What would be the best way for me to boost 3.7V to 12V with 300ma max draw?

I've been looking at this: LT1303 and this MC34063A.

I actually bought an lt1303 last week but I couldn't get anymore than 20mA from it, if I try to draw more, the avg voltage goes down, ~10v at 70mA (hadn't got the chance to hook it up to a scope yet). It looks like the lt1303 limits current draw from my battery to about 200mA.

Now, after digging around digikey a bit, I found the other part, MC34063A and it looks like it can do better. I say that only because it is has this "Current Output Rating" of 1.5A, and I don't completely understand what it means. Do you think I can at least get 100mA 12V from a 3.7V source if I use this and then just wire 3 in parallel so I can do 300mA. Thoughts?

Update: Here's a schematic that's close to what I have. just ignore the extra parts and replace L1 with a 15uH (also tried 30/45uH) and R4 with a 100K. It outputs ~13.6V with no load.

  • 1
    \$\begingroup\$ I can't answer your question directly, but I remember this being a really good video for using the MC34063A: eevblog.com/2010/09/10/… \$\endgroup\$ – Dave Sep 19 '12 at 18:14

The thing you need to consider before going further is the current draw from the source. At 12V and 300mA, assuming 90% efficiency, which is probably generous, you're looking at 1.08A draw. Depending on your battery, that may not work for you. That is probably why the LT1303 didn't do what you wanted. We need to know more about the battery to say for sure, and for good measure, please post your schematic.

And if you choose to go with the MCP34063A anyway, the 1.5A "Current Output Rating" is exactly what it says it is. The datasheet says output switch current up to 1.5A, your 300mA will be fine. You don't need to put these in parallel, you need a source that can source enough power. That IC looks pretty straight forward. Just make sure you go through the datasheet and do the math on the reactive components and make sure they are appropriately sized.

| improve this answer | |
  • \$\begingroup\$ the battery should be no problem, I was able to draw 1.5A+ from it before (its a 2C 2000mAh lipo). so 1.5A is output current? not the draw on my battery? \$\endgroup\$ – Mel Sep 19 '12 at 18:22
  • \$\begingroup\$ Yes, that is the output from the IC switch. The draw on the battery would be over 4A at maximum output. \$\endgroup\$ – Matt Young Sep 19 '12 at 18:24
  • \$\begingroup\$ Where did you get your 875mW draw number? Efficiency = Pout / Pin. In this case Pout = 3.6W, at 90% efficiency you need 4W of input. With an input of 3.7V you need a current of 1.081A for those 4W. \$\endgroup\$ – Gustavo Corona Sep 19 '12 at 18:25
  • \$\begingroup\$ Alright, this isn't working, not ASCII art or something. \$\endgroup\$ – Matt Young Sep 19 '12 at 18:28
  • \$\begingroup\$ Fixed, was multitasking and put the n on the wrong side. \$\endgroup\$ – Matt Young Sep 19 '12 at 18:34

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.