


So. Cal.
Joined Oct 2004
9,398 Posts

Discharge rate is the ability of a battery (or any voltage source, for that matter) to source current. As it relates to lipoly batteries, it's typically a rating based upon capacity. e.g.  a 20C rated battery can be drained at a continuous current draw equal to 20X its capacity.
Voltage is simply the measure of electrical potential. Mark 



To add to what Mark said, the cell count in a Lipo determines the voltage.
A Lipo cell has a nominal voltage of 3.7V. So a 3S pack is 11.1V The 3.7v / cell is an UNCHARGED pack. When fully charged a cell will read 4.2V. So your 3S pack will be about 12.6V That was my 2cents. 


Letchworth, Great Britain (UK)
Joined Jul 2004
13,397 Posts

Power, which is what we're usually interested in, is measured in watts, and is volts x amps. So, to get more power you can either use more volts (more cells in your battery pack) or make the motor draw more amps, or a combination of the two.
How many amps the motor wants to draw is determined by how many volts you give it, and what size prop you put on it. The battery doesn't push amps through the motor, but has to be capable of delivering whatever amps the motor wants. The formula in post #2 lets you figure how many amps a battery is capable of  e.g. a 2200mAh battery rated at 20C is theoretically capable of delivering at the rate of 2200 x 20 = 44000mA, or 44A. 



Of course that isn't technically accurate, but an easy enough way to explain it I suppose. The down side is that this kind of explanation can mess up people's understand of how and why Ohm's law works. (We've had this conversation before. ) At the fundamental level, the voltage of the battery pack is what pushes/pulls the current through. Of course this is also balanced by the resistance/impedance of the setup and what it lets through at a particular voltage. Motors don't have a constant resistance/impedance, and that is also relative to the prop size used, so that tends to complicate things a little.




Quote:
EXAMPLE: Electric outrunner plane draws up to 20 amps at full throttle. If the power output of a 3S pack can generate at full throttle 240 watts(at say 12 volts), then what is the most power this pack can generate at full throttle when the voltage drops to 9.6 volts? How long it generates maximum power at full throttle from 12 volts to 9.6 volts depends on the capacity of the pack. A 2200mAh pack should take at least twice as long to deplete 80% of its capacity compared to depleting 80% of a 1000mAh pack in the above example. 



Letchworth, Great Britain (UK)
Joined Jul 2004
13,397 Posts

Quote:
Until he understands the basics  that too big a prop can kill a motor, ESC, and battery  I think it's not worth going into the technical details of how and why that happens 




Yeah, nothing against you. I think you're doing a great job helping people here from your posts that I've seen. It's just one of those things that bothers me. I guess ideally I'd like to see things explained more thoroughly and accurately when it comes to electricity. I realize though that that isn't always practical and that some people don't want to bother with having a greater understanding beyond what will get them up and running.



United States, IN, Fort Wayne
Joined Oct 2008
401 Posts

Thanks all  seriously not just lip service  THANK YOU... It's funny, abenn and Wrend, that you guys discussed what you did. I can handle a complex explanation, but it was dialog from different sources that 'threw' me!
Is it safe to say discharge rate it a continuous measurement relating to how long a battery can discharge itself at "full" rate? 



Quote:
Using a battery with a higher maximum rate, without increasing the actual discharge rate, will likely provide a more constant, and slightly higher, voltage, resulting in better performance, especially as you near the fully discharged state of the battery. 




Quote:
The actual current rate depends on the voltage of the pack and the resistance/impedance of the circuit. Or in other words, how much current your electrical circuit drains from the battery which also depends on the size of the prop and how fast you spin it. (It's a little more complicated than this even when it comes to the specifics of how ESCs and motors work, but this is essentially accurate.) 



South East Pennsylvania
Joined Jun 2007
204 Posts

Quote:
But I just asked all my 7.4, 11.1 and 14.8 20c 25c 30c 35c 45c Lipo packs if they are discharging themselves and they all said no, but.... my NiCd said ................ 




Quote:
Getting practical, if you buy a "20C" rated battery, how long can you run it at 20C? Being practical, we only want to use 70% to 80% of the batteries capacity. LiPo capacity is rated/listed in Amp/Hours (although regardless of the amps, we are only talking about 1 hour) and there are 60 minutes in an hour. 60 / 20 = 3 or 3 minutes run time (and you will not even get 3 minutes at 20C discharge due to low voltage) and 75% of 3 minutes = 2.25 minutes or 2 minutes and 15 seconds of run time. The point I am getting to is the higher the discharge rate, the less run time you will have. Electrics are all about matching up the components for the required job. Michael (if your not fryin, your not tryin!) 



Letchworth, Great Britain (UK)
Joined Jul 2004
13,397 Posts

I think we're in danger of getting too complicated here Discharge rate (I assume we're talking about C) is a multiplier which can be used to determine how many amps the battery should be able to deliver. Just multiply the battery's capacity in Ah by C and you get its maximum amps.
You can also use C to determine the duration taken to discharge the battery fully (not a good idea, as others have pointed out) if you discharge at its maximum amps  just divide it into 60 and you get the duration in minutes. 
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