0
\$\begingroup\$

I'm trying to imagine a fancy electronic circuit for an electric guitar with lots of customization options. Is it possible to design a guitar pickup analogous to an electro-dynamic speaker? Ie. using electromagnets for the pole pieces instead of fixed magnets? My hope is this would provide an extra dimension of customizability by varying the magnetic strength, but my fear is that the coils of the electromagnets will "short circuit" and induce a current in the outer pickup coils.

If there is induction (transduction?) among the coils, can it be shielded somehow without interfering with the needed magnetic field? Or can it be tuned to a very high frequency and removed later with a low pass filter?

parametric pickup sketch

Please ignore the op-amp portion for the purpose of this question, they are intended to compensate for the varying signal strength from the multiple coil taps.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ I play guitar, experiment with my pick ups, have patents in metal detectors and know a fair bit about inductors and transformers BUT I have no idea what you are trying to ask. \$\endgroup\$
    – Andy aka
    Commented Jan 18, 2021 at 19:06
  • \$\begingroup\$ I must honestly say, I'm confused, and your picture doesn't help that at all (a uncropped, overcompressed screenshot of an unsharp, overzoomed photo it seems?). Maybe you could focus on describing how your approach is (exactly!) different to conventional magnetic pickups. And then, you'd also describe what advantage over the conventional method you hope to achieve through that (hint: I think we'll have bad physics news for you at that point :( ), and lastly, you'd describe the problems you foresee (I don't think any of the concerns you raise make any sense) \$\endgroup\$ Commented Jan 18, 2021 at 19:23

1 Answer 1

4
\$\begingroup\$

All you're asking is whether the magnetizing field in a pick-up can be generated by an electromagnet and thus made variable.

Sure - it's possible, and furthermore the separation of the coils is entirely optional: you could use the same coil both to generate the magnetic field and to pick up the changes induced by the motion of the string.

So, you can use an existing pick-up for that - no extra coils! - and two approaches are feasible:

  1. Keep existing magnets. Use the DC current in the coils to generate a field that subtracts from the magnet's field, decreasing it, or adds to it, increasing it.

  2. Replace existing magnets with powdered iron cores. Use the DC current in the coils to generate desired magnetizing field.

One problem with re-using an existing pick-up is that its coils may not be wound with a wire thick enough to generate sufficient magnetizing field without overheating. You could re-wind the pickup or make your own in that case.

Using a single coil to both generate the static magnetizing field as well as for detection relies on the realization that the magnetizing field is static, i.e. DC, whereas the pick-up's output is AC. The two can thus be separated.

The voltage source used to generate magnetizing current should have high AC output impedance so as not to load down the pick-up output signal. The high AC impedance should extend fairly low - below the frequency of the lowest string. E.g. for a 5 string bass, it should go below 31Hz, so design for a 3db point of about 2/3rds of that, i.e. 20Hz.

Raising the AC impedance can be done by connecting a "low-pass" network between the voltage source and the pick-up. Typical LC or RC filter's cannot be used since they provide filtering but have low AC output impedance. Thus we're limited to a series inductor, in the simplest case.

Now let's see what it really means to have "high AC output impedance" provided by a series inductor. The guitar pickup output impedance at DC is on the order of 5kΩ. The power supply's output network will need to have at least the same impedance at 20Hz. A series inductor would need to have inductance L=Z/(2πf) or about 26H (Henries) - that's not particularly practical. Such an inductor will be not only mechanically large and use many turns of wire, but will also need to be magnetically shielded not to pick up hum.

Instead, to make it practical, the power supply can be designed to have suitably high AC output impedance.

One way to accomplish intrinsically high AC output impedance is to subtract - from the voltage setpoint - the AC component of the output current. Think of the power supply as a power amplifier (PA) that amplifies the current of a voltage setpoint input, at a voltage gain of 1.0. Then, measure the current flowing out of this power supply, convert it into (inverted) voltage with some negative transimpedance Zm (ratio of volts at the output to amps at the input), and sum this output into the input of the PA. The transimpedance Zm of the transimpedance converter will be also the output impedance of the power supply.

Intuitively: We're subtracting a measure of the current from the PA input, thus the larger the current, the more the PA output drops - that's what having an internal impedance would look like.

Given the relatively high impedance of the pickup, the PA in the power supply can be a suitable op-amp, since most types can deal with load impedances in the kΩ range, although it may be a higher power variant to handle the dissipation.

\$\endgroup\$

Your Answer

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

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