Is there a product like this? 2 x AA to 8 x AA "adapter" / booster? ~2.5v > 12v - Page 1
Forum section: electronics / general

author: salil
author_rank: Regular Contributor
timestamp: August 08, 2014, 10:51:19 pm
content: Does a pre-built product like this already exist?Apparently some rechargeable AA batteries (i.e.: Energizer) can handle more than3A discharge rate, so that got me thinking, for some low current devices that take 8 x AA batteries in series (because the devices needs an input voltage of ~12v), is there an adapter with a boost converter that will act like 8 x AA batteries by stepping up the voltage of 2 x AA batteries? I have about a dozen or so devices like this that require 8 x AA batteries but consume very little current. Recharging 96 AA batteries gets pretty tedious. It should be fairly simple to buy boost converters and wire something up like this, but the form factor and such would be convenient to already have so it fits in the battery compartment securely like AA batteries do.So, before building something like this I wanted to see if anyone had heard of a product like this already in existence?

author: Rick Law
author_rank: Super Contributor
timestamp: August 09, 2014, 01:46:14 am
content: Quote from: salil on August 08, 2014, 10:51:19 pmLook up LM2577 boost on eBay you will find plenty of ready-to-go boost boards at about $2-$3 USD. It will boost 2.5V to 12V no problem at about 2Amp max. BUT - it will not be able to deliver the same amps that 8xAA could, and efficiency will be awful.How well it works depends on what you mean by low current: You need to do some evaluation of the power you need."Back of the envelop quick math": At that boost level, you are looking at about 5 to 1 current assuming 100% efficiency. Your 3A@2.5V becomes only 600mA@12V@100%efficiency.At 3A current, your AA battery (even assume it is 3000mAH) wouldn't last an hour.Perhapsnot even 30 minutes at that current.Of course 100% efficiency is not possible. I typically get about 70% efficiency boosting 2AA to about 12V at around 100mA with my LM2577 board. If you have higher draw, you will get less since power lost is proportional to current3A@2.5V gets you 600mA@12V, but it is down to 500mA at 70% efficiency. It will last for may be 30 minutes but surely less than an hour.These are guestimates. You have to do some serious testing on what current you need at 12V and test how long your pair of energizer can deliver 6X that at 2.5V.Good luck,RickEDIT: by the way, 2.5V is just about the cut off for the LM2577. Nominal CUT OFF for the LM2577 is 2.6V I think. I have taken it down to 2.3V (12V output drops to 11V-ish by then) and still working at 100mA-ish output, but not much lower. Eneloops are very good at delivering most of its power at >=1.2V, other batteries may not be as good.

author: mariush
author_rank: Super Contributor
timestamp: August 09, 2014, 02:24:10 am
content: LM2577 only works with a minimum of 3.5v, and then the efficiency will be lousy, at around 65%.You must define first what low current means for you.. because low current could be 1A which may be a lot to me.I would actually recommend going with something like LT1308:http://www.linear.com/product/LT1308High frequency boost regulator, works with tiny inductors and ceramic capacitors and resistors , works as low as 1v (to output 2.5v or 3.3v or something like that, needs more for higher voltage like 12v).As a demonstration, here's the LTspice simulation for LT1308, you can see with 2 rechargeable batteries (2.5v, about two half full AA batteries) can produce about 12v @ 175-200mA but the current pulled from the two batteries will be about 1A, which means the batteries will last about 30 minutes to one hour.Second picture has 12v @ 20mA ... you can see the current pulled from batteries is more reasonable.LT1308 is available in SOIC package, it's easy to solder on a pcb, it may work on a soic to dip adapter but it's really not recommended, ideally you follow the layout from the datasheet as close as possibleYou can download LTSpice and you have the Linear Tech chips "built in", you just need to go to insert a component into the schematic and there you can click "open this part's fixture" and it gives you the circuit ready made and ready for simulation.

author: Rick Law
author_rank: Super Contributor
timestamp: August 09, 2014, 03:13:22 am
content: Quote from: mariush on August 09, 2014, 02:24:10 amGeeze, I must have my boards mixed up.Norminal for LM2577 is indeed >= 3.5V, not 2.6V. I just retested my two LM2577 boards. It did work below 3.5V (but is not recommended working off the margins).Of the two I tested, the "better" one goes down to 2.60Vin (on my UT61e calibrated 5V with DMM Plus) with 100mA@12Vout dropping from 12.3Vout down to 12.1Vout.The other down to 2.65Vin with steady 12.3Vout, but by 2.60V, the 12.3V out drops to 10V out while at the same 100mA-ish current.Both cuts out at just around 2.5Vin generating essentially nothing at the output. With nominal Vin>3.5, 2.5Vin is way off norminal. I would cheat a little but not that much.My 2.3V cutout is incorrect.Sorry, must have the boards mixed up or brain scrambled.Thanks Mariushfor catching it!Rick

