lowering AC heater voltage help?

[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

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?

[chrish]

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]

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.

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]

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?

[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.

[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

[sbelyo]

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

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?

[chrish]

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

[sbelyo]

thanks guys

[tubelab.com]

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).