guitars and electricity

DonovanB

So I understand that a magnetic pickup generates the electricity by electromagnetic induction, but what are the specifics?

Does a hotter pup generate more current? Or is it voltage?

And what does an amplifier amplify? Current again?

MikeM

Man, I've been struggling with this for a while.

One thing to remember is that impedance plays a MASSIVE role in all this, but typically a hotter pup puts out a higher voltage - at least I have seen pickups being measured by voltage. Never seen much with regard to current though.

Gearhead

It really isn't all that complicated ?

The coils in pups pick up a certain amount of power. Put simply this is the product of current and voltage (and then some, but forget that). The coils also form an impedance. Put simply that is voltage divided by current. Now if there is no current (unplugged) the voltage will range between a half and two volts. If there is a current, the voltage will drop. This drop is equal to (give or take some funky factor) the product of the current and the total impedance of the loop, so that is the pup plus the volume pot parallel to the input resistor in the amp parallel to the first tube in the amp.
So now you have as many equations as you have unknowns and you can work it out... Typical values would be such that only half the voltage makes it into the amp.

A hotter pup generates more power and you can measure this as voltage while unplugged. It however also usually has a higher impedance and thus more drop. It will not give your amp twice the voltage if it measures twice the impedance, but it will certainly push it more. Remember that the grid of said first tube also consumes power.

Inside the amp, the preamp multiplies the voltage by about 50-100 and subsequently the power amp multiplies the voltage by 4 (small amp) to 50 (biggun') and the current by 100-1000.

DonRoos

Read http://buildyourguitar.com/resources/lemme/ this article, it gives a really good explanation.