07212020, 12:29 AM
Opus is you want a multibay charger, or a RC/hobby balance charger would be my top pick... and it's also a really nice item to just have around for battery work.
3.2v 26650 LiFePo4 testing

07212020, 12:29 AM
Opus is you want a multibay charger, or a RC/hobby balance charger would be my top pick... and it's also a really nice item to just have around for battery work.
07212020, 10:03 PM
(This post was last modified: 07232020, 05:45 PM by gauss163. Edited 14 times in total.)
(07202020, 02:57 PM)rearden Wrote: What discharge rates do you consider to be high current? If we fully charge an empty cell then the reported mAh is very close to its (current) chemical capacity (assuming that the charge termination current is not set unreasonably high). But the actual usable capacity that we can extract from that charge depends on various factors, esp. the discharge rate (current), health (esp. R = total DC IR) and temperature, because  by Ohm's Law  the voltage while (steadystate) discharging is dragged down roughly by I*R at discharge current I, which causes it to sooner reach the termination voltage. If the product I*V is not too large then we'll be able to discharge almost all the charged capacity (typically over 99%), so the charged capacity yields an excellent estimate of the discharge capacity. But this close correspondence breaks down when I*R gets too large. This is easiest to comprehend by looking at discharge curves at various rates, e.g. below, for two identical NMC Liion cells. For each rate there are two similarly colored curves  one for each cell. His charge curve shows they charged about 1044mAh, and discharged 1044mAh @100mA and 200mA, and 1039mAh at 500mA, all > 99.45% of the charged capacity. But at higher rates the curves 2.8V termination points move leftward, yielding lower capacity, e.g. at 15A (two lowest curves) the stronger cell yields about 923mAh(92%) and the weaker peters out much earlier at around 416mAh(40%). This is because the I*R drop has essentially pulled down the voltage curve so much that its termination point is now very close to the "knee" of the curve  where it has much smaller slope, so even a small voltage change causes a large capacity change. To see this more clearly, suppose our termination voltage is 3.0V instead of 2.8V. Then the lowest 2 curves in the above graph show that at 15A both cells peter out very early  at around 255mAh(24%) and 310mAh(30%) when they hit 3.0V. As we can see from the stronger cell's curve, most of its energy is delivered in the flat part of the curve below between 3.0V and 2.9V, but we cannot access it because it lies below our 3.0V termination voltage. [Note: the reason this curve has flatter shape than the lower rate curves above it is due to selfheating effects. The large 15A rate generates sufficiently more internal heat that it decreases the cells internal resistance I, which decreases the voltage drop I*R. This was enough rise for the stronger cell to deliver most of its energy above 2.8v, but the weaker cell couldn't recover in time]. Those were new cells. If they were more aged their internal resistance might be twice as large, so we'd notice the same effects as above at about half the current, since (I/2)(2R) = IR yields the same voltage drop. But still the discharge capacity remains above 99% of the charge capacity at the lower rates, so we don't need to do a discharge to know the (chemical or lowrate) capacity  we can simply use the charged capacity. Notes For simplicity I have mostly ignored temperature effects above. But these can play a large role  esp. at the extreme ends of temperature ranges. e.g. in cold winter termperatures the selfheating effect may make or break whether or not a cell delivers decent capacity (IR greatly increases at cold termperatures, which is why EV packs often employ heaters). The linked tests use professional equipment. Consumerlevel (dis)chargers are less precise and accurate so they may report slightly more variation between charge and discharge capacity due to their limitations. 
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