12V Battery Backup power supply for GPS Tracker

Hi All,
I'm new to the Forum as its quite a few years since I built electronic projects.
I have a small project I'm working on at home and I would be grateful for some assistance.
I'm fitting a 12v Tracking device to a horse box trailer, and I need to include a long life back up battery for the Tracking unit. The tracking device itself runs off 12v and it does not have an internal battery. My plan is to enclose the tracker and a 12v 1.2AH Lead Acid Rechargeable battery in a Waterproof poly-box. I will then connect the circuit to the trailers rear lights (12v). Therefore when the vehicles lights are on, it will recharge the battery and power the Tracker unit also. When the lights are switched off, the Lead Acid battery does the work. The Tracking device will switch to "sleep" mode when it scenes movement has stopped. In "Sleep" mode it consumes only 2mA. When in Track mode it consumes 150mA.
I must also ensure that the backup battery does not try to power the trailers electrical system also.
I'm sure this type of circuit is very common, but its years since I buit circuits and I'm a little out of date.

All help appreciated.

Barry

Dear Bary,
you can use the circuit like this...


But it is not advisable to charge a 12v battery using 12v DC supply in this method... Because the diode drop will be 0.7V so you have to give atleast 13v or above... You can use 9V battery & 2X series diodes[check the minimum operating of your tracking device]. It is advisable to use charge cut off circuits also to improve battery life...

Regards
Udhay

There is one thing, you have two voltages on input. When car/truck engine is running you have at least 13,8V or more, not 12V, because alternator voltage, and when you turn off engine voltage is from battery 12,6V-12,5V.

Check GPS device specification about input voltage ranges.



Schottky diode have less voltage drop around 0,2V.

Thanks guys,
Udhay,when the engine is running the recharge level will be around 13v, so that should not be a problem charging a 12v battery.
tpetar, the voltage range of the GPS Tracker is 10 - 30v, so no problems there.

My only issue is how to extend the life of the Sealed Lead acid battery. I intend to use a 1.2 Ah type. Am I correct in saying that the max recharge current that should be used in a SLA battery is 10% of its capacity? therefore the max recharge current should be 120mA? If so should I connect a resistor in series with the diode? If so what value and power resistor would you recommend?

Thanks again.

Barry

Lifetime of battery depends from many things, charging voltage should be minimum float charge voltage at 13,5V-13,8V (@25C and reduce each 18mV per one C if temp is above 25C) and can go to 14,2V-14,4V cycling recharge voltage, and most important thing is charging current, for that small battery should be around 0,2A but check manufacturer battery datasheet.

Better use LM317 to limit current.

Its very bad for lead acid battery to stay in empty or below full voltage (12,72V-12,73V), specialy for longer time, and its very bad to discharge it completely, sulfatisation can occure. That small lead acid have around 200 cycles, if you discharge it 100% each time, 400 cycles if you discharge it 60%, around 1000 cycles at 30% discharge level, but check manufacturer datasheet, and expected lifetime is 3-5 years. Capacities of this small lead acid batteries are often declared in C/20 current-time.


Maybe is better solution to use NiMh battery. You can get 12V from 10 NiMh cells or 13,2V from 11 NiMh cells. But you can use AA 3000mAh-3500mAh (3Ah-3,5Ah), or smaller AAA 1200mAh-2000mAh (1,2Ah-2Ah).




Here is one battery datasheet 12V 1,2Ah as example :
**broken link removed**

I have to admit that I've never heard for this company. I take first result from Google search. :smile:

The simple expedient of a diode in series with the vehicle's supply is perfectly adequate to prevent discharge into anything else other than the tracking device.

Your concern about maximum charge current assumes that the battery is well discharged. Is this really going to be the case? If so, let's be optimistic and assume that its terminal voltage is no lower than 10.

With the engine running, the electrical system's voltage may be as high as 14.4, 0.7 of which will be lost across the diode.

The difference between the charging voltage and the batteries initial reading will be 13.7 - 10 = 3.7

3.7V and with 120mA through it means that the resistor's value will be 30.8 ohms. The maximum power dissipated in it will be 443mW.

The figure of 30.8 is not critical. You may use an off-the-shelf 27 ohms one or a 33 ohms one. A half watt should survive quite well (it will dissipate 440-odd mW) for only a short while. The battery voltage will increase fairly rapidly and the disspation will therefore reduce.

If you're of a nervous disposition, use a 1W resistor.

Thank you Syncopator.... and all others..
I think I know where I'm going now.

Barry

We hope that you will share your ideas with us. :wink: