Internal resistance comparison number

When comparing cells or batteries their internal resistance is a parameter. But if the cells/batteries have different capacity or different voltage the number isn't directly comparable. It would be better to use a number which is directly comparable.

My proposal is to make this number by multiplying the cells/batterys internal resistance with it's capacity and divide with it's nominal voltage . That number would be directly comparable between cells/batteries with different capacity and voltage. The lower the number the better.


Example 1:
Type: A123 M1 LiFePO4
Nominal voltage: 3.2V
Capacity: 2.3Ah
Internal resistance: about 10 milli ohm

Internal resistance comparison number: 10 milli ohm * 2.3 Ah / 3.2 V = 7 milli hours
(Yes, the unit will be hours)

Example 2:
Type: Sanyo RC-3600HV NiMH
Nominal voltage: 1.2V
Capacity: 3.6Ah
Internal resistance: about 4 milli ohm

Internal resistance comparison number: 4 milli ohm * 3.6 Ah / 1.2 V = 12 milli hours

So, even though the internal resistance is lower than A123 the comparison number is higher. That is because you will have to put these in series (which increases IR) to make the voltage of a A123-cell, and you will have to parallell "one and and a half" A123 to make the same capacity. (which lowers IR).

Example 3:
Type: Zippy Flightmax 5ah 20C cell LiPo
Nominal voltage: 3.7V
Capacity: 5Ah
Internal resistance: about 5 milli ohm

Internal resistance comparison number: 5 milli ohm * 5 Ah / 3.7 V = 7 milli hours

So, a Zippy pack will have about the same IR as A123 M1:s configured to the same capacity and nominal voltage.

What do you think about the idea?
The number could be called "corrected IR" or similar.

the internal resistance is a derived number, from justin's definition where the drop in voltage with increase in current produced by the cell defines the internal resistance. Ri=(V1-V2)/(I2-I1).

internal resistance is the measure of the battery to produce current at a specified voltage. it is a function of the electrolyte, the separator materials, the anode and the cathode structure and surface interface. many people use the low frequency electrical impedance to evaluate the Ri, but i prefer to consider it as the derived number from justin's definition. how it relates to cell heating is not clear, but i don't consider it to act as normal resistance heating.

you would expect the Ri to be consistent in a particular battery configuration, and if the cell had problems from abuse such as delamination of the separator from the electrolyte and electrodes, or collapse of the cathodic structure, then the Ri should change dramatically, but it must be done on an individual cell level, imo.

I may have been unclear about the purpuse of this thread. What I want to do is to propose a quantity which would make it easier to compare batteries internal resistance.

I too like the Ri=(V1-V2)/(I2-I1) derived number better than the 1kHz AC number, by the way.

Would this number in ( mil/hrs ) represent performance?

Yes, you can say it represents performance. It would basicly be proportional to the voltage sag under load, just as internal resistance is, but this quantity is corrected for capacity and voltage.

The unit is hours. milli is just the prefix, it means 1/1000.

Let's see if I can start from first principles and see if I arrive with a similar expression.

If a cell is added in series, this increases the internal resistance but also increases the voltage so the proportion between the two would represent a constant.

If a cell is added in parallel, this decreases the internal resistance but increases the capacity by the same factor, so the product between internal resistance and capacity would represent a constant.

Adding cell in series:
R/V = K_1

Adding cell in parallel:
R*capacity = K_2

If we want a performance number that's comparable, then any change in adding a cell in series or parallel should mean the "ultimate constant" stays constant. Some mathematical ways of creating this constant is by some mathematical operations on constants, whether this be multiplication, division, addition or so on and to include all relevant information, some mathematical operation between K1 and K2 should take place.

K1*K2 = R^2*capacity/V;

K1/K2 = V*I*t (Expression of energy)

K1+K2 = R/V+R*I*T = R(I*t + 1/V) ... something odd.

So, it seems R^2*capacity/V produces the desired constant that doesn't represent some other unknown(like the total energy).

Your proposal,

"My proposal is to make this number by multiplying the cells/batterys internal resistance with it's capacity and divide with it's nominal voltage"

is R*capacity/V. Since you'd want this comparable number to remain constant for a battery constructed of a given cell(a 24 volt 10 ah and 48 volt 20 ah ping should have the same number), let's test out your formulation and my formulation.

