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Help needed reducing heater voltage

Discussion in 'Shock Brother's DIY Amps' started by jvande7, Oct 14, 2020.

  1. ThermionicScott

    ThermionicScott Poster Extraordinaire

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    The 5C1 Champ ran its 6V6 at about 9 watts, the 5E1 at 11.6 watts, the 5F1 about 12.2 watts, and BF/SF Champs generally over 14 watts, so don't think there is ONE magic number you need to hit or it won't sound good. ;)

    If we're voting, I'll cast another vote for R.G. Keen's bucking transformer over a Variac. A 6V/3A filament transformer is only $15 at AES, and that would be plenty to knock down your voltages into an acceptable range. Overshooting and running the filaments way too cool is also not good for business.

    BTW @King Fan, nice work on your Vintage Voltage box!
     
    Last edited: Oct 15, 2020
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  2. jvande7

    jvande7 TDPRI Member

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    UPDATE:

    Since my B+ is already down to 352v thanks to the zener diodes at the PT CT to ground, I decided to leave the heater voltage as is at 6.8v and just try to set the 6V6 dissipation higher than 11.2w by adding a parallel resistor to the 470R cathode resistor.

    I added a 3k 1w metal film resistor which drops the 464R (actual measurement) down to 402R and brings the dissipation up to 12.5w

    The overdrive now sounds noticably better. I'm going to try a 1.8k in place of the 3k which should bring the dissipation up to about 13.54w to see how that sounds.

    Does anyone suggest going higher? I'm using NOS 6V6GT and I have an old Ken-Rad 6V6G that tests NOS as well. I could use a 1.5k to get 13.76w dissipation, or 1.2k for 14.32w
    I've read that these Champs really shine when dissipation is at or slightly above 100% dissipation, but I still don't know if that means 12w or 14w.
     
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  3. Bendyha

    Bendyha Friend of Leo's Silver Supporter

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    The Ken-Rad 6V6G is the original 6V6, and hence could be 84 years old..doesn't mean it is no longer good, they are in fact excellent. But it does mean that they were built to the original specifications which where subsequently up-graded several times. They have a small plate compared to most things that came later, and pushing the dissipation to its limits could soon see it glowing red and buckling. :(
     
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  4. enorbet2

    enorbet2 TDPRI Member

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    Heya
    If it was mine, I'd just mod it to a DC Heater supply and revel in the quiet :D
     
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  5. shucky

    shucky NEW MEMBER!

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    I'm not sure if you've already wired up a solution, but I think I can help figure out the math.

    The transformer you specified appears to be supplying 6.3V to both the 12AX7 & 6V6GT. 12AX7 heaters in parallel are spec'd 300ma @ 6.3V. Series heaters would be 12.6V @ 150ma. To make sure, measure the voltage separately at pins 4 (or 5) & 9; they should be identical to pins 2 & 7 on the 6V6GT. In parallel configuration, pins 4 & 5 on the 12AX7 are connected together to form one leg, and pin 9 is the other leg.

    Without a schematic I'm only guessing, but your total current draw would be 300ma (for a single 12AX7) + 450ma (for a single 6V6GT) for a total of 750ma (.75A).

    Now your transformer doesn't put out amps; the 2.25A is a rating, ie the maximum amount it was designed it to supply. Not trying to be pedantic of course, just with things like math and electricity, precision is a virtue.

    So anyways, you need to drop .25V across each leg of your heaters. Using an Ohm's law calculator, like this one:

    https://ohmslawcalculator.com/ohms-law-calculator

    putting in .25 in the 'Voltage' box, and .75 in the 'Current' box, it tells us you need ~.33ohm resistor at each leg, and each will dissipate .1875 watts. You could use a 1/4w resistor (.25 watts) but generally you want to avoid exceeding 50% of the power rating, which is being conservative in the interests of safety and reliability, so 1/2W or higher is better:

    https://www.mouser.com/ProductDetail/Yageo/MF0207JT-52-0R33?qs=KUIzHt/e91mUEHnbNvn0ZQ==
     
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  6. robrob

    robrob Poster Extraordinaire Ad Free Member

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    From my website:

    Desired voltage drop / amps = resistor ohms

    .5 volt drop / 2.7 amps = 0.19 ohm dropping resistor

    We want to keep the voltage balanced on each heater pin to reduce 60Hz hum so we need to use two resistors, one on each pin

    We divide the 0.19 ohm by 2 to get 0.095 ohm (I rounded up to 0.1 ohm)

    Note: These same resistors would drop .78v when using EL34 power tubes (3.9 amps * .2 ohms = .78v)

    amps * voltage = watts of dissipation

    2.7 amps * .5 volts = 1.35 watts so to be safe the two resistors should be rated at 2 watts each.

