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Author Topic: Transmitter power output measurements  (Read 460 times)
WA2LTD
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Posts: 12




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« on: January 06, 2006, 09:56:28 PM »

My Johnson Valiant is 275 Watt input for about 165 out. On AM it is rated for 200 in (carrier) and at 60% eff, about 120 out (X4 for peak w/100% modulation?).

When I look at the transmitter CW output across the dummy load with a scope, I see 370 V p-p. that devided by 2 for peak, *.707 for RMS, squared and divided by the 50 Ohm load = 334 PEP Watts?

Is that 334 Watts suposed to be ~3 DB higher than the original figure of 180 for CW output?

It gets worse when I modulate the AM carrier...and get 540 PP Volts from + crest to - crest. That would indicate a PEP output of over 700 Watts!

Am I thinking right here?

73,
Tom  
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AB0WR
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Posts: 77




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« Reply #1 on: January 09, 2006, 05:16:25 PM »

Are you sure you are working into a 50ohm non-inductive dummy load?

It's very doubtful you are getting over 300watts output from this rig.

You need to accurately measure your HV and input current to see what the actual input power is. This will give you a good guess as to what the output power should be.

If this isn't what you are seeing either the measurement equipment is giving you funny readings or the dummy load isn't really 50ohm.

Do you have a wattmeter you trust or can you borrow one? It doesn't have to be a Bird, even one that is plus/minus 10% will let you tell if your measuring equipment has a problem or if the dummy load has a problem.

tim ab0wr

tim ab0wr

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G3RZP
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Posts: 4731




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« Reply #2 on: January 17, 2006, 01:39:01 AM »

Even a Bird is only +/-5% of full scale. So if you have a 1kW slug and you measure 100 watts, the uncertainty is +/-50 watts.

But I'd figure either your load isn't 50 ohms, or there's a pretty big error in the voltage measurement - or both.
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