I want to create a circuitry for switching to backup power (small battery) in case of mains power failure for my ADSL modem (to avoid the internet going down in such scenario).

This is a basic circuitry however I have only very basic knowledge of electrical and electronics and hence need help.

The below diagram shows what I am trying to create. Request help on how to create this "instant switching circuitry".

Modem specs: Current=0.6A Voltage=9V

Modem Power Backup

  • \$\begingroup\$ Just to know, why don't you use UPS battery ? Any specific reason ? They're not expensive and are designed for that. \$\endgroup\$ – Emmanuel Istace Dec 29 '13 at 19:12
  • \$\begingroup\$ @EmmanuelIstace (a) To get a kick out of creating something on my own (b) UPS batteries for supporting only the modem would indeed be costlier (use versus cost). \$\endgroup\$ – iCrus Dec 29 '13 at 19:23
  • \$\begingroup\$ What are the current specs of your modem? \$\endgroup\$ – alexan_e Dec 29 '13 at 19:39
  • \$\begingroup\$ @alexan_e Modem specs: Current=0.6A Voltage=9V \$\endgroup\$ – iCrus Dec 29 '13 at 19:45
  • 1
    \$\begingroup\$ A small 9v battery with low capacity is not a very good plan for backup. I doubt you will get even minutes of power supply because the voltage will drop very fast on load. \$\endgroup\$ – alexan_e Dec 29 '13 at 20:03

The most simple backup source selector involves just two diodes

enter image description here

The problem with this kind of circuit is that there is always a voltage drop on the diode so the voltage that reaches the load is Vsupply - Vf where Vf is the diode forward voltage drop.
Also the backup voltage should be lower than the main supply or both the diodes will be on simultaneously and both sources will share the output current.

A circuit that would work better (untested) would include the addition of a mosfet (it's a P-mosfet) that operates as a switch and connects/disconnects the battery depending on the availability of the main supply voltage.
Note that the main supply voltage shouldn't be higher than the max Vgs which is about 20v (check the datasheet of your specific device). The battery voltage level is not a problem in this case, it can even be higher than the main voltage source.

enter image description here

Apart fro the mentioned solutions you can always design a circuit that involves a comparator and drives a switching device to connect he appropriate source or even use a commercial chip intended for supply monitoring.

  • \$\begingroup\$ Cool, thank you very much. I am going to try the mosfet based solution. \$\endgroup\$ – iCrus Dec 30 '13 at 9:36
  • \$\begingroup\$ @Anupam The diodes are Schotky type for low voltage drop, for example 1N5817. The mosfet isn't going to be a problem (doesn't need to be a logic level model) since the Vgs with 9v main supply will be high enough to turn on any mosfet you use. \$\endgroup\$ – alexan_e Dec 30 '13 at 9:58

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