Message-ID: <003601c81faa$d39e08b0$0d00000a@AGB>
From: "Graham" <g8fzk@g8fzk.fsnet.co.uk>
To: <rsgb_lf_group@blacksheep.org>
References: <007301c81fa0$aabd9a00$65ebfc3e@g3kev> <20071105121359.09bfb671@saluki.ngw.uk.com>
Date: Mon, 5 Nov 2007 12:53:02 -0000
MIME-Version: 1.0
Subject: Re: LF: CLASS D E.... Instrumet induced errors ?
Content-Type: text/plain;
	format=flowed;
	charset="UTF-8";
	reply-type=original
Content-Transfer-Encoding: quoted-printable
Sender: owner-rsgb_lf_group@blacksheep.org
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org


..... The 'I' supply to  the stage dosent look like a sine wave , more like=20
a pulse train ? , thats the case will need a little more than a avo-8 to=20
get the  correct value,  vlaues may be actually higher than conventionally=20
measured, giving lower conversion factors ? supply bandwith may also be a=20
issure, caused by switching -rise times- is there a voltage  collapse on the=
=20
leading edge, there must be some , but how much  , reducing  the 'area under=
=20
the curve' ?

G ..





----- Original Message -----=20
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 =3D 18.7v at 3.4A

Efficiency =3D 78%


The math's is trivial.....

18.7V x 3.4A =3D 63.58W DC input

50W RF output (as measured on a homebrew power meter calibrated against
a =C2=A320,000 R&S Spectrum analyser and precision 40dB power attenuator)

Efficiency =3D Power Out / Power In


50 / 63.58 =3D 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 =3D 73.14W



RF output as measured by your RF ammeter/50 ohm load

1.17A in 50 ohms =3D 68.4W

Efficiency =3D 93.6%



If your RF ammeter reads 5% high...

I =3D 1.23A in 50 ohms =3D 75W Out

Efficiency =3D 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 =3D 67W

Efficiency =3D 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




--=20
No virus found in this incoming message.
Checked by AVG Free Edition.
Version: 7.5.503 / Virus Database: 269.15.19/1106 - Release Date: 11/2/2007=20
21:46