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From: "Fritz Schlunder"
Subject: Re: Looking to make a 56 Amp NiMH Battery Charger
X-Newsreader: Microsoft Outlook Express 5.50.4522.1200
Date: Mon, 6 Jan 2003 07:23:45 -0700
NNTP-Posting-Date: Mon, 06 Jan 2003 08:21:25 CST
"Ian Rotheram" wrote in message
> can anybody suggest methods of making a 56 Amp NiMH battery charger
> for pack of cell either 3,4 or 5 cell packs or multiples of that, with
> the possibility of a car battery as the power source.
> Also as an extra while were on large currents, how about a homebrew
> dummy load one that can handle 40 Amps and turn off when the batteries
> reach a set level and another that can sink 270 Amps (Stop Laughing)
> Ian Rotheram
Well as far as the dummy load construction is concerned I would probably
tend towards constructing a home brewed water cooled resistor (since this is
presumably low voltage 270 amps we are working with).
You don't specify what voltage your battery pack is for the 270 amp load,
but lets assume about 6V (5 cells) since you mentioned that earlier. That
suggests you need about 22 milliohms of resistance for your load, but with a
power dissipation of 270 x 6 ~= 1600W.
If we go here http://www.mogami.com/e/cad/wire-gauge.html we find that 16
gauge solid copper wire has a resistance of about 13 milliohms per meter.
So how about taking about 1.7 meters or so of 16 gauge wire and stuffing it
into a 5 gallon (~20 liter) jug or bucket of reasonably clean water. The
wire need not be insulated (or if so only enamel coated) since reasonably
clean water will have dramatically higher resistance than copper will, and
therefore the water should not significantly alter your load resistance.
20 Liters is 20,000mL and each mL takes 1 calorie (about 4 joules or 4
watt-seconds) to raise one degree C. So it will take about 80 kilojoules to
raise our 20L of water one degree C. Suppose you dumped a fully charged
270Ah 6V battery into your 20L water reservoir. The battery energy storage
is 270A*6V*3600s = 5.8 megajoules. Assuming the water was well insulated
this suggests a temperature rise of (5,800,000J)/(80,000J/K) = 73 deg. C.
Of course that is over a one hour period so the maximum temperature rise
will be less since losses will become signficant at higher temperatures.
Assuming the water was initially at room temperature it should not boil.
A simple comparator with plenty of hysteresis (essential to insure
transistion without oscillations, or worse operation in a linear mode)
controlling the gates (hopefully with a real gate driver) of a set of
paralleled low on resistance MOSFETs could provide the control to switch the
load off when the battery voltage decays to your set value. I might suggest
using 10+ of the IRF3711 MOSFETs since they offer good bang for your buck.
Datasheet at http://www.irf.com/product-info/datasheets/data/irf3711.pdf
Digikey will sell you ten of them for $1.01 a piece. Make sure you've read
(and understand) International Recitifier's application note AN941
http://www.irf.com/technical-info/appnotes/an-941.pdf before construction.
If you parallel 25+ or so of the devices heatsinking may not be necessary.
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