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Subject: Re: Reducing contact resistance for low volt use?
Date: Thu, 12 Dec 2002 17:46:47 -0800
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John Fields wrote:
> "mike" wrote in message
> > There are a whole bunch of practical issues
> > building something in a box
> > that can be carried off the bench and hooked up
> > with real wires.
> > To get the measurements I posted,
> > I had to use my parallel-gap spot welding head and
> > hook it up with
> > #2 cable just to stay within the dynamic range of
> > the 4-terminal
> > milliohm meter. 12 milliohms is a damn small
> > number in a practical
> > pressure contact test configuration.
> > Here's what I suggest.
> > Get a +5V 100A power supply, just cause they're
> > cheap surplus.
> > 2V is probably plenty.
> > Modify the supply to adjust the current limit over
> > the required
> > range of 0 - 100A. DON'T modify it for lower
> > voltage unless
> > you reduce the input voltage. You keep the heat
> > in the supply
> > down by running it at it's max voltage.
> > Sit the battery on top of the +5V.
> > Hook a 50 milliohm 500W+ resistor from the top of
> > the battery to ground.
> > You want the resistor to sink the 100A at as high
> > a voltage as possible.
> > Making the resistor smaller will run it cooler,
> > but the heat will go into
> > the supply instead. Keep it outside in the
> > resistor.
> > Assuming your resistor is stable, you can measure
> > the current as the voltage
> > at the resistor. If not, put a calibrated meter
> > shunt in series.
> > Measure the voltage differentially across the
> > battery with a separate
> > kelvin connection. It's already a floating
> > measurement, so sitting at 5V shouldn't be
> > a problem. You can put the battery "under" the
> > supply, but I'd rather not
> > have to float the whole thing...dpends on what
> > supply you get.
> > Build a comparataor at 5.5V or so to abort the
> > test when the cell voltage
> > hits 0.5V...or pick a number.
> > The low tech solution is to put two 100A diodes
> > from the battery + to the resistor
> > and one 100A diode from the - of the battery to
> > the top of the resistor.
> > Will need a lower resistance resistor to
> > compensate for the diode drop.
> > That will limit the disharge to half a volt or
> > so. If you can't get the parasitic
> > resistances low enough, use one diode in each leg.
> > Probably sufficient if it's an attended
> > measurement and you just want to prevent
> > an explosion if you get distracted.
> > Having 5V to burn dramatically simplifies the
> > interconnect problem.
> > You can use wires that you can flex by hand. But
> > remember, that doesn't
> > relax the requirement for a low resistance contact
> > point. You don't want
> > heat at the contact.
> > And you don't have to build any active circuitry
> > or worry about heat sinks.
> > As for the brass brush connection, watch out for
> > the resistance/heat.
> > To get reliable contact, the bristles need to
> > contact at an angle so each
> > has it's own springiness/compliance. Copper is
> > not good for this cause
> > it's too ductile.
> > I'd worry about too much pressure with current
> > applied.
> > In my welding experiments, I've blown holes in the
> > battery case.
> > Learned very early that I needed more current for
> > shorter time to prevent this.
> > A hot contact area under pressure can bend
> > significantly inward.
> All well and good, but unless I missed something I can't see how your rig is
> going to keep the required constant current coming out of the cell. That
> is, where is the mechanism which will lower the resistance of the load as
> the cell's voltage falls?
The current is constant, because it's controlled
by the current limit
in the power supply. Therefore, the voltage on
the resistor will be constant.
The output voltage of the power supply will change
to keep the total series
voltage of the cell and power supply constant.
Stated another way, the
power supply is run in current limit and is a
fixed current source.
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