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I've got a question about this very simple 2 diode UPS circuit, using a non-rechargeable battery. Basically, the battery takes over the circuit when the external mains power supply is removed (provided there's enough of a voltage difference between the 2)

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If the mains power is never lost over a few years, there's obviously still a small amount of leakage current that's drained by the battery from being attached to the circuit. Is this a significant amount of current consumed, or is it so minor that it's not much different to the natural drain that batteries get when they're not powering anything and collecting dust? Is there any simple wattage calculation I can make for this drain when attached to the circuit?

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    \$\begingroup\$ Depends on the diode and the voltage on the other side of it (D2). What does the datasheet say? How does it compare to the self-drain on the battery? \$\endgroup\$
    – winny
    Commented Feb 21, 2022 at 17:21
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    \$\begingroup\$ It depends what diodes those are and what voltages are over them to find out a value for the leakage current. Then it depends on what batteries those are to compare the leakage current value to battery self-discharge value whether it is significant or not. Without any hard data it's just hand-waving. \$\endgroup\$
    – Justme
    Commented Feb 21, 2022 at 17:22
  • \$\begingroup\$ There would only be a drain on the battery by this circuit if the External power supply voltage is less than a diode drop higher than the battery voltage. If D1 is forward-biased such that the voltage at the D1/D2/C1/Load node is greater than the battery voltage then no current will flow through D2 (excluding D2's reverse-leakage current which would actually serve to charge the battery). \$\endgroup\$
    – brhans
    Commented Feb 21, 2022 at 18:37
  • \$\begingroup\$ @brhans ahhh thanks, that's reassuring. But does that mean though that a non-rechargeable battery would never be suitable for such a system? The only benefit I see in non-rechargeable is that they may have a longer life if they are barely used. \$\endgroup\$
    – Turboduinoz
    Commented Feb 21, 2022 at 19:14
  • \$\begingroup\$ The reverse leakage current is not significant enough to overcharge the battery. What you have is not a “UPS” system per se, but an automated power transfer switch that uses diodes as switching devices. It is a reasonable circuit in most circumstances, unless you’re really using the tiniest of batteries at high temperatures while under power. In that case, reverse leakage could be significant enough to damage e.g. a tiny hearing aid battery. But that really requires the smallest of capacities of batteries available. \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented Feb 22, 2022 at 5:44

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If the external power supply is higher voltage than a battery, then the only current draw will be a diode leakage current. This depends on both diode type and diode's temperature. For a common diode like 1N4001-1N4007 at room temperature, it could be around 0.1 microampere (see Fig 5 in diode datasheet).

I don't know your battery type, so I am going to choose a random one -- say we have D-sized cells, which may have 12 amp*hour typical capacity (according to Wikipedia).

Dividing one by other, I get just over 13,000 years -- which means that battery's self dicharge is going to be the deciding factor.

(One caveat is fresh batteries might have higher than label voltage. For example, a fresh pack with 8x alkaline 1.5V cells might have as much as 13.5V initial voltage. This can be a surprise to some -- if you have "12V" battery pack, and "13V" external power supply, you'd still see some battery discharge current.)

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