dual voltage supply and diode protection

[munman]

Hi,
I have a bus I am converting and want my 12v lead acid battery to be a backup in case my AC/DC supply fails. So I am using the same circuit but different input sources. The load isn't much, just a car audio system and a few led lights. Now I understand I will need to protect the sources and have found a circuit which suggests a 27 ohm resistor in parallel with a 1n5400 diode. I've tried this but voltage isn't being blocked. (Yes, it is inserted correctly). I would guess I should put the resistor in series, but is the value correct?

 

[munman]

Sorry, should have supplied this image

 

[MikeMl]

Dump the resistor. Dump the 1n5400s. Get a commercial battery isolator.

Or get some hi-current Schottky diodes. They have a much lower forward drop than Si diodes.

 

[munman]

Yes a battery isolator will do it but I don't want to manually switch over. I am looking for a resistor/diode combination that will permit the 12v to supply to the load if the 14v fails. On the issue of the Schottky diodes I assume they would need a series resistor, and the above 27 Ω would be a good value?

 

[MikeMl]

Dump the resistor = you dont need any resistors...

 

[munman]

Thanks, I will give this a go.

 

[munman]

Have tried this --- is it usual for the 14v diode to get hot?

 

[picbits]

It depends on how much current you are putting through it.

 

[munman]

OK, I did jump the gun somewhat and wired in IN5400's. Then turned on every light. The IN5400 on the 14v side got very hot. However, the solution will be to go to the electronics store and purchase Shottkeys as suggested by MikeMI.

Below, my revised schematic...

 

[MikeMl]

Some other possibilities:

With the power supply un-powered on the AC side, connect it to a charged lead-acid battery (should have an open-circuit voltage of 12.65V at room temperature). Measure how much current leaks backwards from the battery into the power supply. If it is under 1mA, you do not need a diode between the supply and the loads.

If the power supply voltage is well-regulated, and at no-load its open-circuit voltage is less than 13.20V, you do not need a diode between the battery and the loads. If the supply is adjustable, diddle it down to 13.10V, leave out the diode, and the power supply can be left on 24/7 to act as a float charger for the battery. At this voltage setting, the power supply will not charge the battery, but it will act to keep the battery from self-discharging over the winter months. As a minimum, diddle the supply voltage to overcome the forward voltage drop of the diode if you need it.

I used to have a RV where I used a large 35A switcher to run the trailer loads from AC, recharge the battery, and float-charge the battery during winter. I built a controller that automatically sensed the battery voltage and current, and it would run the battery voltage up to 14.5V while the battery was accepting charge, then it would lower the voltage to 13.1V to float the battery after the battery was fully charged. My add-on controller effectively implemented the three-step lead-acid charging algorithm. You can buy such RV battery charger/power read made. No diodes required.

Another possibility: Add a high-current, AC coil SPDT relay. Use the presence of the AC to pull-in the relay. Use the SPDT contacts to select between battery and power supply. No power loss between the battery and the loads that way...

 

[munman]

Thanks for taking the time. The AC powers a 13.8v DC regulated power supply, and has no way to adjust the voltage. I do like the idea of float charging the battery but the budget doesn't allow for a controller at the moment. I guess that due to fluctuations of current in the load that a cheap but stable solution isn't possible. Otherwise no need for controllers. I have a simple circuit which alerts me to put the battery on charge when the voltage is at 11v. Seems to work OK.

 

[MikeMl]

Repeated discharge of a lead-acid chemistry battery all the way to 11V will greatly shorten its life. I quit at 11.8V, measured a few seconds after the load is removed.

 

[munman]

I will adjust my circuit so it alerts at 12v. BTW, off topic somewhat, but I assume that a clear signal to ditch the battery wouldn't be that it won't take it a charge but that it's discharge rate is much faster than it would be for a good battery?

 

[MikeMl]

I will adjust my circuit so it alerts at 12v. BTW, off topic somewhat, but I assume that a clear signal to ditch the battery wouldn't be that it won't take it a charge but that it's discharge rate is much faster than it would be for a good battery?

A low capacity battery discharges faster, but its terminal voltage comes up much faster during recharge. Ah out ~= 0.8*Ah in