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Date: Fri, 14 Jan 2011 11:25:29 +0100
From: =?ISO-8859-1?Q?Stefan_Sch=E4fer?= <schaefer@iup.uni-heidelberg.de>
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Subject: Re: LF: Re: Re: FET RDS
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Hi Chris,

Of course you are right about the measurement errors. Furthermore one=20
has to think about if the driver "losses", so the power needed to drive=20
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=
=20
i do not care if the losses are 40 W or 50W. It's important that the PA=20
runs stable and the heatsink remains cool, even in QRSS 120 and a SWR of=
=20
1.5 (SWR is an issue in class E)
I am measuring the input voltage and current and measure the output=20
voltage across a 50 Ohm load (this 50 Ohm load consist of series and=20
parallel switched resistors of 100 Ohm +- 1% / 100W / TO247). Maybe=20
there is an error of 5% :-)

73, Stefan

PS: I've got efficiencies above 100% as well. But this was always caused=
=20
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 Sch=E4fer wrote:
>   =20
>> My LF 1 kW class E PA i using a single IRFP360 at>  96 % efficiency
>>     =20
> I'm sure you are getting high efficiencies. But measuring it at these nu=
mbers close to 100% actually makes quite interesting demands on the accura=
cy 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 vo=
ltage measurement.
>
> If you measure the output on a scope, personally I wouldn't trust a scop=
e measurement to better than a couple of percent. The Vout and Iout measur=
ements (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 erro=
rs 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 cla=
ss 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
>
>
>
>
>
>   =20