lowering AC heater voltage help?
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 13th April 2008, 04:17 AM #1 sbelyo   diyAudio Member   Join Date: May 2004 Location: East Coast lowering AC heater voltage help? I've got 7.7 VAC on a 6.3 tap thanks to high ac lline voltage. I obviously need to lower it 1.2 volts I have 2 100 ohm 3 watt resistors and several 500 ohm 3/4 watt laying around What wattage value will I need if 1.8 amps is being drawn from the circuit?
 13th April 2008, 04:36 AM #2 chrish   diyAudio Member     Join Date: Oct 2003 Location: Sydney Make sure that your measured voltage is while under load. Here is a link if you could not be bothered with Ohm's Law http://ourworld.compuserve.com/homep...en/ohmslaw.htm
dsavitsk
diyAudio Member

Join Date: Jan 2005
Location: Hartford
Re: lowering AC heater voltage help?

Quote:
 Originally posted by sbelyo I've got 7.7 VAC on a 6.3 tap thanks to high ac lline voltage. I obviously need to lower it 1.2 volts
Is that connected to the heater, or not? If not, connect it and that will drop some.

Quote:
 Originally posted by sbelyo I have 2 100 ohm 3 watt resistors and several 500 ohm 3/4 watt laying around What wattage value will I need if 1.8 amps is being drawn from the circuit?
W = VI = IIR

Double it for safety.

sbelyo
diyAudio Member

Join Date: May 2004
Location: East Coast
Re: Re: lowering AC heater voltage help?

Quote:
 Originally posted by dsavitsk Is that connected to the heater, or not? If not, connect it and that will drop some. W = VI = IIR Double it for safety.
With .6 amp draw it's at 7.7

with all 4 tubes I expect 1.8 amps

Now the V in the equation, is it the voltage dropped accross the resistor or the voltage of the circuit?

 13th April 2008, 06:07 AM #5 dsavitsk   diyAudio Member     Join Date: Jan 2005 Location: Hartford With 1.8A draw the voltage will be lower. It's the V across the resistor as the resistor does not need to dissipate the heat dissipated in the heater. __________________ http://www.ecpaudio.com :: http://diy.ecpaudio.com
 13th April 2008, 06:19 AM #6 chrish   diyAudio Member     Join Date: Oct 2003 Location: Sydney You are looking at the voltage drop across the resistor. You have not given much indication of what you are doing. If you are building an amp from scratch, you are probably doing the heater wiring first. If this is the case, wire up allowing for the addition of a dropping resistor. Put all your valves in and fire her up. Measure the voltage with the full load, and then use Ohm's law to calculate the resistance required for the voltage drop required. Use Ohm's law again to calculate the power dissipation of the resistor, then choose a conservative rating. Dsavitsk recommends doubling for safety... V=I/R, R=V/I So, to drop 1.2 volts (you may need a different amount when you measure under load) with 1.8 Amps, you will need 0.666 Ohms and dissipate 2.16 Watts, so go with a 5 Watt resistor. Not sure what valves you are using, I have looked up heater voltage ratings and not many show a +/- value. THe Russian valves do, and usually have 5.7-6.9 volts listed, so if it is well inside this I would not worry, though higher voltage may lead to shorter valve life. Chris
sbelyo
diyAudio Member

Join Date: May 2004
Location: East Coast
Quote:
 Originally posted by dsavitsk With 1.8A draw the voltage will be lower. It's the V across the resistor as the resistor does not need to dissipate the heat dissipated in the heater.
Got it

Quote:
 Originally posted by chrish You are looking at the voltage drop across the resistor. You have not given much indication of what you are doing. If you are building an amp from scratch, you are probably doing the heater wiring first. If this is the case, wire up allowing for the addition of a dropping resistor. Put all your valves in and fire her up. Measure the voltage with the full load, and then use Ohm's law to calculate the resistance required for the voltage drop required. Use Ohm's law again to calculate the power dissipation of the resistor, then choose a conservative rating. Dsavitsk recommends doubling for safety... V=I/R, R=V/I So, to drop 1.2 volts (you may need a different amount when you measure under load) with 1.8 Amps, you will need 0.666 Ohms and dissipate 2.16 Watts, so go with a 5 Watt resistor. Not sure what valves you are using, I have looked up heater voltage ratings and not many show a +/- value. THe Russian valves do, and usually have 5.7-6.9 volts listed, so if it is well inside this I would not worry, though higher voltage may lead to shorter valve life. Chris
Yes, building a headphone amp from scratch.

I'll plug in all the valves and turn it on just too see what it drops to. I only had 2 of the 4 in when I measured.

Once I find the value of the dropping resistor, do I put it on one leg of the tap or both?

 13th April 2008, 07:09 AM #8 chrish   diyAudio Member     Join Date: Oct 2003 Location: Sydney You will need to wire the resistor in series with the circuit. That is, take one of the 6.3 taps, take it to one leg of the resistor, then take the other leg of the resistor back to the circuit. The heaters will be wired in parallel (assuming 6.3 volt heaters). Chris
 14th April 2008, 01:11 AM #9 sbelyo   diyAudio Member   Join Date: May 2004 Location: East Coast thanks guys
 14th April 2008, 01:28 AM #10 Tubelab_com   diyAudio Member     Join Date: Jul 2005 Location: West Virginia panhandle Are you still planning to use a TubelabSE board for your headphone amp? If so do NOT add any resistors in the filament circuit. There is a voltage regulator on the board that gives the output tubes exactly 2.5 or 5.0 volts depending on the jumper settings. An external resistor will cause the regulator to lose regulation and increase hum, or worse. There is a resistor on the board to drop the unregulated DC voltage for the 5842's (R3). After the amp is all finished and working you can measure the voltage on the 5842's and adjust its value if needed. This is usually only needed if using a power transformer with a over rated filament winding (especially if it is a Hammond). __________________ Tubelab, it's 5 year mission. To explore strange new tubes, to seek out new circuits and topologies, to boldly go where no tube has gone before......

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