• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

Is it possible to use regulated DC for a rectifier tube?

Status
Not open for further replies.
my main transformer's 5V line is limited to 2A which limits the tubes that I can use. I have a 5V4G (5V@2A) now but it would be nice to have other options like 5U4G (5V@3A).

I have a 2nd transformer (that has 2 secondaries) which will be used to heat up 8CG7 (8.4V@450mA). I am hoping I can use the other secondary to heat up rectifier tubes using 5V.

Thanks for the help 🙂

ps. I know that tube data mentions AC/DC heaters can be used. but my experience with tubes amps is very limited and most schematics I've seen uses AC heaters. thanks again
 
Geek said:
I've used regulated DC for DH tubes and have yet killed one.....

OK, I did blow up a 45 before I realized the meaning of soft-start :bawling:


I am planning to use one of the regulator chips from the LM317 family (higher current rating). would this be suitable? if not, can you share your schematic?

thank you
 
jarthel said:
I am planning to use one of the regulator chips from the LM317 family (higher current rating). would this be suitable? if not, can you share your schematic?

It can work, but you'll need lots of caps before the regulator, probably large heatsinks on the diodes, and a large heatsink on the regulator. How about just using a resistor in series with the filament to drop a few volts?
 
dsavitsk said:


It can work, but you'll need lots of caps before the regulator, probably large heatsinks on the diodes, and a large heatsink on the regulator. How about just using a resistor in series with the filament to drop a few volts?

thinking of doing that and but no more DC. just use resistors to bring 9VAC to 5VAC.
 
I have a 9V secondary.

so I need to drop 4V from it.

4V / 3A (current rating for 5U4G) = 1.33ohms (5x 6.8ohms 5W in parallel)

4V / 2A (current rating for 5V4G) = 2ohms (5x 10ohm 5W in parallel)

Is my calculation correct? thank you
 
Status
Not open for further replies.