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I'm attempting to design a circuit that will be powered from a 12V primary source but if the power fails then a relay switch to battery (which is maintained charged by an UC3906).

http://i49.tinypic.com/1jrcwx.jpg

The idea is that if primary source is present the MOSFET is on, activating the relay. When primary source fails the MOSFET is off and the relay go to the position showed in the schematic, the application is now powered by the battery. Do yo see this reliable?

Update: I wish to ensure that no current is drawn from the battery if primary power is present and I prefer to not put the primary voltage above the battery voltage (which is already slightly above 12V), with the two diodes solution I added a pMOSFET to get it (may be D1 is not necessary now?).

update

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  • \$\begingroup\$ In your schematic you need to add a back biased diode across the relay coil so that at the time the the 2N7000 FET turns off the inductive kick of the coil will not take out the FET. \$\endgroup\$ – Michael Karas Nov 9 '12 at 19:43
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    \$\begingroup\$ If you want to user a relay at all (I would prefer two diodes as Michael suggestst) why not power the relay coild directly from the 12V primary source (without the FET)? You use a FET just to be able to draw the enegising current from the other source, which seems silly to me. \$\endgroup\$ – Wouter van Ooijen Nov 9 '12 at 20:30
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    \$\begingroup\$ What is wrong with two diodes? Pretty much like powerguru.org/wordpress/wp-content/uploads/2012/07/… \$\endgroup\$ – jippie Nov 9 '12 at 21:43
  • \$\begingroup\$ @Tino - You should look at the comment I added to the end of my answer posting. \$\endgroup\$ – Michael Karas Nov 11 '12 at 12:46
  • \$\begingroup\$ @Michael I see. The circuit has improved a lot. Thanks. \$\endgroup\$ – Tino Romero Nov 14 '12 at 22:28
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This circuit idea is OK if the "application" can live with a momentary dropout of its supply voltage during the changeover. There are two things that cause this. First the 12V primary power must drop to less than about 2V or so before the 2N7000 N-FET will turn off. Secondly the relay contact switch over time will take a millisecond or two.

If you want uninterrupted power to the "application" during the changeover then you could consider several alternatives.

1) Add a comparator circuit that detects when the primary 12V has only dropped a small amount (such as to 11.4V) and switch the relay before the primary rail fails. Some charge storage capacitors on the "application side" can minimize the voltage droop there while the relay contacts are switching over.

2) Use a pair of power MOSFETs to switch the voltage instead of the relay. These would switch much faster than the relay and the "application side" hold up capacitors can be much smaller.

3) Use just two power Schottky diodes to OR the two power sources to the application. No switch over control logic required but you do lose some voltage from the sources to the application power rail.

NOTE: In your schematic you need to add a back biased diode across the relay coil so that at the time the the 2N7000 FET turns off the inductive kick of the coil will not take out the FET.


Tino proposed to change his circuit to use the two diode approach as follows:

enter image description here

He questioned as to whether the Diode D1 would still be needed. Unfortunately the diode would still be needed because of you look closely at the data sheet for the IRFU9024 you will note the body diode of the MOSFET....

enter image description here

You can quickly see the problem!

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  • \$\begingroup\$ I edited the post with a new scheme, derived from the two diodes solution. \$\endgroup\$ – Tino Romero Nov 10 '12 at 16:23
  • \$\begingroup\$ @Tino - See above regarding why Diode D1 would still be needed. \$\endgroup\$ – Michael Karas Nov 11 '12 at 12:48
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Further to Michael's answer, I have a similar (but opposite) circuit, for trickle charging a battery from my car electrical system, for my in-car ham radio installation.

This uses a TL431 shunt regulator to switch the relay on or off, when the supply falls below 12V (adjustable). The design is straight out of the data sheet.

A very similar design to mine can be found on the net, here (although that is not my blog)

A similar set up looks relevant here, for you...

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Switching between backup batteries and primary supplies can be done passively with diodes.

The circuit can be fed from the backup battery via a diode. The primary source connects below this diode.

When the primary voltage is present, the diode is not forward biased because the voltage from the battery is equal to the primary voltage. You can ensure reverse bias by having the primary voltage be slightly higher than the battery.

When the primary voltage cuts out, or simply drops to 0.7V less than that of the battery or lower, then the diode becomes forward-biased and current can flow from the battery.

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