I want to model a battery using a capacitor because the simulator gets flaky when you send recharge current to a voltage source. It's only for simulation purposes, so whether the resulting capacitor is physically possible doesn't matter.fechter wrote: ↑Mar 04, 2018 11:31 pmWell, that's about right. I'd compare the energy stored in each one but you get the same result. 38,016 Joules is what the battery holds, and to get the same energy from a capacitor at the same voltage, the capacitance needed is huge, as you calculated. Also keep in mind the voltage on the capacitor needs to go from full to zero in order to use up all the energy, while the battery maintains a useful voltage over the discharge cycle.
Since the energy in a capacitor is a function of V squared, using a higher voltage is generally beneficial.
It's why we don't see capacitor powered vehicles. Yet.
This is the biggest issue with using capacitors in place of or in addition to battery.
I forgot. As if $6000/54kg wasn't enough penalty (amberwolf wrote: ↑Mar 05, 2018 12:49 am
Also remember ... you have to use 10 in series to get the voltage you're after, you then only have 1/10 the capacitance, so you now have to parallel 10 series strings of 10 capacitors to get the same capacitance you would have had with just one capacitor at the lower voltage.
For voltage, you are correct. But for energy, it's 1/2CV^2. So if you are going from 52 to 44 volts, you are using only 15% of the voltage range, but 30% of the energy range.
Could you simulate the pack with a zener diode? Might be closer to what actually happens.
The simple answer to that is I don't know. My electronics fu isn't strong enough to begin to see how that would work.fechter wrote: ↑Mar 05, 2018 10:15 amCould you simulate the pack with a zener diode? Might be closer to what actually happens.
The simulator I'm using, because it is simple enough that I didn't have to look anything up to construct my model, is also too simple to allow me to program bespoke components
That works pretty well. Be sure to include a series resistor to simulate the ESR of the battery.
Thanks. I discovered what happens if you omit it! A huge GA surge when I connected a voltahe source to initially charge the battery. Makes for messy graphsbillvon wrote: ↑Mar 05, 2018 4:43 pmThat works pretty well. Be sure to include a series resistor to simulate the ESR of the battery.
Depends completely on the battery. To get a very rough estimate, you can assume your battery will droop about 10% under full load, then calculate from there (via R=V/I.) For example, a 42V fully charged battery might droop to 38V under full load. If full load is 30 amps, then R=(4/30)=.2 ohms.
Typically under 10 mOhm for the High C-rate stuff, but it also depends on what the SOC is. It's higher at 4.2 V resting than it is at 3.7 V resting. But it goes up again at low states of charge.
I'm specifically interested in the 12V SLA I'm trying to model.
From teh voltages you gave, I'd assume that's per-cell? Then for a 14s pack that would be around 140milliohm?
According to this
.European car maker PSA Peugeot Citroen has been using them in its cars since 2010 - supercapacitor maker Maxwell Technologies says more than a million vehicles now incorporate its products.
That's really intriguing!
Which implies that the SuperCap can accept 25V and they buck convert it down to 12V for the car systems and battery.The alternator, although about the same weight as its 12-V predecessor, is a varying-voltage design that operates in a 12- to 25-V range. So the circuit also requires a dc-dc converter (which weighs 1.8 kg/4.0 lb) to provide 12-V power for electrical accessories. As soon as the driver lifts his foot off the accelerator, the regenerative mode begins, and the alternator uses the kinetic energy of deceleration to produce electricity at maximum possible voltage for efficiency. The “free” electricity goes through the dc-dc converter, and if there is any available electricity beyond the car’s electrical load, it goes to charge the 12-V battery.
?? By the method you just listed - you put them in series.