$ Try a 100M resistor from Vin+ to Gnd - you'll probably find that removes the "bias" too. That shows it's an artifact of simulation - in a real circuit the "bis" could be anything and continuously varying until you provide a DC path (to 4.5V) \$\endgroup\$
    – user16324
    May 20, 2015 at 9:25
  • \$\begingroup\$ You have a 10 or 20 or 30 or 50 or 500 MegOhm resistor from non inverting input to ground, and to supply, and to various other pins - and maybe 1 femtofarad to 5 picofarad coupling to your body,m depending where you are relative to it, and ... . If you don't want all these unintended bits and pieces to influence the outcome then adding a low enough to be significant bias path from inverting input to a point of choice would be a very good idea. Picture: A 100 femtofarad capacitor - YMMV \$\endgroup\$
    – Russell McMahon ♦
    May 20, 2015 at 9:50
  • \$\begingroup\$ Thanks for your answers. Could I do this by connecting the 4.5V reference through a "large enough" resistor to the non-inverting input or would it be advisable to just form a voltage divider at the input with two resistors of the same value? \$\endgroup\$
    – dingari
    May 20, 2015 at 9:57
  • \$\begingroup\$ @BrianDrummond 100M is excessive; on the order of the PCB impedance, quite likely! 1M would be more than sufficient. \$\endgroup\$
    – Nick Johnson
    May 20, 2015 at 10:10
  • \$\begingroup\$ @Nick - absolutely! It's to demonstrate the unrealistic nature of simulation. On the board, either 1Meg to 4.5V or (say) 2Meg to each of 0V and 9V would be reasonable. \$\endgroup\$
    – user16324
    May 20, 2015 at 11:05

1 Answer 1

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Of course it is "biased": you cannot leave amp input "hanging" in the air. In real life your schematic will potentially destroy itself and everything connected to it because opamps usually have very large input impedance and will pick up static electricity from the air, resulting in complete malfunctioning of the device. You have to have the resistor between the input and ground.

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  • \$\begingroup\$ I've purposelessly left op amp pins unconnected in the past. It wont destroy itself that easily and unless Vcc is huge it wont really damage other components connected to the output as they mostly cant amplify past their supply voltage. \$\endgroup\$
    – Password
    May 20, 2015 at 12:28
  • \$\begingroup\$ @Password I do not know what was your purpose (cannot think of any), but the issue is not only about having high voltage going through other components but also about having large DC offset (because opamp goes "nuts" and the output voltage is completely undefined) at the output with an inductive load connected. I once burned an expensive speaker by not having a resistor at the input of my amplifier (based on an opamp). \$\endgroup\$
    – ilkhd
    May 21, 2015 at 3:25

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