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You're almost there.
Think of 50% efficiency like this: half the DC input power is going as useful RF, the other half is going as heat. So if you have 500 watts RF output, then you need 1000 watts going in on the DC. In other words, for 50% efficiency you need to add 100% to get the total DC = 72.4Amps.

Its probably most useful to compare RF watts in RMS with DC watts.

[QUOTE="dudmuck, post: 155575, member: 3904"] You're almost there. Think of 50% efficiency like this: half the DC input power is going as useful RF, the other half is going as heat. So if you have 500 watts RF output, then you need 1000 watts going in on the DC. In other words, for 50% efficiency you need to add 100% to get the total DC = 72.4Amps. Its probably most useful to compare RF watts in RMS with DC watts. [/QUOTE]

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