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Old Apr 09, 2012, 10:37 PM
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What's the difference between discharge rate and voltage?

I thought I was really getting the hang of this.... ;(
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Old Apr 09, 2012, 11:02 PM
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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
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Old Apr 09, 2012, 11:25 PM
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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.
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Old Apr 10, 2012, 03:21 AM
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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.
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Old Apr 10, 2012, 11:06 AM
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Originally Posted by abenn View Post
...The battery doesn't push amps through the motor, but has to be capable of delivering whatever amps the motor wants...
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.
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Old Apr 10, 2012, 02:38 PM
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Originally Posted by Kevinfred View Post
What's the difference(relationship) between discharge rate and voltage?
The rate of discharge and pack capacity determines how long it takes for the voltage of a fully charged Li-Poly cell to drop from say 4.2 volts per cell to say 3.2 volts per cell during flight. The difference in how long it takes to drain 1 volt per cell or 3 volts of a 3S pack from 12.6v to 9.6v depends both on the capacity of the pack (e.g. 800mAh vs 2200mAh vs 5000mAh), and the discharge rate (10amps vs 20amps vs 30amps).

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.
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Old Apr 10, 2012, 03:00 PM
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Of course that isn't technically accurate, but an easy enough way to explain it I suppose. ...
Yes, I know I stated it that way because, so often, beginners think that if you use a battery that's capable of delivering more amps (more mAh and/or higher C rating) it will do so -- and the original poster is obviously a beginner because of his confusion about discharge rate and voltage. He needs to understanding that, all else being equal, the motor will take, or try to take, what it needs to drive the prop that's on it -- no more, no less.

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
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Old Apr 10, 2012, 08:43 PM
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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.
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Old Apr 11, 2012, 04:59 AM
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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.
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Old Apr 14, 2012, 07:04 PM
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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?
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Old Apr 14, 2012, 08:04 PM
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Originally Posted by Kevinfred View Post
Is it safe to say discharge rate it a continuous measurement relating to how long a battery can discharge itself at "full" rate?
The discharge rate is how much current the motor draws from a battery; the C rating is the MAXIMUM safe discharge rate for that battery. If, however, your power system approaches the maximum rating of the battery, odds are, it will drop significant voltage and begin to warm up. This effect becomes more pronounced as the charge in the battery is depleted.

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.
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Old Apr 14, 2012, 08:17 PM
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Originally Posted by Kevinfred View Post
...

Is it safe to say discharge rate it a continuous measurement relating to how long a battery can discharge itself at "full" rate?
If I'm understanding your question correctly, that's what a "C" rating is, more or less. That is the suggested rate limit where it is safe to discharge a battery relative to its full capacity. Rates lower than that are fine.

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.)
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Old Apr 14, 2012, 10:02 PM
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Originally Posted by Kevinfred View Post

Is it safe to say discharge rate it a continuous measurement relating to how long a battery can discharge itself at "full" rate?
No.
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 ................
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Old Apr 14, 2012, 11:55 PM
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Originally Posted by Kevinfred View Post

Is it safe to say discharge rate it a continuous measurement relating to how long a battery can discharge itself at "full" rate?
Assuming we can change "it" to "is" in the above statement, the answer is YES, however.....it might be more correct to say, "that is what it is supposed to mean". Currently many people are suggesting that most manufactures overrate the abilities of their batteries to "discharge" (supply energy). For most LiPo users this is (IMO) a non issue as most users are not hammering their batteries.

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!)
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Old Apr 15, 2012, 03:10 AM
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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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