ANd when good tubes glow red, it is usually the grid telling it to do so. Grid voltage is he first thing to check.
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What makes a preamp tube bias go crazy???
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While we're at it... since I'm in the process of learning how to troubleshoot a problematic tube amp...
The glowing tube had both plate (430V) and grid (0V) voltages in specs, but a much higher than normal cathode voltage (getting higher as long as the plate was glowing brighter). The problem was a bad joint from one grid (pin 7) to ground (via the grid resistor). How did the grid and cathode interact and make the plates glow? Does the cathode voltage increase when the grid looses its connection to ground?Carlo Pipitone
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Originally posted by slidincharlie (Carlo P) View PostWhile we're at it... since I'm in the process of learning how to troubleshoot a problematic tube amp...
The glowing tube had both plate (430V) and grid (0V) voltages in specs, but a much higher than normal cathode voltage (getting higher as long as the plate was glowing brighter). The problem was a bad joint from one grid (pin 7) to ground (via the grid resistor). How did the grid and cathode interact and make the plates glow? Does the cathode voltage increase when the grid looses its connection to ground?Building a better world (one tube amp at a time)
"I have never had to invoke a formula to fight oscillation in a guitar amp."- Enzo
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Nope, your hypothesis is fine tubeswell
The only subtlety is, that if the grid had really been at 0v, the cathode voltage wouldn't have been 30v and the plates wouldn't have glowed.
What probably happened is that when you measured the grid voltage, your meter provided a path for the (tiny) grid current and restored proper operation. (A DMM on voltage range usually looks like an 11 Meg resistor.) If you left the meter there long enough you'd see the plates cool down.
When you went to measure the cathode, the grid would have shot up to some positive voltage, maybe 20v, as soon as you took your meter probe off it, and the tube would have started conducting too hard and overheating again."Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"
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With the grid not connected to anything I don't think there can be any grid current flow, like there would be if the grid were actually sitting at zero or +1 or +2 volts relative to the cathode. So, no direct grid heating from grid current in the usual sense.
If the grid is unconnected (floating), then it will have whatever potential exists in the field at its particular location in the path from cathode to plate (the 20 or so volts suggested). I was puzzled as to how it would "measure" zero volts, but Steve's explanation has cleared that up.
Carlo, when you measured less than 1 Meg to ground, the connection must have been ok at that time. That test should have located the problem, but intermittent failures are always hard to pin down.
MPMLast edited by martin manning; 11-25-2008, 01:53 PM.
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Look at it this way: the voltage on the cathode is there because of the current flowing through it. That's Ohm's Law. COnsider a typical circuit with a 12AX7 triode, a 1.5k cathode resistor and a 100k plate resistor. If 1ma of current flows through all that, then there will be 1.5v at the cathode.
V = I x R = .001 x 1500 = 1.5v
And that 100k p[late load? With 1ma through that as well, there would be 100v dropped across it (.001 x 100,000) Of course we would need to know the B+ to say where that left the plate, but the plate would be 100v less than B+.
The grid controls the current through the tube, and a floating grid does nothing. If the grid is disconnected, the tube will conduct as hard as it possibly can. That means the most current it can. SO really high current means really high voltages across those resistors. The cathode voltage will be really high, and the plate voltage will be a lot lower.
When you have a broken connection at pin 7 - the grid - then the grid is left floating.
How do the grid and cathode then interact? They don't. As far as the rest of the tube is concerned, now there is no grid.Education is what you're left with after you have forgotten what you have learned.
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Okay well, I am yet again enlightened thanks to Steve and Enzo - I can see how that works. Ya see, I got confused when Carlo said "The problem was a bad joint from one grid (pin 7) to ground (via the grid resistor)". But I wasn't taking account of the rising cathode voltage.Building a better world (one tube amp at a time)
"I have never had to invoke a formula to fight oscillation in a guitar amp."- Enzo
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Originally posted by pdf64 View PostWhat makes you think that the cathode resistor is ok? The only sure test is to sub in another one.
Best to replace it with 2k2 whatever, the various SF values of 470-1k are way too low and run the 12AT7 plates around or above their max dissipation. Peter.
The AC outlet is another thing they screwed up; they wired the hot & neutral backwards for many years (incl. BF 2-wire ones).
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Great analysis, Enzo! This makes sense to me. As I was taught, a tube is a "normally on" device, and it requires a negative voltage on the grid to control it from running away. I wasn't aware that the plate voltage would go down, but I can see it now that you explain the voltage drop across the plate resistor.
Now I'd like a little clarification on grid current. Is there any such thing? How is it measured? Or is it better to just look at the voltage?
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Yes, the tube is like a JFET. WIthout its grid, a triode become a rectifier.
Grid current? Grids can conduct current, but I don;t think we invoked grid current here.
For current to flow anywhere - grids or otherwise - there has to be a path for it. Tubes are voltage devices. They work on electron flow. There are electrons flowing in the tube, and electrons have negative charges. The only way for electrons to flow to the grid is for the grid to be more positive than other things in the tube. Let us say a triode has +2v on its cathode. The cathode is what is emitting electrons. For them to be attracted to the grid and cause current, the grid must be at some positive voltage higher than +2v. Otherwise, no electrons go there, and thus no current flows.
Now assume we have a conventional stage 12AX7. signal comes to grid through a film cap of some sort, and a 1M resistor to ground. I got +2 on the cathode. If I send a 6v p-p signal to that grid, the positive side is 3vP. SO each half cycle the grid is more positive than the cathode for part of the peak. All the time that grid is over 2v and up to that 3v peak, the grid will conduct current through that 1M to ground.
Otherwise there is no grid current. That is why your bias voltage can flow through 330k resistors on teh way to the output tube grids without dropping - no current flows.Education is what you're left with after you have forgotten what you have learned.
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