I have the following things:
- 24VDC wires through the house for lighting and more.
- A 24VDC 10A power supply, can be adjusted with a potentiometer between 22V and 28V (more than the specified +-10%).
- Two 12V 7Ah lead batteries.
Now I want to combine them to get an uninterruptible light source.
First the power supply needs some modifications to be able to control the output voltage from an external circuit. It is a TL494 based design. The potentiometer is connected to ground. The stronger it pulls down the higher is the output voltage.
I decided to disconnect the middle pin of the potentiometer and add a FET to pull it to ground. This way the potentiometer can still be used to adjust the maximum output voltage. Then disconnect one of the positive output pins and connect it to the gate of the FET.
That was easy. I feared I have to change more components of the power supply. And I am sure this can be done with other brands too. At least it’s worth a try.
The maximum output voltage is adjusted to 27V. Now the external circuit just has to lower the output voltage when the battery charging current is too high. Additionally it will switch off the load when the battery voltage is too low.
So here is the circuit:
Bottom left is the charging current limiting circuit. I built the shunt using 20 0.1 ohms 0805 resistors (2 series 10 parallel). At 10A I have 2W loss in the shunt, but that’s OK for now. The LM358 is not the best joice for this application since it has an input offset voltage of typically 2.9mV. If I had a better opamp I could make the shunt smaller. I tried with an NJM4560 but it did not work at all. Probably 20mV is too low for its inputs.
Bottom right is a little schmitt trigger disconnecting the load from the battery if the battery voltage is too low. I had to tune it by adding a 100k resistor parallel to R11 since the tolerances of the resistors moved the “on” and “off” voltages a little bit.
At the top is a little 6.8V power supply for the opamp. Q1, R1 and D1 are a 8mA current source for the following Z diode D2 and transistor Q1. D2 adjusts the voltage and Q1 boosts the current. All together a small linear regulator (yes, a 78L09 would have worked fine here).
D4 in the middle is to disconnect the power supply from the battery during a power outage. Without this diode the switching regulator circuit in the power supply would probably continue running and waste power. D3 is to start up the linear regulator after a power outage.
The circuit works fine. I am thinking about doing one improvement: Disconnecting the ground of the potentiometer and the FET and connecting them to the minus pole of the battery. This way the voltage drop over D4 will be compensated too.