On Jan 26, 9:14, dvcorbin_at_optonline.net wrote:
> >Bwahahaha, I've written 10 replies, but canceled them all, because I
can't
> >teach a house
> >plant calculus. You Win.
Yeah, I thought about explaining why the paragraph around the idea that
.5mA kills is drivel, and gave up.
> Seriously the TOTAL lack of regulation on "WallWarts" is quite
common. I am currently (preofessionally) developing a product which
normally operates off of AC power (via a wallwart), but needs to remort
low AC conditions and fall back to battery.
>
> The load on this device is VERY dynamic ranging from under 10mA to
over 850mA depending on what it is doing. The voltage fluctuations out
of the wall wart (which is rated as 11.8V _at_ 1A) will rise as high as
17V when under a minimal load.
>
> Since the device is intended to have a very low production cost, they
really cut some regulation requirements on the board as well, since the
components WILL tolerate this range of voltages.
>
> Unfortunately, the side effect is that the voltage variation based on
load is significantly greater than the voltage variation based on
fluctuations in the AC (eg during a brown out). This has required the
development of software that is constantly monitoriing the "active"
state of many of the devices to "calculate" the current load, and then
going through a transform to estimate the RAW AC that is providing
power to the wart.
Well, there might be a way to deal with the problem. Most of the poor
regulation is down to the following. Most small DC wall warts consist
of a transformer feeding a bridge rectifier with a moderate
electrolytic capacitor across it. Under low or zero load, the
capacitor charges to the peak voltage. As you apply more and more
load, you get more and more ripple, and the average voltage goes down
(in fact, because there is resistance between the transformer and
capacitor, so does the peak). However, you could analyse the ripple
and work out what the input was doing. At moderate loads, the
capacitor will charge rapidly to something near the peak voltage on
each half-cycle, and discharge relatively slowly; at high loads, you'll
get a more symmetric ripple. If you want to know in detail, you could
sample it, but my guess is if you're just looking for brownouts all
you'd need to do is compare the amount of ripple to the average voltage
and apply some rule of thumb. Or compare the slope of the rise to the
slope of the fall. Of course when the ripple goes away completely
you've got no mains input at all :-)
--
Pete Peter Turnbull
Network Manager
University of York
Received on Mon Jan 26 2004 - 14:32:53 GMT