    So we would install two 0.1 ohm 2 watt resistors on V5's heater pins (2 & 7) and connect the inbound wire (from the power transformer or pilot light) to the resistors. These two resistors will drop the heater voltage by 0.5 volt. These types of resistors are available at http://www.mouser.com/ for about 50 cents a piece.

    Just plug in your numbers above to calculate the resistor value. Mouser sells pretty much any resistor value you need including .1 ohm 2 watt resistors used in the above example. Notice only 2.7 watts of heat is generated by the resistors so it's not going to overheat your amp.
     
    Last edited: Oct 17, 2020
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  7. awasson

    awasson Poster Extraordinaire Gold Supporter

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    Hey Rob, shouldn’t we be dropping .5 volts on the actual draw rather than the rating of the heater windings?

    He’s got a power tube, a preamp and a panel lamp. My understanding is that we should be adding those numbers to get the draw and then calculate the resistance required.


    6V6GT - 0.45 A
    12AX7 - 0.30 A (It’s parallel)
    Lamp - 0.15 A (Just a guess)

    Total Draw 0.9 A

    With that in mind wouldn’t we calculate:
    .5 volt drop / .9 amps = 0.556 ohm dropping resistor (or 2 - 0.27 ohm resistors)?

    Then we figure out wattage:

    0.9 amps * .5 volts = 0.45 watt, which is about half a watt but why not just use 1 watt or 2 watt resistors anyway.

    Am I misunderstanding something regarding the rated amperage vs. the actual draw?
     
    Last edited: Oct 18, 2020
  8. Lowerleftcoast

    Lowerleftcoast Friend of Leo's

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    If the internet BTU calculators can be trusted, it will only raise the temperature inside the amp about 50 degrees Fahrenheit higher than it would be without the resistors over a two hour period. Not enough to overheat the amp but it would likely pass the 65C (149F) capacitor rating. Might even pass the 85C (185F) cap rating when the ambient temp is high.

    Tube life vs cap life? Who's to say which life should be shortened?
     
    Last edited: Oct 19, 2020
  9. chas.wahl

    chas.wahl Tele-Meister

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    If you consult Robinette's webpage here:
    https://robrobinette.com/Generic_Tube_Amp_Mods.htm#6.3V
    you will see that he is using the actual draw, of a much more energy-consuming amp:
    "My 5F6A draws 2.7 amps of 6.3v heater current using 6L6GC and 5881 power tubes (3.9 amps with EL34's) so we can use this equation to calculate what resistors to use to lower our heater voltage."
     
  10. awasson

    awasson Poster Extraordinaire Gold Supporter

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    Thanks for that. I didn’t understand the context of the values Rob was using and since his example reduces the .5 volts which is the same amount of voltage that is being discussed here, I though that solution was aimed at the OP’s goal.

    Good enough, easy to achieve. I’d do it if the amp is being tinkered with anyway.
     
  11. robrob

    robrob Poster Extraordinaire Ad Free Member

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    chas beat me to it.

     
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  12. tubedude

    tubedude TDPRI Member

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    2) .47 Ohm 5 Watt sand cast resistors. For less heat, an array of diodes will drop your voltage also. I used 8 in parallel with another 8 paralleled reverse biased on a '66 Bassman, as my mains is regularly 124VAC.
     
  13. robrob

    robrob Poster Extraordinaire Ad Free Member

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    Two .47 ohm resistors (one on each 6.3v line) will drop 2.5 volts in my example amp above. Typically 1/2 volt is all that is needed.

    tubedude, what diode are you using to drop .25v in the heater lines?
     
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