Preface: I am designing a UPS such that if the AC main fail the device can continue to run for at least 10 days. My goal here is to make a design that can last years off a non-rechargeable battery. When starting this I originally thought to use a comparator, but after looking/googling I found that the diode approach was superior.

Design: What: A 3.3 Volt .5 Amp UPS(Uninterrupted Power Supply)//I know you guys like numbers :)

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1) Use a diode or schottky diode?

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schottky diode pros - I will enjoy the efficiency on the low voltage drop when operating from the battery

schottky diode cons - The higher reverse leakage current will decrease the lifespan of the battery which is my biggest design constraint.

2)Is this sufficient with regard to the battery not supplying current back towards the mains? Should I include a diode from the mains towards the load?

3) Would it make more sense for the back up power to come after a voltage regulator? I know that AC mains can fluctuate quiet a bit so this could remove any instance were the battery may engage when its not needed.

4) Are there cheap/simple alternatives to the diode approach that could provide superior results to my proposed method?

Thanks, Josh

  1. Go with the schottky if you can. In your design, the leakage current will actually charge the battery a bit. This prolongs battery life if the battery can handle this extra current. Example: if you have a 10amp-hour battery with 10 year life, its self discharge current is 114 microamperes; if your schottky has less leakage than that, you can use it.

  2. It is sufficient, as long as your Vcc is always > (3.6 + Vdrop) volts

  3. Your regulator will only decrease voltage, so putting the battery after regulator and keeping everything the same will not help. The advantage of putting battery after regulator is longer battery life: regulator drops at least 0.1-1.5 volts, so your load will not see full 3.6 volt if battery is before the regulator.

  4. Diode is quite good. If you are really concerned about voltage drop, two mosfets and a special controller will be more efficient, but will make design significantly more complicated.

  • \$\begingroup\$ So this stems me to another question is it ok to have that small leakage current into a non-rechargeable battery \$\endgroup\$ – tman May 28 '14 at 2:18
  • \$\begingroup\$ I'd say it is OK as long as it is less than battery self-discharge (which can be very roughly calculated as capacity/lifetime). \$\endgroup\$ – theamk May 28 '14 at 2:36

Instead of a diode, use a relay.

Connect the coil across the capacitor, the normally open contact to the + side of the capacitor, the + side of the load to the common contact, and the battery + to the normally closed contact.

That way, with the mains hot the relay will be made and the capacitor + will be connected to the load through the relay's NO and COM contacts, but when the mains fail the cap will be disconnected from the load and the battery + will be connected to the load through the NC and COM terminals.

The main advantage in using the relay in this way is that there will be no voltage drop across the contacts and the load will be able to stay powered up longer than if the diode was used.

  • \$\begingroup\$ Thanks for the answer and I truly agree with your response... ahh but the price of a relay :( .... \$\endgroup\$ – tman May 28 '14 at 1:28
  • \$\begingroup\$ This particular connection (coil across capacitor) is a bad idea. The relay pickup voltage is much higher that relay drop-out voltage. For example, a random 5V relay (sparkfun.com/products/100) will drop-out at 0.6 volt! This means that if your load is a computer/complex electronics of some sort, it will reboot every time the power appears or disappears. \$\endgroup\$ – theamk May 28 '14 at 2:43
  • \$\begingroup\$ Good point, but that problem's easily sidestepped by leaving the diode in the circuit and shunting it with the NC and COM terminals of the relay. \$\endgroup\$ – EM Fields May 29 '14 at 0:05

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