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I want to connect some DC devices(< 1.5 A rating) to the output of a 12V/7Ah battery. Will this system work as a UPS if I connect a battery charger to this battery?

i.e the DC devices, Battery and charger will be connected parallely. Will this work as a UPS?

The charger is auto-cutoff enabled to prevent over-charging. See : http://www.amazon.in/Battery-charger-12V-1A-Adapter/dp/B01GZRBWLW Will the auto cut-off be affected for above setup?

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    \$\begingroup\$ Sure thing. Just make sure you have some under voltage lockout in your devices or by external means in series with your devices from the battery. \$\endgroup\$ – winny Jan 3 '17 at 10:00
  • \$\begingroup\$ Great! Thanks! I'm connecting them using a DC voltage regulator like : amazon.in/Adjustable-Voltage-Regulator-Supply-Module/dp/… \$\endgroup\$ – NS Gopikrishnan Jan 3 '17 at 10:03
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    \$\begingroup\$ That does not have an under voltage lockout, high dropout and consumes at least 1 mA of quiescent current, so your battery will be drained to a very low voltage unesss your devices have high enough undervoltage lockout themselves. That Iq itself will take a few months to destroy the battery if left unchaged so it's up to your application if that's ok or not. \$\endgroup\$ – winny Jan 3 '17 at 10:17
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This sounds like it will work, but there may be some pitfalls. First, ensure the devices are very flexible in their input voltage, lead-acid batteries can vary a bit in their voltage. Put a voltage regulator in between if you're not sure. Second issue from the top of my head could be that the charger puts a load on the battery when turned off, but the auto-cutoff makes this unlikely. This could be fixed with a heavy diode.

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  • \$\begingroup\$ Great! thanks! Yes I am planning to use a voltage regulator: amazon.in/Adjustable-Voltage-Regulator-Supply-Module/dp/… \$\endgroup\$ – NS Gopikrishnan Jan 3 '17 at 10:04
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    \$\begingroup\$ @NSGopikrishnan It is a bit unfortunate to put a linear regulator in this situation. I don't know what output voltage you will be using, but if, for example, it is 5V, it means you waste more than half power as heat, dramatically reducing the runtime when on battery. You should consider using a DC-DC buck converter instead (you can easily find cheap modules with adjustable output voltage based on LM2596, for example). \$\endgroup\$ – dim Jan 3 '17 at 13:09
  • \$\begingroup\$ Thank you! I'm new to this! didn't know about them! Just got some question. My charger is rated 1 AMP. And the cumulative rating of the DC devices exceeds this. Is that a problem? \$\endgroup\$ – NS Gopikrishnan Jan 3 '17 at 15:41
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    \$\begingroup\$ If all the devices are powered full-time, then the charger must be able to deliver a bit more than the total current. If some of the devices are only used part time, then the charger has to provide a little more than the average current. Batteries are not 100% efficient, so the charger has to provide a little more current than is drawn by the loads, in order to keep the battery charged. \$\endgroup\$ – Peter Bennett Jan 3 '17 at 17:26

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