I want to build guitar active preamp inside my bass guitar. It has 2 humbacker pickups (3.4kOhm DC resistance each) for 2 lower and for 2 upper strings, and i want to preamplify each with a separate preamp, so i choose TL072. Let's look at just one pickup.

I draw following schematics, using non-inverting op-amp design:


I have no oscilloscope and i cannot answer how much voltage does produce the coil, but what i want to get is +4dBu (1.737V) line output. So i cannot even estimate required gain. I just suppose it is around 10db, from the external sources. As it also stated in that document, upper scheme has gain of 20*log(1+R2/R1). But there are infinitely many combinations of R1 and R2 which would give me gain of 10dB. You could calculate one if you know the nominal of another.

My question is - how to estimate nominal of R2? I could not find the good basic explanation for that.

Also, being a novice, i appreciate any comment if my schematics is wrong or i miss something, or maybe improve. More passive components introduce more noise and attenuation. And input is well-known. Could we go with a simplier design, soldering the coil directly to the op-amp inputs, since we want just to ampify the difference between coil ends?

  • 1
    \$\begingroup\$ Have you tried a simple bjt configured in emitter follower? The pickups’ can’t drive the line due to impedance matching, not voltage deficit. You have plenty of voltage in your coils, the problem is that when you plug it in a line interface the low impedance drops your voltage, and that’s why you need amplification. \$\endgroup\$
    – PDuarte
    May 20, 2018 at 16:38
  • \$\begingroup\$ I’d consider this topololy fourier.eng.hmc.edu/e84/lectures/ch4/node9.html \$\endgroup\$
    – PDuarte
    May 20, 2018 at 16:40
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    \$\begingroup\$ To answer the question in your title: a single JFET (J201 for instance) common-source amplifier makes a decent guitar preamp. To get to line level with good distortion specs, you'll want to pick a different op-amp that runs at lower voltages, or much wider supply rails. \$\endgroup\$
    – uint128_t
    May 20, 2018 at 19:06
  • \$\begingroup\$ @PDuarte with voltage i need greater current, but puckip is only 3.2k (DC resistance) so cannot get much of it. you re right \$\endgroup\$
    – user149105
    May 20, 2018 at 20:33

2 Answers 2

  1. Instead of using a split battery, I think a virtual ground at 1/2 the supply would be better. That way you are essentially running from a single power supply. That leaves the possibility open, for example, of using a boost converter to make reasonable voltage, and doesn't end up with two batteries that are used somewhat separately.
  2. A TL072 is not appropriate for ±3.6 V. It's not even specified for 7.2 V operation. Take a look at the headroom it needs, both for input and output, with ±12V supply. Or, put another way, 7.2 V is inappropriate for a TL70x.
  3. In your circuit, C1 actually hurts. There is no need to AC couple, and C1 leaves the positive input floating DC-wise.
  4. With a virtual ground, you would AC couple both the input and the output, but that would be a different circuit.
  • \$\begingroup\$ 1. Li-Ion cell has highest voltage among others known cell types, and we cannot go higher, without connecting cells in series. Boost converter involves efficiency, PWM noise, demand of having PCB, controllers, extra inductor, so on.. and i want to avoid it at all costs. Also dual polarity supply is most natural in amplifying audio ac signals and is always considered better that virtual 1/2 ground. There is a special circuit for batteries, balancing them. Also i can go with 4*li ion cells, but why if i just need ±1.7V output? \$\endgroup\$
    – user149105
    May 20, 2018 at 16:16
  • \$\begingroup\$ 2. many people are running TL07x from 9V crona battery, (which could easily go to 7.2V, as the 6 alkaline cells in series goes to 1.2V each, and crona is 6 series alkaline batteries) resulting same 7.2V operation. \$\endgroup\$
    – user149105
    May 20, 2018 at 16:18
  • \$\begingroup\$ 3. Can i just omit C1 and solder that end directly to op amp? \$\endgroup\$
    – user149105
    May 20, 2018 at 16:21
  • \$\begingroup\$ 4. Lets consider that i cannot have virtual ground and we have ideal ±5V voltage source with real ground, which is not the topic. and a question is right about on how to connect pickup to preamp (without pcb, having least wire lenghts, minimal number of passive components and soldering points). \$\endgroup\$
    – user149105
    May 20, 2018 at 16:28
  • \$\begingroup\$ despite it does not answers neither header, nor question(bold) in body i accept this answer to close it \$\endgroup\$
    – user149105
    May 20, 2018 at 20:54

Let's consider that I cannot have virtual ground and we have ideal ±5V voltage source with real ground

The virtual ground doesn't change much, the circuit's gain resistors will be set exactly the same.

C1 is unnecessary. It provides no DC path to an op-amp input. Most op-amps won't work like that. Replace C1 with a short, so that the op-amp has DC bias on both inputs. The bias through the pickup coil is to 0VDC.

My question is - how to estimate nominal of R2?

It's not your question, because the value of R2 in vacuum is rather meaningless. What the op-amp cares about is that the output is not loaded with too low of an impedance. Since the DC loading the output sees is the series connection of R1+R2, we have a lower limit of about 5kOhm. At higher supply voltages it could go down to 2kOhm, but TL072 from 10V doesn't have all that much headroom, so making its output stage do light work is advantageous.

On the other hand, high feedback network impedances create high impedance nodes at the inputs of the op-amp and are prone to picking up interference. The resistor noise also goes up with R1+R2. You probably don't want R1+R2 to be more then 47k-100k.

So those are your limits for the value of R1+R2. I'd shoot for values on the lower end of this range.

Assuming that you want gain of about 3.2 (i.e. 10dB ratio of output to input), R2 can be 3k3, and R1 then is 1k5, for a gain of 1+2.2. R1+R2=4k8 - close enough to 5kOhm to call it good enough.

R2 needs a small parallel stabilizing capacitor, like 10pF.

To isolate the output from the cable capacitance, there should be a 100-600 Ohm resistor between the op-amp's output pin and the input jack. This resistor must not be inside the feedback path, i.e. R2 must be attached directly to the output as you show.



simulate this circuit – Schematic created using CircuitLab


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