I have a system that is powered by a main voltage supply and I want to connect it to a battery for backup, in case of power outage.

I need to know what would happen if I connect the system with the battery in parallel with the source and both the battery and the source give 5 V. Where from will the load take its power? From the source or the battery? And how can I isolate the battery from the system if the source is working? And will the battery power the system when the source stops providing?


2 Answers 2


Paralleling power sources is a Bad Idea™. The simplest solution is to use two diodes to separate them.

Suppose the main power source is 12 V and the battery 9 V. Then the 12 V source will drop, say, 0.5 V (Schottky diode), so the voltage at the cathode will be 11.5 V, which is higher than the battery's 9 V, so that diode won't conduct current. When the main source is off the battery will supply power via its diode.

Note that the highest voltage has priority, that's why I chose 12 V and 9 V as example. If the main power supply would be for instance 8 V then the battery would supply the power all the time.

  • \$\begingroup\$ thanks at first ... but here i have regulator at both so the voltage comes from both is 5v if i use a diode i will use it at the battery so it cant provides when the source is providing and at the source so the battery cant provides other hings that the source is connected to at this case the voltage still parallel and for example 4.7v from both which will provide at this case ?? \$\endgroup\$
    – alone
    Commented Sep 4, 2012 at 10:51
  • \$\begingroup\$ @alone - Sorry, but I don't understand what you're saying here. At least make different shorter sentences. If I get the gist of your comment it's about both voltages being equal, right? If the main supply is from a three-legged regulator like a 7805 you can increase its output by one diode drop by inserting a diode in series with the ground connection. The diode in series with the ouput will bring it back to 5 V, and higher than the battery's voltage. \$\endgroup\$
    – stevenvh
    Commented Sep 4, 2012 at 11:05
  • \$\begingroup\$ yes the voltages are equal and that is my regulator 7805 but i didnt understand how to increase the voltage by the diode thank you \$\endgroup\$
    – alone
    Commented Sep 4, 2012 at 11:20
  • \$\begingroup\$ @alone - I mean this. The diode in the ground lead shifts the reference level and therefore also the output 0.3 V (Schottky) up, so that the 7805's output is 5.3 V instead of 5 V. Then the series diode also drops 0.3 V, so that you ultimately get 5 V; both diodes cancel each other out. \$\endgroup\$
    – stevenvh
    Commented Sep 4, 2012 at 11:38
  • \$\begingroup\$ oh ya... thank u very much this will do the job just one more time the diode terminal (the cathode on the commen leg of the regulator ...and the anode to the ground )is this right? \$\endgroup\$
    – alone
    Commented Sep 4, 2012 at 12:04

enter image description hereI know this was posted half my life ago, but whatever. Seems like a great use of a relay. Connect the relay so that your main power source is connected across the relay trigger and the relay-on output. Then you can connect the batteries to the other relay terminal. If the main source goes out, the relay will switch off, connecting the batteries to the load. The relay just keeps the two circuits separated so you don't have to worry about the sources' interaction. Just put a couple current limiting resistors wherever needed and it should work fine.

  • \$\begingroup\$ Can you embed the image using the image tool instead of using a link? Also there is a circuit tool \$\endgroup\$
    – Voltage Spike
    Commented Apr 1, 2022 at 20:27
  • \$\begingroup\$ This is very useful \$\endgroup\$
    – RichieHH
    Commented Dec 5, 2023 at 8:05

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