purpose of resistor.....

[trem]

I guess it contacts the diodes that are involved with the two Op Amps in the foot-switch circuit.?
Anyway.....what is R97 up to.?
It is the 1k2 resistor right by the foot-switch jack.
Thank You

[Enzo]

It makes the whole circuit possible. There is 39vAC coming through that resistor from the power supply. So ther is that AC voltage on the tip of the FS jack. AC of course means the voltage alternates moving up to roughly +50v, then down to -50v.

If the foot switches or panel switches are open, then those voltages move on to the two op amps through CR23, CR24. Note one is reverse from th other, so essentially the positive half of th AC goes to TP32, and the negative half to TP35.

When the CHANNEL switch is closed, then CR21 is in the circuit and it conducts on the positive half. The drop across that diode is roughly half a volt, so then the positive side is not clipped off over half a volt. And that means the positive pulses going to TP32 are now reduced. And a similar thing happens when teh MORE DRIVE switch is closed, except it affects the negative side, through CR25, and the changes ther go to TP35.

When those changes happen, it changes the outputs of the two op amps, which in turn control various aspects of the switching circuits.

[trem]

Thanks Enzo -
Yeah...I see the note for 39V.
The main reason I ask is.....this resistor gets pretty hot in all these amps. Hot enough to burn your finger and darken the PCB. Do not know if THAT actually causes any problems, but I wonder if a switch to a 3 Watt resistor (5 watts would, physically, be pretty big in that position) might be a prudent step.? ....preventative maintenance and the like.

[Enzo]

I expect it would get hot. If you short the bottom end to ground, and that half a volt of diode drop is close enough to ground, then ther is 39vAC across the 1.2k resistor. Ohm's Law tells me there will be a little over 1.2 watt dissipated there. The part is a 2W resistor. a 3w resistor would dissipate the same heat, but it would be spread through a somewhat larger part. I can't say as I recall replacing that resistor before, I suppose they can fail like anything else. If that circuit works, I'd probably let it be.

[trem]

10-4
Thank You

[J M Fahey]

It's a clever way to control 2 functions, 2 buttons, with a single conductor and ground wire , even a regular guitar cord.
To boot, current is important enough to light LEDs, both on the amp and in the footswitch, that's why the resistor has such a low value, so enough current can properly light those LEDs.
The price is that the resistor works somewhat hot.

[Enzo]

This circuit has only the two control circuits, but the same system can handle 5 or 6 switched functions as in other models. All that is needed is for the switches to include zeners so the resulting voltages are at certain levels. Then in the amp, additional op amp comparator circuits can decode it. All through a single conductor guitar cord.

[loudthud]

It would take 8 zeners to do a four function footswitch. Wouldn't a USB footswitch be easier?

[Enzo]

Well, that would work, but then you couldn't connect it with a common guitar cord. I like that better than a 7-wire DIN cord or some such. And a 20 foot USB cord, well...

8 zeners? WWOuldn't we just need two on the positive and two on the negative? And of course the extra op amps to decode it.

[Enzo]

Here, this one has a four button switch:

http://support.fender.com/schematics..._schematic.pdf