LF: Re: 500kHz Classe-E

["James Moritz" <james.moritz@btopenworld.com>]

Dear Andy, LF Group,

Looking at the Class E design formulas, initially assuming efficiency near 
100%, 55Vdc, then the peak current in the switch is about 24A, which is 
clearly going to drop quite a few volts in the 0.4ohm Rdson. The text I have 
here gives a formula for efficiency where the losses are entirely due to 
Rdson - putting in the relevant numbers gives 86% efficiency, which with 
500W input would result in about 70W dissipated in the transistor. That 
would need quite a big heatsink!
2 in parallel would give 93% efficiency by this reckoning, a more reasonable 
36W would be dissipated across 2 devices, so much more manageable. Actually, 
you would have to increase the DC input power in order to reach your target 
500W RF out, and to overcome additional losses in the feed choke and tank 
circuit, etc.. Then the Rdson is specified at 25degreesC junction 
temperature, and it rises with temperature - about 1.5x at 80degrees, for 
example so you are probably looking at well under 90% efficiency now. Maybe 
3 in parallel...
Cheers, Jim Moritz
73 de M0BMU

----- Original Message ----- 
From: "Andy Talbot" <andy.g4jnt@gmail.com>
To: <rsgb_lf_group@blacksheep.org>
Sent: Thursday, January 06, 2011 9:29 AM
Subject: LF: 500kHz Classe-E


> Starting parameters :
>   Supply voltage anything from 28 - 55V adjustable.  25 A max
>   PA device  - IRFP452 (have a small handful, so can allow a few to go
> pop)  Rated 500V,   0.4 Rdson, 14A,  190 Watts
>   Drive, accurate 50% square wave from ICL7667
>   Resonant Inductor -  air wound with 2mm Litz wire.
>
> I reckon that at 55V 500 Watts might be a value to aim for with a single
> device, with the option to back off by lowering Vsupply.