Hi there,
My tube amp project involves 3 6BQ7A tubes, and I will use a 12VDC supply for the heaters. I'm wondering if the best choice would be to:
A) Use a voltage divider with high Wattage resistors to drop the voltage down to 6.3 and wire the tubes in parallel from that supply.
B) Wire two tubes' heaters in series, and have one tube's heaters in parallel with one of those two tubes. I am unsure of the consequences this approach, and it seems unconventional.
Thanks for your advice in advance!
My tube amp project involves 3 6BQ7A tubes, and I will use a 12VDC supply for the heaters. I'm wondering if the best choice would be to:
A) Use a voltage divider with high Wattage resistors to drop the voltage down to 6.3 and wire the tubes in parallel from that supply.
B) Wire two tubes' heaters in series, and have one tube's heaters in parallel with one of those two tubes. I am unsure of the consequences this approach, and it seems unconventional.
Thanks for your advice in advance!
Just use a 12BQ7A, same tube but uses a 12.6 Volt heater..........
___________________________________________________Rick...
___________________________________________________Rick...
Hi!
If you want to use option B, do it like this: wire two tubes heaters in parllel.
Wire a resistor in parallel to the single tube which is left, which draws the same current.
Then put these two proups in series
Thomas
If you want to use option B, do it like this: wire two tubes heaters in parllel.
Wire a resistor in parallel to the single tube which is left, which draws the same current.
Then put these two proups in series
Thomas
I tried searching eBay, but there don't seem to be any for sale... These I also have in my hand, so I'd like to use them if possible 🙂
http://www.r-type.org/pdfs/6bq7a.pdf
Each heater takes 0.4 amps, so at 12 volts, I should use a 30Ohm, 5Watt resistor, just to confirm:
+12V ---^--^--- GND
+12V ---R--^-- GND
Where ^ = 6.3V heater, R = 30Ohm, 5Watt resistor.
or
+12V ---^--\/--^--- GND
+12V ---R--/\--^-- GND
Where ^ = 6.3V heater, R = 30Ohm, 5Watt resistor., and the wires cross at \/ and /\
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If your 12V is strong and healthy (comfortably sized transformer), you could try to rectify and filter, and feed the three tubes in series.
You should use schottky diodes in the bridge to minimize the losses.
You should use schottky diodes in the bridge to minimize the losses.
The filaments are not strictly identical between tubes. If you wire them in series, one tube may end up at 5.7v and the other at 6.3...
The right voltage if you want the maximum performance is 6.3V, not 6 or less. Yes, it's only a 5% difference, but it does have an influence on the tube's parameters and lifespan.
It didn't matter in the 60's because there were millions of spare tubes around and they didn't have good and cheap voltage regulation... But today we should do better!
Why not doing it the right way??
Buy an adjustable DC-DC LM2596 converter module ($3 on ebay), it's good for 2 Amperes and wire the heaters in parallel.
The right voltage if you want the maximum performance is 6.3V, not 6 or less. Yes, it's only a 5% difference, but it does have an influence on the tube's parameters and lifespan.
It didn't matter in the 60's because there were millions of spare tubes around and they didn't have good and cheap voltage regulation... But today we should do better!
Why not doing it the right way??
Buy an adjustable DC-DC LM2596 converter module ($3 on ebay), it's good for 2 Amperes and wire the heaters in parallel.
Use the first connection in post 4: one leg with two heaters in series, the other leg with heater plus resistor, and no cross-connection. Omitting the cross-connection will reduce the warm-up stress on the heater which would otherwise be placed in parallel with R, and will avoid any similar stress created if R goes open circuit.
Although (unlike 4BQ7A and 5BQ7A) the 6BQ7A does not have a controlled warmup heater it should be OK in short series heater strings.
Although (unlike 4BQ7A and 5BQ7A) the 6BQ7A does not have a controlled warmup heater it should be OK in short series heater strings.
That's....actually a good point. LM317s could also go up to 1.5Amp..... I think I'll go with this method as I have voltage regulator chips lying around, and no 30 Ohm 5Watt resistors (I was planning on using 3 10 Ohm sandbars I already had). Thanks for the help!
I'll do some more research into heater power specifics to understand DF96's proposal to keep the knowledge for when I run out of chips 🙂
Just be careful of relative voltages.
Don't put yourself in a position where one heater could be grossly positive with respect to the cathode, you might end up with conduction between the cathode and the heater.
Don't put yourself in a position where one heater could be grossly positive with respect to the cathode, you might end up with conduction between the cathode and the heater.
Would that be possible if I wire the three heaters in parallel with the output of the LM317 chip? How would I go about minimizing this risk?
Just look at the schematic and look at the relative voltages of the cathodes with each other.
I'm scare mongering a bit, it's unlikely but it's worth looking into especially when some designs have heaters lifted above 0V.
I'm scare mongering a bit, it's unlikely but it's worth looking into especially when some designs have heaters lifted above 0V.
Each 6bq7a uses 400mA each. That is 1.2A total. Lets not for get the inrush current involved with cold tubes, I don't doubt each tube's current draw would exceed 1A when cold.
A LM317 can barely handle 1.5A in absolutely prefect conditions. In practice I only trust them up to 1A. I would recommend if you wish to go this route, you have a voltage regulator for each tube. Each voltage regulator would in turn need a fairly large heat sink.
A LM317 can barely handle 1.5A in absolutely prefect conditions. In practice I only trust them up to 1A. I would recommend if you wish to go this route, you have a voltage regulator for each tube. Each voltage regulator would in turn need a fairly large heat sink.
Or a LM350. Or just buy the module I mentioned. It's cheap, already assembled and needs no heatsink!
Am I missing something? At 12 - 6.3 = 5.7V drop across the resistor and 0.4A current, I get 14.25 ohms (and 2.28 W)
Why not keep it simple. Fit a 15 ohm resistor in series with each heater. Bridge each resistor with a higher value to get 6.3v on each heater. This way, the cold inrush current is limited as well.
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