Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: markmalin on July 27, 2015, 04:43:06 pm
-
Hey guys,
Here's a quote from Paul Ruby Amps: http://www.paulrubyamps.com/info.html#ProblemIsolation (http://www.paulrubyamps.com/info.html#ProblemIsolation)
"A better, but slightly more difficult, solution than grounding the CT is to connect it to some clean, positive DC voltage source. If the power stage in your amp is cathode biased with a nice big bypass cap on the cathode resistor, you're set! Just tie the CT to your power tube's cathode. There's somwhere between 9VDC and 35VDC at this node. It's just as free and easy as ground, so use it instead of ground if you have it. "
I guess I never thought of using the cathode bias as a DC reference for the heaters. The only discussions I've read on this are in TUT3, where he talks about setting up a clean DC supply from the power supply filtering circuitry (adding another cap and resistor). Granted, the amp I'm working on has two separate cathode resistors/caps (one for each set of 2 power tubes) - I'd just be using one side - but I"m curious to know if anyone's used this in a build?
Mark.
-
most of the push-pull CB amp i build that have a heater winding CT, then it's strapped to the output stage bypass.
--pete
-
I have seen that technique in a few PA amps from the late 1950's onwards. It was useful for microphone inputs, as intrinsic hum performance of some valves could vary widely, and age, so it kept the worst valves from being objectionable.
Unless you have knocked every other hum source on the head, and have a poor valve, you probably won't notice any improvement.
Some PA amps even used the input stage valves heaters for the output stage cathode bias resistor - that has its hum advantages, and potential pitfalls.
-
I guess I never thought of using the cathode bias as a DC reference for the heaters.
A number of guitar and Hi-fi amp makers did that back then.
The only discussions I've read on this are in TUT3, where he talks about setting up a clean DC supply from the power supply filtering circuitry (adding another cap and resistor).
Kevin has written 25dcv to 35dcv stand off is not enough, it works but there's more noise reduction to be achieved.
He always references 70dcv to 80dcv in his TUT books.
Fellow member Ed C tested this once and said it was quieter with the higher dc stand off.
-
Some background to the topic, if you like it technical:
http://dalmura.com.au/projects/Hum%20article.pdf (http://dalmura.com.au/projects/Hum%20article.pdf)
-
Thanks guys!
Mark
-
if anyone's used this in a build?
I used it on one problem build(SE). did knock hum down some, elevated +18vdc, but final solution/Band-Aid for that build was DC filaments.
-
Sometimes heater elevation is needed to avoid exceeding a tube's maximum rated heater-to-cathode voltage (Vhk(max)).
If your design uses a cathode follower or cathodyne/concertina phase splitter, or any other circuit that operates with a higher-than-usual voltage on a tube's cathode, elevating the heaters to between 30VDC and 90 VDC will protect the tube and make it last longer, as well as avoiding the "crackling" noises that can be produced using such circuits without heater elevation.
Merlin has much more info on this in his preamp book.
If you use a voltage divider off any convenient point in the B+ chain, this is also useful as a safety bleeder for the power supply.
There are many examples on the web with heater elevation - here is one that yields approximately 70V elevation from a B+ of about 400V:
(http://i514.photobucket.com/albums/t346/jaxmoons04/talon_schematic.png)
-
Ken, that's great. Thanks. Good schematic!
-
Note: the dropping resistor should be rated for 1watt. Calc the dropped voltage across the resistor, square it, divide by the resistance, and square the result. That's your wattage rating that should be used.
-
> Calc the dropped voltage across the resistor, square it, divide by the resistance, and square the result. That's your wattage rating that should be used.
I think you squared once too many.
The last step, I would say "double it".
-
> Calc the dropped voltage across the resistor, square it, divide by the resistance, and square the result. That's your wattage rating that should be used.
I think you squared once too many.
The last step, I would say "double it".
Thanks PRR, good catch! My fingers can do it without thinking but my mind can't do it without my fingers :laugh: