# How to measure the internal resistance of a battery using a digital multimeter?



## Bat (Jul 6, 2011)

is it possible to measure, or at least derive from the readings, the internal resistance of a li-ion rechargeable using a multimeter? 

Many thanks!


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## lctorana (Jul 6, 2011)

Ideally, you need two meters.

First, a DC current meter in series with your test load (bulb, resistor, LED, coil of wire, heater element, whatever).

Second, a DC voltmeter accross (i.e. in parallel with) your battery cell.

Take current and voltage readings at two different load currents.

Don't use the no-load (zero current) measuremrnt.

The internal resistance will be the difference between you two measured voltages divided by the difference between the two current readings.

An example may help.

Say you have a single-die LED, which, when in series with the ammeter, draws 1.1A. The battery voltage holds up at 4.05V 
And a quad-die LED that draws 2.6A, the battery drops to 3.81V
The internal resistance is then (4.05-3.81)/(1.1-2.6) = 0.24/-1.5 = -0.16 ohms, or -160 milliohms. It's always negative, of course, being an EMF.
The figures in my example were plucked out of my @r$e, just to illustrate the concept. They're not intended indicative of any real-world devices.


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## Justin Case (Jul 6, 2011)

You want to select a load resistor with an appropriate resistance value to give you a "safe" discharge current and with an appropriate power rating so that it can handle the wattage that will be dissipated.

Let's say that you want to test for internal resistance at about a 1A discharge current. This current level should be appropriate for larger cells like 17670s, 18650s, 26500s, and 26650s, giving a discharge rate well below 1C. For smaller cells like 16340s, you might want to test at 0.5A, although 1A is still within 2C and is acceptable. For IMR type cells, you can discharge at higher rates.

A fairly typical Li-ion internal resistance is around 100 mohms for cells in top condition. For a 1A discharge current, that would give you a calculated voltage drop of

Delta V = Voc - Vload = I * Rinternal = 1A * 0.1ohms = 0.1V

where

Voc = open circuit voltage of the cell being tested
Vload = voltage under load for the cell being tested
Rinternal = cell internal resistance
I = discharge current through the load resistor

Thus, your hypothetical, fully charged cell would show a voltage drop from 4.2V down to 4.1V.

The load resistor value needs to be

Vload = I * Rload ---> Rload = Vload/I ~ 4ohms

One amp through 4 ohms is 4W. So you should use something like a 4 ohm, 5W resistor. A 3 ohm resistor should also be fine. You'd probably get about 1.3A discharge, which should be very close to 0.5C for something like an 18650. The power dissipation through the load resistor in this case can be slightly greater than 5W, so I'd suggest a load resistor with a higher power rating than 5W.

If you had a smaller 16340 cell that had 100mohm internal resistance and wanted to test it at 0.5A discharge, you can use something like an 8 ohm, 3W resistor.

How to measure:

1) Measure the cell's open circuit voltage when fully charged. Call this Voc. It should be close to 4.2V.
2) Set your DMM to the appropriate DC voltage scale and connect the DMM to each side of the load resistor.
3) Briefly connect the resistor across the cell (Batt+ to one end of the resistor, Batt- to the other end).
4) Read the cell voltage under load. Call this Vload. As soon as you have your Vload measurement, disconnect the cell from the resistor to end the discharging.
5) Use your DMM to measure the load resistor. Call this Rload.

To calculate internal resistance:

Delta V = Voc - Vload = I * Rinternal (from above)

Solving for Rinternal gives

Rinternal = (Voc - Vload)/I

But

Vload = I * Rload

Thus,

I = Vload/Rload

Substituting into the equation for Rinternal, we get

Rinternal = (Voc - Vload) * Rload / Vload

You've measured Voc, Vload, and Rload, so you can calculate Rinternal.

Depending on how you connect wires from the cell to the load resistor, your test may add some noticeable amount of resistance just from the test setup. In addition, any protection circuit will probably add some additional amount of resistance. Thus, the calculated value for Rinternal may not be quite the actual value.

This link gives a table for Rinternal vs cell condition for Li-ions.


```
[B]Milli-Ohm   [/B][B]Battery Voltage    [/B][B]Ranking[/B][IMG]http://www.buchmann.ca/shim.gif[/IMG]
75-150mOhm       3.6V           Excellent
150-250mOhm      3.6V           Good
250-350mOhm      3.6V           Marginal
350-500mOhm      3.6V           Poor
Above 500mOhm    3.6V           Fail
```


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## samgab (Jul 6, 2011)

... My mind just boggled.

Also, you can look for a post by HKJ (check in the reviews section) and look at his sig line. He has a link to some very helpful info on how to use a DMM to do this, among other things, with clear photos etc.


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## badtziscool (Jul 6, 2011)

Wow. Can this be sticky'ed somewhere? Good info.


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## Bat (Jul 6, 2011)

thx all. it'll a lot of time for me to digest


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## jarekdadej (Sep 24, 2011)

Was wondering the same thing about IR of a cell and reading the really helpful post thought of something. One can use a hobby charger to do it without using two multimeters. But please correct me if I am wrong.
The test would look something like that:
1. Charge the cell or cells
2. Once charged run a discharge at set current and observe the voltage for first few seconds of the process.
3. Top off the cell? (not sure about this one so correct me if I am wrong)
4. Set the discharge to the current different from the one in step 2 and observe the voltage.
5. Follow the calculations from previous posts.
OR
Set discharge and start it,my charger starts applying the load from 0.1 amps and in 0.1 increments all the way to the set load. Observe the voltage at every step and follow calculations.


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## Astro (Feb 1, 2012)

Is it necessary to test batteries with a load giving 1A? My torch will draw 1A in its turbo mode, but I usually use it in lower modes - say .3amp. It seems the testing would be safer using a higher resistance/lower current, but will this mean I get an inaccurate reading on the resistance?


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## vectorspace (Oct 7, 2014)

samgab said:


> ... My mind just boggled.
> 
> Also, you can look for a post by HKJ (check in the reviews section) and look at his sig line. He has a link to some very helpful info on how to use a DMM to do this, among other things, with clear photos etc.



Thanks, I'll try to get that link. Sounds very interesting.


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