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In Reply to: Re: For my understanding... posted by drlowmu on February 27, 2007 at 22:38:56:
Jeff, I'm still not understanding. Would that be 5/8 of suggested plate current at suggested plate voltage or 5/8 of suggested plate current at 5/8 suggested plate voltage?Assuming 5/8 suggested plate current at suggested plate voltage.
You would run a 300b (I know you don't like 300b's but if you were going to run one) at 37.5ma at 350 volts. Right?What would the proper plate load impedance be for a 300b running at that operating point?
Follow Ups:
WE 300B max ratings times 5/8 (.625)...400V X .625 = 250V
100ma X .625 = 62.5ma
36W X .625 = 22.5W250V X 62.5ma = 15.625W
So which is it?
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Boo!
I don't think Jeff knows the answer to that question. I think Jeff is just making this stuff up as he goes.Jeff really said 5/8 of suggested not max, so that would be
350V X .625 = 218.75V
60ma. X .625 = 37.5ma218.75V X 37.5ma = 8.2 watts But that's not an operating point for the 300b.
250V X 62.5ma = 15.625W is not an operation point for the 300b.
350v X 37.5ma = 13.125W is also not an operating point for the 300b.Either way, the load impedance would have to be very high to keep things somewhat linear that it would be impractical.
I really don't understand why anyone pays any attention to Jeff at all. He clearly does not know what he is talking about.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
is 300V @ 62ma. That looks fine to me and is likely close to what WE actually used in amps they designed back in the day.That would equal a plate dissipation of 22.2W which is very close to 5/8 max (35W X .625 =21.875W).
About a 5 watt amp after transformer losses and a WE 300B would last for a long time at that dissipation rate.
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Boo!
Yes but Jeff said 5/8 of suggested. Not 5/8 of max. 5/8 of max is the WE suggested operating point (well, very close). But that's not what Jeff is saying. Jeff said "Of Maufacturer's $ugge$ted operating point$, and NOT max points at all"So Jeff must mean
350v X 37.5ma = 13.125W 5/8 of suggested current
or
250V X 62.5ma = 15.625W 5/8 of suggested voltage
or
218.75V X 37.5ma = 8.2 watts 5/8 of suggested voltage and current
not
300V @ 62ma. = 22.3W as this is 5/8 of max. voltage and current
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
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