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Old Mar 26, 2015, 11:24 AM
david.grimes is offline
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Quite enlightening to say the least...
Explains why my 30A PS wouldn't light 3 bulbs or 2 bulbs. I pondered the cause as I arranged a single bulb which did light up so I set mentally set aside the oddity. My assumption now is that either the PS overload protection is inadequate or more forgiving or that bulb didn't require 10x its watt rating to light up (maybe high quality or pre-heated).
I'm still consuming much of the discussion in these threads and the content of the referenced sites. If someone would be so kind as to validate my knowledge thus far to hopefully illuminate any gaps or flaws in what I've learned.
My understanding is that a pack that maintains a higher voltage under load will be faster than the one that sags further. It is also my understanding that iR and C ratings are directly related.
So with regards to quantifying battery performance I should observe a battery with a higher C rating would possess lower iR values and maintain a higher voltage under a given load compared to a battery with a lower C rating and thus higher IR values.
In the event my logic is flawed or I have misunderstood something consider the following choices and rewrite the statement:
I should observe a battery with a (higher|lower) C rating would possess (higher|lower) iR values and maintain a (higher|lower) voltage under a given load compared to a battery with a (higher|lower) C rating and (higher|lower) IR values.

It's funny how hard it is to construct a question or articulate ones understanding to deliberately illuminate what they don't know that they don't know whereas it seems incredibly easy for one to talk out of there ass

Anyways thanks in advance and I'm hugely grateful to those who have taken part in this discussion with me thus far you guys and those on the few pages I've read so far of the these two threads are Rock Stars in my book.

Quote:
Originally Posted by Wayne Giles View Post
David,

When John says 'significantly' he means just that.
I have just quantified this on a halogen bulb and the cold resistance is over 10x lower than the working resistance .
I came across this problem in linear psu design when feeding any load with a proportion of tungsten or halogen lamps in the total load. At switch on the psu 'sees' the load resistance as an overload and will not start up.

So it is unsurprising that you see a huge spark on connecting what should be an 83A load , when in fact the initial current will be around 850A, getting into short circuit territory!

The drop you see in IR values over the first few cycles appears to be general although some packs only drop the value on the first full cycle. It should be stable afterwards apart from the very slow decline through ageing.

A 5Ah pack with the IR figures you quote should comfortably cope with 20C or possibly 25C, but you can forget 50C continuous from any pack .

The heat generated in the pack is a function of the square of the current, so that the heat generated at 50C would be 9 x as much as at 17C for the same IR value.
IR drops with temperature rise, but not that much once the temperature is over about 40*C


Wayne
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Old Mar 26, 2015, 02:58 PM
buckaroo is offline
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Quote:
Originally Posted by david.grimes View Post

It's funny how hard it is to construct a question or articulate ones understanding to deliberately illuminate what they don't know that they don't know whereas it seems incredibly easy for one to talk out of there ass

Me and you would get along good David.

Dean
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Old Mar 26, 2015, 03:24 PM
jj604 is online now
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David see comments in the text.
Quote:
Originally Posted by david.grimes View Post
My understanding is that a pack that maintains a higher voltage under load will be faster than the one that sags further.

Correct. If a resistive load is unchanged then the higher voltage will produce a higher current. From Ohms Law. Voltage equals Current x Resistance. Since the Power is the product of the Voltage and Current then the power will increase.

It is also my understanding that iR and C ratings are directly related.
So with regards to quantifying battery performance I should observe a battery with a higher C rating would possess lower iR values and maintain a higher voltage under a given load compared to a battery with a lower C rating and thus higher IR values.

This is what is observed. Note however that some of us would desribe it in reverse. It is the IR that determines the C rating, not the other way round. C rating is just a description of allowable LiPo performance under load. It is also not clearly defined. The actual C rating you quote for a pack depends entirely on how much heating you will regard as acceptable and how much voltage drop you regard as acceptable. This is why some vendors can make absurd C rating claims like "150 True C".

It is possible a single cell discharged for a brief period and delivering a practically useless low voltage will perform at "150C". The chances of a whole pack delivering a constant 150C with adequate terminal voltage is zero.

There are some specialised cells used in high performance applications like straight line boat racing that have high C rates but it is generally accepted that for commercially available LiPos that ordinary folks can purchase the maximum C rate is about 40C continuous with nearly all packs significantly less than this.


In the event my logic is flawed or I have misunderstood something consider the following choices and rewrite the statement:
I should observe a battery with a (higher|lower) C rating would possess (higher|lower) iR values and maintain a (higher|lower) voltage under a given load compared to a battery with a (higher|lower) C rating and (higher|lower) IR values.

It's funny how hard it is to construct a question or articulate ones understanding to deliberately illuminate what they don't know that they don't know whereas it seems incredibly easy for one to talk out of there ass

Anyways thanks in advance and I'm hugely grateful to those who have taken part in this discussion with me thus far you guys and those on the few pages I've read so far of the these two threads are Rock Stars in my book.
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Old Mar 26, 2015, 03:48 PM
Wayne Giles is offline
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David,

Your understanding is basically correct but the some of the options are reversed of which you may be aware.
I am not trying to be pedantic but it should be packs with higher/ lower C rating should display lower/higher IR and maintain higher/lower voltage under load.

