Battery/DC-DC boost source-select for a low-power application - Page 1
Forum section: electronics / projects

author: horzza
author_rank: Newbie
timestamp: May 27, 2012, 05:09:50 am
content: It's been a while since I've been on the forum, but I had a good look and can't find anything similar to my query.I'm making an LED wrist-watch, using an MSP430G2102, after seeing some low power stuff it had been used for. The idea is that it's in a low power mode most of the time, only leaving sleep mode to increment the time, once a second, or when a button is pressed, activating the display for a few seconds after the last press.I plan to run it off a single 3v coin cell, but as this voltage drops, the blue LEDs it's required to power might not even turn on. They are being Charliplexed, which for this problem, means that all of the LED current for this display is sourced and sunk from/into the MSP430 itself, rather than directly from a supply line. I would require a step-up DC/DC converter, but these aren't terribly efficient at low current draws (like the low power mode the device will spent most of its time in) and so I am planning to use the enable line on the DC/DC (http://au.element14.com/jsp/search/productdetail.jsp?sku=1077152) to turn it on, and power the MSP430 only when the LEDs are required to be on.I'm having difficulty designing a circuit which can use a single GPIO on the MSP430 to both disconnect the battery directly from supplying voltage to the MSP430 for its low-power mode, and turning on the SMPS when required. The more simple the external componentry is, the better, as I've got to solder the board by hand and think smaller/more elegant is better etc..If anyone has done this before, or could help me out, that would be great! Thanks!

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 27, 2012, 06:40:10 am
content: The simplest solution is to use two Schottky diodes on the outputs of the two voltage sources and ensure that the converter outputs a higher voltage than the battery. (These are called OR-ing diodes.) Once this is done all you need to do is use the converter's SHDN# pin to enable/disable the converter. The diodes will do the rest. (See attached for an example schematic.)There will be a voltage drop because of the diodes so this isn't the most efficient solution. A more efficient solution would be to use an IC that is dedicated to the task and that uses MOSFETs instead of diodes. These are called OR-ing controllers and they will handle switching between the two supplies based on which one has the higher voltage. You only lose about 25mV because of the MOSFETs instead of 300mV to 500mv when using diodes.

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 27, 2012, 09:50:13 am
content: Here's another idea. It uses a comparator, an inverter, a couple of p-channel MOSFETs and some resistors. It will (if it actually works) switch between whichever supply has the higher voltage. The MCU simply turns the converter on or off and the circuit will handle the switching.Note how the two MOSFETs are 'backwards.' This prevents current from flowing from one supply to the other through the body diode. The dividers on the comparator's inputs are to prevent SENSE from exceeding the comparator's supply voltage (Vbat) as well as to pull SENSE to ground when the converter is shutdown.Edit: Just realized that the resistor on the bottom MOSFET needs to be pulled-up to Vbat instead of the output of the converter. Although, it may not matter either way.

author: horzza
author_rank: Newbie
timestamp: May 27, 2012, 11:08:08 am
content: Thanks for the speedy reply!I'd considered using diodes, though hadn't considered using Schottky diodes, which makes sense. The thing with the 300-500mv drop is that it also nerfs the operating time of the watch per battery, by requiring the battery to be another 300-500mv above the 0.6v minimum supply voltage for the MSP430. There might be a negligible amount of energy in the battery once it's below 1.8v, but it seems your suggestion of a MOSFET would be more efficient for the lower drop, and allow a lower operating drop for the supply requirement as well.I was then wondering if I could get the best of both worlds, and go with a single MOSFET as a battery switch, with an OR-ing diode behind that for the dc/dc supply. I couldn't figure out how to employ a MOSFET to do this efficiently however, and I am thinking that for now the best thing for efficiency at <10uA seems like just using two Schottky diodes, as you suggested. I can deal with 80% efficiency at full charge, but it's frustrating how this drops to 45% or so once the battery reaches 1.8 volts. Can MOSFETS operate more efficiently than this when being used to switch less than 10uA from a 1.8v source? I guess it's pulling hairs at this point, as the display will draw several mA for however long it is operating, so bigger overall power savings are to be had by not having that on for as long. I guess it became an interesting exercise.Thanks for your help!--Just did a refresh before posting; Your second post would be a sensible way to switch two power supplies perhaps if the current draw wasn't so ow as it is in my situation. It reminds me of the diagram on wikipedia, of a CMOS output, except with Vss and Vdd instead replaced by the two potential sources.http://en.wikipedia.org/wiki/CMOSI'm trying to decide at this point between using a BJT, Diode or FET for switching the battery into series with an output Schottky from the SMPS. I considered using an NC relay for this task actually, but couldn't find one small enough or cheap enough to do the trick.Thanks AGAIN!!

