Yep, will do. Not had a chance to do much with the pack so far since getting it back. Hopefully next week.Are you going to do a capacity check ?
Can you link the thread?Can you make a discharge curve like saneagle's in the other thread?
Ok, but I'll probably use a higher current than 0.5A, otherwise it'll take too long.This thread, post #63 maybe;
Batteries Charging Routine
I will remind you, but remember, I am interested to know what the charger consumes when the voltage at the output is above 41V. My SANS charger does not let me do that above 41.3V I will use a lab power supply to measure current and voltage for charging and battery tester for discharge.www.pedelecs.co.uk
Yes, but it is a check from 4.2V to 4.1V per cell, so for accuracy, done at 0.5 A.This graph from saneagle is very easy to understand
In my experiments around 7 or 8% and most of that is between 41 V and 41.5 V. Above 41.5 V the added energy becomes very small and then when the balancing trips in at around 41.8 V, it becomes hard to tell how much energy is going into charging and what is being consumed by bleed off in the balancing process. It might be better doing such capacity experiments on a single cell, where there is no balancing.I think there is strong interest in charging at 41v in order to prolong life of the battery and reduce fire risk during charging. However, the cost needs to be quantified, how much loss in capacity?
That graph gives you the capacity loss when drained from 4.2V to 4.1V per cell. Which does not tell you what the capacity would be when charged to 4.1V and drained from there.Yes, but it is a check from 4.2V to 4.1V per cell, so for accuracy, done at 0.5 A.
I noticed that point in your experiment but you have not yet repeated the same experiment with Sturmey 's diode idea.That graph gives you the capacity loss when drained from 4.2V to 4.1V per cell. Which does not tell you what the capacity would be when charged to 4.1V and drained from there.
I used a practical drain load of 2A on my 42V 5Ahr battery as that was the current used when on a typical ride, assist level 2 at 14mph or so.
Not yet, waiting for a 1A charger to arrive.I noticed that point in your experiment but you have not yet repeated the same experiment with Sturmey 's diode idea.
Battery Capacity Test | |||||
Time (Sec) | Voltage (V) | Current (A) | Watts (W) | Energy (Wh) | |
0 | 41 | 0 | 0 | 0 | |
1 | 40.9 | 1.6 | 65.4 | 0 | |
100 | 40.8 | 1.59 | 64.8 | 2 | |
220 | 40.74 | 1.59 | 64.7 | 4 | |
300 | 40.71 | 1.58 | 64.3 | 5 | |
1080 | 40.42 | 1.57 | 63.4 | 19 | |
1920 | 40.16 | 1.56 | 62.6 | 34 | |
2700 | 39.92 | 1.55 | 61.8 | 47 | |
3600 | 39.66 | 1.55 | 61 | 63 | |
4680 | 39.36 | 1.54 | 60.6 | 81 | |
8520 | 38.36 | 1.5 | 57.5 | 144 | |
10260 | 37.91 | 1.49 | 56.4 | 172 | |
11820 | 37.54 | 1.47 | 55.1 | 195 | |
14580 | 36.88 | 1.45 | 53.4 | 237 | |
16860 | 36.48 | 1.43 | 52.1 | 270 | |
19680 | 36.16 | 1.42 | 51.3 | 311 | |
21720 | 36 | 1.41 | 50.7 | 333 | |
22560 | 35.91 | 1.41 | 50.6 | 352 | |
25200 | 35.68 | 1.4 | 49.9 | 388 | |
28440 | 35.33 | 1.38 | 48.7 | 434 | |
29280 | 35.21 | 1.38 | 48.5 | 444 | |
29820 | 35.12 | 1.38 | 48.4 | 452 | |
30600 | 35 | 1.38 | 48.2 | 462 | |
31200 | 34.88 | 1.37 | 47.7 | 470 | |
32520 | 34.55 | 1.37 | 47.3 | 488 | |
33300 | 34.29 | 1.39 | 47.6 | 497 | |
33840 | 34.09 | 1.39 | 47.3 | 505 | |
34440 | 34.08 | 1.39 | 47.3 | 513 | |
34980 | 33.98 | 1.38 | 46.8 | 520 | |
36060 | 33.73 | 1.37 | 46.2 | 534 | |
36120 | 33.71 | 1.37 | 46.1 | 535 | |
36240 | 33.66 | 1.37 | 46.1 | 536 | |
36360 | 33.63 | 1.37 | 46 | 538 | |
36480 | 33.57 | 1.37 | 45.9 | 539 | |
36600 | 33.52 | 1.36 | 45.5 | 541 | |
36720 | 33.45 | 1.36 | 45.4 | 542 | |
36840 | 33.38 | 1.36 | 45.3 | 544 |
Your assumption is incorrect. It's impossible to have 10% of the charge between 40.9v and 42v. I alreay showed you that it's about 3%. Any 18560 discharge chart will show the same. That means that your battery lost 7% not zero %. It's impossible to lose zero % after hundreds of cycles. Not even LiFePO4 can do that.So, the capacity check...
I charged the pack to 41V. I left it to stand for a week, then did a discharge test. I found the heating element I had lying around from a heat gun, with the intention of discharging at around 1.5 to 2A, depending on how the resistance of the element changes as it heats up, compared to it's measured resistance at room temperature.
