Battery Pack capacity and C ratings and so on

[Tim Mackey]

Battery Pack capacity and C ratings and so on…….So many people keep asking about this, so here’s a quick explanation which is extracted from a reply I gave in another posting. The question was about C Ratings and duration. 15 - 25C means it can THEORETICALLY be discharged ( used in other words ) at up to 15 -25 times its stated capacity, so if its a 1200 m/a battery ( 1.2 Amp ) that's 18 to 30 Amps

because 15 X 1.2 = 18 ……and ….25 X 1.2 = 30.

The letter "C" simply means its Capacity, so a 1C pack would be only useable at 1 X its capacity, or in this case 1.2A.

What is suitable for your application depends on how much current the motor/ prop combination is going to pull, which is a factor of motor choice, and more especially prop size. The higher "C ratings" claimed by some packs are to be used with caution, anyway if you used that example pack at 25C ( 30 Amps ) two bad things will happen

The pack will have a pretty short and hard life

Your flight time will be just 2.4 minutes - assuming the throttle stick doesnt move much from "GO"

[Tim Mackey]

Part 2

This is calculated thus.....

The pack is rated to supply 1.2 A ( 1C ) for 1 hour ( 60 minutes / 1 = 60 )

So to supply 25 times that amount you again divide the minutes (60 ) by the C number (25) and you get..... 2.4 minutes.

As a general rule, to stay well within the packs capability, prolong life, and get reasonable flight times, I usually aim to use the pack at approx half of its claimed maximum C rate.

Remember also, that by putting 2 x packs in parallel ( and that should only ever be 2 identical packs incidentally ) the C rate of the individual packs does not change, but the current capacity doubles.

So 2 x batteries as above, of 1.2A each, at say 15C maximum, will become a battery of .........2.4A capacity but STILL 15C capable.

EG: 2.4A X 15 = 36A

Hope that helps a little…..

[Fats Flyer]

Hi Timbo,

very useful info that i was wondering about only just getting into lipo's.

One other thing that i was wondering about with lipo's,what is the normal charge rate for a lipo, and does the charge rate increase when the pack increases in size  i.e....2200 ...3300 and so on......

thanks

vince

[Tim Mackey]

The charge rate is 1C so just match the capacity of the pack, and charge will take around 1hour +

[Piper Cub]

The murkiness is starting to clear, slowly!

Thanks for these snippets Timbo.

[Fats Flyer]

Nice one timbo,

cheers

vince

[Terence Kwan]

There are newer quality lipo like the thunderpower, flightpower and etc is capable of charging at 2~3C with no adverse effect on the battery according to the manufacturer.

[Tim Mackey]

The jury is still out on these, and it only applies to the higher capacity versions anyway. I worry that these claims are simply a way to try and retain /increase market share rather than a proven change in the technology. I can find no substantive information as to what is actually different about these cells.

I stick to 1C as it is still the method recommended by the established experts such as Fred Marks of Kokam fame.

[Reno Racer]

Timbo,

It may be sunday night, or I'm a thicky. I've tried to follow your calcs:

I have a 400 Watt motor on 2200mah 3S lipo, pulling 36A max at max throttle (16C?)

If I use a 2200mah Lipo, rated 20C, which therefore delivers 44A, how do I work out flight duration for this battery??, when pulling 36A (if this is correct)

I also assume that If i connect 2 identical batteries togther to get a 4400mah 3S2P lipo, I will therefore get twice the duration??

What does an increase in prop size do to this figure, say moving from a 10" to 11" x6

Apologies for the raft of questions, I am keen to add these calculation formula to my database, whcih currently just calculated Watts required for scale (80W/Lb), sport (100W/Lb) and 3D (200W/Lb) and relates to IC equivielant outputs.

Regards

Chris

[Tim Mackey]

Irrespective of the motors wattage capability, if your battery is 2.2A per hour capable ( which is another way of describing your 2200mahr pack ) and you are using it at 36A, then it will last for 3.6 minutes. THEORETICALLY OF COURSE.

Divide the packs "hourly current ability" (2.2) by the actual current ( 36 ) = .06 of an hour

Then multiply by 60 ( mins in a hour ) = 3.6 minutes. My spreadsheet then adds 50% to this figure, to allow for the fact that I am NOT at wide open throttle all the the time, and the fact that the current pullled drops as the model speeds up with the unloading prop and increased prop pitch speed.

It is a best guesstimate, and I usually then do a couple of timed flights, measuring the amount of ma put back in to the pack, and see how close the calculation was to actual. It varies with so many factors such as weather, wind, flying style, and so on, but has proved to be surprisingly accurate.

It has nothing to do with the C rate.... this merely tells you WHAT the maximum current could be pushed to.

Connecting two identical packs in parallel will provide twice the duration, assuming you change nothing else, however of course, your power to weight ratio changes.

Increasing the prop from 10" to 11" will increase the current drawn by the motor, in an effort to continue turning the larger prop at the same revs....this is pretty well what is meant by KV, and the effect of such a change in prop depends again on many factors, some motors tend to exhibit more current demand with a larger prop than others and the only sure way to monitor the effect, and stay safely within the motor and ESC limits is by use of a whattmeter or similar.

A mere 1" increase in pitch can have a big effect on current drawn ...maybe as much as 30 - 40%

so I repeat..

the only sure way to monitor the effect, and stay safely within the motor and ESC limits is by use of a whattmeter or similar.

