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8th September 2003, 10:33 PM  #1 
diyAudio Member
Join Date: Apr 2002
Location: Belleville, IL.

Help with lowering filament winding voltage.
Hi All,
I've got a Hammond 372JX to use with my SE EL34 amp but Steve (Positron) was kind enough to point out that I may have problems because the filament winding is rated for 8 amps and the filaments in my amp will only put a 3.9 amp load on it. I would like to keep the voltage at about 6.1 volts or so to prolong the life of the heaters of my 5687's (they are getting more costly everyday). How would I go about figuring out how much resistance to use to lower the voltage or at least keep it at 6.3 volts? 6.3 volts at 8 amps means the winding is capable of 50.4 watts of power at full load. I will be putting about 25 watts to use. So do I need to use a high power resistor to dissipate the other 25 watts to get 6.3 volts on my heaters? I know this sounds simple but I can't seem to figure out how to figure this out. Any help on this would be greatly appreciated.
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Gavin 
8th September 2003, 10:53 PM  #2  
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

Hi,
Quote:
It's not because something is capable of delivering more than what you actually load it with that it's going to push this through the load. What Steve meant is that the efficiency is higher with the more powerful xformer giving you excess voltage. Just use Ohm't law to calculate the value and wattage of your dropping resistor or use a regulator and you're done. Cheers,
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Frank 

8th September 2003, 10:55 PM  #3 
diyAudio Member
Join Date: Sep 2002
Location: Oregon, USA

I had a similar problem, and added resistrs to my filament wiring to bring the voltage down. If you want to get really accurate, you'll have to build your amp and measure exactly how much voltage you're seeing across the filaments, and then you'll know what resistor values to use. You could guess by figuring out your transformer's regulation, but that was too much math for me.
Here's more or less how I figured this out. Your tube's ratings chart says it takes 900mA @ 6.3V, so when the filament is hot, it has about 7 ohms resistance (R = V/I = 6.3 / .9). Now, let's say you build your amp and you find that you're getting 6.5V across the filaments. Your target is 6.1, so you need to drop 6.5  6.1 = 0.4V across the resistor. You have 900mA flowing through this circuit, so your resistor is (again using R = V / I) .4 / .9 = 0.4 ohms. You can use series/parallel combinations of resistors to get to this value if you want. What I did was to use 1/2 the resistor value, and put 2 resistors in both legs of the filament circuit. Something like this: Code:
Transformer R 3E/\/\/\/\ 3E  3E  3E \ 3E > Filament 3E / 3E  3E  3E/\/\/\/\ R And yes, watch the power rating. For this example, each resistor is dissipating (I^2*R) = .9*.9*.2 = 0.162W, I would use a 1/2W or 1W resistor here. Hope that helps. Saurav 
8th September 2003, 10:59 PM  #4 
diyAudio Moderator Emeritus
Join Date: Jun 2002
Location: U.K.

Gavin,
rather than a big 25 watt resisitor burning up heat, why not go for a series resistor for each valve. To calculate the series value required, you need the measured voltage (actually at the valve pins), the desired voltage, and the heater current of the valve in question. For operation at nonstandard voltages eg 6.1v, there will be a small inacuracy in the calculation due to the temperature cooefficient of the heater. But it's fudgeable. Cheers, Edit: I wrote this a while ago. Since then others have posted. 
8th September 2003, 11:03 PM  #5 
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

Hi,
A side effect of using the resistors on the secondary side of the xformer is that they reduce inrush current...which in turn extends tubelife... Cheers,
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Frank 
8th September 2003, 11:08 PM  #6 
diyAudio Member
Join Date: Sep 2002
Location: Oregon, USA

And putting a resistor in each leg reduces common mode noise.
Actually, I don't know if it does, but I impress myself when I say things like that 
8th September 2003, 11:32 PM  #7  
diyAudio Moderator Emeritus
Join Date: Jan 2003
Location: Near London. UK

Quote:
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9th September 2003, 12:36 AM  #8 
diyAudio Member
Join Date: Apr 2002
Location: Belleville, IL.

Thank you all very much. For some reason I thought the answer I was getting with the equation R = E / I was wrong because of the limited power ability of the winding. I guess what I'm saying is that just using R = E / I puts no limit on P theoretically. In this instance though there is a limit as to the power the winding can supply. Even so, I now see that P is meaningless in this context except for the power rating of the resistor I use. Thank you for all of the help.
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Gavin 
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