I tried googling but couldn't find anything about Battery 18650 SPICE Model for MultiSim.
Does any one who can help me on this?
Thanks.
Does any one who can help me on this?
Thanks.
Hi,
It depends what you are interested in simulating.
If you just want to do a general check of your charge circuit, then use a voltage source with a small resistor in series. 0.05 ohms would be somewhat typical.
If you want to check charge time, then use a large value capacitor with small resistor like 0.05 Ohms in series.
If i remember right, the value of the cap should be 3600 Farads per Ampere Hour battery capacity with the initial voltage value set to the min cell voltage before a charge.
There are really better models on the web but you probably dont want to go through the trouble of setting all the parameters.
Hi,Hello again,
I went back over the idea and found that the cap value is better placed at:
C=3600*AH/1.4
which means:
C=6686 Farads.
The whole idea is that the cap must charge from 2.75v to 4.15v (or a little less) in a bit less than 2.6 hours with a constant current of 1 amp.
dv/dt=i/C
(4.15-2.75)/(2.6*3600)=1/C
and that is with 0.05 Ohms for the series R. With R=0.07 it would be closer to 6783 Farads but that's not too critical.
Hi,Hi,
I've 2 questions,
1. In "C=3600*AH/1.4", what the magic number 1.4 is?
2. In "(4.15-2.75)/(2.6*3600)=1/C", how the expression "2.6*3600" could get unit of time, i.e. seconds?
Thanks.
Hi,Hi,
They are both because the cap does not charge from zero to max voltage, it chargers from some value like 2.75v to 4.2v. That changes teh overall capacitance of the capacitor itself. The "1.4" happens to come from that.
For the second though we have:
V/Q=1/C
or taking the inverse we get usual:
Q/V=C
with charge A in Ampere Seconds and V in volts (voltage differential) and C in Farads.
The value of 4.15 volts is based on a charge current of 1 Ampere and series resistance of 0.050 Ohms. That is the transition point from constant current to constant voltage. This provides for an estimate of charge time and therefore the value of the equivalent capacitance. It's not that critical though.
by Jake Hertz
by Jake Hertz
by Jake Hertz