HTH

[Piper Cub]

As a matter of interest, for the larger type scale plane, I'd assume a battery pack of 14v would be the norm (at least for the approx 1\6th to 1 \5th Piper Cubs I've been looking at.)

So, this would seem to cause a problem down the field when you want to fly and need to recharge your batteries since most car batteries are 12v unless you get a 24v wagon battery!

Now if I have two motor batteries in parallel say 14v 4000mah packs to give 1 @ 14v 8000mah, it effectively means I have to charge at home with a 240v charger since I'll be lucky to get both these batteries charged fully from a 12v supply.

Alternatively, Im looking at buying 4 14v 4000mah packs so I'd have sufficient batteries for more that 2 flights and I'd have thought that could be costly? Or am I missing something?

How do other flyers address this ie field charging large voltage packs?

TIA

[Tim Mackey]

Modern clever chargers use voltage boosting circuitry to enable them to be able to charge at much more than the 12V input. My Astro 9 Lithium dedicated charger for instance has voltage boost from 12V up to 40V and  will happily charge up to 9S packs, at up to 8Amps. This of course places pretty big demands on the input supply, so it s best to keep the car engine running to avoid a flat vehicle battery !

The cost of packs is obviously relevant to several things...more cells = £ more capacity = £

and of course your chosen supplier My latest packs cost me £35 ( 5S / 2200 /25C ) and these are Enerland cells which are the dogs danglies.

[Geoff Neilsen]

Timbo

I'm new to electric and am very confused with the methods of labelling things. What does the 3S or 4S mean when talking LiPos? For the Typhoon in the RCM&E special they recommend the AXI 4120-14 or equivalent, but how do you determine an equivalent? This is all very confusing to a slope nut getting into power. If this is too much for this forum, can you suggest a Web site that can explain things for me. I am an electrician so I understand the theory, just confused about the labelling.

Cheers

Geoff

[Former Member]

[This posting has been removed]

[JC]

Timbo

Just read this chain and explanation of C ratings ect was again very usefull. One point however(and you will probably get fed of trying to clear this up for me) how do they arrive at a btys C rating? Also when I used to teach bty theory at the RAC signals school we had a formula for working out the safe charge/discharge rate for a bty of a given amp hour capacity(can't remember it now I'v got it in a manual somewhere) would it apply to lipos.

Jon

[Tim Mackey]

Generally speaking most if not all lipos should be charged at 1C ( 1 x the capacity ) only, which should take around just over an hour.

How manufacturers arrive at a C rating is something only they can tell you and is a topic of much debate with some wildly extravagent claims by some !! I always treat them with sceptism, and if it says max Crate is 20 for instance, then I use it at no more than about 10 to 15 max.

[JC]

Timbo

Thanks for the help.

[Craig Jones]

Could a model using a 480, 960Kv outrunner brushless motor powered by a 11.1v 2100mAh 20(c) LiPo be setup to use a 14.8v 2100mAh 20(c) LiPo...?

I know the Voltage is different but could it work? 

[Tim Mackey]

Well yes possibly, but you would need to prop down a lot to stay at the same current consumption.

It depends also on the motor specs - IE what is the maximum wattage or current capacity of the motor? Once you know that, the only safe way to check that you are not going to burn it out with the higher voltage pack is by using a whattmeter or similar.

[Myron Beaumont]

Timbo    Are you saying basically that voltage controls rpm and the current is controlled by an ESC to prevent things getting hot!? Unlike ic engines where the smaller the prop the higher the speed (light load) the smaller the lekky prop the less the current or have I got it all wrong again? I know I'm not the only dummy around  but one day maybe with your help we'll end up knowing just what we're doing.

[Tim Mackey]

Yes the voltage controls the rpm ( KV), but current is not controlled by the ESC - current is "consumed" relative to the resistance in the circuit EG the motor and its workload- the harder it is asked to work ( by turning larger props ) the more current will be consumed.

Our old friend Ohms law steps in here, and says of course, if the resistance ( load ) stays the same, and you increase the volts, then the current goes up proportionally.

The ESC is just a commutating  switch, albeit a very rapid one. When on low throttle, FULL current is still being drawn by the motor, just for VERY short periods of time... milliseconds. As the throttle increases the ESC merely applies the same full voltage to the motor for slightly longer bursts, until at full throttle it is almost like a continuos connection. There is a common misconception that ESCs feed different voltages or current to motors, but its as I say above. Thats partly the reason they are more likely to get hot under partila throttle than full....they are working harder !!

[Myron Beaumont]

I see BUT Why the figures denoting what the ESC can do are apparently nothing to do with "throttle " settings at anything below flat out? when you say it is working harder and getting hotter?Or have I got it wrong again? Myron

[Tim Mackey]

dont understand your q mate.

[Myron Beaumont]

ITimbo       I think what I'm saying is Does the ESC "absorb" unnessesccary! current which would normally go straight to the motor if it wasnt there?(the ESC that is) Bit like a resistor with a sort of built in heat sink ?Like a thyristor in a dimmer switch?

[Bruce Richards]

Excuse me for butting in but let me try to help

Yes it is similar to a thyristor in a dimmer switch that a good comparison.

This is a simplification to help with this specific question it may not hold true for all electronic circuits but is the way I think about it and I hope it helps.

1.  The motor pulls the current from the battery through the ESC.

2. The ESC switches the electrical supply on and off quickly and increases the amount on time to increase the speed.

3. When at WOT the electrical supply is on almost all of the time and hence the ESC is doing very little work.

4. At  partial throttle settings the ESC is switching the current on and off at a high rate and this switching is what produces the heat.

Bruce