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From: "Da Man"
Subject: Re: Looking to make a 56 Amp NiMH Battery Charger
X-Newsreader: Microsoft Outlook Express 5.50.4807.1700
Date: Tue, 07 Jan 2003 13:16:36 GMT
NNTP-Posting-Date: Tue, 07 Jan 2003 09:16:36 AST
"Fritz Schlunder" wrote in message
> "Fritz Schlunder" wrote in message
> > "Ian Rotheram" wrote in message
> > news:email@example.com...
> > > Help,
> > >
> > > 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)
> > >
> > > yours
> > >
> > > 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
> > presumably low voltage 270 amps we are working with).
> > You don't specify what voltage your battery pack is for the 270 amp
> > but lets assume about 6V (5 cells) since you mentioned that earlier.
> > suggests you need about 22 milliohms of resistance for your load, but
> > power dissipation of 270 x 6 ~= 1600W.
> > If we go here http://www.mogami.com/e/cad/wire-gauge.html we find that
> > gauge solid copper wire has a resistance of about 13 milliohms per
> > So how about taking about 1.7 meters or so of 16 gauge wire and stuffing
> > into a 5 gallon (~20 liter) jug or bucket of reasonably clean water.
> > wire need not be insulated (or if so only enamel coated) since
> > clean water will have dramatically higher resistance than copper will,
> > 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
> > 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
> > is 270A*6V*3600s = 5.8 megajoules. Assuming the water was well
> > this suggests a temperature rise of (5,800,000J)/(80,000J/K) = 73 deg.
> > 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
> > load off when the battery voltage decays to your set value. I might
> > using 10+ of the IRF3711 MOSFETs since they offer good bang for your
> > 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
> > (and understand) International Recitifier's application note AN941
> > http://www.irf.com/technical-info/appnotes/an-941.pdf before
> > If you parallel 25+ or so of the devices heatsinking may not be
> Just thought of something that is very important that I forgot to mention
> earlier. Design the load resistance to be as non-inductive as possible.
> This means do NOT coil up the ~1.7meters of 16 gauge wire (or whatever you
> select). Instead construct the resistor by taking the wire and looping it
> back over itself over and over (like a ribbon, or like a hiking trail
> up a steep mountain that switches back and forth on itself) to lower the
> entire high current loop inductance.
> The energy stored in a 2uH indunctance (an easily obtainable figure with
> meters of wire with an air core) is 0.5LI^2 (where L is inductance in
> henries, I is current flowing through inductor, and the result is the
> stored in joules).
> At 270 amps 2uH stores 0.5(0.000002)(270)^2 = 73millijoules, a definitely
> non-negligible figure.
> If the 270 amps causes your MOSFETs to avalanche (which it certainly would
> since the output capacitance of 10+ IRF3711 MOSFETs is very small and
> nowhere near 73 millijoules at under 20V), it is probable that the
> will not be evenly distributed over all MOSFETs equally. As such any one
> MOSFET could easily exceed (massivly at that) the maximum avalanche
> capability of the device (30amps), and your MOSFETs may (likely will) be
> So, what you need to do is snub this energy. You could use a massive RC
> snubber (or RCD clamp), or what I would be more inclined to do would be to
> use about three 1.5 kilowatt transient supressor diodes in parallel with
> small series resistors (to insure even sharing of current). Perhaps three
> or more 1.5KE10A (aka 1N6271A) (datasheet at:
> http://www.onsemi.com/pub/Collateral/1N6267A-D.PDF) transient supressors
> (with say 0.05 ohm resistors in series) could be paralleled across your
> MOSFET bank. Alternatively if you can find one perhaps a single 5KP9.0A
> (http://www.littelfuse.com/PDFs/Products/5KP.pdf) transient supressor
> (without any resistor in series) could be used instead.
> The 1.5KE10A devices are available from digikey for around 70 cents a
> My question for other people suggesting using a relay to switch off the
> current. Where do you find a relay suitable for switching greater than
> 270Amps DC? Also, how much does it cost? Although I've certainly never
> seen one I'm sure they exist somewhere, but I suspect they cost
> significantly more arms and legs than a MOSFET solution would...
Ford style Starter Solenoid! They cost about $10 Canadian. 12V drive (in the
range of 1A) , and most will tolerate long duty cycles.
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