I want to implement typical guitar preamp scheme, like following: enter image description here

I see, that it will have input impendance =2.2M1.1M and Gain = 1 + R4/R3 = 3.136. So its easy to find R4, knowing R3, and vice versa. But selecting both actual values R3 and R4 is the impossible task.

What is the difference between R4=4.7k; R3=2.2k and R4=47k; R3=22k? AFAIK, no one does ever questioned this topic in details, so I think this is a good question.

How to properly select order of magnitude for those two resistors in feedback circuit?

Upd. I know basic theory how OA works.

  • OA is a comparator, sticking its output to V+ or V-, dependent on "if In+ greater than In- or less", accordingly.
  • These two R3+R4 makes a voltage divider between output and a ground, for negative input terminal. And output is is working on R3+R4 circuit, establishing some level at negative.
  • First plus second facts leads to behaviour, that OP amp is trying to euqalize voltage at its inputs: to make" negative terminal's voltage equal to positive's.
  • And divider's ratio determines gain (how much more voltage should OA pop out of its output to equalize negative input to positive)
  • Inputs does not virtually draw any current to OA, any output current goes from power rails.

What i dont understand is reasoning while choosing magnitude of R3+R4 sum. According to ohms law, total divider resistance will determine value of unwanted current(and power draw) that goes to ground. To save battery I better put there higher resistors, like 2M2 and 4M7, is that correct? Or maybe should I set this current exactly to 3.1415 mA? Or 2.71828mA? I know that humbacker coil DC resistance is 2-3 kOhms, and line level input voltage (what i want to get at output) at next stage ranges from 0.3v for consumer to 2v for profesional hardware. But nowhere current is mentioned: at next stage (the mixer) current is also not coming into op amp.

  • 1
    \$\begingroup\$ Q1: Why do you imagine the input impedance is \$2.2\:\text{M}\Omega\$? (Curious how you arrived at that figure and I stopped reading further as the answer you provide to this may affect how I read the rest.) \$\endgroup\$
    – jonk
    Commented Jan 2, 2019 at 20:32
  • \$\begingroup\$ @jonk, Because I've accidentally messed it with dual-supply scheme, which I would use (series of several li-ion batteries with a balancer charger and a midpoint) \$\endgroup\$
    – xakepp35
    Commented Jan 3, 2019 at 0:50

1 Answer 1


First, the input impedance is 1.1M because the two input biasing resistors are effectively in parallel (the battery impedance is virtually zero ohms).

Beyond that, you are correct - there are many feedback resistor combinations that will yield the same gain, and there is no perfect value set. Some thoughts:

The higher the resistance values, the more susceptible the circuit is to noise pickup.

The lower the resistor values, the harder the opamp output stage has to work to drive them; this increases distortion.

The lower the resistor values, the larger C2 must be for the same cutoff frequency.

Back to R1 and R2 - the frequency response profile of an inductive guitar pickup changes depending on the load resistance it sees. (The pickup is basically a small AC power generator). The datasheet on your pickup should have information on this.

  • 1
    \$\begingroup\$ Higher resistance values don't just make the circuit susceptible to noise pickup. They generate voltage noise by themselves. That resistance combination generates about 22\$\mu\$V RMS of noise over a 20kHz bandwidth, if I'm getting my numbers right. That's about six times less than the quantization noise of a 16-bit ADC scaled for a 9V swing. So it's probably in "don't worry now, but don't make it much worse" territory. \$\endgroup\$
    – TimWescott
    Commented Jan 2, 2019 at 20:49
  • \$\begingroup\$ Tim makes a very good point. A low-noise JFET input opamp, like this one, has significantly less RMS noise voltage equivalent than these two resistors provide. (Though I'm not getting exactly the same result from the combined Johnson noise (\$\sqrt{4 \, k T \, B \, R}\$) voltages of both input resistors. [The \$\frac{k T}{C}\$ noise is negligible.]) \$\endgroup\$
    – jonk
    Commented Jan 2, 2019 at 21:11
  • \$\begingroup\$ Yes, 1.1M, sorry, I've messed it with dual-supply scheme. 1M is a typical advice for average pickup, I think I will just stick this value. As per R3 and R4 - I often see like 100k resistor for a mixer input op amp. Should I make R3+R4==100k, in order to "match preamp output and mixer input impendances"? Really dont quite get what does mean to "match impendances". \$\endgroup\$
    – xakepp35
    Commented Jan 3, 2019 at 0:46
  • \$\begingroup\$ @TimWescott So do you think this cannot be used with studio 24-bit ADC, as resistor noise would be 40 times greater than quantisation noise? \$\endgroup\$
    – xakepp35
    Commented Jan 3, 2019 at 0:56
  • 1
    \$\begingroup\$ Many golden-eared audio nutbags swear by the Burr Brown (now part of TI) OPA134/2134. It is supposed to have a less harsh "feel" than the first true audio opamp, NE5534/5532. \$\endgroup\$
    – AnalogKid
    Commented Jan 6, 2019 at 4:09

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