Lithium Charging Load Resistance?

[johnywhy]

I'm trying to understand the load-resistance of an A123 Nanophosphate High Power Lithium Ion cell.

https://www.buya123products.com/uploads/vipcase/844c1bd8bdd1190ebb364d572bc1e6e7.pdf

My understanding is that resistance increases as the cell charges.

The datasheet says "Internal Impedance (1kHz AC typical, mΩ) 8". I don't understand what that means.

Here's my (probably incorrect) attempt to apply Ohm's Law, at the low-SOC and full-SOC of the cell:

Low SOC
charging at 4A
R = V/I
R = 2.5/4
R = 0.625

High SOC
charging at 0.00001A
R = 3.7/0.00001
R = 370K

[RoGeorge]

Impedance is for alternative current (AC).
Resistance is for direct current (DC).

Impedance is not the same thing as resistance, though they both measures in ohms ().

The 8m\$\Omega\$ from the datasheet means if you pass an audio signal of 1kHz through that battery, the battery will appear almost as a short-circuit to that AC signal.  That is for passing AC currents only, not for DC.

[johnywhy]

But we don't pass audio to a battery cell. We pass DC current.

Does the AC impedance tell us something about DC charging behavior?

[Siwastaja]

Here's my (probably incorrect) attempt to apply Ohm's Law, at the low-SOC and full-SOC of the cell:

You are calculating the resistance of your load. That has nothing to do with the cell.

The interesting parameter about the cell is its internal equivalent series resistance, ESR. This acts in series with your load resistance, and "causes" a voltage drop. For example, if the cell is at 3.3V and you apply a load of 10A, and cell voltage drops to 3.2V, cell ESR is 0.1V/10A = 0.01 ohms. You can also then calculate that I^2*R = 10^2 * 0.01 = 1W of heat is generated in the cell. If the load is supplied with 3.2V and 10A, you can further calculate that the efficiency of the cell is 32W / 33W = 97%.

Now the rest of my post discusses this equivalent series resistance of the cell.

My understanding is that resistance increases as the cell charges.

Usually not. While internal resistance is a function of the state-of-charge, it's usually fairly constant between maybe 30%-100%. It is typical though that resistance increases near empty.

The only ways to get actual data instead of such generalization are: measure for yourself; find reliable data published by others; or commit to buy millions of cells and get the manufacturer to supply the data you need (under non-disclosure agreement, of course).

The datasheet says "Internal Impedance (1kHz AC typical, mΩ) 8". I don't understand what that means.

Many don't. It's a nearly useless number for a battery user, i.e., the system engineer. It is somewhat usable as a cell quality control number.

If you apply a load which changes its load current in a sinusoidal waveform at 1kHz, you will see a varying voltage drop compared to the ideal cell output voltage. This impedance is dU / dI.

The cell physically acts like an electrolytic capacitor, which helps source (and sink) current at short timescales. This makes the AC impedance number lower (better) than the DC resistance. But if you try to load the cell for 10 seconds, capacitively stored charge is consumed and the chemical reaction thing has to supply all the current, making the voltage drop larger, and lower-frequency impedance larger. Usually maybe something like double the 1kHz figure. The problem is, while the AC impedance number is optimistic, it's hard to predict how much exactly, so it's nearly useless if you want to calculate things like cell efficiency, how much the cell heats up, how much the output voltage sags, etc.

1kHz AC impedance is also more constant regardless of state-of-charge, which some people consider an "advantage" for some weird reason. Of course, really it's only an indicative of this number not representing the real world, because we mostly care about the DC characteristics of the cell. (Not to say 1kHz performance is totally useless - for example mobile phone applications need short large burst currents and benefit of this capacitor-like behavior of cells.)

[RoGeorge]

But we don't pass audio to a battery cell. We pass DC current.

Does the AC impedance tell us something about DC charging behavior?

Yes, it gives an idea of what response to expect from the battery when the current varies.

For example, let's say a load has an average current of 10A, and the measured voltage on the battery is 3.700V.  Now, if that load current starts to oscillate +/-1A around the average 10A, then we would expect a deviation on the battery voltage with +/-8mV (assuming those 8m\$\Omega\$ from the datasheet the voltage will vary between 3.692V...3.708V).  So a load current of 10A average, but oscillating between between +9 and +11A will produce a very voltage variation of only +/-8mV around the 3.7V average voltage of the battery.  (We assumed the battery is capable of 10A/3.7V DC)

Now, if the battery is capable or not of delivering 10A DC and still have 3.7V at its terminals, that's a completely different problem.  A small size battery might have the same 8m\$\Omega\$ impedance for AC variation as a big battery, yet the small one might not be able to deliver more than 1A, while a big one might be able to deliver100A.

For DC, there are other specifications for the max current.  If not, the capacity of the battery may give an idea of how much DC current a battery is able to sustain.  As a thumb rule, the mAh capacity is measured at 0.2C current.  For example if a battery is marked as C = 2300mAh, this means that battery was measured to last 5 hours at a continuous load of 460mA DC.  That gives an idea about what average current can such a battery sustain for hours, 0.46A.  That battery might keep it's 3.7V at 0.5A, but at 10A I would expect the voltage to be much lower than 3.7V.

[tunk]

No expert, but I think I would do this to get an idea about the internal resistance:
- measure cell/battery voltage unloaded
- discharge it with e.g. 1A (and/or charge it)
- wait a few seconds to let the voltage stabilize and then measure voltage again
Use current and the voltage difference to calculate a resistance.

[Siwastaja]

My preferred DC ESR measurement method is this, usable during cycling test, with example numbers,

Let I0 = 3A (discharge or charge current used for the cycle)
Let I1 = 0.75*I0 = 2.25A
Let I2 = 1.50*I0 = 4.5A

During constant current discharge at I0=3A, change current to I1=2.25A for 10 seconds. At 9.99 seconds into this, measure voltage V1.
Now change current to I2=4.5A for another 10 seconds. Again at 9.99 seconds, measure voltage V2.
Now change current back to I1=2.25A for another 10 seconds. At 9.99 seconds again, measure voltage V3.
Change back to normal operation I0=3A.

Repeat the cycle every now and then at interesting points, e.g. near 100%, 75%, 50%, 25% and 0% or whatever.

Voltages V1 and V3 were measured at 2.25A. Voltage V2 was measured at 4.5A. Because the cell was discharging or charging slightly during the test, one can linearly interpolate what the voltage would have been at V2 sampling moment, if the current were 2.25A: (V1+V3)/2.

Therefore, the dynamic DC ESR is:
R = dU / dI = ((V1+V3)/2 - V2) / (I2 - I1)

Note that the multipliers 0.75 and 1.5 were chosen so that average current over the 3-step pulse equals I0 when the timesteps t1 = t2 = t3 = 10 seconds are equal.

10 second step is more than "DC enough" for most practical purposes, i.e., the capacitor characteristics of the cell disappear. It's still worth noting that if one was to make a full discharge cycle at higher current, the cell would heat up more and its DC ESR get lower thanks to that.

It's finally important to understand that resistance number is nothing more than modelling of how the cell voltage sags under charge/discharge current, put in one number. How the measurement is done affects the result, and it's best to choose a measurement procedure which is closest to the final application profile.