Coupling Cap Leakage?

[bobloblaws]

Hello,
I have a Gibson Skylark GA-5 here. Fortunately this schematic lines up perfectly with the specimen I have in front of me (apparently some of these vintage Gibson amps and schematics are all over the map specifications-wise). The amp was humming like crazy and I replaced the filter caps, much quieter now. Anyway, I went to check the coupling caps for leakage. The .0047uF cap between V1A plate and the section which I am guessing is the tone stack (followed by the volume control and V1B grid) is reading 58VDC on the V1B grid side. I measured its capacitance out of circuit and it seems fine. I also tried swapping in a brand new cap and got the exact same thing. Is that amount of DC voltage to be expected there? The other coupling caps show close to zero volts of DC leakage, the highest being .025 VDC at the grid of V4.

- B.L.

[xtian]

There should be no DC at the V1B grid. Try replacing that tube.

[bobloblaws]

There should be no DC at the V1B grid. Try replacing that tube.

Oh, I see, so you think the DC is leaking from the V1B plate to grid as opposed to the coupling cap? I did test that tube on my tube tester and it didn't show any inter-element leakage.

[The Dude]

It's easy enough to pull the tube and see if the voltage is still there.

[Enzo]

And maybe a coincidence - those happen - but while I expect zero DC on V1b grid, I DO expect such a voltage on the grid of V2.

[g1]

while I expect zero DC on V1b grid, I DO expect such a voltage on the grid of V2.

Good point. Easy enough to check the cathode resistor values, that will tell what's what.
No way you will have 58V on the grid and 1.5V on the cathode.

[bobloblaws]

And maybe a coincidence - those happen - but while I expect zero DC on V1b grid, I DO expect such a voltage on the grid of V2.

It makes more sense after having a closer look. V1 was wired backward compared to the pin numbers on the schematic. In other words, the pin numbers at V1A on the schematic should actually be 9, 8 and 7 rather than 4, 5 and 6 for the cathode, grid, and plate, respectively. That means that I measured 58V downstream of the coupling cap after pin 6 on the first tube, but that is actually connected to the V2 grid and not the tone/volume section as shown on the schematic. So obviously that jibes with your point about expecting such a voltage on the V2 grid.

It's not immediately apparent to me how that specified 68.5V is formed on the cathode (and the 58V I measure on the grid) of V2. Can anyone give me a brief explanation?

[The Dude]

V1's cathodes are elevated from ground by 1.5K and 2.2K. V2's cathode has 104.7K to ground. For this purpose, think of the tube circuit as a voltage divider.

[bobloblaws]

V1's cathodes are elevated from ground by 1.5K and 2.2K. V2's cathode has 104.7K to ground. For this purpose, think of the tube circuit as a voltage divider.

Ok, I follow that to a point. There is a greater voltage drop on the V2 cathode than on the V1 cathodes due to the higher resistance. As for the voltage divider part of it I'm guessing that you mean the 100K and 4.7K resistors form the voltage divider with the +68.5V being Vin and the +65V being Vout.

What I'm really not clear on is the relationship between the on the plate, cathode and grid voltages. If I pull V2 I measure 221V on pin 1 (plate). When I put the tube back in I measure 132V on the plate, 76V on the cathode and the aforementioned 58V on the grid. So can we say 89V (221-132) is being dropped by the tube and cathode resistors/voltage divider and grid? When the Vout of the voltage divider (I hope I'm using correct terminology) hooks back around to the grid is that some kind of feedback loop or is just a way of regulating the voltage on the grid so that is slightly more negative than the cathode?

[The Dude]

The "voltage divider" consists of the plate resistor, tube, and cathode resistor (in this case, 2 of them in series). When you pull the tube, the plate voltage goes high because there is no conduction. You've basically taken something out of that series. The voltage on the grid is taken from the junction of the two cathode resistors and passed through R12 to the grid. Notice in the V1 circuits there is no resistor from cathode to grid, so there is no direct relationship. I'm not sure I explained that very well, but hopefully, you get my drift.

Edit: The reason it's done that way in this amp is so that V2 (6C4/single element tube) can drive the push pull amp by itself. If you look, the cathode feeds the lower output tube and the plate feeds the upper output tube.

[bobloblaws]

The "voltage divider" consists of the plate resistor, tube, and cathode resistor (in this case, 2 of them in series). When you pull the tube, the plate voltage goes high because there is no conduction. You've basically taken something out of that series. The voltage on the grid is taken from the junction of the two cathode resistors and passed through R12 to the grid. Notice in the V1 circuits there is no resistor from cathode to grid, so there is no direct relationship. I'm not sure I explained that very well, but hopefully, you get my drift.

Edit: The reason it's done that way in this amp is so that V2 (6C4/single element tube) can drive the push pull amp by itself. If you look, the cathode feeds the lower output tube and the plate feeds the upper output tube.

Good stuff. Thanks, Dude!

[The Dude]

If you want to learn more about how it works, Google "cathodyne phase inverter". I'm sure you'll find better explanations than I just gave you.

[bobloblaws]

If you want to learn more about how it works, Google "cathodyne phase inverter". I'm sure you'll find better explanations than I just gave you.

I will do that, thanks again.