I set up the battery with a new Watt meter. Open circuit, I was getting 41V although it is probably about 40.9 as the Watt meter has an an error of about 0.1V compared to the calibrated DVM.
Initial current was around 1.6A and I took readings at intervals over a 10 hour period.
View attachment 53417
After 10 hours, I had to go to bed (it was 1am by this time). The pack was down to 33V or so, but it wasn't exactly falling off a cliff at this point and I still think more could have been taken out of it. So from 41V to 33.38, I got 544 Wh.
Here is the raw data:
Battery Capacity Test Time (Sec) Voltage (V) Current (A) Watts (W) Energy (Wh) 0 41 0 0 0 1 40.9 1.6 65.4 0 100 40.8 1.59 64.8 2 220 40.74 1.59 64.7 4 300 40.71 1.58 64.3 5 1080 40.42 1.57 63.4 19 1920 40.16 1.56 62.6 34 2700 39.92 1.55 61.8 47 3600 39.66 1.55 61 63 4680 39.36 1.54 60.6 81 8520 38.36 1.5 57.5 144 10260 37.91 1.49 56.4 172 11820 37.54 1.47 55.1 195 14580 36.88 1.45 53.4 237 16860 36.48 1.43 52.1 270 19680 36.16 1.42 51.3 311 21720 36 1.41 50.7 333 22560 35.91 1.41 50.6 352 25200 35.68 1.4 49.9 388 28440 35.33 1.38 48.7 434 29280 35.21 1.38 48.5 444 29820 35.12 1.38 48.4 452 30600 35 1.38 48.2 462 31200 34.88 1.37 47.7 470 32520 34.55 1.37 47.3 488 33300 34.29 1.39 47.6 497 33840 34.09 1.39 47.3 505 34440 34.08 1.39 47.3 513 34980 33.98 1.38 46.8 520 36060 33.73 1.37 46.2 534 36120 33.71 1.37 46.1 535 36240 33.66 1.37 46.1 536 36360 33.63 1.37 46 538 36480 33.57 1.37 45.9 539 36600 33.52 1.36 45.5 541 36720 33.45 1.36 45.4 542 36840 33.38 1.36 45.3 544
View attachment 53418
So, it's got me scratching my head a bit. These are supposed to be HG2 cells at 3 Ah each, giving a total of 18Ah (648 Wh). I have always been sceptical about this as they were bought from Hong Kong on eBay for £2 each and I never felt that their capacity was that large and the pack seemed to perform similarly to my 15Ah packs. I assumed they were LG but 2.5 Ah versions which had been re-sleeved.
Anyway, if the pack is at 40.9 V, you can expect to lose approx 10% there (compared to a fully charged pack to 42V), so say 64Wh. I would guess there is still probably another 50 Wh to be had if the pack is driven right down to 25 V, which would give a figure just over 648 Wh. Which would mean zero loss in capacity for cells that have done hundreds of cycles and are several years old. However, if you look at Lygate's review of LG HG2 cells:
View attachment 53419
At 1 to 2 A discharge rate the capacity achieved is less that 2.8 Ah/cell, so 605 Wh for a 6p pack. Hence these cells of mine seem to be giving better results than brand new cells. This makes me wonder if they are actually LG MH1 cells that have been re-sleeved as HG2 cells.
Test of LG 18650 MH1 3200mAh (Cyan)
lygte-info.dk
View attachment 53420
These are nominally 3.2Ah, but were found to give just over 3 Ah when discharged between 1 and 2 A.
So there you go, discuss...
And that is the real World way of doing the capacity comparision;Anyway, if the pack is at 40.9 V, you can expect to lose approx 10% there (compared to a fully charged pack to 42V), so say 64Wh.
Exactly. You have to let the battery rest to get a true picture. When discharging and you hit 41 V, if you stop and let the battery rest it will go back above 41 V after resting. In my discharge experiment, the next day the voltage of the pack had rebounded from 33.38 to 34.2 V.And that is the real World way of doing the capacity comparision;
Stop charging at 41V, measure the capacity.
Wait a bit
Stop charging at 42V, measure the capacity.
Compare the capacities.
As I said above, you have to let the battery rest in order to find out how much capacity is between 4.1 V/cell and 4.2 V/cell. I have found it to be typically 7 to 8% and this is in line with what is widely reported online.Your assumption is incorrect. It's impossible to have 10% of the charge between 40.9v and 42v. I alreay showed you that it's about 3%. Any 18560 discharge chart will show the same. That means that your battery lost 7% not zero %. It's impossible to lose zero % after hundreds of cycles. Not even LiFePO4 can do that.
Can you confirm that with your method of reducing the charging voltage (was it NealH who suggested using a diode inline?), the effective reduction in full charge voltage is from 41.7V to 41V? That would giive about 0.7V difference. Saneagle did not give much details how he charged the pack to 41V so it's difficult to interpret his 3% reduction.As I said above, you have to let the battery rest in order to find out how much capacity is between 4.1 V/cell and 4.2 V/cell. I have found it to be typically 7 to 8% and this is in line with what is widely reported online.
There are a lot of issues that have to be considered before it can be claimed how much capacity has been lost.