Guitar pickup current?

[peskywinnets]

Just wondering if anyone has an idea of how much current flows throw a guitar's pickup wiring?

If say (for arguments sake) an average peak to peak signal voltage of say 200mV (ie not the initial peaky 'pluck', but a couple of hundred milliseconds afterwards - this is typically what I see on a scope for my test guitar), then that works out at 70mV RMS....but to calculate the current, I need the load.

I realise their is no 'set in stone' load a pickup sees - as this would depend on the guitar vol pot & guitar amp - but given that most guitars use something in the order of 250K for a Vol pot (ok, ok, I know some use 500k!) ...then this in parallel with the guitar amp's input impedance (say 1M?) - must mean the load that a pickup sees is under the 250k Vol Pot value ....would 200k be a reasonable figure? (ie 250K & 1M in parallel)

If so then that suggests about 0.00000035A? (350 nanoamps)

Are my figures correct?

[Steve Conner]

Yes, seems like a reasonable figure

[peskywinnets]

I had this hairbrained idea to detect which pickup was active (selected) by using a current sensor around the exposed pickup wire (ie internal to the guitar - don't ask...'Project X' is top secret!), but detecting 350 nanoamps is probably pushing my envelope a little.

[FunkyKikuchiyo]

I've got a bad feeling about this...

[StarryNight]

i'm not great with math, but what about the load from the coil wire itself, and also the patch cable load?

[salvarsan]

i'm not great with math, but what about the load from the coil wire itself, and also the patch cable load?

You can get a rough estimate using the average output voltage and the pickup DC resistance.

[Mike Sulzer]

....would 200k be a reasonable figure? (ie 250K & 1M in parallel)

If so then that suggests about 0.00000035A? (350 nanoamps)

Are my figures correct?

Don't forget the tone control. That puts another 250 K in parallel at most frequencies. It rises a bit a the lower frequencies from the capacitor.

[peskywinnets]

So - to all you heavy (techie) hitters - should the pickup's own DC resistance be included here? (if so, then that's going to mean I'm out by a factor of 25!)

If the DC resistance of the coil *should* be included, is the resistance considered to be in parallel or series wrt 'load'?

Not taking Pickup's DC resistance.... load is derived from 1Meg (amp input impedance) + 250k Vol Pot in parallel (Mike whilst the tone pot will affect things, let's assume a one pot guitar!) = 200k

Including the Pickups DC resistance (in paralell) 1M (amp impedance) + 250k (Vol Pot) + 8K (DC resistance of pickup = 7.69k which would translate into a current of 9.6uA (Still tiny, but positively national grid like vs the first calculation!)

[Steve Conner]

So - to all you heavy (techie) hitters - should the pickup's own DC resistance be included here? (if so, then that's going to mean I'm out by a factor of 25!)

I don't think it should be included, certainly not in parallel. That would only be the case if the pickup were shorted.

I'd go with 1 microamp as a ballpark figure.

[salvarsan]

I don't think it should be included, certainly not in parallel. That would only be the case if the pickup were shorted.

I'd go with 1 microamp as a ballpark figure.

If you specify the voltage across a resistance,
then you've specified the current flowing through it.

[peskywinnets]

If you specify the voltage across a resistance,
then you've specified the current flowing through it.

I can relate to that.

So, from a load perspective, should the pickup's internal resistance be considered in series with the load, or in parallel with the load.

[Steve Conner]

The voltage generated by a pickup is not across its internal DC resistance. If that were the case, there would be none left for the output.

In other words, the internal resistance is in series with the load.

[salvarsan]

The voltage generated by a pickup is not across its internal DC resistance. If that were the case, there would be none left for the output.

In other words, the internal resistance is in series with the load.

Modeling the pickup as a lumped RLC assembly, the Vac source goes anywhere you like but once you define a voltage across it, you have defined a current flow through it.

At this point, someone should post a rational circuit model to disambiguate the whole mess.

[Ronsonic]

I can relate to that.

So, from a load perspective, should the pickup's internal resistance be considered in series with the load, or in parallel with the load.

Source impedance is in series with the source. Sorta by definition. We clump it into one thing as an approximation despite the fact that the capacitance and resistance is distributed throughout the coil(s).

Really messy to model. And I can't say I've ever bothered.

[Plucky]

I had this hairbrained idea to detect which pickup was active (selected) by using a current sensor around the exposed pickup wire (ie internal to the guitar - don't ask...'Project X' is top secret!).

If you are only trying do ascertain which pickup is active on the switch, just look for DCR fluctuations when each pickup is independantly 'agitated'.