From: Chris Carlen
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Subject: Re: 87% -- measured inductor current finally
References: <firstname.lastname@example.org> <email@example.com>
Date: Wed, 25 Dec 2002 05:09:27 GMT
NNTP-Posting-Date: Tue, 24 Dec 2002 21:09:27 PST
Tony Williams 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.
I was a bonehead and measured the supply current with an ammeter, and
with all the noise in the line, it read too high.
I put the scope on there, and the current went to 1.2A in 2.67us with
45.1V supply, so that gives 100.uH like I expected :-)
But that also means my efficiency calcs. were way off, and it seems that
with the current sense resistor installed, I got 320V into 10k ohms,
with 45.1V in at 1.2A*0.4*0.5=0.24A (add 0.02A for the PWM), oh that's
I pushed the output to 400V, and got 1.8A peak inductor current, with no
trace of upward curling. So I am well clear of saturation. At 1.8A,
that's about 0.25mT! The frequency was such that the on time was 4.0us,
so that's still 100.uH.
The power end is 98.5% efficient at this level! (That's just based on
the inductor current triangle, you can tell me what I might be missing???)
I'll try to get updated circuits and waves on my website, but I might
not be able to do so until after I get back from Thailand Jan 18.
Christopher R. Carlen
Suse 8.1 Linux 2.4.19