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Message-ID: <00a701c8200c$be6d7540$0900a8c0@Lark>
From: "Alan Melia" <alan.melia@btinternet.com>
To: <rsgb_lf_group@blacksheep.org>
References: <007301c81fa0$aabd9a00$65ebfc3e@g3kev> <20071105121359.09bfb671@saluki.ngw.uk.com> <003601c81faa$d39e08b0$0d00000a@AGB> <005101c81fcb$fc19cab0$4bd0fc3e@g3kev> <000b01c8200a$2e1885e0$0d00000a@AGB>
Date: Tue, 6 Nov 2007 00:34:02 -0000
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Subject: Re: LF: CLASS D E.... Instrumet induced errors ? Misunderstanding ?
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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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