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LF: Re: 500kHz Classe-E

from [James Moritz] [Permanent Link][Original]
To: <[email protected]>
Subject: LF: Re: 500kHz Classe-E
From: "James Moritz" <[email protected]>
Date: Fri, 7 Jan 2011 00:13:48 -0000
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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" <[email protected]> 
To: <[email protected]>
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.
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