author: DarkPrince
author_rank: Regular Contributor
timestamp: May 27, 2012, 04:11:52 pm
content: I would like to dump my 2 cents into this. I am in the same boat as you, designing a project which is powered by a single coin cell battery, and also uses a blue led. Different function though.I am looking at these: MCP1258 or MCP1259 (charge pump ICs). Unfortunately there are a couple short comings, one is the output voltage of 3.3v (though it may work), and it has a maximum input voltage of 3.6v (abs-max 3.8v). What is interesting in these two versions of these devices though is that they have a bypass pin, which allows Vin to connect to Vout and disables the converter. Perfect for this kind of design.I just wanted to put those devices out there, maybe it will give some ideas or whatnot. I am still looking around for my own project.

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 27, 2012, 07:22:51 pm
content: The low current requirements make this a bit tricky. Here's another idea. The circuit is simpler and more efficient in the normal operating mode--with the converter off.The optional Schottky diode across the MOSFET is to reduce the sag on Vccto at worse Vbat- Vf(schottky)during the switchover from Vbatto the converter's output. Otherwise the sag will depend of the Vfof the MOSFET's body diode.The sag will also depend on the capacitance on Vccand the current being drawn during the switchover. Because of this you will likely want to introduce a delay in your code before you start driving the LEDs after de-asserting SHDN#.The MOSFET will need a very low worse-case Vgs(th), obviously.

author: NiHaoMike
author_rank: Super Contributor
timestamp: May 28, 2012, 04:39:38 am
content: You can drive the MOSFET from the microcontroller itself. Or if one Schottky diode drop is acceptable, just use a Schottky diode from the input to the output of the boost converter.

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 28, 2012, 05:46:55 am
content: Oops. I guess it helps to read the datasheet. I didn't before I drew-up my schematics so they probably won't work.It looks like there is a parasitic diode in the IC from the input to the output so the IC doesn't really shutdown when SHDN# is asserted. This means the IC's output doesn't go high-Z like I assumed.NiHaoMike's idea about the Schottky would likely work.

author: free_electron
author_rank: Super Contributor
timestamp: May 28, 2012, 06:45:21 am
content: Mcp1400..Or, you use an io pin of the cpu to simply make you own boostpump !Code:Bat ---- inductor----+---|>|------ vcc|mos|gnd[/codeIn normal operation the cpu gets power through the inductor and diode. It runs at 2.7 volts then and is in low power mode.When you need more juice you simply apply a pulse train at the gate of the mosfet. You can use a pwm channel or a timer output. The pulse on the mos gate should be very short with a longer delay between pulses.This is the same effect as having a step-up converter. The inductor will charge when the mos is on and when the mos turns off the back ems in the inductor will be added to the battery voltage. Since the diode only conducts one way ... The vcc of the cpu is being lifted.Youll have to experiment a bit to get the right output voltage but it is perfectly doable. If you cpu has a comparator you can even monitor the produced voltage and use that info to modulate the pwm train.

author: horzza
author_rank: Newbie
timestamp: May 29, 2012, 07:55:42 am
content: Thanks for all the great input, guys! Even if I only go with one approach, in other situations, having the other options in the back of my mind will likely be useful in the future.I think in a mass-produced environment, I might go with the the cpu-driven boost-converter, for low part count and moving the cost from the parts to the program, but at the moment I'm not confident enough with the code to get that running both with/without feedback, and the G2102 which I already posses (along with the converter chip in question) doesn't have an internal comparitor, making this even more difficult/require more external components. I think that I'll just go with the dual OR-ing Schottky approach, as I can't see the overall current draw changing much when the device is in such a low power mode, as a result of just a few missing mV. ~not much impact on efficiency. I guess that leaves the MOSFET as somewhat of a middle-ground, which is possibly the most useful in other perhaps higher-power applications?Really, I just wanted to say thanks for the advice!

author: TerminalJack505
author_rank: Super Contributor
timestamp: May 29, 2012, 09:18:05 am
content: I thought free_electron's idea was pretty cool. Sounds like a fun circuit to play around with.I also think NiHaoMike had a really good idea. You should be able to drive the MOSFET with the same signal that you use to shutdown the converter. And--here's the good part--because of how the converter is designed, you don't need a diode on its output.You should be able to do something like the attached. Once again the optional Schottky diode is to prevent excessive voltage sag during the switchover. And, once again you will want to introduce a delay in your code after switching on the converter before you drive the LEDs. You'll want to give the converter time to start up.You might try it on the breadboard.