Measuring amps on 9 volt battery

:One thing that's interesting about it is that :this way schematics can get archived as text files :by outfits that don't archive binary files, like Google, :and yet the data's there, and worth a thousand words. : John Fields

Yup. What he said.

The name of the program Fields mentioned is: Andy's ASCII Circuit by Andreas Weber

* As you hang around, you will see this appended to some drawings.

Text schematics can be done with any text editor (monospaced font--Courier). A text editor that does columnar blocks can make life easy. I use QEDIT, an old DOS editor.

BTW, are you really using OS/2?

John, R1 = battery internal resistance R2 = meter internal resistance

Isn't the current increase due to the lower value of R2 (meter internal resistance)?

Why is R2 (meter internal resistance) not utilized here? The meter is still connected. The Load is a new type of resistance. Shouldn't it be represented as R3?

More questions,

Your example with 20 ohm load shows R1 (battery internal resistance) remaining at 1 ohm. Your earlier post indicated R1 changes according to the load.

I may have missed something, but how do you measure R1?

Thanks

Thanks for the program. I'm not familiar with the language, but I'm guessing "bne 120kR 12k" means branch to 12k if not equal to 120kR?

Does "bra end" in calci not terminate the program?

Wouldn't the program need to consider the factors you described in the other post (R1:battery internal resistance and R2:meter internal resistance)?

*

Thanks for those links. I will be checking them out.

Yep. VIrus and spyware free since 1992 :-)

John

:) And that thousand word savings in bandwidth leaves us plenty of room for off-topic stuff.

*

No he won't, it's:

| ||-+ || BTW, are you really using OS/2?

--
Best Regards,
Mike

resistance)?

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Well, in a manner of speaking.  Before the meter was connected to the
battery there was _no_ current flowing through R1, so battery + was
sitting at 9.0V. Then, when the meter was connected, the infinite
resistance across the + and - terminals of the battery went to 200k
ohms, so current stared to flow out of the battery, through R1,
through the meter, and then back to the battery.  That current flowing
through R1 caused a voltage drop across it, so the meter could only
measure the voltage from the end of R1 not connected to the battery to
battery minus, and that's why it didn't read 9.000000V.

--- You don't, really. You measure the voltage across the load, the current through the load and then, using Ohm's law, calculate what the resistance should be for that current if you had the full battery voltage across the load. The difference between that resistance and the actual load resistance will be the internal resistance of the battery. Also, the difference between the measured voltage and the full battery voltage multiplied by the current should be the internal resistance of the battery. That is, both readings should be the same. I think...

The example was just that, an example. If you calculate the internal resistance with different load currents it'll change.

-- John Fields

--
Yes.  It's just some Motorola 6800 assembler instructions and some
sloppy pseudocode.

Watch those pesky periods! That should be 19.999 960 ohms

And that is 19.998 000 ohms - both such a tiny bit under 20 ohms, that we may as well call it 20 ohms.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

The next to last sentence "Also, the difference between the measured voltage and the full battery voltage multipied by the current should be the internal resistance of the battery." Shouldn't "multipied" be "divided by"? I assumed that modification to agree with Ohm's law where E is divided by either I or R. I'm calling this sentence "Method B".

On my first test, E full = 8.18 volts. R load = 118800 ohms. The meter shows I readings of .05, .06, and .07 mA alternating. Do better meters show more precision?

Using .05mA, the Battery IR is 44,800 ohms by method A, 400 ohms with method B. Using .06mA, the Battery IR is 17,533 ohms by method A, 333 ohms with method B. Using .07mA, the Battery IR is -1943 ohms by method A, 286 ohms with method B.

Considering that an alkaline 9v battery's nominal internal resistance is 1.7 ohms, are these numbers reasonable?

Why is Method A very different than B?

I made a simple Excel spreadsheet with good visibility, with these values in each row starting with row 1: A1 STEP B1 FORMULA C1 CELLS D1 VALUE E1 UNIT F1 VARIABLE

A2 Measure Battery Full B2 C2 2:input D2 (ENTER VALUE HERE) E2 volts F2 E full

A3 Measure Resistor Load B3 C3 3:input D3 (ENTER VALUE HERE) E3 ohms F3 R load

A4 Calc Expected Current B4 E full / R load C4 4:2/3 D4 =D2/D3 (format with 10 decimals) E4 amps F4

A5 Measure Current B5 C5 5:input D5 (ENTER VALUE HERE) E5 mA F5

A6 (milliAmp divisor) B6 C6 6:constant) D6 1000 E6 F6

A7 ...converted to amps B7 C7 7:5/6 D7 =D5/D6 E7 amps F7 I load

A8 Calc R full B8 E full / I load C8 8:2/7 D8 =D2/D7 E8 ohms F8 R full

A9 Calc Battery IR (a) B9 R full - R load C9 9:8-3 D9 =D8-D3 E9 ohms F9 R batint.a

A10 Measure Voltage load B10 C10 10:input D10 ENTER VALUE HERE E10 volts F10 E load

A11 Calc Battery IR (b) B11 (E full - E load) / I load C11 11:(2-10) / 7 D11 =(D2-D10) / D7 E11 ohms F11 R batint.b

(Note: value calculated in row 4 is not utilized here.)

I color the constant and calc cells different to remind me not to enter values there. The spreadsheet makes it easier to calculate and focus on formulas.

Is there a better way to describe spreadsheets in a text file?

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Ooops...

Yup, I straightened it out later on in the post but never came back
and fixed it here.  Thanks. :-)

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Yup...

At

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)?

My favorite "help me learn electronics" site on the web is

When you click on each volume, scroll down and there's a pdf file you can download and read offline. Or print (watch out, Volume 1 - DC is

538 pages). The graphics and diagrams are clearer and better in the pdf files, compared to the online html version.

The author, Tony Kuphaldt, doesn't assume that you know anything and explains things very clearly, which is what I need. I've found them immensely helpful.

Any time some> Have a look at

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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 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

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