Question About Guitar Pickups - RE: Current

[Paul Marossy]

I know that guitar pickups produce a voltage somewhere between 100mV and nearly 1 volt, depending on the pickup. I am assuming that some amount of current is also produced. What kind of current would we be talking about? Is there a way to calc that out?

(I tried doing a search on the internet about it, but I just found the usual stuff on how they work, etc.)

[The Tone God]

You have the resistance of the pickup and how much voltage the pickup puts out ? Try Ohm's Law.

Andrew

[Paul Marossy]

Well, I don't know the exact DC resistance of the pickup. But, I guess I could get an idea doing that. Sometimes I ask some really dumb questions...  :oops:

Although, your suggestion seems a little too simple. Wouldn't inductance also be involved?

[The Tone God]

Well, I don't know the exact DC resistance of the pickup. But, I guess I could get an idea doing that. Sometimes I ask some really dumb questions...  :oops:

There is no such thing as a dumb question. Only dumb people who don't ask questions.

If its just for interest sake and you don't need exact numbers then IIRC Seymour Duncan's site listed the DC resistance of their pickups. They might have the output too if not throw in an arbatrary figure in and do the math to give you an idea.

Andrew

[niftydog]

DC resistance is no good. You'll need to know the impedance of the pickup - and that is likely to vary with frequency.

Hook your guitar up to a large resistor to ground. Measure the voltages accross that resistor and use ohms law to calculate the current. (Make sure you know the exact resistance first, don't rely on the colour bands!) It'll be close enough to get an idea.

[brett]

The few pickups I have encountered (3 or 4) seem to have impedances of about 10k.  Given outputs of 100mV when first plucked, the current is likely to be about 0.01mA.   Not a lot!!  That's why high-impedance stages usually follow the pickup.

[WGTP]

The DiMarzio site lists DC resistance and voltage for it's pickups.  I have seen impedance figures listed for various frequencies, but don't recall where.  

[Paul Marossy]

Thanks for the replies guys.

niftydog- I like your idea. It probably would be a pain to do it, though.  That would be a good use for an Ebow, if only I had one. Maybe I can get my wife to help on that one... :wink:

brett - 0.01mA?! Wow. That would be a lot less than I would have thought.

WGThickPresence- I have some DiMarzio pickup literature somewhere, I'll have to dig it out. I know that they list DC resistance and impedances at some frequency. I don't remember seeing any specs on current, though.

[DDD]

Hi men,
IMHO pickup current calculation is extremely simple:
1. Let's assume that the PU's impedance (10 kOhm active plus 30 kOhm reactive) is negligible relatively to the input impedance of widely-spread amps/preams/effect boxes e.t.c (about 1 MegaOhm).
2. Then PU current is to be the PU voltage divided by 1 MOhm (approximately, but accurately enouogh).

[Paul Marossy]

DDD-

Well, I hear what you're saying, but I was wanting to know the amount of current coming directly from the guitar, before any effects...

[Mike Burgundy]

Which, approximately (cable losses, that kind of stuff) is what the input impedance of the first effect sees. Good point, DDD.

[DDD]

Well, I'm sorry. I have forgot the Volume pot that is 250k or 500k and the tone circuit - let's think it's 300k.
Also let's consider that the string frequency is low enough not to be affected with pickup/cable capacity - something about 150 pF and 300 pF.  
So the pickup is loaded with 250(500)k in parallel with 300k and in parallel with 1M as described above or without that 1M amp input impedance.

[dolhop]

For what reason are you interested in the current?  Guitar pickups are voltage devices - current must be minimized in order to retain full signal....if you put too much load on the pickup your tone will change radically.  Because of the long windings, the more current, the worse the high frequency response will become.

[Paul Marossy]

For what reason are you interested in the current? Guitar pickups are voltage devices - current must be minimized in order to retain full signal....if you put too much load on the pickup your tone will change radically. Because of the long windings, the more current, the worse the high frequency response will become.

dolhop-

Well, I wanted to know what kind of current a guitar pickups produces because I want to mess around with some LM3900s (quad norton opamps), and I wanted to get some idea of what kind of current we were looking at because the LM3900's inputs operate off of a differential input current rather than the usual opamp which operates off of a differential voltage. I don't know that I need to know any of this, but I thought it would be good to know in case I do need to know...

[dolhop]

Looks like these opamps have been discontinued.  Nevertheless,  I believe that they should work okay with a guitar signal.  It's worth a try anyways.  Are you expecting some kind of advantage over a typical opamp such as the TL072?

[dosmun]

Couldn't you just stick your ammeter in line?

[dolhop]

No, an ammeter only tells an attempt at the RMS current.  It will not give the real picture.

[Paul Marossy]

dolhop-

Well, I just have some of them and I want to goof around with them. I don't expect anything special, but, who knows? Maybe I'll get some really interesting noises out of the thing...