fechter wrote:One of the major issues with most BMS/cell balancing schemes is the amount of heat given off by the shunt resistors.
liveforphysics wrote:Everyone of the de-coupleing caps would need to be rated for the full pack voltage+20% or so, and to pass meaningful current, the freq needs to go way above 100hz unless you plan to use giant low ESR caps.
I can visualize how the cells could all be charged with out removing the series connections from one 4.2v high freq AC supply, but I'm unclear how it would balance during charging.
fechter wrote:Whichever cell had the lowest voltage would present the path of least resistance, so would take the most current.
number1cruncher wrote:fechter wrote:Whichever cell had the lowest voltage would present the path of least resistance, so would take the most current.
Well, I could understand the schematic without any help, so it must be really simple. Plus it follows one of life's major rules...
How would this setup prevent overcharge of the cells? Would it be based on the power supply setting? Or is there some sort of governor in the circuit?
swbluto wrote: Presumably, 5v - 2 diode drops = 3.7 volts. I don't understand how it's supposed to limit the voltage at lower currents, however, since the diode drop is smaller at lower currents and so it seems like it has the potential to creep up to the supply voltage.
number1cruncher wrote:swbluto wrote: Presumably, 5v - 2 diode drops = 3.7 volts. I don't understand how it's supposed to limit the voltage at lower currents, however, since the diode drop is smaller at lower currents and so it seems like it has the potential to creep up to the supply voltage.
Is the forward drop proprotional on a diode? I thought this was a characteristic of a resistor, but not a diode. However, I did have this argument in my head before posting earlier...
fechter wrote:In this case, the capacitors allow AC to pass, but block the DC.
I did some initial testing using a junk switching power supply. I attached a pair of wires to the output of the switcher transformer to get an AC signal to work with.
This particular switcher runs at about 50khz. Feeding the AC through a capacitor, I measured how much power came through and tested the cap for heating.
With a 4.7uf capacitor, it looked nearly identical to a direct connection. This is encouraging. I also tried a 0,1uf, but there was a significant loss in output, though it was nearly 1/2 of the direct connection.
Another thing I figured out is the output at the cell end will be the peak-to-peak voltage of the AC waveform minus the diode drop. As the switcher regulates, the duty cycle of the on pulse increases, but the peak-peak voltage remains nearly the same. This means the AC source needs to be a bit more sophisticated than the output of a switcher transformer. My guess is you'd want to take a 5v supply and run it through a FET bridge to change it to a high frequency AC. This way, the voltage of the 5v supply could be adjusted to get the desired output at the cell. You could also crank the frequency up higher to reduce the drop in the capacitors or allow smaller capacitors and have a nice 50% duty cycle square wave.
This circuit won't suck power from the high cells and feed it to the low ones. It will only be able to charge the lower ones at a higher rate than the high ones. I think in the end, this will have about the same effect during a charging cycle. If the total imbalance is corrected during the same time that the bulk charge takes, then there won't be any need for additional balancing time.
Toorbough ULL-Zeveigh wrote:
where does the top of the uppermost cap ultimately hook to?
cuz it doesn't connect to anything right now, or is that one cap left off entirely?
sorry if that's a previously established given assumption i'm not aware of & would just like a little clarification.
with say a 4 cell balancer as pictured, what would the completed circuit look like, that might help me out.
also what kind of load did u run the test at?
can u really pull one amp thru a cap, if that's ur target?
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