I'm using a desktop computer over a power inverter that's connected to a 12V lead acid battery. When the main power supply fails, the inverter switches to battery backup and can power my desktop for 4-6 hours. However, during the transition, my desktop shuts down (the transition time from mains to backup is too large for the desktop).

Is there a circuit / device that'll power the desktop for a few 100 milliseconds till the inverter backup takes over (something like a big capacitor).

Additional info:

An alternative is to connect the desktop to a UPS but that'll cost another $70-$100. I'm wondering if there's a cheaper option at a fraction of that.

Some inverters have a "UPS mode" that has a faster switchover time. The downside of this mode is that the operating input voltage range is narrowed i.e. the inverter switches to backup if voltage is outside a narrow range. This isn't a great solution when power quality is low.


closed as off-topic by PeterJ, Transistor, JRE, Nick Alexeev Mar 5 '16 at 22:17

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  • \$\begingroup\$ Define expensive. A non-expensive UPS could be the trick. In any case, I don't think there's anything that you can do to solve the problem with switching time by adding a device between the computer and inverter. \$\endgroup\$ – AndrejaKo Mar 8 '15 at 7:53
  • \$\begingroup\$ @AndrejaKo I've added a price now. \$\endgroup\$ – user Mar 8 '15 at 12:07
  • \$\begingroup\$ Define what is "narrow" for you. My UPS can be configured to accept 100V dips. \$\endgroup\$ – PlasmaHH Mar 31 '15 at 12:38
  • 2
    \$\begingroup\$ You should include a circuit to show how the power is being "switched". Does the battery connect in any way to mains (ACDC to charge it)? Does your computer use DC? I couldn't suggest anything if I don't know the circuit. \$\endgroup\$ – Dave Dec 25 '15 at 8:13

Put a decent sized cap on the 12V line. Calculate what the estimated load is by the power supply rating. If your computer is \$(12V)^2/300W = ~2Ω\$ because \$ V^2/P = R\$. Then calcualte the 1/RC time constant, if you use a big capacitor like 1Farad then you would be protected to 60% of the voltage for 1/(1Farad*2Ω) = 0.5 seconds, you would probably want to be protected to 90% of the voltage so you would really be protected for much less than that but your switching time is probably in the msec, so a 1Farad cap would do great.


simulate this circuit – Schematic created using CircuitLab


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