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|12/21/1999 5:00 AM|
|RANDY||impedence of guitar?|
What is the output imp. of a guitar?
|12/21/1999 6:57 PM|
|J Fletcher||Don't know, but...|
I'll take a guess...Depends on the type of pickup. If it's active I'd guess lower than 10K ohms, maybe as low as 1K depending on the active circuit. Passive, greater than 10K ohms. Maybe somewhere around 25K for a Srtat. Would also depend if it was single coil or a humbucker, and whether you're using only one pickup at a time. Two pickups in parallel would be about 1/2 the impedance of one (he guessed). Series connected would mean a doubling of impedance. Would also somewhat depend on the value of the volume and tone controls, and probably where they're set. And maybe even what note you're playing...good question, I wonder too. I was thinking that to test the impedance, you need to measure the signal voltage coming from the pickup, and then connect a variable resistance, a pot, in series with the pickup. When the voltage across both is equal, then the resistance of the pot equals the impedance of the pickup. I suppose you'd have to also have a high value resistor as a load resistor in series also, say 1 meg. You could use an Ebow to drive the whole thing, to get a constant AC voltage from the pickup. I don't know how accurate this would be, and I wouldn't be surprised if there was a much easier way...Jerry
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|12/21/1999 8:08 PM|
||Re: impedence of guitar?|
Strat (typical) around 6-7K
Humbuckers maybe 11K.. they vary a bit
but these are good working numbers.
|12/21/1999 8:12 PM|
Thought the Z would be higher than the DC resistance...Jerry
|12/21/1999 9:54 PM|
The AC impedance is of course a complex quantity that's frequency-dependent. At say, 1 khz, the impedance would be close to the DC resistance of the parallel combination of whatever pickup is selected and the value of the volume control pot. So for a typical strat with a 250k volume control and 6.5k coil resistance, the parallel combination (when volume control is on "10") is appx. 6.3k ohms. For two pickups in parallel, the resistance/impedance would drop to about 3.2k.
|12/21/1999 10:05 PM|
You live..you learn. I've wondered about this before, and figured that since the value of the volume pot is 250 k on a Fender, that the Z of the pickup would be somewhere around 1/10th of that, for best voltage transfer. Why make the value of the input stage grid resistor so high, 1 meg, if the output Z of the guitar is only 6K? I would have thought 100K would be fine, and it'd be quieter to boot...Jerry
|12/21/1999 10:31 PM|
The problem is this: what if you have a 500K volume pot, for example (a common value). Now, at top of rotation, the output impedance is equal to the impedance of the pickup in parallel with 500K, so if the magnitude of the pickup impedance were only 10K, the output impedance would be very low, slightly less than 10K. However, if you adjust the volume control to midpoint, the output impedance changes from 10K to 127K. If you used a 100K input impedance on your amplifier, you would lose about half your signal. The effective output impedance changes throughout the rotation of the pot.
In addition, this 100K load is effectively connected to the wiper of the guitar pot, so it would radically change the taper of the volume pot. 1 Megohm is a good compromise between loading and noise. I certainly wouldn't go below 500k.
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