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Effect of switching current on Li-ion battery packs

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fcb:
OK. Lets try again.

Is this statement correct: 3A is your average (RMS) current draw from the battery and this is being drawn from the battery in short pulses (PEAK) of perhaps 30A.


fcb:
C is a ratio, and is calculated thus: (dis)charge divided by rated capacity of the battery.

So 0.4A discharge current of a battery rated at 4Ah would be called a "0.1C discharge" and should exhaust the battery after 1/0.1C=10hours.
Likewise charging a the same battery at 2C would mean charging the same battery at 8A and result in a 30 minute charge time.

My tests where on various 18650 cells rated at 2 to 2.8Ah.

ogden:

--- Quote from: fcb on February 07, 2020, 06:19:50 pm ---With sufficient capacitance (and the right type of capacitors) between the battery and the converter you will 'smooth out' the AC ripple - reducing the AC component you are presenting to the battery and then you are treating the battery more as DC device.

--- End quote ---
Exactly. Next thing to do - redo simulation. Introduce real capacitor(s) between battery and converter, use real components with real impedances everywhere including battery and wires.

VEGETA:

--- Quote from: fcb on February 08, 2020, 01:02:29 pm ---OK. Lets try again.

Is this statement correct: 3A is your average (RMS) current draw from the battery and this is being drawn from the battery in short pulses (PEAK) of perhaps 30A.

--- End quote ---

When I set the final voltage at 30v and final current limit to a little bit more than 3 amps (to allow 3 amps) + at 10 ohms of load... I get 30v and 3 amps of output through that 10 ohms load. this is supposed to be the absolute maximum rating ever of this design. Normal uses are way below this.

At this statuation... switching current through the pre-regulator taken from the battery is about 20-30 amps of pulses as seen in the picture and attached complete ltspice simulation file.

____


--- Quote ---My tests where on various 18650 cells rated at 2 to 2.8Ah.
--- End quote ---

so these 2000 mah batteries gave you +10 amps of continuous current? no severe voltage drop or something?




--- Quote ---Exactly. Next thing to do - redo simulation. Introduce real capacitor(s) between battery and converter, use real components with real impedances everywhere including battery and wires.
--- End quote ---

I have attached the simulation so you can check yourself. it is very well done and compensated through help from a certain member in this forum long time ago... I spent sleepless nights to learn and modify it until we made it stable and working.

kindly check it, it is slightly modified to use consolidated parts like 40k resistors (in real-life would be 4 of 10k in series) instead of 43.2k which gave 300khz switching frequency while 40k gives slightly more than 300khz. this has no major effect on the design.

Also, the design uses not-orthodox type of grounding and referencing since it is a floating regulator... so when you probe, keep holding mouse left button and drag to get the drop voltage needed. You'll know when you see.

thanks!

ogden:

--- Quote from: VEGETA on February 08, 2020, 03:53:01 pm ---I have attached the simulation so you can check yourself.

--- End quote ---
Yes indeed you have zero impedance battery, ideal (and small) capacitor after battery, no battery wires "modelled", L3 & L4 also are ideal. You would want to update circuit and run simulation again.

That "not-orthodox type of grounding" is  |O. What's the problem to make OUT- as simulation ground?

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