Nothing is that simple of course and the two complicating factors here are C ratings being wildly exaggerated by sales blurb and the fact that IR varies with temperature.
We can't do anything about the first because salesmen will never be truthful (don't believe any C claim of over 40C) but the second can be quantified and John Julian has plotted a selection of IR variations with temperaure for a selection of packs.

http://static.rcgroups.net/forums/at...06.26%20am.png

If you take this on board and consider a lipo as a perfectly stable voltage source with a small resistor in series (the IR) then you have a simplified but adequate model of reality.

On load the voltage will be the open circuit value less the volt drop across the IR.
The voltage drop is equal to the the current in amps x IR in ohms (Ohms law) and the power ( in Watts) lost and dissipated as heat in the Pack is equal to the current squared x IR .

The power loss at the motor is approx. double the % voltage lost. So if the voltage drop at the lipo is 5% you can expect a power loss at the motor of just under 10%.

Hope this helps without being patronising,

Wayne
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Old Mar 26, 2015, 04:12 PM
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I wholeheartedly echo John and Wayne's comments.

I have been using measured internal resistance to monitor the performance of my lipolys and to also predict their in-flight performance for about 6 years now and have empirically determined that IR is a near-perfect 1:1 predictor of expected performance in my particular models that have typical flight times of 6-10 minutes. i.e. - Of two 2200mAh 3S packs, the pack with lower measured internal resistance has always demonstrated better performance in my particular applications.

As one moves into extreme high performance applications that can deplete a lipoly in 1-1/2 to 2 minutes, heating effects and deltas in temperature coefficient can become a more significant factor. For these extreme applications, replicating actual discharge conditions (current, airflow over packs, etc.) become more important if one desires absolute accuracy. For the rest of us, my extensive testing has demonstrated to my satisfaction that knowing IR, weight, and actual capacity of a lipoly pack are more than adequate to obtain a very solid predictor of expected performance within our models.

Mark
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Old Mar 26, 2015, 04:35 PM
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Quote:
Originally Posted by CelloGuy View Post
I think the tool is designed to give an estimate of the current that can safely be drawn from a pack rather than how well it will hold voltage. The max current and voltage sag are related but I'm not sure how close the connection is. The tool just uses a couple of the main factors that determine max current output (IR and capacity) - there are most likely other things that can affect it, and I imagine there are other things that can affect voltage drop too. The IR (combined with capacity) still seems to usually work as a useful way of estimating how a pack will perform under different types of loads and whether there is possibly a problem with the pack. It overcomes the problem that C ratings are used as a sales pitch and are unreliable for estimating output but it's never going to be foolproof and probably wasn't meant to be.
Gotcha. IR has been my next thing to start researching. Just recently picked up a reactor 300 and till now I had no way of testing my packs.
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Old Mar 26, 2015, 06:59 PM
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Slightly off topic but not exactly as some may not know this...I find the DPS 800GB power supplies, and some others that I have converted, have a delay on reporting a dead short and get through this surge from the inrush of amps to our light bulbs and chargers.

Rick
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Old Apr 06, 2015, 08:27 AM
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Quote:
Originally Posted by jj604 View Post
It is entirely concevable that for some LiPo chemistries the electrode/electrolyte transfer affects the overall behaviour more than the electrolyte conductivity - and vice versa.

So you can imagine that for someLiPo chemistries as the temperature increases the conductivity goes up faster than the reaction at the the interface (due to Arrhenius equations) increases - and vice versa.

So for some LiPos internal heating due to IR drop increases the breakdown of the cell and for others it increases performance (measured by voltage drop).

Again I emphasise this is all a generalised data-free heuristic (guesswork). I have no evidence for any of this.

John
Hi John,

Thanks for the reply. Certainly seems plausible.

My experience has been that packs when they're stressed will either sag fairly rapidly (small packs particularly) or will give good voltage (presumably with the bump in voltage from internal heating) but at the expense of heating and puffing. The packs that sag straight away don't seem to heat up much if at all and I'd assumed that that meant that they weren't being damaged nearly as much as the ones that were puffing. I guess the sagging means that they're not exposed to high current for long.

I'd been critical of the packs that puffed up because they seem to get damaged more easily but if it's true that the lower IR packs will heat up and puff more easily then puffing could actually be a sign of a better pack (albeit one that's been mistreated) at least relative to a similar pack that was reacting to the same current draw by sagging quickly rather than heating/puffing.

I have one pack that under a high load used to sag very quickly but now gets a voltage bump and heats and puffs a little. The load on it is less than it was (balanced a very unbalanced prop) so that might explain it but it almost seems like the pack has been rehabilitated a bit. Not sure how likely that is.

Anyway, thanks again to you and Wayne for your replies. Interesting stuff to get a handle on and your work makes it much easier. Cheers, Guy
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