The original question is why there is a voltage drop of 310V rather than the 150V he experienced with the pre-amp stage.
The circuit is likely fine, the problem is that the 6dj8s draw more current than the 6n1ps so there is more sag (voltage drop over the filter resistors). You can use ohms law to calculate what you need.
V=IR
310 = I(66000) so I = about 5mA
150 = .005(R) so R = about 30000
So try two 15K resistors and see what you get.
The circuit is likely fine, the problem is that the 6dj8s draw more current than the 6n1ps so there is more sag (voltage drop over the filter resistors). You can use ohms law to calculate what you need.
V=IR
310 = I(66000) so I = about 5mA
150 = .005(R) so R = about 30000
So try two 15K resistors and see what you get.
His diagram details the 6dj8 current at 5ma. each. A quick look at a 6N1P data sheet shows typical current of 8ma for each section. 2 sections per tube x 2 tubes. If I read that Russian correctly. Perhaps the entire circuit design with the 6DJ8s pulls a higher load.
thanks a lot for that. that is what i was looking for. but now im not sure if that is what the problem is. the capacitors loose their charge very quickly when the tubes are plugged in so there must be some kind of short some where, im thinking it could be one of the smoothing capacitors that are soldered near the sockets.
just a process of elimination i guess. but yeah, the load draws current which causes a voltage drop across those resistors and with this the B+ voltage level is determined. thanks for that
5VCT transformers were (and are) quite commonly used for heating older 5V direct heated rectifiers like the 5U4/5R4/5Y3. Supposedly this significantly reduces modulation of the raw dc by the ac voltage drops across the two halves of the filament - theoretically the two should cancel each other out since the ac filament voltage is balanced relative to the common mode dc which is the argument for doing it this way. Whether it is worthwhile in a relatively low current 5V rectifier is debatable - I haven't measured very significant differences in either case.
This connection should not be used with tubes like the 5AR4.
It does definitely make a measurable difference in rectifiers with high filament currents. (866A etc)
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lol
can you check my understanding?
if i use two e88cc tubes and all 4 triodes and the anode current is rated 15mA that means the total current drawn is 60mA
so if i have 400v supply voltage and i want to bring this down to 250v then i want to drop 150v across the resistors
E=IR
150=.060R
R=2500
since i have two resistors between two smoothing caps i need two 1.25k resistors.
P=EI
P=150(.06)
P=9
i need two 1.25k 9watts resistors?
OK. 2 6dj8 with 2 sections each. Your diagram says they run 5ma per section = 20ma. Not sure where the 60ma is coming from?
really? ok. allright. i seee hmm. ok you are right.
it says here that B+ is 250v and then it says the tubes operate at 100v, 5mA. right. so how is it that B+ is supposed to be 250v but the tubes operate at 100v. i dont get it.
can help me to pls figure out what i need to do to get this doodad working properly?
i need to have only 100v at the plates? that is about what i got now but it gives very little gain only. i dont get it
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Those type of low power preamp and other driver tubes never run at high B+. There is always a dropping resistor in series to the plate on those. Or the voltage is divided out to different circuits at the filters with dropping resistors.
If you have near 100v at your plate, you're good. Maybe a little higher or lower, 5%-10%. Not critical, usually. So low gain may be your tube characteristics when you get the voltages sorted out. Not familiar with those tubes.
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He's just saying that the dropping resister is 30K in order to take the 250v down to 100v for the tube operating voltage. You just need to have the 250v coming out of your supply. That 30K resistor is actually listed as a part. R4 and R9.
And then as soon as I say "never" and "always" I find a preamp diagram with 250v on the plate of a 12ax7... (slinks away in shame, while flogging self, and repeating to self, try not to say the words: never, always, can't, ..etc., etc..)
never say never.
i still dont understand what is going on. without anything on the B+ rail is 400 volts. when i wire it up it drops down to about 100v but at the plate its about 50v after those resistors. so should i just ought to by-pass those 30k resistors?
why is there such a major voltage drop? can you polease help meeeee. i need some attention
I would suggest looking back at my previous note. It is a simple Ohm's law thing. The current draw causes too much voltage to drop over the PS resistors because they are too big for this application. You just need to use smaller resistors in place of the 30k ones. Calculate approximate values and plug 'em in. You may or may not need to recalculate the plate load resistor to get the right voltage on the plate.
you will sacrifice some filtering for less voltage drop. Another option would be to use a different regulator (SS for example) or a bigger PS transformer, but those are more complicated.
If you need more filtering with less voltage drop, you may replace one or both resistors with inductors, but then it is more complicated as well and you may want to run it through PSD II.
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