John Fields said:
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I don't think so. What's happened is that you have fixed in your mind
what you want to hear and you don't understand what you've being told
so you've concluded that since it doesn't sound like what you want to
hear it must be bogus.
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Maybe your right. I have locked my mind on a possible solution and refuse
to accept "No" for an answer. I'm a computer programmer by trade and
find that other developers (and me sometimes) say that something cannot be
done, only to find that with a little more thought and creativity, that
there is a way (all-be-it complicated) to accomplish whatever is was.
I do see electronics as being any different to a computer program, after
all they're not that unrelated. A program is, to put it simplistically is
basically a sequence of flipping switches. (ok ok, that's probably too
simple)
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If you increase the number of devices demanding current from the
battery it won't "overrate" the rest of the stuff, all it'll do is
reduce the amount of time until the battery voltage drops to some
arbitrary point.
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oops, sorry. I should have been talking Amperage, the leads that run from
the battery can carry about 40Amps (Not sure the exact thickness, but it's
the stuff that's used in model power boats to connect from the battery to
speed controller. I think it's about 4 or 5 mm thick)
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Yes, it will. There is a voltage below which current shouldn't be
taken from the battery, and once the battery voltage decays to that
point it should either be disconnected or recharged.
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OK, I need to get more litteral about this. At the moment, the circuit
that I'm using measures a voltage range between 11v and 14v. I know from
previous experiments that I want to recharge my battery when the voltage
drops to about 11v with no load (other than the monitor circuit). Also I
know that as I increase the load the voltage that the monitor reads is
forced lower by .x of a volt depending on the load. The higher the load,
the higher that this .x is. At the moment my circuit, when under high
load, shows the 11V reading pretty quickly when the battery is under a
high load. When I shut off the load, the voltage returns to a higher value
and gives it's true reading, which match the chemical thingy in the
battery (which I can't see in normal use as it's not illuminated, and shut
in the boot of a car)
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Yes, it does. If you look at the discharge curves for _any_ battery
you'll find that as the rate of discharge increases (as more current
is drawn from it) the smaller its capacity becomes. The capacity of
most batteries (C) is rated in Ampere Hours, but full capacity can
only be achieved if some fraction of the one hour rate of current is
drawn, ususally C/10 or C/20 for lead-acid batteries. That means that
if you have a fully charged 100AH battery rated for C/10 and you draw
10 amperes from it, its voltage will decay to the cutoff point (say
10V for a 12V lead-acid battery) in 10 hours. However, if you take
100A from it its voltage will decay to 10V in substantially less than
1 hour. Also, since the battery's internal impedance will cause its
voltage to fall more and more as more and more current is drawn from
it, that will futher shorten the time until it reaches cutoff.
Let me put this another way. Say I've been using my battery for a while.
I will consider it completely flat when is reaches 10v either under load
or not. At the moment, my battery is reading 11.3v and is under load. My
circurit, is showing 11.3v. When I turn off the load, the battery voltage
immediatly jumps up to 11.7v. When I turn on the load it slowly returns
to 11.3v. One of the devices that I'm using is a laptop. When the hard
drive is working, the current rapidly changes as the drive heads move
across the disk and data is read/written. This turns the LEDS on the
circuit into quite a good light show, which is really annoying and I want
to stop it doing that.
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Well, you should, and that's precisely why I said that you want to
hear what you want to hear, not what's at variance with what you
believe.
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hmm, I shouldn't have said that. It's why I built the circuit in the
first place. What I want is a steady reading, not one that makes the
LED's look like a reject from a bad SCI FI Film.
I asked for that.
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Won't anyone give me a straight answer on how the hell to build an
Ammeter
circuit, so that I can get the "BATTERY LEVEL MONITOR" to give a correct
reading?
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Yowzah boss!!!
+-------------------+
/ |
A--+ [FUSE]
| |
[AMMETER] +----------+
| | |
+-----------+ [DRIVE] [VOLTMETER]
| | | |
[BATTERY] [VOLTMETER] +----------+
| | |
B--+-----------+ |
\ |
+-------------------+
Everything else connects (just like the drive with its own set of
wires and its own fuse) to points A and B. That is, directly to the
ammeter and the battery.
But... That's still only going to give you voltage readings and
current readings, so you'll still have to disconnect loads depending
on battery voltage.
Or maybe you want something to let you know how much charge is still
in the battery or how much time you've got left until it goes flat? A
battery "gas gauge" kind of thingy?
Yes, that's exactly what I'm trying to achieve.
Well, boss, if that's what yuh wants, just ax fo' it an' ahm sho' some
of us ol' niggers'll jump at the chance to serve yuh.
Colin Dawson
www.cjdawson.com