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I have a powerful system (can use up to 20A in max load when using batteries) that can be either powered by batteries, or when possible, by another power supply with the same voltage, so it stops draining the batteries. The problem that I encountered is I don't know how to swap between these sources without a huge power-loss (I don't want to use oring diodes because with 20A can produce a lot of heat).

enter image description here

Right now I have this setup being 11VU the auxiliar power supply, and 11VS the battery suply. I can use a diode with the auxiliar because it doesn't drain too much while in this state, but I'll be glad to accept any suggestions to change it. But I'm afraid that when 11VU connected, the power can return to 11VS by the MOSFET diode. What can I do?

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  • \$\begingroup\$ A very common is to use a mechanical solution like barrel connector that disconnects the batteries when power supply is connected. Electrically you can do it with relay or back-to-back mosfets. \$\endgroup\$
    – Michal Podmanický
    Commented Jun 23 at 2:16
  • \$\begingroup\$ Look at an ideal diode controller to reduce power loss through a diode. \$\endgroup\$
    – Eric Johnson
    Commented Jun 23 at 12:33

1 Answer 1

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Here's the schematic using Schottky diodes having a low forward voltage drop.

enter image description here

A 14 V power supply is used, presuming a 12 V system backed up by a 12 V sealed lead acid battery. It doubles as the predominant source and battery charger. Resistor R1 is the charging current limiter.

An electromagnetic relay K1 may be used instead of the Schottky diodes.

enter image description here

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    \$\begingroup\$ A diode across R1 may be required to ensure the battery holds up the load if PSU1 (14V) fails. The contactor may not be fast enough to prevent loss of power to the load. I am not sure if that was a requirement of this application. \$\endgroup\$
    – Fabio Barone
    Commented Jun 23 at 7:46
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    \$\begingroup\$ Let me specify the power supply characteristics. It has a external power supply, which is a 3S8P li ion battery, and a 3S2P backup li ion battery. This is for a rocket. We can spend more than 8h in the launchpad without launching, that's why I need the auxiliar. When it is time to launch, we disconnect the auxiliar, and the power must be on always. Also the heat can accumulate, that's why I want the minimum power loss. \$\endgroup\$
    – Gonzalo García
    Commented Jun 23 at 11:11
  • \$\begingroup\$ Thank you. Please include this data in the original question to obtain appropriate answers / solutions. \$\endgroup\$
    – vu2nan
    Commented Jun 24 at 2:31

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