Say i have a circuit that runs on 3 v. if the circuit consumes 114 uA. If the circuit is powered by 2 AA batteries with a 1000 mah rating than the circuit would have a lifetime of about 1 year. If i replaced the batteries with a 12 v battery with the same capacity would the lifetime be extended to 4 years?
The lifetime of the product would most likely be measured in microseconds if you connected it directly. If you have a circuit inbetween, the results depend on the circuit. Four years would be a theoretical maximum.
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
why do you say microseconds?
Spehro Pefhany wrote:
3V device run off 12V. Try sticking a cruise missile up your tail pipe.
No - the lifetime would likely shrink to approximately zero.
If the circuit is intended to operate on 3 volts, and you apply 12 volts, the circuit will probably be destroyed.
-- Peter Bennett, VE7CEI peterbb4 (at) interchange.ubc.ca new newsgroup users info : http://vancouver-webpages.com/nnq GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca
say the 12v is converted to 3 v via a dc to dc converter at 90 % efficiency. what i am asking is does the fact that 12 v is 4 times 3 volts equal an increase of 4 times the life of the device.
You would do all of that for a 114 uA load?
That's a lot of stuff. Why not 2 C cells or 2 D cells?
Tam
If things were ideal-- the converter was 100% efficient, if the batteries were the same A-h capacity measured under the same conditions and if the batteries had zero self-discharge.
There's self-discharge current in the batteries which (depending on type and conditions) might be of the same order of magnitude as your load. You're also losing 10% in your hypothetical 90% efficient converter (might be hard to achieve 90% efficiency at such a light load). There are a few other factors, but those are likely the most important ones. It could be a lot worse than 4:1.
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
./................................................................................
Even if he could build a DC/DC converter that only drew 114 ua at 12V, the efficiency would already be down to ~25%.
Tam
The simple answer is no (it will be somewhere between less and much less, assuming that the circuit has a regulator that can handle the higher input voltage!).
I am assuming that since you have given 1.5V for the cell voltage you are currently using Alkaline AA cells. At such a low continuous current load, converter inefficiency and battery self discharge are going to become big battery draining factors. Also, you haven't mentioned what the cutoff voltage for the circuit is. This will greatly effect how long the circuit is capable of running before the supply voltage gets too low.
Since it sounds like you have the space (if a 12V battery is an option) you are better off changing to C or D cells to extend circuit life. If you want to get a bit more technical, a small solar cell and associated converter/regulator may be a better long term option (although regulator and battery choice will effect life).
I've used a MAX1674 step-up converter/LDO with a 20uA draw, and that's only because I didn't want to buy a full reel of the one that only draws 1.5uA.
So what's your point? Think a switcher necessarily draws >114uA?
OK, maybe I was wrong, but the OP wants 12V in and 3V out; the 1674 has a max input voltage of 6V. The 114 ua is his load current. I was merely suggesting that he may not be able to get a 12V input current any lower than that.
Tam
I concur with the larger cells at 3V idea. But since the device is going to be running near the cell's shelf life (larger cells will just self discharge over roughly the same time period), the OP might want to consider some cells with longer shelf life, like lithium cells.
-- Paul Hovnanian mailto:Paul@Hovnanian.com ------------------------------------------------------------------ "A doctor can bury his mistakes but an architect can only advise his client to plant vines." -- Frank Lloyd Wright
Sure ... 4 years is your theoretical maximum assuming zero inefficiencies and losses in your dc-dc converter. Shelf life shouldn't affect the lifetime as it is already incorporated in your earlier calculation of 1000mAh/114uA*24 ~365days. If you think hard about it you'll realize the converter plus the 12V battery may not be worth your time. How do you get the 12V battery? By putting in series 4 of your 2 batteries. Put in parallel 4 files of
2AA series connected batteries and you've got your 4 years of capacity. The math is quite straightforward but you can easily see why it works for 4 yrs because you are actually letting each cell deliver less current in a parallel combination or another way is you are actually reducing the output impedance of your battery combination by paralling the internal resistances of the batteries. Fiddle around with such combinations and your options are limitless. Larger cells? Ummm remember the internal cell resistances go up with larger cells.exskimos
and if one cell doesn't go bad there by, killing the others! you'll be fine.
-- "I am never wrong, once i thought i was, but i was mistaken" Real Programmers Do things like this. http://webpages.charter.net/jamie_5
1000mAh rating at what discharge rate? If it's a typical 10 hour or so rate, you're liable to get quite a bit more out of them if you discharge them at a constant low rate for a year. But another part of the answer lies in how low a voltage you circuit can tolerate and still operate properly. If it's designed to operate down to, say, 2.2V input, then you'll get practically all the energy out of your AAs. But if it dies just below 2.99V, you'll waste a lot of available energy, and you'd be better off with an efficient switching regulator, driven by whatever--a single cell or three or more cells. You should research energy versus weight and volume for some typical battery voltages and cell chemistries, if you want to optimize operating life for a given volume or weight. The info is out there on the web if you go looking.
Also, decent fresh alkaline AAs at low rates should be able to deliver more than twice the 1000mAh you suggested, at least if you can let them discharge down to 1.2V or less. See, for example,
People have mentioned self-discharge. I would expect with a good alkaline at room temperature (20-25C), you won't notice self-discharge in a year. If you design for ten year battery life, you should consider a different chemistry, though.
Note that it wouldn't take a very big capacitor to keep running this device for several seconds while the batteries were being changed...at least assuming it doesn't draw big pulses of current occasionally.
Cheers, Tom
Huh? Not sure what kind of cells you are talking about, but that's generally NOT true of the cells I'm familiar with. A good "D" alkaline will have lower internal resistance than a "C" of the same series, which in turn will have lower than an "AA" of that series. There are big differences in internal resistance of a given size with changes to different chemistries, though.
Cheers, Tom
Speaking of long shelf life, I have an HP 32SII calculator that I bought a long, long time ago--maybe 15 years ago. I don't use it a lot, but I do use it from time to time. It still has the original batteries in it. I have never replaced them.
Based on a Google search this calculator uses 1.5V, silver-oxide, Duracell D 303/357 button cells with a capacity of 165 mAh. The calculator uses 3 of them in series.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.