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I have little knowledge on electronics apart from following a few guides. I have made a few 7 segment displays, nothing complicated.

Recently my girlfriends car was stolen and she has an insurance style GPS tracker, but only gave the road that the tracker was down. We spent hours looking for clues and eventualy we found the car and it had been chopped up into bits with the tracker striped out.

Now these GPRS trackers are good but useless if the battery is pulled out.

So my question is:

How do I create a circuit so that when the car battery is dead/removed, a small 12V battery hidden away kicks in and powers up the tracker.

Now I have done my research and all I can come across is complicated circuits that involves the 2nd battery being charged up aswell.

I don't need this, all I need is a few good hours out of the second battery so charging the second battery is not needed.

This is the tracker: http://www.amazon.co.uk/Tracker-Vehicle-Theft-Protection-System/dp/B003XDN58K/ref=sr_1_2?ie=UTF8&qid=1389875239&sr=8-2&keywords=gps+tracker

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  • \$\begingroup\$ Can you provide a link to the complicated circuits to see what you class as complicated? How much current and voltage does the GPS tracker need? Maybe a link to it would also help. \$\endgroup\$ – Andy aka Jan 16 '14 at 13:49
  • \$\begingroup\$ A circuit like this would work fine. \$\endgroup\$ – alexan_e Jan 16 '14 at 14:13
  • \$\begingroup\$ This is the gprs tracker: amazon.co.uk/Tracker-Vehicle-Theft-Protection-System/dp/… \$\endgroup\$ – user35643 Jan 16 '14 at 22:56
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Use two diodes, preferably Schottky diodes in this case due to the low voltage. The diodes are arranged such that either the car's 12 V line or the small battery can run the GPS, whichever is putting out a higher voltage.

Almost certainly the GPS unit doesn't use 12 V directly and has a power supply inside that makes regulated and filtered lower voltage from the car "12 V". Most likely, the GPS will still run fine from 9 V. 6 or 7 AA primary cells in series should work well anough to keep the GPS powered up for hours if the car battery is disconnected. Get batteries with a long shelf life so that they are still ready to provide power after sitting there doing nothing for years.

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  • \$\begingroup\$ Thanks for the reply, What type of Schottky diode would i require? \$\endgroup\$ – user35643 Jan 16 '14 at 22:54
  • \$\begingroup\$ So i could this: electronics.stackexchange.com/questions/96059/… \$\endgroup\$ – user35643 Jan 16 '14 at 22:59
  • \$\begingroup\$ @user: Any Schottky that is rated for enough current and enough reverse voltage. I doubt the GPS needs all that much current, so a generic 30 V 1 A Schottky should be fine. Those are cheap and available from many sources. \$\endgroup\$ – Olin Lathrop Jan 16 '14 at 23:17
  • \$\begingroup\$ Thanks, So two 30v 1a schottky, a p mosfet, and do i still use a 10k resistor like in the provided link? \$\endgroup\$ – user35643 Jan 16 '14 at 23:24
  • \$\begingroup\$ @user: Huh? I didn't say anything about a MOSFET. Read my answer again. Just two diodes. \$\endgroup\$ – Olin Lathrop Jan 17 '14 at 0:14
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It can be as simple as a relay that connects your backup battery.

schematic

simulate this circuit – Schematic created using CircuitLab

The resistor can be as high as needed to trip the relay when the car battery is hot. This draws power to the GPS unit from the car battery. When that fails, the relay trips and power is drawn from the backup battery. No charging, subject to relay lifespan as it will be energized perpetually.

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  • 2
    \$\begingroup\$ This is not a good idea since it will draw current to keep the relay energized all the time. The current is small compared to the car battery's capacity, but it could start to matter to a weak battery in a car unused for a few weeks, especially in winter. You don't want to add extra load like that to a car battery when the engine is off. The extra complexity of a relay is also unnecessary. \$\endgroup\$ – Olin Lathrop Jan 16 '14 at 14:47
  • \$\begingroup\$ I agree its not a good idea. But the OP did ask for a simple solution with little electronic knowledge. A high enough resistor and a low power relay could draw less power than the car alarm system - which stays on 24/7 and car battery is fine. \$\endgroup\$ – Ron J. Jan 17 '14 at 14:13
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    \$\begingroup\$ How is this "simple" compared to two diode? \$\endgroup\$ – Olin Lathrop Jan 17 '14 at 14:15

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