I have a 65 mAh thin Li-Po pouch cell. I tried measuring the internal resistance of the same using the foll. steps: (The battery was fully charged before this process)

  1. Battery open circuit voltage measurement: 4.0662 V
  2. Connect a 560 ohm resistor across the battery
  3. Voltage measured across the resistor: 4.0562 ohm
  4. Current through the circuit: 4.0562/560 = 0.00724 A
  5. Voltage drop across battery's internal resistor: 0.01 V
  6. Battery Internal Resistance: 0.01/0.00724 = 1.38 ohm

I repeated this process with 150 ohm and 47 ohm resistors and got the internal resistance to be 1.55 ohm and 2 ohm respectively.

From what I have read in the literature, battery internal resistance is generally in double-digit milliohms whereas the resistance of my battery is 2000 milliohms.

  • Is such high resistance possible?
  • Is this because I'm measuring the resistance at a fully charged state?
  • I know the SoC affects the battery resistance, but would it make the resistance so large?
  • If not what is the mistake I'm making in the resistance measurement process?
  • 4
    \$\begingroup\$ you're misreading the literature. You're dealing with a low-current battery, these have other design criteria than low internal resistance. 2Ω doesn't sound unrealistic at all. \$\endgroup\$ Commented Mar 9, 2020 at 14:52
  • \$\begingroup\$ I agree with Marcus. I read the description about this product. It states an internal resistance of about 50 milli ohms. I think you should approach that value if you redo the test with lower resistors. Allow the supply to provide at least 200mA of current, in my opinion. \$\endgroup\$
    – a360pilot
    Commented Mar 9, 2020 at 14:55
  • 3
    \$\begingroup\$ @BenFM I don't think you're right! This is a 65 mAh battery. If you draw 200 mA, it'd last 20 minutes, at most (battery capacity is reduced when current is high). This battery wasn't designed for applications where an internal resistance of 2Ω "matters". \$\endgroup\$ Commented Mar 9, 2020 at 14:58
  • 1
    \$\begingroup\$ One possible source of error : as you discharge the battery, its open circuit voltage will change. Quite rapidly until it gets down to its "nominal" 3.7-ish volts. If you aren't re-measuring that after each experiment your resistance measurements will be artificially high. \$\endgroup\$
    – user16324
    Commented Mar 9, 2020 at 15:22
  • 3
    \$\begingroup\$ The internal resistance is quite temperature dependent (and when you load the battery it will self heat), an effect you are seeing (higher output current, higher internal resistance). The values you have computed seem perfectly reasonable for a 65mAh battery. The test conditions for internal resistance are often no load or a specified load. \$\endgroup\$ Commented Mar 9, 2020 at 15:22

1 Answer 1


A battery ESR test depends on vendor choice of test method. The most consistent is Hioki’s low-frequency AC-IR method. It cuts testing time down to 10 seconds while still maintaining measurement precision and stability.

The battery ESR determines the C rating for charge and discharge which varies widely upon dimensions and chemistry. Thus a 1C battery has an ESR 10x bigger than a 10C battery. if it has no rating, assume it is 1C with high ESR.

Although 1kHz ESR Test gives the lowest result 1Hz IR drop gives the same results as DC ESR = IR drop test results in 15 seconds without full charge and using only 1% of max rated current of the DC ESR Test.

The equiv circuit is R1+R2//C1 so 1Hz Test gives same results as DC with ESR= R1+R2. But the 1kHz Test takes 1 second and ESR=R1 for production speed.

The equiv cct. Is more complex for testing mAh for C1//C2 with separate ESR’s for each C. This also describes the short term memory effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.