I am trying to create an op-amp amplifier that would work from single 5V supply, and would be able to amplify -100mV to +100mV audio signal to around a 1V peak-peak or so. I've came across this circuit from this article, that could seem to work, but am having trouble calculating the actual values:


simulate this circuit – Schematic created using CircuitLab

From the article I read that R1 and R2 should both be the same and around 42kOhm for 5V power supply. R4 should be R3+(0.5*R1) and thats about it...

So how would I go about actually calculating the capacitor, resistor values needed for a varying frequency signal with maximum frequency at around 20kHz and gain of about 5?

Thank you for helping me!


In the article the author wrote by the ground symbol: "*STAR GROUND". Is it really imporant that I combine all ground trances in the schematic to one point, or can I use a ground plane across the whole circuit?

  • \$\begingroup\$ What is the load on the op-amp output? \$\endgroup\$
    – Andy aka
    Feb 12, 2015 at 19:12
  • \$\begingroup\$ @Andyaka I am not quite sure, this will get connected to an actual audio amplifier. Is there any way I could measure the input impedence of the amplifier? \$\endgroup\$
    – Golaž
    Feb 12, 2015 at 19:16
  • \$\begingroup\$ If it's going to an audio amp it'll be fine. Just wanted to make sure you weren't driving a speaker or headphones. Star grounds are probably best for audio rather than groundplanes but a combo of both (providing you know what you are doing) is better. \$\endgroup\$
    – Andy aka
    Feb 12, 2015 at 19:27
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    \$\begingroup\$ @Andyaka Haven't really read that much about this subject. I guess I'll read up some articles and go with star grounding. Thanks! \$\endgroup\$
    – Golaž
    Feb 12, 2015 at 19:42
  • \$\begingroup\$ Good going. You've effectively created a "virtual earth" at the junction of R1, R2, R3, C2, getting around the single supply rail problem. It may be best to connect the "grounded" end of C3 to that point rather than the power supply ground, to prevent noise in the power supply being injected there. \$\endgroup\$ Feb 13, 2015 at 11:43

2 Answers 2


You seemed to have actually found a reasonable circuit on the internet. I heard there was one out there somewhere.

The equations you cite are overly strict. Instead of just telling you the values, it's better to explain what each part does.

R1 and R2 are a voltage divider to make 1/2 the supply voltage. This will be the DC bias the opamp will operate at. C2 low pass filters the output of that voltage divider. This is to squash glitches, power supply ripple, and other noise on the 5 V supply so they don't end up in your signal. R3 is needed only because C2 is there. If R3 weren't there, C2 would squash your input signal too, not just the noise on the power supply. Ultimately, the right end of R3 is intended to deliver a clean 1/2 supply signal with high impedance. The high impedance is so that it doesn't interfere with your desired signal coming thru C1.

C1 is a DC blocking cap. It decouples the DC level at IN from the DC level the opamp is biased at.

R4 and R5 form a voltage divider from the output back to the negative input. This is the negative feedback path, and the overall circuit gain is the inverse of the voltage divider gain. You want a gain of 10, so the R4-R5 divider should have a gain of 1/10. C3 blocks DC so that the divider only works on your AC signal, not the DC bias point. The divider will pass all DC, so the DC gain from the + input of the opamp to its output will be 1.

C4 is another DC blocking cap, this time decoupling the opamp DC bias level from the output. With the two DC blocking caps (C1, C4), the overall amplifier works on AC and whatever DC biases may be at IN and OUT are irrelevant (within the voltage rating of C1 and C4).

Now for some values. The MCP6022 is a CMOS input opamp, so it has very high input impedance. Even a MΩ is small compared to its input impedance. The other thing to consider is the range of frequencies you want this amplifier to work over. You said the signal is audio, so we'll assume anything below 20 Hz or above 20 kHz is signal you don't care about. In fact, it's a good idea to squash unwanted frequencies.

