Lipo
More MAh, More Power. The race to increase capacity.
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), 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 (short 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.
As always, that last 10% of performance costs as much as the initial 90%.
The voltage will always drop for batteries under load. When we say the battery should be able to deliver, we mean that it won’t drop to 6 volts while providing “X” amount of current. Here’s another example to help clear it up:
For this example, I’m simplifying the “C” rating to the idea that the 6000 mah 10C pack can deliver 60 amps continuous before it drops below 6 volts. This doesn’t mean that when it’s fully charged it will supply 60 amps at 8.4 volts and then all of a sudden drop to 6 volts when you try to pull 61 amps. The voltage drops roughly proportional to the percentage of power demanded from it. If you demand 30 amps, a full pack will probably be around 7.5 volts. If you demand 50, it will probably be close to 6.5 volts. 60 amps will put it right at 6 volts. If you remove any current demands, the pack will jump back up close to 8.4 volts.
The “drops quickly, levels out, and then dumps” voltage curves you are thinking of are for constant current discharge rates. You take a pack, discharge it and plot the voltage. This just tells you how the pack will react under constant currents, not varying ones. If you discharge one pack at 30 amps and another at 60, the voltage curve for the 60 amp run will always be below the 30 amp run when plotting it not only against time, but against how many mah have been discharged.
Voltage dips that you see from in car telemetry devices or other on track recording systems are caused by high current demands at low rpm. The voltage quickly recovers primarily because the demand diminishes as the rpm of the motor increases. At low rpm there is very little impedance and the back emf is also very low. This creates a high delta voltage and a very low overall resistance, so high current spikes and battery voltage drops are the result.
