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#1
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Since its so trivial for you, how about a basic indication of the
components hookup or reference to the circuit elsewhere? Anybody (else) know of a finished product source of this trivial product? Kit? RST Engineering wrote: Trivial to design and build. Jim |
#2
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abripl wrote:
Since its so trivial for you, how about a basic indication of the components hookup or reference to the circuit elsewhere? Anybody (else) know of a finished product source of this trivial product? Kit? RST Engineering wrote: Trivial to design and build. Jim Nope. That's mostly a roll yer own. Sounds like a relay circuit to me. Off the shelf automotive supplies? Wally World? |
#3
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This is starting to smell very much like the upcoming November Kitplanes
column. Jim "cavelamb" wrote in message k.net... abripl wrote: Since its so trivial for you, how about a basic indication of the components hookup or reference to the circuit elsewhere? Anybody (else) know of a finished product source of this trivial product? Kit? RST Engineering wrote: Trivial to design and build. Jim Nope. That's mostly a roll yer own. Sounds like a relay circuit to me. Off the shelf automotive supplies? Wally World? |
#4
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You mean I started a new idea? It sure would be handy to plug in
portable devices into the cigarette lighter socket source and not drain the battery with engine off. OR in case of charging system failure some non-crucial devices could be turned off automatically. We need this before November. RST Engineering wrote: This is starting to smell very much like the upcoming November Kitplanes column. Jim |
#5
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What do you want the cutoff voltage to be? Personally, I'd pick 13.0 volts,
but it is child's play to do any voltage you want. There will be about a milliampere of quiescent current for the sensor that will still be on-line when the output shuts off. Jim "abripl" wrote in message ups.com... You mean I started a new idea? It sure would be handy to plug in portable devices into the cigarette lighter socket source and not drain the battery with engine off. OR in case of charging system failure some non-crucial devices could be turned off automatically. We need this before November. RST Engineering wrote: This is starting to smell very much like the upcoming November Kitplanes column. Jim |
#6
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"RST Engineering" wrote in message
... What do you want the cutoff voltage to be? Personally, I'd pick 13.0 volts, but it is child's play to do any voltage you want. There will be about a milliampere of quiescent current for the sensor that will still be on-line when the output shuts off. Jim One-chip solutions designed for this purpose that can detect both undervoltage and overvoltage are available from several vendors, such as Maxim. Example: http://pdfserv.maxim-ic.com/en/ds/MAX6457-MAX6460.pdf These offer very low quiescent current when off (couple of uA). The output can be used to activate a relay (maybe needs an extra transistor). I think you can request free samples. Rob |
#7
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A. Maxim is very poor on production delivery.
B. Joe Sixpack in East Undershirt Ohio can't go down to the corner Maxim store and buy onesie-twosies on Sunday afternoon. C. One milliampere out of a 25 amp-hour battery will run the battery down in 25,000 hours (1040 days, or a little less than 3 years). If you don't fly your airplane but once in 3 years, that's not my problem. D. Why would you use an unreliable electromechanical device like a relay when a FET is half the price and a thousand times more reliable? Jim "Rob Turk" wrote in message news:Y7_gg.1536$YI3.268@amstwist00... "RST Engineering" wrote in message ... What do you want the cutoff voltage to be? Personally, I'd pick 13.0 volts, but it is child's play to do any voltage you want. There will be about a milliampere of quiescent current for the sensor that will still be on-line when the output shuts off. Jim One-chip solutions designed for this purpose that can detect both undervoltage and overvoltage are available from several vendors, such as Maxim. Example: http://pdfserv.maxim-ic.com/en/ds/MAX6457-MAX6460.pdf These offer very low quiescent current when off (couple of uA). The output can be used to activate a relay (maybe needs an extra transistor). I think you can request free samples. Rob |
#8
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Yeah,
13V cuttoff sounds right. Most charging systems go to at least 13.5 volts and batteries seldom settle near 13 when not charging. 1 ma is fine. FET? is that what it takes? Its gota work for low and high loads same... RST Engineering wrote: What do you want the cutoff voltage to be? Personally, I'd pick 13.0 volts, but it is child's play to do any voltage you want. There will be about a milliampere of quiescent current for the sensor that will still be on-line when the output shuts off. Jim |
#9
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![]() "RST Engineering" wrote in message ... What do you want the cutoff voltage to be? Personally, I'd pick 13.0 volts, but it is child's play to do any voltage you want. That's about right, but here's a couple lessons learned the hard way: I'd like to see some hysteresis in the switching -- say power on at 13.1V and off at 12.9V. It would also be a good idea to have some limitation on inrush current or at least a delayed response. I've seen too many cases where it goes into oscillations when a device with a hefty inrush pulls the detector voltage below the switch point which shuts down the power at which point the voltage recovers and it turns back on and the cycle starts over. Gerry |
#10
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![]() "Gerry Caron" wrote in message .. . "RST Engineering" wrote in message ... What do you want the cutoff voltage to be? Personally, I'd pick 13.0 volts, but it is child's play to do any voltage you want. That's about right, but here's a couple lessons learned the hard way: I'd like to see some hysteresis in the switching -- say power on at 13.1V and off at 12.9V. I never do a comparator without hysterisis. It would also be a good idea to have some limitation on inrush current or at least a delayed response. I've seen too many cases where it goes into oscillations when a device with a hefty inrush pulls the detector voltage below the switch point which shuts down the power at which point the voltage recovers and it turns back on and the cycle starts over. I was sort of figuring on about a ten second loop constant. Thanks. How you been, Gerry? Long time no type. Who you working for now? Jim Gerry |
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