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Tweakers' Asylum Tweaks for systems, rooms and Do It Yourself (DIY) help. FAQ. |
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Tweakers' Asylum Tweaks for systems, rooms and Do It Yourself (DIY) help. FAQ. |
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In Reply to: Re: Resistor Value posted by cantskienuf on January 22, 2004 at 11:39:31:
regardless. I will assume you set each tube at 330 millivolts and that your amp uses a pair of 6L6's operated in push pull class ab1.So each tube is set to 330 mv across a ten ohm resistor. Volts=current times resistance. So .33/10 = 0.033 amps or 33 milliamps. Wattage = current squared times resistance. So 0.033 squared times 10 = 0.0109 watts. Double that for a safety factor (0.0218 watts) and pick the next largest standard value. You are way under 1/8 watt....under 1/32 watt for that matter. Or maybe since it IS millertime I messed up on the math but hopefully you got the idea. Post again or email if you didn't. I think you can see that a one watt resistor is way beyond what you need even with the tube working at maximum.
In addition you can then measure the voltage between the plate and the cathode of the 6L6 and calculate how many watts the tube itself is doing at idle conditions. Lets say you measure 350VDC. This time lets use watts = volts times amps. 350 times 0.033 = 11.55 watts at idle. As a rule of thumb a class a amp would run around 70% of the tubes rating and a class ab amp around 50%. I didn't bother to check but off the top of my head I think 6L6 is rated for about 20 watts maximum plate dissipation so it looks like you are where you need to be.
Russ
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- Re: usually bias is in miliamps not milivolts but... - Russ57 14:35:00 01/22/04 (0)
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