Say we have a cell with initial resistance, voltage and capacity of R_o,V_o, and Ah_o. Their "constant" is supposedly R_o*Ah_o/V_o or R_o^2*Ah_o/V_o. Adding a cell in series doubles the resistance but doubles the voltage, so the first formulation results in (2R_o)*Ah_o/(2V_o) = R_o*Ah_o/V_o whereas the second one would result in 2R_o^2*Ah_o/V_o which clearly isn't the same as the original one. So it seems your formula passes the "series test" and mine doesn't. :lolMust have messed up my mathematical reasoning somewhere)

If we added a cell in parallel, its capacity would double and the internal resistance would halve(checking... r^2/(r+r) = 1/2*r - check). Your formula would result in the same initial constant whereas my formula, which has already been discredited, would result in a halved initial "constant". My formula is again disproven and yours is supported.

Congrats, it seems your formula makes theoretical sense. The meaning of the dimension of "time", though, seems kind of weird as it doesn't seem to have any immediate meaning. Oh well, that's like the meaning "frequency" with its unit of (1/time), right? Well... sort of, maybe.

By the ways, you shouldn't put this units of time. Since it's clearly your invention and it's a simple way of directly comparing different cells' internal resistance, it should be named after you. So, what's your last name or does "This battery has 100 bearings" sound good enough?

On another tangent, this seems sooooooooooooo weird but yet it makes soooooooooo much sense! Why does it seem like this induces a brain-warp?

Anyways, bearing, I say this idea is genius. Let's define a millihour as a bearing, ok? :wink: "millihour" just sounds awkward since a base 60 unit doesn't readily mesh with base 10 prefixes.

swbluto said:

......
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Your formula would result in the same initial constant whereas my formula, which has already been discredited, would result in a halved initial "constant". My formula is again disproven and yours is supported.

Congrats, it seems your formula makes theoretical sense. ...
....

Click to expand...

You basically claim that the fault in your own thought proves that some other thought is correct!

That makes no sense to me.

Could they not both be wrong?

Yes strange logic for supporting bearings idea. Can you not take a different approach to deriving his expression from first principles that comes to the same conclusion Swbluto?

i think bearings idear is sound from my limited experience. what is ping's bearings?

=) guys, this is funny, naming a unit after me. But I'm not sure we should name it after me, because when I woke up this morning I just realized why the unit is time. The number is the time it would take to discharge the cell by shortcircuiting it with a superconductor. That is in theory of course, because the cell can (probably) not supply the current required.

So, 7 milli hours is the time it would take to discharge one shortcircuited A123-cell, and also the time it would take to discharge any A123-pack shortcircuited. (If all conductors are superconductive.)

What about naming the value: Shortest Discharge Time, or SDT in short?

And, we should use the unit seconds instead, by multiplying with 3600 (or 3.6 if the prefix was milli)?
Then 26 seconds is the SDT for a 10 milli ohm A123.

Again, genius! That makes a lot of sense, I guess the late night and strange units might have thrown me out of whack.

Mr. Mik said:

swbluto said:

......
...
Your formula would result in the same initial constant whereas my formula, which has already been discredited, would result in a halved initial "constant". My formula is again disproven and yours is supported.

Congrats, it seems your formula makes theoretical sense. ...
....

Click to expand...

You basically claim that the fault in your own thought proves that some other thought is correct!

That makes no sense to me.

Could they not both be wrong?

Click to expand...

That's not what I was trying to suggest. I was trying to suggest that a given formula should result in a constant that stays truly constant no matter the quantity of cells and the configuration of a pack's cells, and since bearings' formula remained constant when paralleling another cell and adding in series another cell(In the example calculation, one cell to begin with, but the result would be the same no matter how many cells one started with assuming symmetry.), this would directly suggest that any pack/configuration-of-cells would have the same constant since a pack is nothing more than combining cells in series and parallel. So, it seems the idea is sound. A precondition is, though, that the pack is symmetric which virtually all practical packs are.

Anyways, I think I found the flaw in my reasoning. Those two constants are constant under their own corresponding operation and to combine them means you'd have to leave common variables as is... so it isn't really "true" multiplication, but like a "grouping together" type of multiplication. Doing so would result in bearing's formula.

voicecoils said:

Yes strange logic for supporting bearings idea. Can you not take a different approach to deriving his expression from first principles that comes to the same conclusion Swbluto?

Click to expand...

Oh yeah, I definitely believe there's more than one way to skin a cat! Especially when it comes to ideas and theory. I wouldn't be surprised if there was a direct and easy approach from first principles to show that this is just a calculation of the amount of time it theoretically takes a short-circuited standardized cell to drain - the short circuit being a superconductor of some sort. And the internal resistance staying constant even as the cell significantly heats, of course.

