Automatic switching, dual power supply - Page 1
Forum section: electronics / beginners

author: braincat
author_rank: Contributor
timestamp: July 19, 2020, 04:45:21 pm
content: Hi All,I am working on a device with a primary power source about 27VAC (think of common wire on the AC furnace). There is a full wave rectifier, and the final output is 3.3V using LM2596. The output current requirement is about 1.5A. I would like to have an option of powering the project from a secondary power source of 5V DC (connected to the input pin of LM2596) so that it could only supply power when the primary source is absent. I need to protect that secondary power source but cannot use a diode because it will drop the voltage below LM2596' minimum (4.5V); in addition, the drop across LM2596 is more than 1V if I am reading the documentation correctly. I did some research, and I feel that LM74700 (https://www.ti.com/lit/ds/symlink/lm74700-q1.pdf) could do what I need. Do you think that's correct, and are there better solutions? I also found an alternative with a built-in MOSFET (LM66100) but could not figure out from the spec what is its max reverse voltage.Thank you for reading.

author: Zero999
author_rank: Super Contributor
timestamp: July 19, 2020, 06:22:30 pm
content: The LM2596 is only rated to 40V maximum, so 27VAC with a peak voltage of 38V, is marginal design.

author: Peabody
author_rank: Super Contributor
timestamp: July 19, 2020, 06:30:57 pm
content: What about using Schottky diodes. You could put one in each line, and the source with the highest voltage would always be selected. The voltage drop would be about half what a regular diode produces.

author: braincat
author_rank: Contributor
timestamp: July 19, 2020, 07:18:44 pm
content: I am just afraid it would be too close for comfort. Suppose diode's voltage drop is 0.4V; dropout voltage of LM2596 could be as high as 1.3V under certain conditions.

author: Peabody
author_rank: Super Contributor
timestamp: July 19, 2020, 09:52:06 pm
content: If you will need the mosfet anyway, perhaps you could just use a standard load sharing circuit. I'll attach a schematic of that where the two supplies are the 5V USB source and the lithium battery, but you would be using 40V and 5V. When the 40v is present, it brings the gate high, which turns off the mosfet. If there's an outage on the 40V, the pulldown resistor turns the mosfet on. Notice the orientation of the mosfet. It's kinda backwards from the usual orientation, but needs to be that way so the body diode points the right way and protects the 5V from the 40V when the mosfet is off. You would need to find a mosfet and diode more appropriate for the voltages you have, but I think this would work.

author: braincat
author_rank: Contributor
timestamp: July 19, 2020, 10:41:23 pm
content: I see. Thank you for the suggestion. Is the boost converter there just to boost the battery voltage?

author: bdunham7
author_rank: Super Contributor
timestamp: July 20, 2020, 12:15:14 am
content: Why not use the 5-volt version of the LM2596 to feed an LDO like the LM1085IT-3.3? Then you can use one or two Schottky diodes with less than 0.5 forward volts to isolate the two sources and you'll still have a bit of margin.

author: Peabody
author_rank: Super Contributor
timestamp: July 20, 2020, 04:36:20 am
content: Quote from: braincat on July 19, 2020, 10:41:23 pmYes. The battery will range from about 3.5V to 4.2V, so you need the boost converter to get to 5V. But in your case all you need is 3.3V, so you would use your LM2596 instead.

author: braincat
author_rank: Contributor
timestamp: July 20, 2020, 02:16:18 pm
content: Quote from: Zero999 on July 19, 2020, 06:22:30 pmThanks, I might look for an alternative. Peak should be about 36.6V because of a diode bridge.