author: salil
author_rank: Regular Contributor
timestamp: August 09, 2014, 03:37:30 am
content: Thanks for your insights and findings, mariush and Rick.For my particular application, the nominal current would be: 150mA @ 12v DCMax current: 200 mA

author: Rick Law
author_rank: Super Contributor
timestamp: August 09, 2014, 04:44:51 am
content: Quote from: salil on August 09, 2014, 03:37:30 amYou could try 3x AA or 4x AA. Either will take you above the 3.5V of the LM2577. Those boards are indeed cheap.200mA in doesn't sound like a lot, but when you boost, that is a lot. Particularly when the more current draw (above a certain minimum), the less total energy you get.3x AA is 3.6 to 12 is 3.33x boost. The 200mA@12 needs 667mA@3.6.4x AA is 4.8 to 12 is 2.5x boost. The 200mA@12 needs 500mA@4.8VIf you get 65% out of 100%, the 667mA in will need about just over 1000mA in to account for the efficiency lost. 3x 2000mAH AA battery will die in about 2 hoursbest casebut likely much much less when you are drawing that high. Many AA's are not very good at pumping >1A sustained except the Eneloops. (and I am not about to go by memory and say how long... I will likely mix up again given it was done around the same time when I testing the booster.)If you boost less with 4x AA (4.8V to 12V = 2.5x boost), it will actually last more than 1/3 longer than 3xAA. At that lower current, you can draw more total power.With the boards being so cheap, perhaps you should get one just for experimenting. It is fun to play with. Note also, in that eBay cheap board arena, there is a new XL6009 which claims to be lm2577 replacement. It may be better but it may be worst. I have not try that one so I can not say.Good luck!

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 09, 2014, 11:43:33 am
content: There are many clever ways to skin this cat. It is, indeed, a tricky problem owing to the combination of a high step-up ratio and low input voltage. Two things I would consider are using a tapped-inductor boost (maybe 1:3 ratio), and, perhaps a bipolar transistor for the switch.However, there are switchmode controller & regulator ICs that can operate efficiently on 2V or less. For example, the LM2621 regulator (ie - the switch is integrated) from National/TI and the LT1619 controller (which uses bootstrapping to reliably drive an external MOSFET).http://www.ti.com/product/lm2621http://www.linear.com/product/LT1619Not sure how to embed the links into the part numbers above with this forum software... oh well.

author: Rick Law
author_rank: Super Contributor
timestamp: August 09, 2014, 06:27:29 pm
content: Quote from: MagicSmoker on August 09, 2014, 11:43:33 amToo bad we can't fight Physics. Physics dictates thatloss is proportional to currentandpower_in > power_out.The OP is looking for a "ready made product", that reduces option further.Now that I refreshed my memory... 3xAA was how I made my set up. When 3xAA NiMH norminal is 3.6V (which is above the LM2577 nominal cutoff). When it goes below 3.5V, there isn't much power left anyhow. That was my logic then.RickBy the way: I remember why I thought it was cut_off = 2.3V. To avoid a cell going negative in my 3xAA+LM2577, I had installed a LiIon cell protection circuit which cuts off at 2.3V. Over time, I got the numbers scrambled.

author: MagicSmoker
author_rank: Super Contributor
timestamp: August 09, 2014, 10:11:05 pm
content: Quote from: Rick Law on August 09, 2014, 06:27:29 pmHuh? I suggested the tapped inductor boost because it uses an autotransformer instead of an inductor to provide some of the voltage multiplication in a boost converter. Volt-seconds and amp-turns are conserved; the universe is happy.And the suggestion to use a bipolar transistor in a circuit that has to run off ~2V minimum should be self-evident.

author: Rick Law
author_rank: Super Contributor
timestamp: August 10, 2014, 01:04:31 am
content: Quote from: MagicSmoker on August 09, 2014, 10:11:05 pmI am not saying your suggestion is not good. I don't know enough about it to say it is or isn't anyhow.I am just expressing that the fight is up-hill:- Anything less than 100% efficiency, there is lost. Laws of Physics said you can't get 100% efficiency. The lost is higher when boosting because of these other two Laws:- Laws of physics said you can't get more wattage out of it then you put in so you need more current.- and Laws of Physics said the higher the current, the more the wattage lost.I trust your suggestion is good. Just that it is still an up-hill fight.Rick