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In Reply to: RE: phasing issue posted by vinnie2 on June 29, 2017 at 12:21:40
Are both 6bq5 cathodes connected to ground through one 130 ohm resistor?At 320 vdc plate to cathode and 8.8 volts bias you should have about 67ma per tube.
I don't get it.
If you're looking for 33.5ma per tube the common cathode resistor should be about 223 ohms. 15 volts / 223 ohms = 67ma.
If you look at the triode plate curves in the link below you will see that about 15 volts bias with 320 volts plate to cathode gives about 33.5ma of current per tube. 2 tubes, 2 times 33.5ma is 67ma.
2 X 33.5ma is 67ma flowing through the one cathode resistor.
Could your tubes be worn out/weak?
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 06/29/17 06/29/17 06/29/17Follow Ups:
Isn't that what the schematic shows, both to ground through one resistor? That's what I did.
The tube manual says 67mA for the 6bq5. Should I be trying to have that through each tube for a total of 134 mA? I am confused.
No Vinnie, I'm not trying to tell you what to do.I'm saying that if both cathodes are connected to the one common cathode resistor (BTW that is the right way to do it) and there is ONLY 8.8 volts at the cathodes then (according to the plate curves) there should be a lot more current flowing through the tubes.
Something is wrong or your tubes are worn out.
Please look at the triode plate curves that I posted a link to.
Look at 320 volts and 8.8 volts grid.....do you see where the current is?
"The tube manual says 67mA for the 6bq5"
That's Max current and not where you want to operate each tube.
Also, where the data sheet says 130 ohms cathode resistor for 2 tubes that's pentode connected. This is not what you are doing.
The data sheet shows a cathode resistor of 270 ohms for one tube at 250vdc plate.
That would be 135 ohms for 2 tubes but you plate voltage is higher so that's not what you want.
Again look at the triode curves. Everything you need to know is there.
The red dot is at about 8.8 volts bias and 320 volts plate and it's 80ma.
That would be for one tube but you have two tubes.
The way you have it now there should be 160ma of current total.
That's not what you want to do.
Again, the triode plate curves show that at 320 volts plate the bias should be at -15 volts to get about 33.5ma from each tube.
That should take a 223 ohms resistor to give the 15 volts bias and if it doesn't then something is wrong.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 06/30/17
That's what my question was from the start, do I use 67mA or 33.5 mA per tube. Now that I know the answer I will adjust the value of the cathode resistor to give me about 30 mA per tube so I am not on the ragged edge. Thanks.
Edits: 07/03/17
That sounds about right.320vdc times 30ma is 9.6 watts
The 6bq5 will dissipate 14 watts (12 for the plate and 2 for the screen grid)
You could go a little higher safely.
320vdc times 35ma is 11.2 watts and that would be 80% of max dissipation (still in the "safe zone").
BTW with 8.8 volts at the top of the 130 ohm cathode resistor each tube IS running at 33.5 ma but that just shows that something is wrong.With only a 130 ohm cathode resistor those tubes should be running 80ma each.
I think your tubes must be worn out or the plate voltage is not what you said it is or something else is not what you say it is.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 07/03/17
I have to go back and start at the B+ and work forward from there. I want to try the unit staying within the recommended limits in the tube amp, so I want to try 250 vdc on the plates. I will report back with some new numbers when I think I have everything right and see what you think then.
The GE data sheet shows 250vdc plate, 34ma., 270 ohm cathode resistor with a 3.5k plate load for single ended triode connected Class A.So for PP triode connected Class A that would be 250vdc plate, 34ma per tube, 135 ohm common cathode resistor with a 7k plate to plate OPT.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 07/03/17
Thanks. I will start there.
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