0
\$\begingroup\$

I do understand that similar questions and maybe even the same question has been asked before, I've found them but unfortunately as a result of inexperience on my part I have been unable to adapt the posted solutions to my design specifically.

I am currently working to build the power side of my overall circuit that uses an 18650 (3.7v, 2600mAh) battery that is charged through an MCP73833 IC. The source power for charging will be a 5V wall wart power supply. What I am trying to accomplish now is to create a path or switch that disconnects the battery when the 5V power supply is connected, while preferably allowing the the 5V to be used to power the project and charge the battery at the same time.

Based on prior answers here on EE, I've heard that two Schottky diodes can be used with a little forward voltage drop; I've also seen recommendations for FET's. I'm trying not to add any more IC's to the project if I do not need too especially since I'm hand soldering all of this (not meaning FET's, but actual IC's).

Forward voltage drop is not critical since I'll need to step-up the battery voltage anyway after this circuit (to 5V for MAX7219). So it wouldn't be a big deal to run both power sources through the same booster.

My problem is that I have zero experience with Schottky diodes and even less with Mosfets. I've been reading and trying to understand but just when I think I do, I look at my circuit and my brain turns to mush.

This is what I have so far:

Obviously if you see any errors within the circuit and wouldn't mind pointing them out, I'd appreciate that as well.

\$\endgroup\$
  • \$\begingroup\$ How is the battery powering the project? \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 31 '16 at 16:21
  • \$\begingroup\$ I may not understand the question (or designed it wrong) but as I see it now, the solution to this question would allow me a common ground and both a +5V and +3.7 lead. Whichever is active would go through a booster after this circuit to provide +5V to load. Unless you just mean where, which is off the + and - of the 3.7V battery. I just didn't add a "to load" net because I do not yet have a load. \$\endgroup\$ – Robert Seal Jan 31 '16 at 16:23
0
\$\begingroup\$

Microchip's AN1149, "Design A Load Sharing System Power Path Management with Microchip's Stand-Alone Li-Ion Battery Charger", describes how to use 3 components to allow switching between off-line and on-line operation without interruption of power.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ @RobertSeal: The exact charging chip in use is irrelevant. The only important part is the power steering. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 31 '16 at 16:35
  • \$\begingroup\$ @RobertSeal: Since you only have a single supply input, yes. The two diodes on the left of the circuit are only to prevent the wall wart and USB connection from flowing into each other. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 31 '16 at 16:41
  • \$\begingroup\$ @RobertSeal: It needs to be able to handle the current being pulled through it. See the AN text for sizing of the resistor. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 31 '16 at 16:46
  • \$\begingroup\$ I thank you very much for your help on this. I did reference the AN for the pull-down resistor (100k in this case). I am having a lot of trouble however finding a P-Channel Mosfet with the Schottky diode; they use (FDFMA2P857) but the packaging makes that part impossible for me to hand solder. Is there a common FET used in these applications that lends itself better to hand soldering? \$\endgroup\$ – Robert Seal Jan 31 '16 at 18:52
  • \$\begingroup\$ @RobertSeal: Not really. I personally would look for a suitable SOT-23, but I don't know how capable you are at soldering. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 31 '16 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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