• Register
  • Login
  • this message might be for you!

    This battery backup circuit can be added to surveillance systems like alarms and others to power the circuit during mains failure. The battery backup will immediately take up the load without any delay.

    The circuit is simple to construct. Regulator IC 7812 gives 12 volts regulated DC for powering the circuit as well as to charge the rechargeable battery. LED indicates the power on status. When the mains power is available, diode D1 forward biases and passes current into the battery through R2. Value of R2 is selected to give 90 mA current (12/100 = 0.1A) for slow charging.

    When the mains power fails, D1 reverse biases and D2 forward biases and backup the circuit. The same circuit can be used in circuits having 6 volt 7.5Ah battery. For 12 volt battery, use 7812 regulator IC and 14 Volt input.

    Battery Backup Circuit Schematic

    battery backup circuit schematic

    battery backup circuit schematic

    attentionThe author D Mohankumar is not an active member anymore. Please take into consideration that the presented information might not be correct.
    ask a question

    26 Responses to "Battery Backup Circuit"

    1. Its an absurd circuit just to mislead ameature hobbysts.
      The charging votage requred to cahrge the 9V battery should be aleast 10.5v ,whereas here 9v -drop across R1(100ohm) -.6v of diode D1 barrier =around leass than 6v for a discharged battery to get charge.i advice hobbysts not to try the circuit.

    2. sir , you have to add the “voltage reg 7812 instead of 7809” and “another section of voltage reg of 10.5v between part1 ( consist of ” 7809 , ri , led , d1 ” ) and part2 (consist of ” d2, 100r ,9v battery “)”.
      but thanks for circuit.

    3. sir can you show me another circuit with the additionals that you mention..thnx

    4. sir,thanks alot for this useful circuit,really its working good.

    5. “The same circuit can be used in circuits having 6 volt 7.5Ah battery. For 12 volt battery, use 7814 regulator IC and 14 Volt input.”

      By this logic the IC should be 7812 and not 7809 as shown and from my very limited knowledge the charge voltage should be higher than the battery voltage . Is D2 Necessary ???

    6. A good circuit!DO YOU HAVE AN INVERTER CIRCUIT WITH 6V,4.5Ah sealed lead-acid rechargeable battery and has an output of 220v,100watts pure sinewave?

    7. Can i use this circuit to charge the Energizer 9 volt battery( Energizer 9 volt Rechargeable NH22N 150 mAh ).

    8. please give me circuit diagram of battery charger circuit for 12v and 7.5ah battery used in ups….

    9. Simple Battery Basics 101
      A lead acid or gel acid battery has a terminal voltage of 2.1V per cell nominal, 1.8V per cell discharged and requires 2.3-2.4V to properly charge.
      A 12V battery relates this to 12.6V, 10.8V and 13.8-14.4V respectively.
      So you MUST charge a rechargeable battery with a voltage higher than the system operating voltage or you will not get a proper charge and quickly destroy the battery.

    10. R2 and D2 are part of a slow charger of the backup battery. The IC 7809 needs 2 capacitors @Vin and Vout. D1 and D2 creates a 0.6 V drop, so the source voltage would be 8.4 V instead of 9V, and will match 9V only if the current will be near to 0 Amps. It is a very simple circuit that provides 8.4 V power, with a modest amperage.

      • i have 9v supply circuit at the input IC cap = 1000uF and output cap = 10uf. Actual i’m using rechargeable battery NIMH 7.2v 150mAh

      • Jim Keith says: on November 2, 2012 at 1:15 am

        What I would do in this case would be to change the 7809 to an LM317 adjustable regulator. To set the voltage, place fully charged battery across the regulator and increase the voltage until the current is in the 10 to 20mA range –that will be close to the proper voltage.

        You can adapt the regulator from this schematic:

    11. What is the purpose of the 100ohms resistor series with 12v battery?
      If u can please email me the answer.
      thank you.

    12. Narendrapal singh says: on January 8, 2015 at 6:58 pm

      Sir we require 9 volt 1 amp dc supply with 6 volt 4.5 ah rechargeable
      battery backup, when mains off output remain 9 volt without unintruped.
      We have to use this supply to operate weighing scale.

    13. Pls how would my project switch to the battery from the main supply and back to the main suppy in case of power failure

    14. Good circuit but I am interested to know why 1N4007 diodes are used when 1N4001 could be used and also have smaller dimensions?

    15. Can i have a load of 3A?


    16. We have built a backup battery system as shown above but instead of the 7812, we power the circuitry and charge the backup battery directly from the aircraft alternator and main battery. Also, there is no need for R2 since the aircraft voltage regulator maintains the main battery voltage at close to 14 volts while running. Given the presence of diode D1, the backup battery sees approx. 13.3 volts, which as it turns out is not sufficient to charge the backup battery. Have you found your backup battery actually charges with this circuit?

      BTW, your assertion that R2 will charge the backup battery with 90 ma is not correct, as the only way 90 ma would flow through R2 is if the backup battery voltage were zero!

    17. hi,

      what changes do I have to do in order to adjust the circuit to one with input of 30VDC, 24VDC output, and 24VDC 7AH battery?


    18. Don’t bother trying to use this circuit. It won’t charge your battery! You need much more than 24 Volts to charge a 24 Volt battery. I am sure the author who has now disappeared never used this circuit, because it won’t work and I am sure he never actually tried it.

    19. Sir… I used two mobile phone li-ion battery’s with voltage 7.4.. Bt thus circut cant recharge di battrys…. Plz give me reply

    20. Sir… I used two mobile phone li-ion battery’s with voltage 7.4.. Bt this circut cant recharge di battrys…. Plz give me reply… I used 7809 ic instead of 7812….

    21. You will charge the batteries, but you will also be feeding 8.4V (9 – 0.6) to the circuit; not 7.4V. Forget this circuit, it just won’t work as advertised, regardless of which components you use.

    You need to log in if you need to post comments on ElectroSchematics.com or register if you do not have an account.