Measuring amps on 9 volt battery

[George]

At http://www.kadentech.com/ they offer Model 3365 to actually measure
battery Internal Resistance. It costs about $500.

The maximum reading is 39.99 ohms.
Why would such an expensive meter not be able to read the IR ranges in
our tests?

I noticed that in my test, there was a difference of only .02v drop
(8.18 to 8.16) when load was connected.

Your battery showed a larger .82v drop (9.00 to 8.18).
Any idea why this difference?

As a beginner at this, I'm a little confused.

My meter seems to have fresh batteries, but I'm not sure if it's
providing accurate info.

Do any meters provide more decimals on the mA reading (eg. .1234 mA)?

 

[John Fields]

On 5 Nov 2004 14:24:00 -0800, tek1940@hotmail.com (George) wrote:

At http://www.kadentech.com/ they offer Model 3365 to actually measure
battery Internal Resistance. It costs about $500.

The maximum reading is 39.99 ohms.
Why would such an expensive meter not be able to read the IR ranges in
our tests?

---
Because a couple of things were assumed which shouldn't have been; one
being that the ammeter you used has zero internal resistance and the
other being that the voltmeter you used has in infinite internal
resistance. Neither assumption is true.
---

I noticed that in my test, there was a difference of only .02v drop
(8.18 to 8.16) when load was connected.

Your battery showed a larger .82v drop (9.00 to 8.18).
Any idea why this difference?

---
I had thought that the 8.18V you mentioned was the voltage across the
load and that the unloaded voltage of the battery was 9V.
---

As a beginner at this, I'm a little confused.

---
Well, then, do you want to start over, from the bottom?

If you do, how about posting what you have for test equipment,
including the values and tolerances of the resistors you've been
using.

And, spring for a fresh alkaline 9V battery. (It should read about
9.5V with only the voltmeter across it.
---

My meter seems to have fresh batteries, but I'm not sure if it's
providing accurate info.

---
The only way you can find out if it is or not is to use it to measure
known quantities or send it out for calibration, which is what the cal
shop will do.
---

Do any meters provide more decimals on the mA reading (eg. .1234 mA)?

---
Yes.
---

--
John Fields

 

[George]

I found several other ways to calculate the battery Internal
Resistance, in addition to the special $500 meter that directly
measures.

Some are more complicated and apparently need special equipment most
people don't have. See http://www.eveready.com/ Technical Info,
Application Manuals, Alkaline, pages 7-9.

The formula in Radio Shack book "Using Your Meter" (1994), page 4-24
is
shown below as METHOD C. It's similar to John Fields' METHOD B but
uses
Calculate I instead of Measure I.


Variable/Formula
-----------------
METHOD A (John)
Measure No-Load Voltage Enl
Measure Resistance R
Measure Current…
…convert to amps I
Calculate No-Load Resistance Rnl = Enl / I
Calculate Internal Resistance Ri = Rnl - R

METHOD B (John)
Measure No-Load Voltage Enl
Measure Load Voltage E
Calculate Voltage Difference DE = Enl - E
Measure Current…
…convert to amps I
Calculate Internal Resistance Ri = DE / I

METHOD C (Radio Shack book)
Measure No-Load Voltage Enl
Measure Resistance R
Measure Load Voltage E
Calculate Voltage Difference DE = Enl - E
Calculate Current Ic = E / R
Calculate Internal Resistance Ri = DE / Ic

I completed tests with a "9v" battery and seven single resistors in
this sequence:
680, 470, 390, 330, 180, 150, and 100.

(I in mA)
680: Enl=8.15, R=665, I=12.00, E=7.98, DE=.17
470: Enl=8.16, R=461, I=17.20, E=7.93, DE=.23
390: Enl=8.13, R=384.9, I=20.69, E=7.88, DE=.25
330: Enl=8.10, R=327, I=24.28, E=7.84, DE=.26
180: Enl=8.10, R=178.4, I=43.30, E=7.67, DE=.43
150: Enl=8.09, R=148.7, I=51.60, E=7.6, DE=.49
100: Enl=8.02, R=98.9, I=76.00, E=7.44, DE=.58

Ri=
680: A=14.1667, B=14.1667, C=14.1667
470: A=13.4186, B=13.3721, C=13.3707
390: A= 8.0435, B=12.0831, C=12.2113
330: A= 6.6079, B=10.7084, C=10.8444
180: A= 8.6670, B= 9.9307, C=10.0016
150: A= 8.0829, B= 9.4961, C= 9.5872
100: A= 6.6263, B= 7.6316, C= 7.7099

As mentioned in a prior post, I'm using an inexpensive MM that shows
only two decimals for voltage and current.
Using Resistors in this low range (100 - 680) provides good comparison
results, whereas Resistors in higher values require more decimals
because the Differences (DE) become much smaller.

Observations:
1. Methods A, B, and C produce results fairly close.
2. As R decreases, I increases as expected (I = E / R).
3. As I increases, Ri decreases.
4. As I increases, DE (voltage difference) increases.

This provides some insight into battery operation with a small range
of
resistance loads.

The above-mentioned Everready web page document (page 7) states:
"While
the absolute Ri will vary with the load, ... The Ri of a cylindrical
Alkaline battery remains relatively constant until it approaches end
of
service life and then increases rapidly as shown in the following
diagram:"

This indicates possibly two kinds of Ri:
1. Absolute Ri that varies with the load.
2. Ri that starts at a certain level for each cell type and only
increases.

The same Ri cannot operate in two different ways. Why isn't there a
different name for each, and formulas to explain them?

BTW, the RS book page 4-23 "Testing under load" suggests if a
battery's voltage is measured outside of its normal load, it should
have these minimum load resistors:
D cell - 10 ohms
C cell - 20 ohms
AA cell - 100 ohms
9-volt battery - 330 ohms