Originally Posted by
Krio
Example as to why a higher capacity pack will give more current:
Going with a 3000 mah 10c and 6000 mah 10c as example packs and assuming at 10mph the resistance (and impedance) of the electrical system is .1 Ohms and the back emf is averaging 4 volts.
With this setup and assuming a pack voltage of 8 volts (almost full and an even number to work with. lol), the system will pull 40 amps if this voltage is maintained. The math I'm using for that is very simplified for what the whole system is, but it is close enough for this discussion:
Delta(V) = IR
VBattery - Vmotor = I(0.1ohms)
8volts - 4 volts = I(0.1ohms)
4volts/0.1ohms = 40 amps
This is higher than the 3000 mah pack can safely give, so the demand for current pulls it's voltage down to 7 volts. At this point the equation works out to 30 amps:
7volts - 4volts = I(.1ohms)
3volts = I(0.1ohms)
3volts/0.1ohms = 30 amps
30 amps x 7 volts gives you 210 watts.
Same idea, but with the 6000 mah pack. The setup wants 40 amps. The voltage will dip a touch under load, but not as much as the smaller pack. This is a function of the packs internal voltage, but to make it easy lets just say it drops to 7.5volts.
7.5volts - 4volts = I(0.1ohm)
3.5volts/(0.1ohms) = 35 amps
35 amps x 7.5 volts = 262.5 watts
That's 25% more power!
Obviously no one is running packs only capable of 30 amps, but the idea still applies. It's just a diminishing point of return as the voltage drop under load becomes smaller and smaller with each capacity or C rating increase. The biggest arguments about packs are centered around this... sure you can buy those amazing new cells that just came out for $100, but what percentage of power are you gaining over a "sportsman" pack for $60? What about even cheaper packs straight from asia? With no standards when it comes to labeling how a pack will perform (near term as well as long term) it all comes down to the arguments you see every other week on various forums about the pack of the week and personal experience.
Personally I get "enough" power from my inexpensive nano-tech batteries. Would an orion or smc pack provide more punch? Probably, but the small percentage in power gains with large percentage in cost increase can't be justified against my diminishing racing budget.
As always, that last 10% of performance costs as much as the initial 90%.
Thanks for writing this up. Of course, I am confused.
For the first case, the system "wants" 40 amps, but a 3000 mAh pack with 10C rating can only give 30A, so there is a voltage dip (I'll avoid the word "drop" here). For the second case, the system wants 40 amps, but a 6000mAh, 10C battery should be able to deliver, no? Which leads to no voltage dip? This would give deltaV of 4V, leading to 40A X 8V to 320 Watts!
I think I prolly should not take the numbers too literally. I guess your point is, there will always be a momentary voltage dip, but the higher the capacity (
for the same given C rating) the dip will be lower, leading to a larger deltaV, leading to larger I.
I may need to learn a little more about batteries and the overall electrical system, and particularly, how it all works as a function of time, to be fully on board.
That said, I am learning a fair bit while at work. Maybe I should tell my boss about my time well-spent . . . or maybe not.