Timeline for Two power sources, auto-switch?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 10, 2014 at 10:49 | vote | accept | Dean Thomas | ||
| Apr 10, 2014 at 10:48 | comment | added | Dean Thomas | Thanks all, this solution worked perfectly. What didn't help was iCircuit doesn't have a Schottky diode in it, so i was using the wrong thing. Working perfectly in another application, now i can order some bits. Next up is the charging circuit :) Thanks again! | |
| Apr 10, 2014 at 7:12 | comment | added | user39962 | @gwideman; Well crap. Ass.u.me. and all that. I promise to read the byline next time. Maybe. | |
| Apr 10, 2014 at 0:00 | comment | added | gwideman | @NickHalden: That's an interesting question, I'd like to see that asked as a proper question to attract some answers! Sean: NickHalden doesn't appear to be the OP. | |
| Apr 9, 2014 at 23:56 | comment | added | user39962 | They would work extremely well, but now that I know the kind of currents your dealing with, may not be warranted necessarily. Or at least you may not particularly need a high frequency one. They are switched mode power supplies that essentially start and stop charging a capacitor to maintain a specific voltage that is lower than the input. Thanks to modern power BJTs and MOSFETs, they can be wildly more efficient than linear regulators. You might even be able to find what you need off the shelf. | |
| Apr 9, 2014 at 23:14 | comment | added | NickHalden | @SeanBoddy Can you provide some details on the chopper style converter you're referring to? I have an application where one of the power supplies is batteries (needs to be efficient when running off this) and the other supply is rectified mains power so it doesn't matter the efficiency. The currents are typically extremely low (10s of microamps) but can jump kinda high (up to a couple hundred milliamps for about a second) and I'm wondering if those converters you mention would work well... | |
| Apr 9, 2014 at 22:12 | comment | added | user39962 | The node would be where the diodes meet, which would be connected to the input of your regulator. Speaking of regulators, I really hope you don't mean a linear regulator. That eats power pretty well. Olin is spot on; after the diodes what you really want to have is a chopper style switch mode buck converter. | |
| Apr 9, 2014 at 22:00 | comment | added | alexan_e | @DeanThomas by internal power node he means the supply input of your circuit (that is the regulator input according to your description), the diodes should be placed between each supply output and the circuit supply input | |
| Apr 9, 2014 at 20:34 | comment | added | Dean Thomas | Sorry, i'm not sure what you mean by internal power node? Thanks though, i'm just a bit new to this! | |
| Apr 9, 2014 at 20:24 | history | answered | Olin Lathrop | CC BY-SA 3.0 |