0
\$\begingroup\$

I am designing a 12 V circuit to provide 12 V DC (500 mA). I would like to have 12 V battery but also be able to plug in 12 V adapter to automatically override the battery without a switch and without removing the battery.

I saw a 12 VDC with 9 V battery circuit on this site with two diodes but I’m afraid this may not work for me as the voltage will be the same for both diodes so it won’t override? Thanks for your help.

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Power path controller does the job, or you may use higher PSU voltage than batt. voltage. \$\endgroup\$ Commented Mar 12, 2020 at 22:01

2 Answers 2

0
\$\begingroup\$

The easiest way is to use a jack with a built-in switch in addition to the diodes, since you cannot predict which will supply current when they are similar voltages.

From the CUI datasheet for their PJ-202A:

enter image description here

The switch would be in the minus side if the adapter, as is customary these days, has the center pin positive.

\$\endgroup\$
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Using a relay to disconnect battery when mains present.

  • BAT1 supplies the load when the mains is off.
  • D1 prevents backfeed from the battery to the relay coil (which would disconnect the battery).
  • When the mains supply, V1, turns on RLY is energised, BAT1 is disconnected and the load is powered via D1.

Be aware that when the power fails that RLY1 may not switch off until V1 falls to 4 or 5 V. This means that the load will be pretty much unpowered until the relay drops out. There are two solutions:

  • Add some big electrolytics in parallel with the load to maintain voltage during the transition.
  • Use a 5 or 6 V relay fed via a Zener diode (D3). The relay should drop out at about 3 V which, with the Zener diode, will cause it to drop out when V2 falls to about 9 V.
\$\endgroup\$
2
  • \$\begingroup\$ Thanks. I appreciate the answers , but really don’t have the space for a relay and I’m trying to make this as simple and cost effective as possible. Would it work if I added a resister to the battery side to lower the voltage slightly (.5v) to cause increased bias to the adapter diode to override? \$\endgroup\$
    – TommyZ
    Commented Mar 12, 2020 at 23:10
  • \$\begingroup\$ If there are constraints such as size and cost then these need to be edited into your question to avoid us wasting our time. A resistor will only work for a fixed load. On a variable load the voltage drop will vary with the current. \$\endgroup\$
    – Transistor
    Commented Mar 12, 2020 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.