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Message-ID: <000901c82095$5f3cdfa0$9b0f7ad5@w4o8m9>
From: "James Moritz" <james.moritz@btopenworld.com>
To: <rsgb_lf_group@blacksheep.org>
References: <007301c81fa0$aabd9a00$65ebfc3e@g3kev> <20071105121359.09bfb671@saluki.ngw.uk.com> <003601c81faa$d39e08b0$0d00000a@AGB> <005101c81fcb$fc19cab0$4bd0fc3e@g3kev> <000b01c8200a$2e1885e0$0d00000a@AGB> <00a701c8200c$be6d7540$0900a8c0@Lark> <004301c8207c$b17be720$0d00000a@AGB>
Date: Tue, 6 Nov 2007 16:52:09 -0000
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Subject: Re: LF: CLASS D E.... Instrumet induced errors ? Misunderstanding ?
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Dear Graham,

For a CW output, as Alan said, it isn't a square wave - or any other wave
really, apart from a tiny amount of RF ripple. For the class E, and the
usual "current switching" Class D circuits, the DC current is fed via an RF
choke - in order not to knock the circuit out of tune, the choke needs a
reasonably high reactance at the operating frequency, which means that the
RF current in the choke must be relatively tiny compared to the DC current.
The small RF ripple current will be further soaked up by the supply
decoupling capacitors, so what is coming out of the DC supply should be
nearly pure DC.

It is slightly different with "voltage switching" class D stages, such as
the one used in the 500kHz EER system. There, the output devices draw
current direct from the supply rail, and large, low-impedance decoupling
capacitors are an essential feature of the PA. But in such a design, you
will always want to have additional DC supply decoupling choke/capacitor
filtering to prevent RF currents circulating in external parts of the
circuit, so again the DC feed to the PA stage will be nearly pure DC.

So unless a switching PA (or any other PA for that matter) has awful
decoupling, you should not need to worry about measuring anything other than
DC votage and current from the power supply.

Measuring the input and output powers under dynamic, modulated conditions is
another thing - how you do it will depend on what you are trying to
measure - peak power, mean power, mean efficiency, instantaneous efficiency,
etc. But since the modulation level varies relatively slowly compared to the
RF signals in the PA, the efficiency of the PA at any particular output
level will not be much different from what you measure under CW conditions
at that level - which is certainly much easier to do.

Cheers, Jim Moritz
73 de M0BMU

----- Original Message -----
From: "Graham" <g8fzk@g8fzk.fsnet.co.uk>
To: <rsgb_lf_group@blacksheep.org>
Sent: Tuesday, November 06, 2007 1:55 PM
Subject: Re: LF: CLASS D E.... Instrumet induced errors ? Misunderstanding ?


Hi Alan,

.. Thats the point ,

>>>>(((((Note that it is not a 50% squarewave)))). Conversely if the design
> criteria are not followed the efficiency will be compromised. These PAs
> are
> really just switching power supplies and those achieve very high
> efficiencies at similar frequencies.<<<<

Its a good system ,



I was noting a lot of 'posts' referring to  , P=I*V , , even if it was a 1:1
square wave, you would still need either a form factor , or better still , a
'true rms' meter . to make the final measurement , assuming it has
sufficient bandwidth ,



A scope with a floating differential input , measuring across a very low
'passive' resistance would do  the trick .. ..



Its a bit like measuring the peak power out of a ssb set , do you  measure
the peak envalope power , by peak votatage and calculate the power and then
mult by 0.707 to  get the rms level .or just take the peak ... or just dont
ask ?



Graham ..

















----- Original Message -----
From: "Alan Melia" <alan.melia@btinternet.com>
To: <rsgb_lf_group@blacksheep.org>
Sent: Tuesday, November 06, 2007 12:34 AM
Subject: Re: LF: CLASS D E.... Instrumet induced errors ? Misunderstanding ?


