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cathode bias resistor ratings

Discussion in 'Shock Brother's DIY Amps' started by guitjopicka, Feb 27, 2014.

  1. guitjopicka

    guitjopicka Tele-Meister

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    So I have done a bit of searching and seem to find every possible answer ranging from 5-25 watts. How high if rating do I need on the resistor to cathode bias a pair el34's? Shared bias resistor. I could do them separate if I have to, but it will look prettier if I don't.
     
  2. tubeswell

    tubeswell Friend of Leo's

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    The usual practice is to have them rated to at least 2 x the actual dissipation that is to be experienced in operating the resistor in the circuit.

    So work out the cathode voltage divided by the number of ohms in the resistor and you get the current through the resistor. The current x the voltage dropped across the resistor is the power the resistor is 'seeing'. Double that to get your power rating.
     
  3. laird

    laird Tele-Holic

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    If you want to do all the math, Merlin has a detailed biasing example of a single cathode-biased EL34 on his site.

    The short answer is that 5W is the BARE minimum for a push-pull pair, but using a higher-rated resistor is the safer choice. A basic rule of thumb is that your cathode resistor wattage rating should be at least 15% of your plate dissipation. Two EL34s = 50w dissipation x .15 = 7.5 watts.

    -Laird
     
  4. guitjopicka

    guitjopicka Tele-Meister

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    thanks guys, I will do some math today!
     
  5. Keyser Soze

    Keyser Soze Tele-Holic

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    Power = Voltage x Current

    But since Ohm's Law tells us that Current = Voltage x Resistance you don't even need to know the current.

    All you need to do is measure the voltage drop across the resistor, square that value and multiple by the resistor value to get dissipation.

    Then, as noted, double that for a safe operating margin.

    And you can always choose a higher power rating if it is more economical.
     
  6. zook

    zook Friend of Leo's

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    This is not correct. Current = Voltage / Resistance.
     
  7. Fred Mertz

    Fred Mertz Tele-Afflicted

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    zook is correct:

    Ohm's law:

    Voltage drop across a resistor = Current times resistance (V = I x R)

    Power dissipated in watts = Voltage drop times current (P = V x I )

    Since V = I x R, power dissipation can also be calculated as I x I x R
     
  8. tubeswell

    tubeswell Friend of Leo's

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    Keyzer Soze probably knew that but was in such a hurry he didn't see his typo
     
  9. Fred Mertz

    Fred Mertz Tele-Afflicted

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    He may have known. But there is more wrong with his post other than a typo. Voltage drop squared times the resistance does not equal the power dissipated.
     
  10. Keyser Soze

    Keyser Soze Tele-Holic

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    Yes, thank you - I was multitasking, and poorly!

    I stand corrected. (I should probably stick to symbols and avoid the terms entirely.)

    But you lost me Fred,

    P = IV

    I = V/R

    therefore

    P = (V/R)V

    (Oh, Hell, never mind, I screwed up both functions on the first go.

    Yep, I'll stick to speaking in symbols!)
     
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