From: Tony Williams
Subject: Re: 87% All that vector calculus paid off
Date: Mon, 23 Dec 2002 07:56:56 +0000 (GMT)
NNTP-Posting-Date: Mon, 23 Dec 2002 08:32:42 +0000 (UTC)
User-Agent: Pluto/1.14i (RISC-OS/3.60)
In article ,
Chris Carlen wrote:
> >>At the moment the thing is driving a 10k resistor with 351.5V,
> >> pulling 0.315A from a 45.1V supply, for 87.0% efficiency.
> >> I'm running at 150kHz with 40% duty.
> > A very quick WAG on those numbers is that Lp= 70uH
> > and Ipk= 1.5A, approx.
> I don't understand why you calculate only 70uH? Are you using some
> formula that attempts to deal with the non-linearity of the
Nothing to do with the magnetics Chris. The guesses are
straight off your input/output figures above.
0.315A at 40% duty cycle gives a 0.8A mean current during
the ON-time. For a triangular current, Ipk= 1.6A.
150KHz, 40%, is an ON-time of 2.7uS.
V= LdI/dT. 45V = L*1.6A/2.7uS gives L= 76uH.
Or from another direction.... P-out is 12W, and guess
that there is another 1W lost on the Sec side (and in
the core). P-in to the core is 13W. 13/150KHz = 87uJ.
At 1.6A, 0.5*L*I-squared gives L = 68uH.
A useful WAG to do because it points to scoping the
ON-current as an early thing to do.... verify that
apparent 1.6A, check that it is reasonably triangular
(and not curling up as the core saturates), etc.