I'm told that guitar signal does not cross a resistor, I'm curious to why they are used on the input?, whats the significance ?
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Guitar Input resistor sizing.
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Does not cross a resistor where?
If you are talking about the series resistance at the input grid of an amp, its there to filter RF, so you don't get your local radio station blasting through your amplifier.
Most amps use one in the 10k-68k range. Some amps don't use them at all. I've had a few of those and its the 1st thing I add. Nothing like a high gain amp blasting NPR at a rock show. Its a real Spinal Tap moment.
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The 68k stops RF (in this particular case). It's called a grid stop resistor.
The 1M is there to provide a DC path to ground for the grid. It's called the grid bias resistor (in this particular case). http://www.freewebs.com/valvewizard1...Gain_Stage.pdf page 14.-Mike
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The 68k is a series input resistor that filters RF. Its optional but highly recommended. Value is subject to requirements and other circuit considerations.
The 1M is a resistor to ground. Its the grid leak resistor for your amp's first triode, and works in parallel with your guitar's pickups and volume pot.
Most tube amps use 1M there. If you start to drop from that, you start to lose dynamics and high end from your signal.
Two different purposes entirely.
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When you see the high/low pair of jacks, it is not the series resistor making the difference. It is the voltage divider TWO resistors make.
Look at a typical circuit, like a Fender 5E3.
http://www.schematicheaven.com/fende..._5e3_schem.pdf
If you plug into the top jack, upper left, it not only ungrounds the input, it allows the signal to flow through a 68k resistor to the grid. But at the same time it can also flow through the second jack and its 68k resistor to the same grid. SO essentially you have two 68k resistors in parallel, and in series with the signal.
But if you plug into the second jack, your signal now flows through a 68k resistor to the input grid, but the upper jack has something to say about it. The tip contact of that jack now remains grounded. SO that places its 68k resistor from the tube grid to ground. Altogether then, you have the signal coming through a 68k to the grid and a 68k from grid to ground. That forms a voltage divider. Look at it a litle different, and you find the signal is applied to a 136k resistance and the signal is tapped out of that at the mid point and fed to the rest of the circuit. Exactly the same as if it were a 136k volume control set to mid-point. So the signal voltage is cut in half.
So note that within reason, the resistance of the resistors is not the determining factor. If the two resistors have the same value, this relation would remian at 68k 47k, 100k, whatever. That is not to say a 68k, 34k, 16k, resistor will sound alike, but the high/low volume difference is not due to the resistance. It is due to the voltage division.
That straight 1/2 voltage pad is 6db. SOme amps do use differeing value resistors so the division is not half, they made some other ratio for their purposes. Works the same though.Education is what you're left with after you have forgotten what you have learned.
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give a look here
http://www.el34world.com/Forum/index...0;attach=11212
there are some explanations I received in this my thread in another forum, some time ago
Input circuit (1Mohm resistor) differences in Vox AC30 schematics - Why ??
hope this can help
Kagliostro
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