Calculating Battery Impedance

[Zooty]

Hi all, i came across this on the internet. Most of you probably know this but for those who don't, here is a method for calculating the impedance of a battery without special equipment. I don't know how reliable or accurate it is but apparently it is a popular method from what i have seen.

What you need to do is measure and record the open circuit voltage of the battery, Then you need to load it with a resistor. Use one that draws a good bit of current (50 ohm, 100 ohms, etc) and measure and record the voltage again. The difference between the two readings is the number we need for the math below. (You might have to do this second measurement fast so as not to burn up your resistor unless you have a big power resistor. Remember that P=V^2/R which for 50 ohms would be 12^2/50 or almost 3 watts; so be fast and careful if you use a resistor with a smaller power rating).

OK now the math. The internal resistance of the battery actually froms a resistive voltage divider with the external resistor you load it with. This is what causes the voltage drop. The standard equation for this voltage drop is:

Vdrop = Vbatt*(Rbatt/(RL+Rbatt))

Where:
Vdrop is the difference in your two readings above
Vbatt is the open circuit voltage measured above
RL is the resistor you load the battery with
Rbatt is the battery internal resistance you're looking for

If you do a bit of algebra and solve the equation above for Rbatt, you get the following equation:

Rbatt = Vdrop*(RL/(Vdrop+Vbatt))

Just plug the numbers you measured above into this equation and you'll have your answer.

Here's an Example:

Vbatt = 12V
When loaded with 500 ohms the battery voltage drops to 11.9 V ... or Vdrop = 0.1V

Plug the numbers in:

Rbatt = 0.1*(500/(0.1+12))

Rbatt = 4.1 ohms

[Sephiroth]

A bit trickier with the the batteries we are using

It may work in theory, but in practice I don't think you can get an accurate measurement using this method.

My 24ah batteries are rated at 9.5 milliohms (according to the datasheet for the batteries) which is a very hard figure to measure accurately since it is so close to zero. Most lead acid batteries' internal resistance is in the milliohms.

[Zooty]

oh well Expensive equipment it is!

[Ted Ewert]

"Battery impedance" is somewhat of a misnomer. Charging impedance and internal resistance are two different things. This is because a battery's charging impedance is calculate as that in which optimal charging power is matched to. Internal resistance is usually calculated as an output capacity.
I've asked the question of how to calculate this number to Bedini, as well as to other knowledgeable people, but have never received a satisfactory answer. Perhaps the above equation is correct, I don't know.

Ted

[sucahyo]

Here is my way. Put the battery in a radiant charger load, measure the voltage, and then replace it with resistor and change it's value until the measured voltage same as when battery used as load.

If anyone already post this, please let me know.

It has been mentioned often that the coil must have the same impedance as the battery. Matching imedance improve circuit efficiency. But some of us can't think of a way to measure battery impedance, atleast I do.

Here is my idea for impedance matching procedure. We know that with high impedance load, the output current will be lower, with low impedance load, the output current will be higher. We can use this to match our coil impedance with our battery impedance.

Since using multimeter to measure impedance may give conversion error, the best way is to use the coil directly as load and measure the current output. It would be better if the multi meter show more precise current value, but I think any meter will do since radiant circuit is sensitive to load.

We do the comparative test by replacing the charged battery with our radiant coil candidate. The output current has to stay the same if both has the same impedance. reduce the wire length if output current is higher, increase if lower.

For the best result I think the circuit has to has stable radiant output. This means mechanical radiant oscillator may not be good for this kind of test. Solid state is the viable option. But since this field is new, any other type may be tried.

The test circuit frequency or duty cycle or input current should be the same with the candidate circuit, maybe anything else too. Same circuit would be best. We use another coil for test coil.

[ksmarke]

I was checking the math of the Rbatt equation and I found an error.

From the standard Voltage Drop Equation:
Vdrop = (Vbatt * Rbatt) / (RL + Rbatt)

You rearrange the equation as follows until you solved for Rbatt.

Vdrop = (Vbatt * Rbatt) / (RL + Rbatt)

Vdrop * (RL + Rbatt) = (Vbatt * Rbatt)

(Vdrop * RL) + (Vdrop * Rbatt) = (Vbatt * Rbatt)

(Vdrop * RL) = (Vbatt * Rbatt) - (Vdrop * Rbatt)

(Vdrop * RL) = Rbatt * (Vbatt - Vdrop)


So Rbatt = (Vdrop *RL) / (Vbatt - Vdrop).