If 9-volt Power Was Inadverdently Run Thru Pups...

  • Thread starter Thread starter stevieb
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stevieb

Just another guy, really.
Would it burn out the pups? Would it make a difference if the pups were active, or passive?

I know, there is more to electricity than just voltage- but that is the only thing that would be constant- amerage would vary.
 
Depends.... Pickups could handle 9v at 11 milliamps (~.1 watts) no problem. if It was 11 amps (~100 watts), your pickups would fry. The wiring in a pickup is some pretty damn small guage wire, it will not take a lot of current. Your pickups (coils) will melt and/or poof if you push too much current thru it, as would any wire too small for the load.
 
suprstar said:
Depends.... Pickups could handle 9v at 11 milliamps (~.1 watts) no problem. if It was 11 amps (~100 watts), your pickups would fry. The wiring in a pickup is some pretty damn small guage wire, it will not take a lot of current. Your pickups (coils) will melt and/or poof if you push too much current thru it, as would any wire too small for the load.
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You do not "push" current through a wire, the wire draws whatever current it does at a particular voltage based on its resistance. Ohm's Law: I = V/R. We know that V = 9V. What's the resistance of the pickup? The current it draws will be dependent on that resistance unless it exceeds the current delivering limit of the power source.
 
ggunn said:
You do not "push" current through a wire, the wire draws whatever current it does at a particular voltage based on its resistance. Ohm's Law: I = V/R. We know that V = 9V. What's the resistance of the pickup? The current it draws will be dependent on that resistance unless it exceeds the current delivering limit of the power source.
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Passive pickups generally range from around 5.5K ohms up to 13k ohms, correct? So the current could range from 1.6mA to 0.7mA. Whether pickup wire would get toasted at that amperage, I don't know.
 
ggunn said:
You do not "push" current through a wire, the wire draws whatever current it does at a particular voltage based on its resistance. Ohm's Law: I = V/R. We know that V = 9V. What's the resistance of the pickup? The current it draws will be dependent on that resistance unless it exceeds the current delivering limit of the power source.
Click to expand...

That's true, and I could be wrong, but I figure the resistance of the coil is so low that the battery/power supply would be running at full capacity. A typical 9v battery may not hurt it, because it probably couldn't produce enough current to meet the 'I' in ohm's law. A much larger power supply would be more capable of frying a near-zero ohm load at 9v.
 
Zaphod B said:
Passive pickups generally range from around 5.5K ohms up to 13k ohms, correct?
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If that's the case then I guess I'm way off.. Isn't a pickup basically a coil wire wrapped around a magnet(s)? I thought the string's vibration in the magnetic field induced a small current in the coil? I thought the coil itself would have very low resistance.
 
suprstar said:
If that's the case then I guess I'm way off.. Isn't a pickup basically a coil wire wrapped around a magnet(s)? I thought the string's vibration in the magnetic field induced a small current in the coil? I thought the coil itself would have very low resistance.
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Correct. The string's movement in the coil's magnetic field induces current within the coil - think of it as the same phenomenon as the way a solenoid works, where current induces movement of an object surrounded by a coil - just backwards.

Anyhow, there's lots of very small-gauge wire wrapped around those bobbins. You can check at any of the large pickup manufacturers' web sites for impedance values, you'll see I am correct.
 
I can't say for sure, as I'm not about to do a test on a customer's pickup, but my guess is you would be fine.


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