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On 24 Feb 2004 11:48:17 -0800,
Norm Freedman wrote: Any electronics experts out there? Would appreciate help with a simple circuit: to light an LED when current in a 12vdc circuit goes above approximately 10 amps (adjustable from about 8 - 13 amps would be nice) and drop out when current goes below that set point. Maximum current is about 20 amps. Any ideas? Thanks for the help Try Maxim, http://www.maxim-ic.com/appnotes.cfm/appnote_number/2015/ln/en They are usually quite happy to send a sample of an IC. Or you can order them from the usual suspects like Mouser. Note that the appnote above, also cuts off the load in the event of overcurrent, but you can cut that part out and ignore it. If you'd rather roll your own, you'll need a (couple of) precision low value resistor(s) about 0.01 Ohm or so, then rig up a voltage drop measuring circuit around that, adjust for expected current levels. Most of the circuit cookbooks will have examples you can copy. With 0.01 Ohm, passing 10A through it, results in 0.1V drop, which is a bit small for some circuits to reliably measure, but only dissapates 100mw, and if the current goes to 20A, your resistors are only dissapating 200mw, which is probably ok for most things. Although it's a couple of amp hours a day if it runs that long. You might be able to get away with using a magneto-resistive system, but with DC, the results are likely to vary widely depending on what other magnetic fields are around, and are likely to change when you start or stop the engine, or other unrelated magnetic sources. -- Jim Richardson http://www.eskimo.com/~warlock If you can modify it via software, then it doesn't count as hardware. -- Lionel, in the Monastery |
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