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It's just occurred to me that the model is based on the assumption that the gauge supply is a fixed 12V. Is the gauge supply the same as the battery you are monitoring? If so, what battery voltage can we take as 'empty' and what as 'full'? The model resistors will need adjusting to cater for a varying gauge supply voltage.
Is the "R27" current now impervious to transient flucuations within that range, or are you saying that what's shown on the graph is dependent upon the supply voltage being 10.5 V @ empty, 12.0 V at 50%, and 13.5 V at 100% state of charge?I've remodelled things assuming the input source goes from 0V to 5V as the '12V' bus goes from 10.5V to 13.5V.
What are the implications if R4 isn't adequately tweaked to produce that exact voltage?R4 may need tweaking so that the input voltage on IC pin 5 is a tad over 2.5V when the source is 5V.
What do I need in the way of capacitors, and what nodes should they be attached to?Not shown on the schematic are essential decoupling capacitors for the IC supply pins.
Yes. The model uses the equation Vsupply = 10.5 + 0.6 * Vsource.are you saying that what's shown on the graph is dependent upon the supply voltage being 10.5 V @ empty, 12.0 V at 50%, and 13.5 V at 100% state of charge?
Step 1 may be at the same level as step 0 (i.e.24mA), or slightly shifted along the x-axis.What are the implications if R4 isn't adequately tweaked to produce that exact voltage?
The datasheet recommends 10uF tantalum or 22uF otherwise, connected as close as possible to the chip supply pins (2,3).What do I need in the way of capacitors, and what nodes should they be attached to?
Oh, yes, I see that now. Hmm. That may not work very well. The supply voltage will never be more than 12.7 V, and shouldn't be less than 11.5 V steady state, with excursions as low as 10.5 V only under heavy acceleration with a depleted battery. (I've always assumed that the brief drop in gauge reading under such conditions was essentially unavoidable. If you've got some trick up your sleeve to provide some sort of truly regulated [as opposed to limited in the less-than-or-equal-to sense] 12 V supply, I'm all ears, although I unfortunately didn't think to add provisions for such a thing when I had the instrument panel out of the car over the summer.)Yes. The model uses the equation Vsupply = 10.5 + 0.6 * Vsource.
A fully charged common lead acid car battery is normally around 13.8V under light load. Why 12.7?The supply voltage will never be more than 12.7 V
There's no voltage headroom to put in a regulator to give the gauge a constant 12V supply, but if you can get at the gauge +V wiring it might be possible to run it from, say, a regulated 9V.If you've got some trick up your sleeve to provide some sort of truly regulated [as opposed to limited in the less-than-or-equal-to sense] 12 V supply
You're probably thinking of a traditional car battery application wherein the alternator is providing a constant 13.5-14.3 V charging voltage. Open circuit voltage of a '12V' lead acid battery is never more than 13 V, in this case 12.5-12.7.[/QUOTE]A fully charged common lead acid car battery is normally around 13.8V under light load. Why 12.7?
Battery voltage is the only thing it measures directly. If I understand correctly, it monitors transient changes in voltage, and the duration of those reduced voltage levels to infer state of charge by some kind of integration technique. It's a Curtis model 933 battery monitor if you want to look it up. (I believe the monitor is made in the USA, but I salvaged it from a British forklift, so I thought it was a rather interesting coincidence that both of you who responded to this thread are in the UK...)I'm unclear how your 0-5V source operates. Is it merely monitoring battery voltage, or does it monitor charge/discharge current and duration to give an output representing charge state?
Couldn't find any useful info on the 933.
Can we assume then, for modelling purposes, that the gauge supply drops uniformly from 12.7V to 11.5V, while the monitor (source) drops from 100% to 0%, going from 'full' to empty'? Or do you have an alternative suggestion?