Replacing AAA Battery with a super Capacitor  - Page 1
Forum section: electronics / beginners

author: GoatZero
author_rank: Regular Contributor
timestamp: January 07, 2014, 07:36:30 am
content: I have a device that requires around 1.541v DC and from 16uA to 30uA to work as inteded, normally it uses 1 AAA battery, originally i wanted to skip the battery and replace it some nokia phone 5v AC adapter 350mA, i ended up making a voltage regulator using an LM317 to set the voltage as low as i needed (see picture)Since i have already come this far i was thinking for a way to make it work with both the AC adapter and the 1.5v battery in order to stop it from resetting the time (yeah it resets without energy) then i found out about super capacitors, i have been reading a bit and im thinking about placing one in parallel at the entrance of the voltage regulator circuit, this way it should be always charged and if power runs out it should slowly discharge and keep the device on for a few hours or if possible days.... however across all this proyect many questions have arised1) When regulating 5v Ac adapter voltage, i get the reading of 8.57v DC (thats the reading from the multimeter without load)2) When regulating the AC adapter makes a some buzz noise, its barely noticeable but it can be heard, why is this happening? (temperatures forcircuit and ac adapter are ambient )3) How can i calculate the Farads and voltage needed for my circuit in order to get a supercapacitor of those values? I have been looking at newark and there only seem to be supercapacitors from 2 volts and above , my device resets itself above 1.8vThanks for advance for any kind of response

author: hagster
author_rank: Frequent Contributor
timestamp: January 07, 2014, 08:44:52 am
content: I think it would be best to put the capacitor after the regulator. You can get higher capacity super caps at the lower regulated voltage and you don't need to worry about leakage through the regulator.You also probably need to limit the inrush capacitor current. Look at it data sheet to see what it can reliably provide. You can probably just put a series resistor inline with it. At the uA discharge rate it won't create much voltage drop.Here are some formulas to work out the discharge curve.http://www.electronics-tutorials.ws/rc/rc_2.html

author: lapm
author_rank: Frequent Contributor
timestamp: January 07, 2014, 08:46:37 am
content: First things first. Capacitors voltage drops as you take current from it. Its energy storage, not battery. Voltage level depends energy left in capacitor and capacity(Farads).http://en.wikipedia.org/wiki/CapacitorSecond get supercap with at least twice the voltage you need. This is to give you safety margin and prolonged capacitor lifetime. Most capacitors don't like to operate near their rated maximum voltage.In case 1, sounds like its not really regulated at all. Idea of regulated output is that it gives same voltage even if output is left open and you don't overload it.2. Some badly designed switching power supplies (usually "Hung Low" Chinese types) can use switching frequency that some conditions can be in in range of human hearing. Sometimes its just badly made inductors, where separate winding's in coil get to move to magnetic forces.

