Pre-amp circuit with virtual ground

Question

I am working on a pre-amp circuit for my guitar, for testing I am currently utilizing a 12 volt power supply. It is to my knowledge that since this power supply is not bipolar, and the output of the guitar is an AC signal, I need to use a virtual ground, with a voltage half that of the power supply, as a reference point, allowing the output to swing above and below this point and then the signal can be passed through a capacitor to provide a suitable AC output.

I am using a LM324N op-amp, the circuit is completely barebones just for testing. Because of the virtual ground, if the input were tied to 0 volts, then the output should be 6 volts, however, instead the output is 0 volts. And with the input left floating, the output hovers above 6 volts by up to 3 volts. Why is this so and how do I rectify this?

"if the input were tied to 0 volts" ... if you're tying the input to the 0V marked in your circuit, the op-amp calls that -6V. — jonathanjo
– jonathanjo, Commented Feb 25 at 13:53
You would need to tie the input to your virtual ground to get 6V out. — Finbarr
– Finbarr, Commented Feb 25 at 14:10
What is meant with ""virtual GND"? Yes if you tie the input (+) to GND, then your opamp is saturated to GND. — Tyassin
– Tyassin, Commented Feb 25 at 14:12
Your virtual ground is too high-impedance for your feedback network. This is a separate issue from what you're asking about, but it's also going to cause you problems. — Hearth
– Hearth, Commented Feb 25 at 14:21
Using an inverting regulator is also a possibility. You could build it yourself or buy a module. youtube.com/watch?v=r49BwY0eFPA — Hörður Mar Tómasson
– Hörður Mar Tómasson, Commented Feb 26 at 0:15

Kuba hasn't forgotten Monica · Accepted Answer · 2025-02-25 15:41:03Z

No op-amp inputs can be left floating.
For amplification of AC signals, both inputs must be biased to virtual ground.
The virtual ground should be VCC/2 - 1V, because the LM324 input and output voltage range is between 0V and VCC-2V approximately.

For DC applications, the feedback network should be use a dedicated virtual ground, configured to be a Thevenin equivalent with the required source impedance. There's no power supply rejection possible, so the supply should be a clean reference voltage for this purpose.

For AC applications, the Thevenin equivalent network isn't necessary, and a capacitor can be used to provide low impedance for AC signals and to filter out noise from the power supply.

For example, suppose you want the gain of 11 non-inverting, and the feedback resistor network uses 10k and 1k resistors:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

schematic

^{simulate this circuit}

C2 above lowers the impedance of virtual ground for AC signals. It needs to be suitably large for the lowest AC frequency of operation. In this particular case, the equivalent high-pass time constant is \$(R_3||R_4)C_2 = 5k\cdot47\mu\$, equivalent to \$ \approx 4{\,\rm Hz}\$ high-pass corner.

Since you're using LM324, you have an op-amp available to buffer the virtual ground before it's used. The circuit will then look as follows:

schematic

^{simulate this circuit}

R1 and R2 provide a 7V virtual ground voltage.,
R4 and R5 set the gain of OA2,
R6 biases the non-inverting input of OA2 to virtual ground.

Op amps need a DC path on both inputs, so you can't just attach a coupling cap to op-amp input, there needs to be a DC biasing resistor to ground - whether real or virtual.
C2 and C3 AC-couple the input and output of OA2, respectively.

Note that LM324 is unsuitable for use with audio at gains over 11 anyway, even that is a stretch. It doesn't have enough gain-bandwidth product for such high gains, so it'll muddle the sound.

Ideally you'd use it at gain 5 or 6 and no more. The gains of consecutive stages multiply, so you'll need several stages in series. Those stages can be DC coupled to each other for the most part.

So, a gain-of-100 amplifier, like you planned, would use 3 stages, each having a gain of \$\sqrt[3]{100} \approx 4.6 \$.

This would be the schematic:

schematic

^{simulate this circuit}

You will find that all those four op-amps are conveniently provided by in the single LM324 package.

Andy aka · Accepted Answer · 2025-02-25 14:31:25Z

You ought to be doing something more like this: -

Using a 220 kΩ to bias the non-inverting input ensures that you'll get reasonable high frequency response from the very inductive guitar pick-up. The input and output capacitors are there to block DC. The 1 μF across the lower right-hand 10 kΩ keeps noise down on the mid-rail generator.

An improvement would be to use a FET input op-amp (maybe like the good old fashioned LF347) and use something like a 2M2 resistor instead of the 220 kΩ shown above.

Tyassin · Accepted Answer · 2025-02-25 15:38:08Z

This is one suggestion with a single power supply. I have not put in any values. But you add a DC offset in the input and for the output. If you analyse it for DC, you can set a DC value at the input to the opamp. This DC value is NOT amplified by the circuit because of C1. Instead it acts as a voltage follower and you also get the same DC voltage at the output. AC wise you can consider C1, C2 and C3 as shorts and you set the gain with R1 and R2.

A few things, C2 make up a high-filter with R3 and R4. Also be aware of capacitive loading for the opamp with C3.

Simon Fitch · Accepted Answer · 2025-02-25 15:45:11Z

The resistor potential divider, consisting of the two 10kΩ resistors, have exactly the same behaviour as this equivalent, shown right:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This is Thévenin's theorem, which you should study to understand this equivalency.

If I replace the divider in your circuit with this equivalent, you get this:

schematic

^{simulate this circuit}

You expected this gain \$A\$:

$$ A = 1 + \frac{100k\Omega}{1k\Omega} = 101 $$

Hopefully you see why \$R_{TH}\$ and \$R_4\$ in series produce 6kΩ, and so gain will actually be:

$$ A = 1 + \frac{100k\Omega}{6k\Omega} \approx 18 $$

The divider itself would need a Thévenin equivalent resistance of \$R_{TH}=1k\Omega\$. That's easy, just use \$R_1=R_2=2k\Omega\$:

schematic

^{simulate this circuit}

Now you have gain \$A=101\$, but there are still problems.

You expected the signal at \$V_{OUT}\$ to be centered on +6V when the input is \$V_{IN}=0V\$, but that won't happen either. The circuit now considers +6V to be the "center" for both input and output. To this amplifier, an input \$V_{IN}=0V\$ looks very negative with respect to the new +6V centre, and the op-amp would try to produce a very large negative potential. It would fail, of course, because there's no negative supply. That's why you see your output stuck at 0V.

If you tell the op-amp to output a signal centered about +6V, then the input too needs to be centered on +6V. You could adapt the above design to "bias" the input signal about this +6V level, using a capacitor and another resistor divider:

schematic

^{simulate this circuit}

This design uses R5 and R6 to bias the signal at Y, so that it is centered at +6V. C1 "AC couples" IN to Y, bridging the potential gap between \$V_{IN}\$ centered at zero, and \$V_Y\$ centered at +6V. Potential changes (AC) at IN will be copied to Y, but the DC levels of IN and Y can be very different, thanks to C1.

This works, but it relies on very precise resistors. The ratio \$R_1:R_2\$ must exactly equal \$R_5:R_6\$. This is difficult to do, and so we generally stick to the oldest biasing technique in the book (for non-inverting arrangements like this):

schematic

^{simulate this circuit}

We employ the same input biasing technique at the input, using R5, R6 and C1, but we add C2 to do the same kind of DC offset compensation for the output, allowing \$V_{OUT}\$ to have an offset exactly equal to the average potential at Y. You'll notice that I went back to the original 1kΩ for R4, because C2 takes care of the 6V offset, rendering all that Thévenin stuff unnecessary.

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