Battery operation solution for dual voltage - Page 1
Forum section: electronics / projects

author: @rt
author_rank: Super Contributor
timestamp: November 01, 2016, 06:26:57 am
content: Hi GuysI have made a device that is powered from a 12 Volt bench supply at the moment,that uses a 7805 to regulate for 5 Volt for a GLCD and some 7400 series logic,and an LM317 for 3.x Volt supply for SD Card, GPS module, MP3 decoder, and a microcontroller.I’d prefer to power the device from a number of AA batteries, and wondering what the best way to go is.I imagine the 5 Volt side could be powered directly with 4xAA batteries in series, but then is the LM317 the best idea forthe 3.3-3.6 Volt side with efficiency in mind? The bulk of the current would be consumed on the 3.x Volt supply.Cheers, Art.

author: t2kv
author_rank: Contributor
timestamp: November 01, 2016, 06:35:08 am
content: What about using a small buck converter to efficiently step the ~5v down to ~3.3v, so you can power the whole thing from one set of batteries?Probably overkill, but these are pretty neat:http://power.murata.com/en/oki-78sr-3-3-1-5-w36-c.htmlEDIT: or, use a LiPo as the primary power supply, and a small boost converter to power the 5v section?

author: @rt
author_rank: Super Contributor
timestamp: November 01, 2016, 07:05:00 am
content: HiI'd really like to stick to AA batteries.The converter you linked to, why is it overkill?The SD cards do have fine tolerance so someregulator is really required, but the 7 Voltminimum input screws me over.4xAA batteries that would probably be ok todirectly supply the 5 Volt side wouldn't run the converter.I don't care if the overall supply is 4,6, or even 8 AA batteries,but two of the converters seems overboard.... or Maybe not, I'm really new to this.

author: cprobertson1
author_rank: Frequent Contributor
timestamp: November 01, 2016, 07:53:16 am
content: Right - just to confirm:- AA-battery IN (any number >> any voltage)- 3.3V OUT- 5.0V OUTWere I doing this I'd use a simple switching regulator likethis LM78xx Pin-compatable 3.3V Regulator- all you need is two smoothing caps on the input/output and you're good to go, with a dropout voltage of 4.75V.The5.0V versionhas a dropout voltage of 6.5V - so you'd probably want 5 or 6 AA batteries powering it.The downfall with these switching regulators is the ±3% regulation on them - in the case of the 3.3V line that'd be in the region of 3.2 to 3.4V - which is within the 2.7-3.6V specified byToshiba,SandiskandTranscend- you'll need to decide for yourself if they're accurate enough for your needs though!These particular regulators have 97% efficiency - so they shouldn't hurt battery life too much (but as I said, you'd still need 5 or 6 AA batteries to provide the initial voltage: once the batteries discharge a bit you'll get below the 6.5V dropout, IF I recall correctly, the 5.0V regulator will start falling off. It will still work for a little below that dropout voltage - but how far below that voltage I can't say - depends on a lot of factors.You might also consider either using enough batteries that the voltage will be above the dropout voltage even when the batteries are 90% discharged (which I think would be upwards of eight batteries? I'd need to consult some datasheets to get a better figure) - OR - using a step-up buck converter to get the same effect, at the expense of deep discharging the batteries and having lower current handling capabilities (if there's no energy left in the battery, it doesn't matter what tricks you use to boost the voltage, it won't be able to provide much current!)The best advice I can give is to check out the datasheets regardless of what you do!I'd be inclined to try and figure out what happens as the batteries discharge - for instance, in this case the 5V rail will cut out before the 3.3V rail will - can you use that to safely shut the device down?

author: @rt
author_rank: Super Contributor
timestamp: November 01, 2016, 01:12:21 pm
content: Quote from: cprobertson1 on November 01, 2016, 07:53:16 amAlmost. 8 is a maximumbut that does still sound feasible, thanksIt wouldn’t hurt to try them because it look like they are designed to drop in place of both existing regulators,albeit a slight change to the LM317 pinout is probably needed.If it turns out the 5 V supply always fails first, that might result in behaviour detectable by the micro,or maybe a spare input of a shift register could look at a resistor divider as input for one of it’s bits.

author: cprobertson1
author_rank: Frequent Contributor
timestamp: November 04, 2016, 09:42:31 am
content: Quote from: @rt on November 01, 2016, 01:12:21 pmLol - when I said "any number" I really meant "any number within reason"I *believe* the 5V reg will stop first (as it has a higher dropout voltage than the 3.3V reg) - you may find, however, that it's performance just deteriorates below the dropout voltage instead of actually stopping altogether. I'll need to check the datasheets again!Since you have a micro, I'd say "hey! let's put it to good use!" - and just use it to monitor the rails (with a big resistor in series, the current draw will be minimal) and do what you need to do from there. Need to test it I'm afraid! Bench power supply to the rescue (or use a pot as a voltage divider and work from there if you don't have a bench supply!)

author: setq
author_rank: Frequent Contributor
timestamp: November 04, 2016, 11:24:54 am
content: There are some interesting tips in here for dual 3.3 and 5.0 volt systems which may cover your requirements:http://www.microchip.com/stellent/groups/techpub_sg/documents/devicedoc/en026368.pdfI'd probably use 4x AA's to get 6v (from 6.6v-4.6v over lifetime - remember that batteries die slowly!) then use a buck/boost converter to get the 5v line. This will preserve a fixed supply voltage higher than your source supply and cope with battery drain and your system will most likely last longer than a simple linear regulator chain. Then use an LDO to drop that to the 3.3v for the rest. It does depend on how much current your device is pulling. If it's a rather hefty by 2016 standards 100mA you'll lose 170mW of output in the LDO. If your load is peaks of power occasionally this makes sense. If it's constant drain, a second buck converter is probably more efficient.You can use a cheap IC to do it. Even the ancient and crappy MC34063 has some mileage here.Also once you have this topology configured then you can sample the battery voltage with the ADC on the microcontroller periodically with a high impedance divider. No worries about brownout detection etc.Also read Art of Electronics 3rd edition on low power design - loads of tips in there.