Peavey Roadmaster disappearing lead signal

[glebert]

Working on a Roadmaster. Schematic here: https://www.thetubestore.com/lib/the...-Schematic.pdf

Right now I am only working on the preamp. Normal channel is working fine. Lead channel passes basically no real signal. If you crank up the input volume you will get some really distorted garbage. Probing through the circuit the signal is clean up until it gets to V2. Signal is clean and strong at pins 1, 2, and 7. Almost nothing at pin 8 unless the signal is very big and then I get very small bumps at the peaks of the signal. Voltages all line up with the schematic. Have tried multiple known good tubes in V2.

Anyone have any idea what is going on here?

TIA,
Greg

[The Dude]

If you measure resistance from pin 8 cathode to ground, what do you get?

[Enzo]

Is your ribbon connector OK? Is it on the right pins? You got 200v on pin 8?

[glebert]

If you measure resistance from pin 8 cathode to ground, what do you get?

Bingo! Totally open. I didn't think I could get the DC voltage at pin 8 if there wasn't a current path, but I admit the cathode follower circuit has always confused me more than my usual confusion.

Thanks!

[minim]

Well thanks glebert, because I had never really analyzed the CF, and after reading this thread I wondered "how the hell does that 200-odd volts actually get there?" So I did some digging...

This is from John Wilder, whom I believe used to frequent these pages, and he summed it up pretty well I think:

"Basically what happens is that on a cathode biased stage, you have a ground referenced grid and the cathode is elevated above ground by some small amount...usually in the 1-2V range. Since the grid is at ground potential (i.e. 0V) and the cathode is at +1-+2V, this presents a -1 to -2V potential at the grid relative to the cathode.

For example, let's say our cathode is at +2V. This would present -2V at the grid relative to the cathode. By "relative to the cathode", I mean that the grid appears to be at -2V as the cathode sees it.

If we were to raise the grid voltage up by +1V, this would make the grid appear to be at -1V as the cathode sees it. This initially presents a -1V difference between the cathode and the grid. However, the increase in grid voltage also causes an increase in cathode current through the cathode resistor (Rk). The increase in cathode current through Rk causes a corresponding increase in the voltage drop across Rk, which raises the cathode voltage. The cathode voltage rises by a factor of 1V also, which maintains the -2V difference between cathode and grid."

[glebert]

Replaced R19, lead channel now works. This one is kind of a paradox, in that normally the cathode voltage is developed by the current going through it, but in this case with an open cathode resistor I guess it was more that the cathode voltage was being pulled down from the plate voltage (?). Actually, I guess that is the normal CF operation.

OK, now on to the power amp. Anyone got a crappy sextet of 6L6GCs they want to donate?

[nickb]

Bingo! Totally open. I didn't think I could get the DC voltage at pin 8 if there wasn't a current path, but I admit the cathode follower circuit has always confused me more than my usual confusion.

Thanks!

You are overlooking your meter. As soon as you connect it to from cathode to ground you provide the required DC path. Due to the tube characteristics the measured voltage will be only a volt or so away form 'normal'.

[glebert]

You are overlooking your meter. As soon as you connect it to from cathode to ground you provide the required DC path. Due to the tube characteristics the measured voltage will be only a volt or so away form 'normal'.

Good point! Probably didn't help that I was using my crappier meter which has a lower resistance.

[glebert]

Put a couple output tubes in it and got some output and no fire, so good progress!