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12th December 2008, 09:07 PM  #1 
diyAudio Member
Join Date: Jun 2008

filament power supply question
Gentlemen,
I'm trying to slap together a DC filament power supply from junk drawer items and questionable bits of knowledge. I have a few boneheaded noob questions: Here's the deal. I have a 14 watt wall wart transformer together with a bridge rectifier and a 22 uf cap followed by about 60 ohms of resistance and another 200uf of capacitance for the 12.6 volts I need. I still have a quarter volt of ripple. So I suppose I should substitute a choke for the resistance, right? I pulled a small dime sized toroidal inductor (I think thats what it is, it lookes like a copper wire minidoughnut) from a old circuit card and put it in the circuit to absolutely no effect. The impedance is only 1.8 ohms! Is it fair to use the formula for inductive reactance to back out the inductance? If so it's 1.8/(2*pi*120) or .0024 Henries. Can that be correct? I know I need about 3 Henries. I don't get it, what value would such a small inductor have in any circiut? For that matter, is this thing even an inductor? Thanks 
12th December 2008, 11:29 PM  #2 
On Hiatus

You're probably saturating the poor thing. To be effective at that low frequency and that high current, you'll need a core the size of a softball at the least to get an effective 3 henry inductor.
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13th December 2008, 12:13 AM  #3 
diyAudio Member
Join Date: Jun 2008

Yup. I get it now. PSUD2 and the Antique Electronic site clearly illustrate that my approach is not practical due to the size and cost of the inductor due to the current requirements. I'll need to use a regulated supply. A good lab experiment though.
The low voltage regulator kit from GlassWare looks like a real bargan. I think that is the ticket. 
13th December 2008, 12:23 AM  #4 
diyAudio Member
Join Date: Jun 2008
Location: SoCal

Load resistance divided by 1200 will get you the Henries needed for a particular circuit. If you know the load resistance then you know the current, in fact you probably need to know the current to figure out the load resistance. Any way you figure out the different values this formula is real easy to remember and works pretty well.
Craig 
13th December 2008, 12:35 AM  #5 
diyAudio Member
Join Date: Jun 2008

Thanks craig. The load is a 6PN1  600ma at 6.3 volts. That should be 10.5 ohms. Using your formula that is only 0.005 henries. Is that correct?? PSU2 is not too happy with it.

13th December 2008, 01:00 AM  #6 
diyAudio Member
Join Date: Jun 2008
Location: SoCal

To be honest with you I've only used that formula for high voltages, ie 400 @ 15 ma, which ends up being a 22 henry choke. It shouldn't make any difference though. Look at the chokes in solid state amps, they too are in the mH range. For your circuit I got almost 9mH. Try googling for other formulas and see what you get, there are many out there. My friend is a transformer engineer, he came up with the same figure I did for my 22 Henry choke but used a different method. There are more ways to skin a cat but the cat always ends up skinned.
Craig 
13th December 2008, 01:30 AM  #7 
diyAudio Member
Join Date: Jun 2008
Location: SoCal

Check out Hagerman Technology, using that formula I got 9.3mH. There you fill in blanks so not sure what formula is. But the cat in still nekked!
Craig 
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