This is indeed a very good idea, because there is no easy way of rapidly comparing cells or batteries. Not a huge deal for tech-heads, but for people less prone to use math when comparing things, having such a number would be universally helpfull for cell/battery selection.

This does however raise the issue of what "watt-hour" rating we use (1C, 2C, ?), and also what voltage do we use? The best would be to average the voltage slope during the discharge at any given C rate, and state the rate used for the test along with the resulting number.

Like: 7milli-bearings@1C; 8milli-bearings@5C; etc

ZapPat said:

This does however raise the issue of what "watt-hour" rating we use (1C, 2C, ?), and also what voltage do we use? The best would be to average the voltage slope during the discharge at any given C rate, and state the rate used for the test along with the resulting number.

Click to expand...

Hmmm.... Hmmm... I think he was stating just using the "capacity" in amp-hours, but that might not be a perfectly accurate rating - I guess a "20 hour" rating or other rating that entails a low percentage of internal resistance losses would be approximate to its "true" Ah rating. As far as voltage, it seems it'd need to be standardized for a chemistry but ensuring it's standardized so different chemistries are comparable seems to be more difficult. I'd be inclined to say the "average" voltage which can be calculated using... (1/AH)*Integral(V_oc(ah) d_ah) but I'd imagine this would need to be the 0pen-circuit voltage function and that function can be found by V_oc(ah) = V_(ah) + R(ah)*I where I is the constant current corresponding to the "20 hour" or whatever slow rating, which should be used for all chemistries to find the Ah and average voltage of a given chemistry.

Even though that's not really difficult to calculate, I suspect many are scared of calculus so we could just take the open-circuit voltage at 50% state of charge. That should approximate the above.

I've taken LiFePO4 to have a 50% state of charge Open-circuit voltage of 3.2 volts. LiPo has one of 3.7 volts. And nimh has 1.2 volts. I'm not sure those are right, though, since I haven't measured it. But, the point is that the voltage per cell should be some constant value for a chemistry so that one could accurately compare one cell to another of that chemistry, and the voltage per cell should be determined by some "practical formulation" to ensure comparability between different chemistries. Does the open circuit voltage at 50% state-of-charge sound good?

Also, zappath, the total energy(as you mention watt-hours) should remain approximately constant no matter the load. This is assuming the total energy dissipated on the load and cell's internal resistance is added together. The total "capacity"(in Ah) would vary depending on the load, but a really low load should be used to get an idea of a battery's "total" ah if bearing's formula is going to use ah.

bearing said:

=) guys, this is funny, naming a unit after me. But I'm not sure we should name it after me, because when I woke up this morning I just realized why the unit is time. The number is the time it would take to discharge the cell by shortcircuiting it with a superconductor. That is in theory of course, because the cell can (probably) not supply the current required.

So, 7 milli hours is the time it would take to discharge one shortcircuited A123-cell, and also the time it would take to discharge any A123-pack shortcircuited. (If all conductors are superconductive.)

What about naming the value: Shortest Discharge Time, or SDT in short?

And, we should use the unit seconds instead, by multiplying with 3600 (or 3.6 if the prefix was milli)?
Then 26 seconds is the SDT for a 10 milli ohm A123.

Click to expand...

Oh I see great brainstorming about Ri here

You guys have great idea and the way you think clarely show you understand how cells work.

Theire is just one think i would like to correct here:

The "Then 26 seconds is the SDT for a 10 milli ohm A123".. can't be right.

You will need to introduce another inportant factor!!!.. Yes.. one more parameter to the formula :lol:

This is the max time the cell can tolerate the heat generated bu the max current discharge capability!!

And what limit that is the max temp the cell can take without damaging it.. in teh case of a A123 the tmax temp limit to 10sec.. and not 26 sec.
cause at 26sec.. let say it's 100% dod.. the temp will be so high that the cell could have irreversible dammage.

I like the swbluto's formula that include the voltage of the cell and the capacity


Doc

what do you think is happening under the heating regime you mentioned.

do you think the electrolyte begins to boil and the gas phase pushes voids into the separator or maybe inside the cathode itself? maybe the jelly roll delaminates between electrode and separator and the gas is trapped there, or the electrolyte changes or maybe the anode grows some sort of surface interface that blocks current transport because it is more of an insulator than the anode itself?

