# Series resistance of battery pack?

Discussion in 'Electronic Basics' started by Joe, Feb 25, 2004.

1. ### JoeGuest

Is there a way to measure the series resistance of a battery, or battery
pack? I am using LTSPICE to try and duplicate the results of a breadboard I
have constructed, but I have no idea what the series resistance of a battery
is. Maybe there's a ball park figure I can use? OR an equation?

TIA,
Joe

2. ### CFoley1064Guest

Subject: Series resistance of battery pack?
Choose a resistor which will load the battery pack to something near
recommended maximum current. After charging the battery pack, measure no-load
voltage with a DMM (Vnl). Place the resistor across the battery pack, and
resistor

(I = V/R).

Then calculate internal resistance by

R = (Vnl - Vfl) / I.

Good luck
Chris

3. ### Costas VlachosGuest

Measure the battery's voltage without any load. Call this value V1. Connect
an external load resistor across the battery and measure the battery's new
voltage and the current flowing through the load. Call these values V2 and
I. The battery's internal resistance is given by:

Ri = (V1 - V2) / I

Make sure you use a resistor that generates enough current to cause a
significant drop in the battery's voltage.

This assumes a battery model consisting of an ideal voltage source and a
resistor. For a more accurate measurement you'll need to plot a graph of
battery voltage vs. load current and evaluate the slope of the graph at the
operating point you need.

cheers,
Costas
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Costas Vlachos Email:

4. ### JoeGuest

Chris,

Thank you for the reply. I am not sure what is meant by recommended maximum
current. I have a battery chart that shows the mAH for different types of
batteries at a certain current drain value, eg, for AA alkaline manganese
dioxide batteries, the chart shows 2000mAH at a typical drain of 50mA.
Is that what I use (50mA)?

Joe

5. ### JoeGuest

Thank you Costas, a significant drop in battery voltage, like maybe it will
drop the voltage by one half? Would that be enough?

Thanks,
Joe

6. ### John PopelishGuest

That is a reasonable load, especially if it is close to what your
future load will be. You might measure the voltage of the pack before
during and after applying that load, and average the before and after
average and divide that difference by the load current (in amperes,
..05 in this case) to calculate the approximate battery resistance.

7. ### Costas VlachosGuest

One half sounds a bit too much. How about setting the current to the nominal
current of your circuit? That would give you a more accurate Rin masurement
as the relationship may not be linear depending on the type of battery, etc.

Costas

8. ### David WoodGuest

It might be interesting to compare your results with reference data
from the cell manufacturer. For example, the datasheet for the
ubiquitous Duracell Alkaline-Manganese AA shows 120 m-Ohm at 1kHz:

http://www.duracell.com/oem/Pdf/MN1500.pdf

Why would they reference a frequency rather than DC in their chart?

9. ### JoeGuest

Thanks John, I measured a rechargeable 9V (actually 7.2Volt) NiMH battery
expect? I measured it with 70mA (approx 100 ohm) load. It is not fully
charged for probly a month or so, just been sitting on my bench, but it has
not been used much either.

I am going to be using a 12volt lawnmower battery to power one of my
projects and that is on the charger right now, so I can measure it when it
is at full charge and keep track of it as it discharges.

Joe

10. ### JoeGuest

Hi Costas,

I did that and came up with about 6ohms for a rechargeable NiMH (7.2V). I am
using a different battery for my application so I will measure that one
when it is fully charged and possibly plot the values for a learning
experience. Thanks for the tip

Joe

11. ### JoeGuest

Hi David,

I was wondering that myself. I went to the site, thanks for the link. The
graphs they show are very instructive, but I don't know what's up with the
impedance at 1Khz, maybe someone else on this forum can explain it...

Joe

12. ### John PopelishGuest

That sounds quite reasonable to me. That would correspond to a >1 amp
short circuit current.
That one should measure well under and ohm.