Battery with boost converter vs USB power - Page 1
Forum section: electronics / beginners

author: gus789
author_rank: Contributor
timestamp: May 17, 2016, 11:34:56 pm
content: Hi, I am trying to design a battery powered arduino-based board running on 3.3V only. I plan to power the board with two AA batteries and would like for the battery power to be disconnected when I connect the USB. Does the attached schematic seem reasonable? Should I put a pull down resistor on the EN pin as well? USBVCC will also go to a 3.3V LDO regulator which I have not yet included in the schematic. Current draw for the board is expected to be less than 100mA. Thanks!

author: KL27x
author_rank: Super Contributor
timestamp: May 17, 2016, 11:54:09 pm
content: QuoteI didn't see anything about a pullup/down on the EN pin in the datasheet. You don't have anything to bias the EN pin when the FET is off.Also, maybe I'm dyslexic, but it looks like you have the logic reversed. That when USB 5V is present, the device will be enabled.Thirdly, FYI, a boost circuit is disabled, the Vbat will still be passed, usually. It looks like you put in a rectifier diode to take care of that part, though.

author: rs20
author_rank: Super Contributor
timestamp: May 17, 2016, 11:56:32 pm
content: Wouldn't it be a lot simpler to just use two diodes and a buck/boost converter like this?Vbatt ---->|-----.|Vusb ---->|-----+------( Buck-boost converter )----- 3V3 outIf only Vbatt is present, the input to the buck-boost converter will be that voltage (minus voltage drop). Once you connect Vusb, the input to the buck-boost converter rill rise to 5V (minus voltage drop), thus preventing any more current being drawn from the battery.

author: KL27x
author_rank: Super Contributor
timestamp: May 18, 2016, 12:00:31 am
content: ^ I concur, since you have a diode in there already...A boost converter wouldn't waste much energy when the USB takes over, anyway.But for good measure, sure why not?Just in case you didn't realize it... if you want to be able to digitally turn the device on/off when the USB is not plugged in, for instance, you can't use the EN pin for the reason stated in my previous post. If you need to do that, put the FET between the battery and the boost circuit. And switch it on/off from there. You will still need the rectifier diode, though.

author: rs20
author_rank: Super Contributor
timestamp: May 18, 2016, 12:05:16 am
content: Quote from: KL27x on May 17, 2016, 11:54:09 pmI believe this is incorrect. When USB 5V is present, the Vgs of the P-channel FET goes to zero (actually slightly positive), and thus switches off. Then the pull-down resistor on EN (which needs to be added) pulls EN low, thus disabling the boost converter.When USB 5V is not present, Vgs of the P-channel FET goes negative due to R28, switching the FET on (assuming the FET will be turned on by a Vgs of -(Vbatt - diode drop), which requires some careful MOSFET selection).So the MOSFET logic is not reversed. A simpler way to think of it is that transistor logic is basically always inverting (any digital buffer is actually two (or four..) inverters in a row). The correct behaviour here is high Vusb --> en low, which is an inverter -- it'd be impressive if the OP had accidentally made a single-transistor digital buffer! (I suppose a follower would count, but I digress..)

author: gus789
author_rank: Contributor
timestamp: May 18, 2016, 01:29:49 am
content: Thanks guys, I think I will adopt rs20's suggestion after all and use a TPS63000 as it will prevent me from having to add another regulator for the USB power.