Page **1** of **1**

### Working out battery internal resistance.

Posted: **Sat Jan 08, 2011 7:24 am**

by **retepsnikrep**

Maths help chaps please.

OK we know to work out IR we measure unloaded voltage, then measure voltage with a load attached. If we know the current flowing we can work out the resistance of the combined load + battery. Then we can dedcut the load R and work out battery IR. Fine

But I don't know the R of the load as i want use the IMA motor to provide the load.

So can I still work out the IR of the battery with known accurate current and voltages but no data for the load?

Any good ideas.

### Re: Working out battery internal resistance.

Posted: **Sat Jan 08, 2011 10:53 am**

by **dillond666**

So can I still work out the IR of the battery with known accurate current and voltages but no data for the load?

Somehow I would doubt it. That lets you work out power but you still won't know the load and supply impedances.

Using a motor for the load makes words like impedance, inductance and saturation pop into my head.......scary.

Bearing in mind I only know enough to be (very) dangerous, I would be tempted to abort mission and use some cooker or kettle elements

for your load as a resistive load sounds easier to me.

Derek

### Re: Working out battery internal resistance.

Posted: **Sat Jan 08, 2011 12:09 pm**

by **Jeremy**

If you can measure the current flowing from the battery and the battery terminal voltage (right at the battery terminals) then that is all you need to work it out.

Here's the procedure:

Connect an accurate voltmeter directly to the battery terminals and an accurate ammeter in series with battery power cable. It's important that you measure the voltage at the actual battery terminals, and not at the connecting lead side of them, because there may well be a small resistance at the lead connection.

Apply a load to the battery that draws a typical operating current, just for a few moments to bleed off any surface charge.

Turn off the load and record the battery voltage.

Apply the load again and record both the battery voltage and the load current at the same instant.

Do the following calculation:

Ri = (V1 - V2) / I

where:

Ri = the battery internal resistance in ohms

V1 = the initial voltage with no load

V2 = the voltage under load

I = the current under load

Here's a worked example, with some made up values:

V1 = 102.0 V

V2 = 99.85 V

I = 55.6 A

Ri = (102.0 - 99.85) / 55.6 = 0.0387 ohms

You'll find that Ri will vary with temperature and state of charge, BTW, so a range of measurements made under different conditions will give you a feel for the typical value.

Jeremy

### Re: Working out battery internal resistance.

Posted: **Sat Jan 08, 2011 1:17 pm**

by **dillond666**

Jeremy,

Easy when you know how

Another useful nugget learnt on this excellent forum.

Derek

### Re: Working out battery internal resistance.

Posted: **Sun Jan 09, 2011 6:07 am**

by **retepsnikrep**

Cheers Jeremy. It's like having a wise old man nurturing his offspring. No offence intended

### Re: Working out battery internal resistance.

Posted: **Sun Jan 09, 2011 9:10 am**

by **Jeremy**

Not that wise, I'm afraid, and not that old either (although I am now retired.........).

This method isn't super accurate, as it doesn't take true account of the effect of pulling pulsed power from the battery (to do that needs a measurement with a super-imposed AC component) but it's good enough for practical purposes and has the advantage of being easy to do.

If you want to do automated measurements on the fly, perhaps using a data logger or microcontroller measurement system, then you don't need to always do the off-load voltage measurement. The method will work by comparing the voltage and current drawn under different conditions, for example, if you have two measurements, one at 50A and 98.5V and one at 75A and 97.7V, then the formula is:

Ri = (V1 - V2) / (I2 - I1)

where:

V1 = voltage at current I1

V2 = voltage at current I2

For the example given this gives:

Ri = (98.5 - 97.7) / (75 - 50) = 0.032 ohms

Jeremy