Hi Doc, nice of you to come in. The value is only theoretical, and it's sole purpose is to make comparison of different cells or batteries easier. I never meant anyone should test a cell by short circuiting it. (It seems, though, that some likes to make hazardous experiments with similar risks involved :wink: (excavator))

The current formula:
SDT = 3.6 * Ri * Capacity / Nominal Voltage
Ri in milliohms
Capacity in Ah
Nominal Voltage in Volts
SDT in seconds.

dnmun said:

what do you think is happening under the heating regime you mentioned.

do you think the electrolyte begins to boil and the gas phase pushes voids into the separator or maybe inside the cathode itself? maybe the jelly roll delaminates between electrode and separator and the gas is trapped there, or the electrolyte changes or maybe the anode grows some sort of surface interface that blocks current transport because it is more of an insulator than the anode itself?

Click to expand...

Check this site out... really good tech info about cells!

Sorry, I actually meant "amp-hours", not "watt-hours"!

SO I guess that the 50% open circuit SOC voltage would be OK to use for manual calculations. However the most accurate source for this would be to periodically sample the voltage to find the average during the whole discharge (like your integral formula example), but would require a micro or other data logging device to do right. As for the capacity rating used (Ah), I wonder about using the "20 hour" result, since it would likely be high when compared to the 1C rating that is often used in modern chemistries. Even lithium suffers some peukert effect, although it is small specially when compared to lead.

Also, I think we might want to drop the multiplier (3.6), since this number is theoretical anyways and no matter what scaling we apply to it, it still is just a number to compare cells/batteries between each other. I figure KISS.

This is getting messy with three different units circling around.

But I agree with you about skipping the 3.6 constant. The value is easier calculated without it, and also easier to use "backwards".

I also want to make a comment about the proposals of calculating integrals and stuff. My view of this value is that it supposed to be simple and rough, just to give a fast indication. That is also the reason I gave the values in my examples without decimals. Everything in a battery is significantly influenced by temperature, age, state of charge and so on. If one battery is 11.2 and the other is 11.4 doesn't really matter because at another temperature the decimals could indicate a favor for the second battery. In the other thread we see batteries differ by factors of 2, 5, 20 and more. If the difference is just a few percent then other factors (like price, lifespan, weight and so on) should be of more importance when choosing the battery.

You maintain simplicity at the expense of exactness and accuracy?

Yah, I joke. If someone wanted to have an exact underpinning, there it is, but it's more practical to take the 50% SOC measurements.

ZapPat said:

As for the capacity rating used (Ah), I wonder about using the "20 hour" result, since it would likely be high when compared to the 1C rating that is often used in modern chemistries. Even lithium suffers some peukert effect, although it is small specially when compared to lead.

Click to expand...

The idea is to get the "true" amount of energy that the cell contains so you can calculate the SDT. If people were to base the AH based on a current where a much higher percentage of energy is lost on the cell's internal resistance, then the AH rating would change, and then this would be kind of redundant when you calculate SDT which takes into account the internal resistance and there's no universally accepted standard rate at what to discharge at - it'd essentially have to be arbitrary. This is a just a theoretical construct anyways, and you don't want imperfections to be repeated in the calculations, so it'd be best to make the percentage of internal losses "standard"(i.e., 0% - 20+ hour rate) in measuring things so you can cleanly incorporate the internal resistance in the SDT formula. Also, I hope you aren't taking measurements of output voltage during discharge - you're talking about the open circuit voltage right? If not, again, we run into that "standardization problem" and the internal resistance redundancy in the chain of calculations. Also, the average output voltage would depend on the internal resistance of the battery if you were to standardize the discharge rate, different batteries of the same chemistry would have a different voltage! This would make things even more complicated as you entrust people to measure the average voltage at some standard discharge which means more room for error, more complications and larger potential deviation from the comparability ideal.

Also, I think we might want to drop the multiplier (3.6), since this number is theoretical anyways and no matter what scaling we apply to it, it still is just a number to compare cells/batteries between each other. I figure KISS.

Click to expand...

Well, I think if you're going to call it the "Shortest Discharge Time" which makes sense of the units, I think seconds is more intuitive than millihours and that'd be effectively "simpler". Also, the crappier cells can be more accurately delineated by their 3+ digits left of the decimal. :lol:

On another question, what do you mean by that lithium has a peukert effect? I was under the impression that lithium's internal resistance is approximately constant in relation to current, ceteris paribus(temperature, age, SOC, etc.), whereas lead's internal resistance increases (I assumed exponentially having read it on a web page, but I think I'm going to mathematically investigate that later because I think a non-exponentially rising one could possibly result in an exponential reduction in AH) and this increasing internal resistance is the basis of the peukert effect. The reason why you see less AH is because a greater percentage of the energy is dissipated on the internal resistance(since the load resistance has to decrease for the load current to increase, so the battery's internal resistance makes a greater proportion of the total resistance) at higher currents. One that's based on an increasing internal resistance is called the "peukert effect".

