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filaments on D.C.

7/24/2000 1:35 AM
filaments on D.C.
Anyone know how to convert an A.C. filament supply into a regulated D.C. supply ?
7/24/2000 3:19 AM

Well, you have to rectify the AC voltage to DC.  
You will need a bridge rectifier, and a rather large capacitor, say 10,000uF/16V.  
You will get 1.414 * the AC voltage you put into your bridge. which will be probably 8.9V if you use a 6.3V tap. This needs to be dropped back down. There are a couple of ways to do this. You can use a very small value of resistor, and fiddle with it until the voltage and the current balance out to what you need. -- Say a 12AX7 takes 300mA at 6.3V, so use V=IR and figure out what size resistor you would need. If you have other tubes, you will draw more current in parallel, so account for that.  
You can also use Diodes. Si will drop about 0.6V depending on how much forward current you put through it. As long as you don't exceed the forward current rating of the diode, you can string them up until the voltage is within 10% of 6.3V.  
You could probably use a zener diode too. a 3V one in series would get you within the ballpark, but the wattage you run may make this impractical (read expensive)  
I wouldn't run DC heaters on your power tubes, only the preamp ones. You will probably stay under 1.5A this way, which is easier to design for-- smaller and cheaper components.  
Hope this answers your question,  
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7/24/2000 3:40 AM

Hey Brian.. Thanks for the info...any reason in particular you recommend NOT putting power tubes on D.C., other than exceeding current ratings? I've got a 4 amp filament tap, 2x6L6 and 5 x 12AX7's. If I do this, does it REALLY decrease hum and noise in the amp ?
7/24/2000 4:24 AM

DC on the power tubes won't do anything for you. Your signal voltages are significantly large by then that the small magnetic interactions really aren't a big deal.  
You can run AC to the power tubes and DC to the preamp tubes on the same transformer tap. Just run the bridge rectifier off in parallel. -- I think-- I am not sure about having different references-- Say the CT of the winding is grounded, can you still run a bridge off of that winding and ground the negative of the bridge???? -- I dunno never done it.  
Does it really decrease noise? It depends on how crappy your wiring job is. A properly wired amp won't have any noticable noise from the filament hum. It may be worth it for you-- It's your call.  
The only stage that really needs careful heater routing is the first preamp stage. The small guitar signals are very suceptible to the magnetic coupling.  
Hum can also be grounding issues, if you have some in your amp. Play with the lead dress first.  
7/24/2000 4:51 AM

I agree with Brian... for the most part guitar amps with DC filaments is a waste of your time  
Use good wiring technique, follow known lead dress patterns and raise the center tap or virtual center tap on the filament winding up from ground with a DC bias derived from the existing power supply by about 20vdc-35vdc.  
Very quiet!  
7/24/2000 9:59 AM
jason d

How do you raise the center tap? I've seen something like this on filament supplies without a center tap, are they really very much quieter?
7/24/2000 9:41 PM

If you have a filament CT, then you float it up (more on this in a second). If you don't, use a 100 ohm resistor from each leg as a pseudocenter tap.  
Now for the floating up part, Yes it will make your tubes run quieter. Not because of hum reduction, but it drops the hiss down quite a bit.  
As for how to float the CT up;  
a) if you have a cathode biased amp, just run the CT to the cathode of one of your power tubes. This will usually give you between 10 and 15V of float.  
b) if you have fixed bias then you have to mod the power supply with a dropping resistor and another filter cap with some kind of load resistor. try to get in the ballpark of 30-50V. Mind your tubes specification for max heater to cathode voltage-- don't exceed this.  
Like so;  
where the first filter cap shown is the existing last filter cap in your power supply. Use a resistor divider formula to find out what value of resistors to use. Any old electrolytic value should do for the second cap, as long as it can take the voltage.  
V=IR, so say you have 250V on your last fiter cap, and you want to pull 0.5mA. so R=V/I -- 500Kohm resistor for the total. now you want to get 50 volts out of that, so since you have 0.5ma running through the chain, you get R=V/I, 50/0.0005 = 100K  
so 500K-100K = 400K, which doesn't exist so use 390K or 470K to get within the ballpark.  
so if you use 470K as R1 and 100K as R2 you will get 44V off of the supply, and pull 0.43mA which shouldn't load your power supply AT ALL.  
Does this make sense to you? Sorry, I had a nice ASCII picture, but the text translator neglects to leave spaces so it didn't make any sense in the preview. This drawing kinda sucks, but I think it gets the point across. If my explaination left you in the dust I can try and simplify it.  
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