Hello guys,
I have a question which I think is very useful for people who fly bigger models.

Can we have a lipo and li-ion in parallel?
Assumptions made --> lipo can discharge a constant current of 90A where as li-ion battery can only discharge 20A.
Few of my points:
1. I wanted to use li ion becuase of its weight, cost but at the same time I have to compromise because of its discharge is too low. If I choose
    lipo then both weight and cost would increase but I need not compromise about discharge.
2. What I have seen is the discharge graphs of li ion battery is little steep compared to lipo.

Advatages of using them in parallel?
1. Simple we can have advantages of both batteries i.e
1. Weight can be reduced.
2. When there is current spike lipo may kick in and supply the load and next for long run i.e when the discharge is low the li-ion and lipo share the load equally.

However we may need experimental data before we can comment on anything. As so many factors need to be considered and I need your points as well before someone can proceed with any experiments.

it does not work like that - i leave it to the more inclined to elaborate

if I ask PT usha and my grandmother to hold hands and run will they reach an avg speed between the two max's ?

ans. no ... my grandmother will collapse


better you buy cheap lipo and charge/ discharge with care.

@balakrishna reddy

Without elaboration, here are two points for you to ponder over, that will clarify the matter:
1. What will happen if two batteries of different voltages are connected +ve to +ve and -ve to -ve (even before they are connected to any circuit)
2. When the battery combo is connected to a load/circuit, will the current through each battery be the same or different

Regards

@Saikat.

Why don't you consider lipo as a capacitor and li-ion as normal battery and then try to overlap your view.
My point is a capacitor will charge to battery voltage and can discharge at very high rate when load is increased immediately. The capacitor will supply as long as it has juice in it. However we even need to consider the terminal voltage during discharging.

@KK Iyer,

1. Battery terminal voltages initially will be same when they put into circuit. However their terminal voltages are subjected
    to change when load is applied.
Terminal voltage drop is directly proportional to load,discharge rate of battery.
So for a given load the voltage drop of lipo and li ion is not same. However when in circuit the terminal voltages of both the batteries remain same as long as they are in circuit.

2. Current sharing will not be equal as their internal resistances are not equal. The load contribution is a factor of internal resistance and terminal voltage.

However if there is change in terminal voltages inside the circuit there will be circulating currents to make the voltages equal.

yes bala ... your assumptions are correct on the surface
but during discharge since the lion is not current limited its voltage sag will be greater and the lipo
will end up charging (at very high current ) the lion as well ..essentially the two will behave like a parallel battery with one bad cell

basically the lion will not contribute anything

better would be a microcontroller with a current sensor which will switch mosfets (pwm) to allow the lipo and lion to discharge according to each threshold. have a combo of p channel and n channel to prevent
current of one charging the other .

@ Balakrishna Reddy
Hi,
Did you finally try it out?
If so what did you find?
Was looking forward to developments...
Regards

@ K K Iyer
No sir. Not yet. However I will conduct a load share test for which I am getting the required tools after which I will perform it.
Indeed we may not really need to conduct the experiment to give rough results but knowing internal resistance will suffice.
Since we cannot obtain the internal resistance I have to choose this method.
Anyway I will share the test results once I do it.

A few years/months ago, i had done some experiments to estimate internal resistance and posted results in this forum. Will provide a link if i can find it!

We can use very same load test and by putting various readings of voltage and current onto a graph and obtaining slope will give us internal resistance but that again requires access to battery.

@balakrishna reddy,
Looking forward to developments/updates...
Regards