tube dissipation - am i getting this right???

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers' started by avwalker, Jan 25, 2006.

  1. avwalker

    avwalker Member

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    I finished my "octal fatness" amp with a single-ended el-34 power section on sunday. I took some measurements last night just to see where I was sitting at as far as plate voltage, current draw, and output. I was kindof expecting this to be around 10-12 watts but these are my numbers.....

    Plate voltage: 370V
    Cathode voltage: 18V
    Plate Current: 51ma

    So the formula for dissipation is Power= (plate voltage-cathode voltage) * plate current........correct????

    This means I'm outputing just under 18 watts??? that seems a little high. I know an el-34 will handle 25 watt dissipation......but my OT (hammond 125ese) is only rated at 15 watts (but it says it will handle 80ma max dc bias). Is it safe?
     
  2. Shea

    Shea Member

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    Plate dissipation does not equal power output. So just because you're dissipating 18 watts of heat doesn't mean it'll put out 18 watts. It's probably not more than 10 watts.

    A single ended, Class A amp should be biased pretty close to the max plate dissipation, so if anything that's a little low.

    Shea
     
  3. avwalker

    avwalker Member

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    so i'm cool as long as i keep the bias under the 80ma OT limit, correct??

    I've got a 330ohm cathode resistor right now.....might try dropping in a 280 or so.
     
  4. VacuumVoodoo

    VacuumVoodoo Member

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    At 370V plate voltage the plate current should not exceed about 65mA but this will bias the EL34 tube into class B, for class A you'd need to bias at ca 150mA. This will unfortunately melt the plate...

    You can get around 10W output power with 250V/100mA (Rk=100Ohm, Ra=2kOhm) or 300V/83mA (Rk=180Ohm, Ra=3.5kOhm) Ra is ofcourse defined by your Output Transformer and speaker impedance, but I trust you know that. Above figures come from Mullard datasheet.
     
  5. VintageJon

    VintageJon Member

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    avwalker:
    I read this "So the formula for dissipation is Power= (plate voltage-cathode voltage) * plate current........correct????" and don't see a response correcting it.

    This is NOT correct. For a single-ended amp, assuming cathode bias, it goes thusly:

    Measure cathode resistor, write it down as Rk.
    Measure Screen resistor, write it down as Rs.
    Power amp up and let it warm up.
    Measure voltage across Rk and write it down as VRk.
    Measure voltage across Rs and write down as Vrs.
    Measure plate voltage and write down as Vp.
    Turn amp off, and get your calculator out.

    VRk / Rk = Ik or cathode current, write it down.
    VRs / Rs = Is or screen current, write it down.
    Ik - Is = Ip or plate current, write it down.

    Now, Ip X Vp = IPD or Idle Plate Dissipation. For a single-ended-cathode-biased-amp it must not exceed the Max plate dissipation.

    -Jon
     
  6. Shea

    Shea Member

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    I don't think you got to the crux of what he was asking. You walked him through the process of calculating plate current,but his question presupposes that he already knows what the plate current is.

    I think he might be using the term "plate voltage" to mean the voltage from plate to ground. Understood that way, his statement "power = (plate voltage - cathode voltage) * plate current" would be correct.

    But the term "plate voltage" refers to the voltage between the plate and the cathode, not between the plate and ground. You can figure that out by measuring the voltage from plate to ground, and the voltage from the cathode to ground, and subtracting the latter from the former. So I think he has the right idea, he just didn't use the term "plate voltage" correctly.

    Shea
     

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