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From: John Woodgate
Subject: Re: Doubling wallwart power rating
Date: Tue, 8 Oct 2002 09:27:38 +0100
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Tue, 8 Oct 2002 10:20:00 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Phil Allison
wrote (in ) about 'Doubling
wallwart power rating', on Tue, 8 Oct 2002:
>"John Woodgate" wrote in message
>> How did you measure it, exactly?
[snip interesting account of meter design]
> You did say exactly !!
Yes; it matters.
>Did you disconnect the rectifier, so as
>> to measure the *transformer's* no-load current?
> ** Oh come on, to remove filter capacitor leakage current!! Bit
Well, while you might know that the cap wasn't faulty, it does happen.
> Did you use a true
>> r.m.s. meter, because the magnetizing current is very peaky? If that
>> transformer drew around 1 A (12 V x 500 mA x 1.6) on no-load, there was
>> something seriously wrong with it.
> ** I think you meant 10VA not 1 amp. Yep, the off load current was
>about 45 mA. The primary was about 350 ohms which gives a copper loss of
Well, 12 VA. I did a secondary-side calculation and inadvertently left
out the conversion to 240 V.
Your copper-loss calculation is correct for no-load, i.e. no copper loss
in the secondary.
> Most wall warts I see are similar - when loaded at the rated
>current the magetising current peaks drop down to about 60 % of the off load
>value and charging peaks appear at slighlty less amplitude. The rms value is
>the same +/- 10%.
With 12 V at 500 mA d.c. output, the secondary voltage must be about
13.5 V and the current must be about 800 mA, assuming a bridge
rectifier. That's a secondary side power of 10.8 W. Only if the
transformer were 100% efficient would that translate to 45 mA (your
measured off-load current) at the primary side. In fact, the copper loss
must be about (not exactly) 1.4 W and there is some iron loss, likely to
be less than the copper loss unless the transformer uses semi-scrapless
laminations. Say 2 W total losses. Primary current 53 mA, plus
magnetizing current in quadrature. That was 45 mA offload. Suppose it's
30 mA on load, due to the primary IR drop reducing the voltage available
for generating induction. That's 61 mA total.
>> Remember our discussion about mains leads? When you gave me ALL the
>> facts about your tests, we agreed.
> ** Not quite the way I remember it.
We initially disagreed because I did calculations based on the wire
having the normal resistivity of copper. When you disclosed that in fact
the wire had a much higher resistivity, we agreed. I passed your warning
on to the British Standards committee responsible for 60065 and 60950,
with advice to members to increase goods inwards checks on mains leads.
I wonder if any others did that in their countries.
>> > You have to do tests to know - theory will not tell you.
>> Ah, well, that depends on whether you know how to apply the theory, of
> ** Believing that is your problem John. Therory will not tell you what
>some maker in a factory decided to do.
Applying the theory when evaluating products *includes* knowing how
other designers apply it. No, the theory won't tell you what design
practices are *used*, but it will tell you how to interpret the results
I do tests; it's what I do to while away the long hours between e-mail
discussions with you. (;-)
If you design a transformer so that the core is deep into saturation at
nominal mains voltage (which is what is implied by the input current
being the same on no-load as on full load), it WILL overheat under
safety testing at high mains voltage (+6% or +10% depending on the
standard; one even requires +15%!).
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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