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In Reply to: Re: My first OTL Circlotron posted by Neon on January 26, 2006 at 15:56:50:
Output power is P=R*I^2 where I is the RMS current in the load. If we assume that only one tube is sourcing the current at peaks which is the case for class AB then the peak current must be I*SQR2. Note also that each tube will only source current for half of the period as it will be more or less cutoff for the other half. Given this assumptions we can calculate the necessary peak current for a given power which will be Ipk = SQR(2P/R) and the average current through each tube which will be IRMS/(2*1.1) (2 due to what I wrote above that each tube don't conduct current during more than half of the period and 1.1 which is the difference between the RMS and average of a sinewave current).An example: 25W at 8 ohm give peak current of SQR(2*25/8) = 2.5A and the average current is SQR(25/8)/(2*1.1) = 0.8A
Normally tubes are not specified for peak current but only for the maximum average current so it is not given that any tube can be used in this manner. Max peak current depends on cathode emission and differs between different tubes, some are very marginally designed when it comes to cathode emission and some other have very overdimensioned cathodes, 6C33C seem to be one of these.
Regards Hans
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- Re: My first OTL Circlotron - tubetvr 04:32:18 01/29/06 (0)
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