R1 and R2 only need to be equal to make 1/2 the supply voltage. You mention no special requirement, like battery operation where minimizing current is of high importance. Given that, I'd make R1 and R2 10 kΩ each, although there is large leeway here. If this were battery operated, I'd probably make them 100 kΩ each and not feel bad about it. With R1 and R2 10 kΩ, the output impedance of the divider is 5 kΩ. You don't really want any relevant signal on the output of that divider, so let's start by seeing what capacitance is needed to filter down to 20 Hz. 1.6 µF. The common value of 2 µF would be fine. Higher works too, except that if you go too high, the startup time becomes significant on a human scale. For example, 10 µF would work to filter noise nicely. It has a 500 ms time constant with the 5 kΩ impedance, so would take a few seconds to stabilize after being turned on.

R3 should be larger than the output of R1-R2, which is 5 kΩ. I'd pick a few 100 kΩ at least. The input impedance of the opamp is high, so lets use 1 MΩ.

C1 with R3 form a high pass filter that needs to pass at least 20 Hz. The impedance seen looking into the right end of R3 is a bit over 1 MΩ. 20 Hz with 1 MΩ requires 8 nF, so 10 nF it is. This is a place you don't want to use a ceramic cap, so lower values are quite useful. A mylar cap, for example, would be good here and 10 nF is within the available range.

Again, the overall impedance of the R4-R5 divider doesn't matter much, so lets arbitrarily set R4 to 100 kΩ and work out the other values from there. R5 must be R4/9 for a overall amplifier gain of 10, so 11 kΩ works out. C3 and R5 form a filter that has to roll off at 20 Hz or below. C3 must be 720 nF or more, so 1 µF.

Note one issue with this topology. Frequency-wise, C3 is acting with R5, but the DC level that C3 will eventually stabilize at is filtered by R4+R5 and C3. That is a filter at 1.4 Hz, which means this circuit will take a few seconds to stabilize after power is applied.

C4 forms a high pass filter with whatever impedance will be connected to OUT. Since you may not know, you want to make it reasonably large. Let's pick 10 µF since that's readily available. That rolls off at 20 Hz with 8 kΩ. This amp will therefore function as specified as long as OUT is not loaded with less than 8 kΩ.

  • 1
    \$\begingroup\$ Wow, I've learned so much from reading your response, thanks! But I still have some questions: Why does R3 has to be larger that R1-R2? And why is a ceramic capacitor not OK to use as input decoupling ? \$\endgroup\$
    – Golaž
    Feb 12, 2015 at 18:41
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    \$\begingroup\$ "You seemed to have actually found a reasonable circuit on the internet." :) Good one. \$\endgroup\$ Feb 12, 2015 at 18:53
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    \$\begingroup\$ @Gola: A resistor divider isn't perfect. It's output will change depending on how it is loaded. By making R3 large compared to the parallel combination of R1 and R2, loading effects are minimized. Note that this is really to keep the frequency response flat of the desired 20 Hz to 20 kHz range. Due to C1, the divider can't be loaded with DC. \$\endgroup\$ Feb 12, 2015 at 19:53
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    \$\begingroup\$ If you wanted unipolar output (let's say, to drive an ADC that expected 0-5V input) would you just leave off C4? \$\endgroup\$ Feb 12, 2015 at 22:43
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    \$\begingroup\$ @Russ: Yes. --- \$\endgroup\$ Feb 12, 2015 at 22:49

There is no need to make R4 that particular value since this is a CMOS op-amp (no matching of input bias currents).

Capacitor values are determined by the desired lower corner frequency. C = \$\frac{1}{2\pi f_c R}\$

So if fc = 20Hz, and R1, R2 = 39K. Let's arbitrarily make R3 100K. Then C = 100nF is about right.

C2 depends on what's on the power supply that you're trying to attenuate, but let's say 1uF for that.

Let's pick R4 = 100K just to keep two resistors the same. R5 will then be 11K for a gain of +10.1

Finally, C3 can be calculated from R5 to be about 1uF (using the above equation).

That's it!


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