2
\$\begingroup\$

I'm looking to build a preamp for a dynamic microphone (SM57, rated at 150 Ω with -56 dBV/Pa). I've just finished a course in circuits and I'm having a bit of trouble approaching this design problem. My end goal is to use this signal for DIY guitar effect pedals and other analog signal processing.

How would I determine how much gain I would need to drive a chain of guitar pedals, especially if I'm using a dynamic microphone?

As for designing preamps, I can't find many high-quality resources online that actually explain the circuit they use and the stages that should be included.

I think I can use a rail-to-rail op-amp (I've got the TLE2141 on hand) and then some passive filtering to reduce noise and to get a flatter frequency response. Would this yield reasonable results?

\$\endgroup\$
8
  • \$\begingroup\$ This is an advanced task, IMO. My SM57 generates 2-3 mVrms at bedroom level. To make it more sensible to analyse and to feed other equipment the output should be at line level (~1 Vrms). So the amplifier's gain should be at least 50 dB. Also it should have low noise, low THD, high input impedance, wide bandwidth. \$\endgroup\$ Jan 10 at 16:34
  • \$\begingroup\$ It needs to be a balanced differential input too. You also need to specify what DC supply voltage you are going to use. Plenty of things to reveal not least a spec data sheet for the SM57 (or maybe @RohatKılıç has one he can link). \$\endgroup\$
    – Andy aka
    Jan 10 at 16:38
  • 1
    \$\begingroup\$ @AndrewLi hm, a preamplifier isn't "complicated" per se, a high-quality preamp just might. Because signal-processing-wise, the functionality is "boring", if you want to buy something and build something else, the preamp would be what I'd buy, and things like effects, equalizers,… would be what I build for fun. \$\endgroup\$ Jan 10 at 16:45
  • 1
    \$\begingroup\$ in principle yes, as your high-level description doesn't rule out that you have a differential input stage. But it doesn't say that, either ;) In the end, a single TLE2141 with AC coupling and a high-impedance biasing of the result into the middle of your input range will not do wonders, but give you some audio. Don't think you'll achive the full 50 dB gain reliably, though, that would require a lot of open-loop gain. To solve the issue that a single-Opamp difference amplifier has different impedances on inverting and non-inverting input,you could add a unity-gain input buffer to both inputs \$\endgroup\$ Jan 10 at 17:12
  • 1
    \$\begingroup\$ which would turn your circuit to an instrumentation amplifier; that's a pretty common circuit for preamplification of sensitive inputs. Use a fourth opamp (or some other method of producing a stabilized center-supply voltage) to giver you a reference point to which you pull both inputs – with impedances high enough to not distort the input through the C-R highpass filter that forms with the AC coupling capacitors, but low enough to be less sensitive to EM noise; your mic says 150Ω, so 300Ω (or 470Ω) from each branch to that middle pseudoground would be a start. \$\endgroup\$ Jan 10 at 17:15

1 Answer 1

1
\$\begingroup\$

I think this is a good learning experience.

  1. Forget about noise!

For now, don't think so much about noise. It is never a problem. Even the worst opamps have maybe 2uV RMS noise which is tiny compared to audio signal levels. Your enemies are distortion and mains interference. Once you have solved these, you can think about low noise amplifiers and resistors.

Also filtering noise is not a concern. For such a mic, the entire audible range is the signal bandwidth. Of course there will be signal and noise at higher and lower frequencies, but you can't hear these anyway. And in any recording interface, they will be cut out by its anti-aliasing filters.

  1. Low distortion

For most basic op-amps that means an inverting configuration. Another advantage of this scheme is that you get to set the input impedance very deterministically. You can AC-couple the input if you want to use a single supply voltage. In this case, use a capacitor without voltage dependence, e.g. electrolytic or foil. But don't use Class II/III ceramics, because again they can lead to distortion and are themselves also (piezoelectric) microphones.

Finally make sure that the amp has enough slew-rate, or distortion will increase.

A good gain value to start of is ~100, which would require at least 5 MHz gain bandwidth to have some headroom. If you want to use much higher gain towards 1000, it is probably better to cascade two stages.

A basic inverting amplifier doesn't have balanced inputs, but that is not so critical.. the important part is that the input is differential. See below..I suggest that you postpone buffers to a possible second iteration. Because non-inverting buffers might introduce distortion again. Inverting buffers add noise, but less distortion. Omitting buffers will add neither, but might be more susceptible to interference.

  1. Mains interference

This is mainly about two things: The most important one is understanding where signal currents flow and where they shouldn't flow. XLR mics are a good case because they use proper differential signaling to become highly immune to interference pickup. That also means that you shouldn't mess up the cabling.

The other aspect is making sure, that your supply voltage has very little differential voltage in the audio frequency range, e.g. mains harmonics. Ideally, the supply is absolutely stiff. For audio frequencies, this can be either achieved by lots of electrolytic capacitors or by linear regulators. Batteries are also an option.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the answer! I just had a few general questions–why does using an inverting amplifier mean lower distortion (but higher noise)? And what do you mean by differential voltage? \$\endgroup\$
    – Andrew Li
    Jan 11 at 3:43
  • \$\begingroup\$ @AndrewLi Inverting opamp has a constant operation point at its input, while non-inverting hasn't. inverting buffers needs resistors which add noise, while non-inverting doesn't. Differential means that you have two wires that have the signal current and you measure the voltage between them, instead of measuring the voltage between 1 wire and ground \$\endgroup\$
    – tobalt
    Jan 11 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.