> Well the thermo RF current meters measure RMS current, but have you looked
> at the current from the PSU on a scope?? It shouldn't vary much if the
> choke
> is doing its job properly. The FET switches the current to ground and when
> it opens the current is diverted to flow into the shunt capacitor. The
> choke
> tries to maintain the current flow by generating a "back emf"....this is
> what chokes do. If the design is done properly (the choke and the shunt
> cap
> are the right size) there should ve very little "ripple" in the current,
> so
> the load seen by the psu should be fairly constant and it should not show
> "sag". Note that it is not a 50% squarewave. Conversely if the design
> criteria are not followed the efficiency will be compromised. These PAs
> are
> really just switching power supplies and those achieve very high
> efficiencies at similar frequencies.
>
> Alan G3NYK
>
> ----- Original Message -----
> From: "Graham" <g8fzk@g8fzk.fsnet.co.uk>
> To: <rsgb_lf_group@blacksheep.org>
> Sent: Tuesday, November 06, 2007 12:15 AM
> Subject: Re: LF: CLASS D E.... Instrumet induced errors ?
>
>
> Thats the point, the out put is 'pure' no problem to  measure , but the dc
> feed looks to be a 'complex' function ...any allowance made in the calc's
> ,
> past  P = I*V ?
>
> Same problem, pulsed radars are very sensitive to psu ripple causing
> 'spurious side bands'
>
>
> ----- Original Message -----
> From: "hamilton mal" <g3kev.ham@virgin.net>
> To: <rsgb_lf_group@blacksheep.org>
> Sent: Monday, November 05, 2007 4:49 PM
> Subject: Re: LF: CLASS D E.... Instrumet induced errors ?
>
>
> It looks like you have a serious problem, sounds more like a Radar tx.
> The
> output from my class E is a beautiful sine wave, this is followed by a 5
> pole LPF.
> For testing it is preferable to use a spectrum analyzer, and a good
> calibrated scope and a quality DL
>
> G3KEV
>
> ----- Original Message -----
> From: "Graham" <g8fzk@g8fzk.fsnet.co.uk>
> To: <rsgb_lf_group@blacksheep.org>
> Sent: Monday, November 05, 2007 12:53 PM
> Subject: Re: LF: CLASS D E.... Instrumet induced errors ?
>
>
> ..... The 'I' supply to  the stage dosent look like a sine wave , more
> like
> a pulse train ? , thats the case will need a little more than a avo-8 to
> get the  correct value,  vlaues may be actually higher than conventionally
> measured, giving lower conversion factors ? supply bandwith may also be a
> issure, caused by switching -rise times- is there a voltage  collapse on
> the
> leading edge, there must be some , but how much  , reducing  the 'area
> under
> the curve' ?
>
> G ..
>
>
>
>
>
> ----- Original Message -----
> From: "John GM4SLV" <gm4slv@sighthound.demon.co.uk>
> To: <rsgb_lf_group@blacksheep.org>
> Sent: Monday, November 05, 2007 12:13 PM
> Subject: Re: LF: CLASS D E
>
>
> On Mon, 5 Nov 2007 11:39:54 -0000
> "hamilton mal" <g3kev.ham@virgin.net> wrote:
>
>> Hi all
>> Tnx to all for the information, but some of the figures do not work
>> out. How are the % figures measured.
>> One simple method is to calculate the DC input PWR in watts (I x V)
>> then using an RF current meter calculate the PWR out (Isq x R) let R
>> be the norm 50 ohm D/load. Compare the two figures and calculate the
>> Eff % 73 de Mal/G3KEV
>>
>>
>>
>
> Again...with my TX
>
> At 50W output :-
>
> Dc supply = 18.7v at 3.4A
>
> Efficiency = 78%
>
>
> The math's is trivial.....
>
> 18.7V x 3.4A = 63.58W DC input
>
> 50W RF output (as measured on a homebrew power meter calibrated against
> a £20,000 R&S Spectrum analyser and precision 40dB power attenuator)
>
> Efficiency = Power Out / Power In
>
>
> 50 / 63.58 = 0.786
>
> Hence 78.6% efficient.
>
> Of course if you can't accurately measure any of the necessary
> quantities then the end result will be meaningless.
>
> Measuring RF output using an RF ammeter and 50 ohm load is all well and
> good...but...how accurate is your RF ammeter? How accurate is the 50
> ohms. How is the 50 ohms measured... at DC? Is it the same at RF?
>
> With a good Class-E amp and 90% efficiency it only takes one of your
> measurements to be out by a few percent to make the efficiency
> calculated to be really way out.
>
> Say (plucked out of thin air example):-
>
> DC input (most accurately measured parameters?)
>
> 13.8V at 5.3A = 73.14W
>
>
>
> RF output as measured by your RF ammeter/50 ohm load
>
> 1.17A in 50 ohms = 68.4W
>
> Efficiency = 93.6%
>
>
>
> If your RF ammeter reads 5% high...
>
> I = 1.23A in 50 ohms = 75W Out
>
> Efficiency = 103%
>
> Contact the Nobel Prize Committee... you've invented the prepetual
> motion machine.....
>
>
>
> If your 50 ohm load is really 49 ohms however...
>
> 1.17A in 49 ohms = 67W
>
> Efficiency = 91.7%
>
>
>
> Don't assume there's some black magic mumbo-jumbo Emperor's New Clothes
> syndrome with MOSFET amps just because it appears that you can't tie
> down a reliable efficiency figure. Little errors make big differences.
> The important thing is that they work, don't get hot and can produce
> serious power for little engineering. Try getting a 2N3055 to produce
> 150W on 500kHz.
>
> Cheers,
>
> John
>
>
>
>
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