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Re: LF: Re: Re: FET RDS

To: [email protected]
Subject: Re: LF: Re: Re: FET RDS
From: Stefan Schäfer <[email protected]>
Date: Fri, 14 Jan 2011 11:25:29 +0100
In-reply-to: <[email protected]>
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Hi Chris,

Of course you are right about the measurement errors. Furthermore one has to think about if the driver "losses", so the power needed to drive the PA is included in the losses or if it is ignored. But all that isn't really important i think. Actually, when running 1 kW i do not care if the losses are 40 W or 50W. It's important that the PA runs stable and the heatsink remains cool, even in QRSS 120 and a SWR of 1.5 (SWR is an issue in class E) I am measuring the input voltage and current and measure the output voltage across a 50 Ohm load (this 50 Ohm load consist of series and parallel switched resistors of 100 Ohm +- 1% / 100W / TO247). Maybe there is an error of 5% :-)

73, Stefan

PS: I've got efficiencies above 100% as well. But this was always caused by a non-sinusoidal waveform, measured with an ordinary analog wattmeter.

Am 14.01.2011 01:01, schrieb Chris Trayner:
Dear All,

Stefan raises an interesting issue.

On 2011 Jan 13, at 21:36, Stefan Schäfer wrote:
My LF 1 kW class E PA i using a single IRFP360 at>  96 % efficiency
I'm sure you are getting high efficiencies. But measuring it at these numbers 
close to 100% actually makes quite interesting demands on the accuracy of the 
instrumentation.

You presumably measure DC input power from the PSU as Vin * Iin, and RF output 
power as Vout * Iout.
The efficiency is then (Vout * Iout) / (Vin * Iin). The percentage error of the 
result (the efficiency) is the sum of the percentage errors of the four 
measurements.
You might instead measure the output power into a dummy load as Vout^2/R, but 
you still have a percentage error of twice twice the error of the voltage 
measurement.

If you measure the output on a scope, personally I wouldn't trust a scope 
measurement to better than a couple of percent. The Vout and Iout measurements 
(or two Vout measurements) will amount to a few percent error each, so your 
efficiency of 96% could really be 96% +/-4% or even 96% +/-6%.
(If you measure Vin and Iin on a multimeter, they could easily have errors of 
0.1% or better, and not contribute appreciably to the overall error.)

John G3PAI brought up this interesting problem when he first built a class E 
amplifier and measured efficiencies of over 100%. It was immediately obvious to 
him that the measurement accuracy was a limitation.

73,
Chris G4OKW

-----------------------
Dr Chris Trayner
School of Electronic&  Electrical Engineering,
The University of Leeds,
Leeds LS2 9JT, United Kingdom
Tel: +44 113 34 32053
Fax: +44 113 34 32032






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