0
\$\begingroup\$

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?

Improve this question
\$\endgroup\$
3
  • 1
    \$\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
  • 1
    \$\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
2
\$\begingroup\$

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.

Improve this answer
\$\endgroup\$
4
  • \$\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
  • 2
    \$\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
  • 1
    \$\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
0
\$\begingroup\$

For a DC UPS, you need the following components:

  • A battery charger (ideally a smart charger that will float charge the battery when full to prevent over-charge).
  • A battery (ideally deep-cycle)
  • A DC power supply
  • A switch-over circuit that switches the load from the DC supply to the battery when AC power is lost
  • A low voltage cutout to protect the battery from over-discharge (typically at 10.5-10.8V, depending on load).
  • Switch mode voltage regulators for the various voltage outputs you need.

I built such a DC UPS for my networking devices using a CCTV battery backup unit to take care of the first 5 points. The final point was handled by a boost converter to 17V, followed by a buck converter to 12V. This arrangement ensures you always get 12V output regardless of where the battery is in the 10-14V range, plus putting the boost to 17V first ensures the input to the buck converter is always high enough for reliable operation. A single buck-boost converter could be used too.

Improve this answer
\$\endgroup\$
0
\$\begingroup\$

In theory it can,

But, many chargers charge the battery with 14.2 ~ 14.4V, and when they detect that the battery is nearly charged, (the charge current drops to lets say 200mA), they lower their voltage to 13.8V.

The 13.8V is the ideal float charge voltage for many lead based batteries. the 14.2V is to much, if you charge your battery for too long with this voltage, they will dry out. And will be damaged.

If you add load to your system, your charger will see always a current greater than 200mA (or whatever your charger trip point is). and always will supply 14.4V.

This can be fixed, to add a diode in series. the diode will drop around 0.7V so your battery will never be overcharged.

But it takes significantly longer to charge.

secondly you need some sort of low voltage cut-off. depending on how long you want your battery to last, somewhere between 10.8V an 11.8V. (the higher the cut-off voltage the longer you battery will last, but the less run-time you have) (also you can use around 40~60% of your rated battery Ah if you want to use your battery more than once).

Improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.