author: mariush
author_rank: Super Contributor
timestamp: January 07, 2014, 01:42:11 pm
content: As you were already told, supercapacitors are a bit (well, quite) different compared to batteries.Batteries, as they discharge, very slowly lose voltage, going from about 1.65v when fully charged to about 1.35v when completely discharged. When the voltage gets near 1.3-1.35v, a battery can no longer provide current to a device, even though it still shows voltage when measuring it with a multimeter.In contrast, supercapacitors can completely empty themselves, they can go from 2.75v or 5.5v (two common maximum voltage ratings for supercapacitors) down to 0 volts. Unlike batteries, the drop in voltage as they discharge is directly proportional to how much current you take out of them.The rechargeable kind of batteries need to be charged slowly, otherwise the internal chemistry is affected, the battery overheats and gets damaged, the battery can even blow up. Supercapacitors don't have this kind of problem, they can be charged in minutes or even less, but that also means that the power source needs to be able to give them so much energy to charge fast. When they're completely discharged, supercapacitors act like short circuits so the power supply gives its maximum into the supercapacitor. As they charge up, the capacitor will start to draw less current.AC adapters use inductors and transformers inside that buzz differently at various degrees of usage, it depends on how much current a device takes from them. Some ac adapters are unregulated, meaning they say they can output 7.5v for example, but without anything serious to use that power the adapter may output 9-10v.Phone chargers like that Nokia are also somewhat unregulated, because it's easier and cheaper to make an adapter that's less than perfect and it's perfectly safe because the phone itself has a chip inside that wil further convert that voltage into something that can charge the battery.Now let's get back to your circuit. Remember what I said about voltage on supercapacitors going down very fast to 0v as they're discharged?In your circuit, you have a problem because you're using that LM317 linear regulator to produce 1.53 volts.The LM317 regulator is a good chip but you have to remember all linear regulators work in a way that make them need a bit of voltage above the configured output to work properly. As the LM317 is a very old design, it needs about 2v above the output voltage.So, if you have a 5.5v supercapacitor before LM317 and you charge that up to 5.5v, the LM317 will produce 1.53 volts just until the voltage on the supercapacitor reaches 3.53 volts. You're losing about half the charge the supercapacitor holds due to this weak LM317. You won't even be able to use 2.7v rated supercapacitors, because the LM317 can't output 1.53v with so little voltage.There are other linear regulators which don't need as much voltage above the set output voltage to work, some only need 30-40mV (0.04v) and they're just as easy to use as LM317.For example, I could suggest* MIC2941 :http://uk.farnell.com/micrel-semiconductor/mic2941awt/ic-reg-ldo-1-25a-adj/dp/1556737or* MCP1825http://uk.farnell.com/microchip/mcp1825-adje-at/ic-ldo-adj-500ma-to-220-5/dp/1578398(this one only supports up to 6v at input so don't use supercapacitors larger than 5.5v if you use this)Such linear regulators will allow you to use all the charge in a supercapacitor, from 5.5v or 2.7v all the way down to about 1.6v before they can no longer output 1.53v you set, giving you longer runtime from supercapacitors.You also need to be careful about how much voltage you put on the supercapacitors, because they have that maximum voltage rating of 2.7v or 5.5v. There are some other higher voltage ratings, or lower (2.5 and 2.3v are other common ratings), you can't just put a supercapacitor at the adapter input because you'll blow it up.You also need a regulator to give the supercapacitors a voltage just below their maximum voltage rating. For example, if the supercapacitors are rated for 2.7v you may wish to configure the regulator to output 2.65v to the supercapacitor.You could use a LM317 here, because the fact that there's a 2v loss on the regulator doesn't matter that much as you have the adapter producing 9-12v. A lm317 would also work to charge a 2.7v capacitor from USB , because USB has 5v and that's large enough to give the LM317 room to output 2.65v or around that value.But if you have a 5.5v rated capacitor... it won't do. The MCP1825 would again be great to get 5-5.5v from USB and output around 5v in the supercapacitor rated for 5.5v maximum.Remember what I said about supercapacitors acting like they're a short circuit when they're discharged? When they're empty they can draw 3-4 Amps from a regulator. LM317 is designed for 1.5-2.5 amps so it will be able to give that much to supercapacitor, but it's also important that the input of the LM317 is able give that much current.Phone chargers, some power adapters, usb jack on your computer are not really designed to output more than around 0.5-1 amps so it's not really a good idea to let the supercapacitor suck so much current.The easiest way to do this would be to use a linear regulator that's designed to output a lower maximum current and which will have internal current limit, not allowing the supercapacitor to exceed a certain current amount.For example, the 1117 regulator :LD1117 :http://uk.farnell.com/stmicroelectronics/ld1117v/v-reg-adj-1-25-15v-1117-to-220/dp/9755829TLV1117 :http://uk.farnell.com/texas-instruments/tlv1117ckct/ldo-2-7-15v-1-3vdo-0-8a-adj-3to220/dp/2322054LM1117 :http://uk.farnell.com/texas-instruments/lm1117t-adj/ic-v-reg-ldo-adj-1-25-13-8v-1117/dp/9485805is just like LM317 but is designed to only output about 0.8 amps and has a current limit set around 0.9-1.2 amps so the supercapacitor won't be able to draw more than that from the input. This 1117 also needs just about 1v above the output, which means you could use it to charge a supercapacitor rated for 5.5v from usb, charging it up to about 4v.so recap ...[ input dc, at least supercapacitor voltage rating + voltage drop of linear regulator ] ----> [ regulator to output less than supercap rating and possibly limit current at same time ] ----> [ supercapacitor ] -----> [ regulator with as low voltage drop as possible for longer life ] -----> [1.53v]Also keep in mind that those battery coin cell type of supercapacitors are not ideal for this type of application. You should look at the supercapacitors that look like regular electrolytic capacitors.

author: GoatZero
author_rank: Regular Contributor
timestamp: January 08, 2014, 06:07:28 am
content: Thanks for the help i will loook into these regulatores in order to complete this proyect, i will post results if they were as intendedthnx again