But... I need to verify what I've just typed because I haven't done so already(It's just been a huge hunch).

( Ah*V = I*V*t = P*t = energy. A lower amount of output energy entails either a lower voltage, AH or both. Since obviously the output voltage decreases with increasing current, it might be possible that Ah doesn't change with a constant internal resistance. So...

(1) Ah*V = I*V*t = P*t = energy (Output current, average voltage, energy)
(2) V = V_oc-I*R

Ah*(V_oc-I*R) = output energy. If V_oc's average and R is a constant, and I increases, then the output energy decreases but yet Ah could still possibly remain constant as increasing I decreases the right term. It seems like measurements might be needed to be taken as it seems my math hasn't proved anything.)

swbluto - check some lithium discharge curves floating around here to see just how they can be non-linear. Specially between open circuit voltage and the 1C curve vs the 1C-2C curve differential. It even changes with temp and SOC. That's why I think some well thought out test algorithms might be required to get a complete analysis of the cell under test, giving us a more robust comparative between cells.

This might or might not be realted to the "peukert effect", I have yet to read anything definitive about this relating to lithium. One thing I do remember is that the peukert effect is a complicated notion, and is not necessarily just explained by a non-linear Ri. Cells tested under heavy loads do lose Amp-hours, and I wonder if this is only the premature triggering of the LVC due to the Ri*I voltage drop, or if it also partly the peukert effect making the cell less efficient at heavier loads.

Of course approximated comparisons would be usefull for sure and much easier to do manually, but wouldn't be as complete as an automated test procedure carried out by a processor and using multiple Ri and Voc samples combined with the cell's real capacity.

ZapPat said:

swbluto - check some lithium discharge curves floating around here to see just how they can be non-linear. Specially between open circuit voltage and the 1C curve vs the 1C-2C curve differential. It even changes with temp and SOC. That's why I think some well thought out test algorithms might be required to get a complete analysis of the cell under test, giving us a more robust comparative between cells.

Click to expand...

Can you please quote what you're directing this at? I must request this because that's one of things that I already knew but yet I presume my wording didn't quite convey. They're multiple points when I'm talking about "open circuit voltage", and I'm actually talking about the open-circuit voltage's average(The average over the entire discharge), and I believe that the open-circuit voltage's average should remain approximately constant for a given internal resistance at any current(Everything else being equal, like Temp, which they aren't in "real tests" where they just let the cells heat up and thus have a decreasing affect on internal resistance. Tsk Tsk. They need to control their parameters!). This presumes that the internal resistance, in relation to load current, is relatively constant holding everything else equal which a normal discharge test doesn't do but it's sufficiently approximated by lower loads for practical purposes(I.e., a 5c a123 only heats up to, what 30-40 degrees above ambient temperature?).

ZapPat said:

One thing I do remember is that the peukert effect is a complicated notion, and is not necessarily just explained by a non-linear Ri. Cells tested under heavy loads do lose Amp-hours, and I wonder if this is only the premature triggering of the LVC due to the Ri*I voltage drop, or if it also partly the peukert effect making the cell less efficient at heavier loads.

Click to expand...

Oh, yeah sure, Peukert's original analysis wasn't just simply based on a non-linear Ri or not-so-constant Ri in relation to load current as lead has multiple sources of internal resistance, but I think the notion has to be with some such simplified association if you're going to liberally apply it to fundamentally different chemistries. But, let's get this straight, are you suspecting that the internal resistance increases at higher load currents holding everything else equal contributing to the loss of observed Ah and you have already considered the effect of a greater percentage of the total energy squandered on the cell's internal resistance due to the greater percentage of internal resistance made by a lower total resistance(load+Ri)? If you have, please share your deductions! And, as to the internal resistance increase with higher load current suspicion(Is that the type of "less efficiency" you're thinking about?), that's particularly easy to test. Just do the internal resistance formula between difference pairs of currents, one appreciably higher than the other and you should be able to see a persistent difference in internal resistances if this is true at different SOCs(And same temps), suggesting that there is at least some similar mechanisms behind the peukert effect and lithium's own "peukert effect". You have the testing equipment, right? I'll be building my own digitized testing equipment soon enough since I'm pretty much forced to test my ping cells now , but I think the data might be redundant by the time these questions are answered.