Support information on internal resitance.
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The internal resistance seems very high: voltage difference divided by current is resistance round the circuit I think, and unless there is something else in it, will be almost entirely the internal resistance of the two batteries.Let's go one step further in our model of DIY extender.
We have already seen that we can connect two "same" battery banks in parallel as long as both banks have the same charge. How charge and voltage are related according to the cell manufacturer's curves. So we can say "equal volages = equal charges".
Now, if bank A has a higher charge than bank B, when they are connected in parallel, charges from A will pass to B through an electric current limited only by the internal resistance of the battery banks. these internal resistances are of the order of milliohms and that is why large currents would be caused depending on the level of charge imbalance of bank A and B. Destruction could occur. For this reason, battery chargers limit the maximum current delivered to the load.
Here comes the first alert with the DIY extender.
In our lab-scale model, we connected two banks with different charge levels in parallel and measured the current that was produced. See attached figures.
Bank A = 83% charge volt=8.10 v
Bank B = 72% charge volt=7.94 v
Charge imbalance = 15%
At paralleling , I = 44.4 mA from A to B charging rate 0.02C
After some minuts , I = 30.7 mA from A to B charging rate 0.01C
If we let the banks be connect , they will equalize their charges.
One thing you should pay attention to is that the basic cell's voltage curve is almost flat before reaching the point of discharge ( from 4.2v to 3.0v). In the DIY extender and main batt ( from 42v to 30 v).
So small voltage differences between bank A and B can mean large charge differences. Don't forget this point.
I will continue...
What you write down is correct. I would say that it is not the expected result. But it is real for that test fix. The only explanation I have is that these banks are not new and have too many charge and discharge cycles.The internal resistance seems very high: voltage difference divided by current is resistance round the circuit I think, and unless there is something else in it, will be almost entirely the internal resistance of the two batteries.
(8.11 - 7.91) / 0.0444 = 4.73 ohms, or 1.18 ohms per cell. The data sheet indicates 0.1 ohms per cell, so is there a factor of 10 error creeping in somewhere?
Something doesn't look right here.
The behavior of the currents and voltages of the Extender and Main Batt, were equal to the measurements in the static test with route simulation in Rouvy and using the Wahoo smart trainer, therefore the project is finished.
So, Antonio, have you concluded whether the extender is working as a charger?
Looking forward to it.Hi Yak, more than that. The mistery now is complete clear. Diy Extender has almost 90km road real test and I am putting all my tests and results in a report.
Hi. How did you measure the current. The meter itself can introduce an error when reading current at such small voltage differences and you need to take into account the resistance of the measuring device, leads and probes at the scale that you were using. For example, some Fluke meters introduce a 'Burden' of well over an ohm on the current scale.The internal resistance seems very high: voltage difference divided by current is resistance round the circuit I think, and unless there is something else in it, will be almost entirely the internal resistance of the two batteries.
(8.11 - 7.91) / 0.0444 = 4.73 ohms, or 1.18 ohms per cell. The data sheet indicates 0.1 ohms per cell, so is there a factor of 10 error creeping in somewhere?
Something doesn't look right here.
I was commenting on the measurements in a previous post, not making my own. It may well be down to the instruments used. I've not got down to that level of detailed analysis in my experiments!Hi. How did you measure the current. The meter itself can introduce an error when reading current at such small voltage differences and you need to take into account the resistance of the measuring device, leads and probes at the scale that you were using. For example, some Fluke meters introduce a 'Burden' of well over an ohm on the current scale.
Can you live with the burden? | Fluke
One of the realities of electrical testing is that you can't measure something without changing it in some way.www.fluke.com
I have a battery case for my iPhone XS which work in this way. The battery in the phone is kept charged at 100% by the battery in the case.Hi Chamzamzoo, have you come across this, it might do what you want:
The extension battery feeds this regulator which you adjust to give up to 2A at up to the fully charged voltage of the original battery. Then to the original battery it looks like a charger and supplies the first 2A of {charging current + motor load}, if any.
Unlike the 'batteries in parallel' method no mods are needed to the original battery, no diode in other words, which I'd guess is why the bike's makers chose the 'extender' route.
The battery voltages needn't be matched, the extender battery's voltage can safely be higher, or it can be lower and will be boosted as needed (up to a point).
This does rely on the external battery cells having a generous current rating, and on its LVC protection, and on the original BMS working with an extender (tick). You should gain about 80% of the extender pack's capacity.
I'd fit a diode in the regulator's output, just in case, and choosing similar battery voltages should make for a stable, efficient regulator. I've not used this one myself but maybe it's worth a punt.
Anthony
Excelente Antonio. You have found that the extender provides more than the 2a charging current when the motor is drawing more than 4a under load. This is what I had hoped. But would this be possible if the extender acts as a passive charger, and as the main battery is depleted more quickly when under +4amp load, that the extender is simply putting more charging current into the main battery? Have you tried to establish whether the extender will equalise slowly (‘charge’) the main battery if the extender voltage is the higher of the two, or whether, as Volabike claim, current is taken solely from the extender under load until the two batteries are at equal voltage?Hello,
I am attaching a draft of the test report that I carried out on my the DIY Extender.