Hi,
The following battery is
Thanks
John
Given that, by definition the capacity is the total charge that you'll get from the cell if you draw current at a moderate rate (70mA is moderate for this capacity) from full charge voltage to shutoff.
2: That depends on the cell. Dig around -- the information is out there. Basically, figure that with a good cell it'll take an hour plus (capacity)/(current). Properly charging a Lithium-ion cell requires that you limit the maximum current and the maximum voltage -- so when you get to the end of the charge cycle the battery is controlling how much charge it accepts, not the charger. 0.2A sounds pretty low for that capacity, but maybe not -- maximum charge current varies a lot from design to design, and newer cells can accept more current safely than older ones. Brand spanking new designs (not product -- design) can take their capacity / 1 hour, or even more. Older cells are much more lame in this respect. You need to find the data on the cell, or just charge conservatively. 1: Discharge it at the rated current from full charge down to 3V, and calculate the total charge.-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Hi,
1: Discharge it at the rated current from full charge down to 3V, and calculate the total charge.I am assuming to use a resistor to do it. But how should I determine the value.
John
Because it says so on the packet +/- a bit...
0.650/0.2 = 3.25 hours+ a bit , assuming it's totally flat. 0.650/0.07 about 9 hours + abit.It's always + a bit...
You need to integrate the current over time to get the charge. In practice discharge it though a resistor and write down the current every five minutes or so, then piece-meal integrate. (I(t=3D0)
*5minutes +I(t=3D5)*5 minutes... etc.)George H.
I am _really tempted_ to tell you that if you can't do the calculation, you shouldn't be messing with lithium batteries. Other batteries -- sure, no problem. Expensive batteries that are sensitive and tend to burst into flame when unhappy -- maybe not. But, I assume you're grown up, or at least that your parents know what you're doing. Just don't burn your house down*.
Ohm's law. The Most Basic Electrical Engineering Equation Ever.
E = I * R,
where E = electromotive force, in volts, I = current (intensity) in amps, and R = resistance, in Ohms.
Do a bit of algebra, and you see that
R = E / I.
So, 3.7V / 0.07A = 52 Ohms. Note that as the battery discharges, the current will change. Note too, that if you're doing this test yourself you need to watch the thing like a hawk toward the end of charge -- the cell voltage will quickly descend from 4.2V down to 3.8 or so, then very slowly go down to 3.4 or so, then it'll hit a 'knee' and start going down rapidly. _You_ have to shut it off at 3V -- if you don't the cell has a built-in low-voltage protection circuit that'll permanently disable the cell (but will hopefully save you from a fire).
** Probably because she didn't see the fire that motivated me to put the extinguisher there. And I never needed it.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Until the battery gets old or abused. Then it's - a bit, or more.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
LOL
I think those commercial cells will have protection built in- over/under voltage shutoff, over current shut down. Unlike Radio Control LIPOs..... 12S 30C, 5000mAHr, no protection and go like a bomb.. literally :)
I think the radio control LiPo cells must have some of that, now -- the word that I've heard is that the newer stuff is less likely to torch. Of course, the old cells were 10C, and everyone ran them right at the limit. Now they're capable of higher discharge, but few people run them over 10C max, because if you do then your flight times suffer.
Between that and newer charging techniques, the only time you need to worry about torching a battery is when you do physical damage to it (like, running a plane into concrete at full speed).
Abuse aside, cells do degrade over time, and when they do the capacity goes down. Even when you take good care of them, rechargeable batteries eventually lose their capacity and have to be retired.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
I kind of think the OP is a student and the questions are from the teacher. But of course it's hard to be sure about that.
Or might take the midpoint voltage, 3.35V, or some other nearby midrange value, and use that to define the resistor value.
....
What about this, though?
Should present a higher dynamic resistance, maybe roughly one order of magnitude larger than you'd get with a resistor determined entirely by the design load current.
But maybe there is a better way to use two active devices here? This came to mind quickly, is all.
Jon
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