Measuring amps on 9 volt battery

[George]

Thanks John and Peter for this great info.
I hope others will benefit as I have.

 

[JeffM]

You guys actually have the patience

to do these great "schematic" sketches in your mail reader,
or is there a secret I don't know about that makes it easier ?
John (uvcceet)

: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
http://groups.google.com/groups?&q=andy's-ascii-circuit&meta=group=sci.electronics.*
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?

 

[George]

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

snip

Now to see how the internal resistance of the battery affects the
battery output voltage, let's connect a 20000 ohms-per-volt analog
multimeter across the battery and see what happens. Let's say the
meter has a 10 VDC range. That would make its internal resistance
200000 ohms if we selected the 10VDC range and, with a 1 ohm internal
resistance for the battery we wind up with a reading of:


E1R2 9V * 200000
E2 = ------- = --------------- = 8.999955V
R1+R2 1R + 2000000R

See the voltage falling? That's because more voltage is being dropped
across the battery's internal resistance as it's required to supply
more current.

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

So far, with the voltmeter loads, the drop has been small, but it
starts to matter when you start drawing significant current from the
battery. for example, assume you have a 20 ohm load and that because
you've figured out that

E 9V
I = --- = ----- = 0.45A
R 20R

you expect the 9V battery to deliver 450mA into the load.


Well, if you look at

E1R2 9V * 20R
E2 = ------- = ------------ = 8.57V
R1+R2 1R + 20R

you'll find that, because of the battery's internal resistance you can
only get 8.57V across the 20 ohm resistor, which is only going to
allow about 429mA to flow through the load.

That's also borne out if you do:

E 9V
I = ------- = ----- = 0.42857...A
R1+R2 21R

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?

 

[George]

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

 

[George]

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)?

 

[Active8]

: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

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

Yup. What he said.

The name of the program Fields mentioned is:
Andy's ASCII Circuit by Andreas Weber
http://groups.google.com/groups?&q=andy's-ascii-circuit&meta=group=sci.electronics.*
As you hang around, you will see this appended to some drawings.

No he won't, it's:

|
||-+
||<-
-||-+
|
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

 

[[email protected]]

The name of the program Fields mentioned is:

created by Andy s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

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

Yep. VIrus and spyware free since 1992

John

 

[John Fields]

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




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

---
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.
---

Why is R2 (meter internal resistance) not utilized here?

---
Because it's insignificant compared to the internal resistance of the
battery and the load it's in parallel with, but that's an excellent
observation and you bring up a valid point which we'll get to in a
second.

The meter is still connected.

Yes, it is, and it's in parallel with R2, so the circuit now looks
like this:

+-----+---->E1
| |
| [R1]
+| |
9V +------+--->E2
| | |
| [R2] [R3]
| | |
+-----+------+--->0V

where R3 is the internal resistance of the voltmeter.

R2 and R3 are in parallel, and to determine the total resistance of
the pair we can write:



R2 * R3
Rt = ---------
R2 + R3


Where Rt is the total resistance of the pair.


For your digital multimeter, with a 10M input resistance we'll get:



R2 * R3 20 * 10E6
Rt = --------- = ---------- = 19 999 960 ohms
R2 + R3 20 + 10e6


and for the 20,000 ohms-per-volt analog multimeter we'll get


R2 * R3 20 * 200000
Rt = --------- = ------------- = 19 998 000 ohms
R2 + R3 20 + 200000

The Load is a new type of resistance.
Shouldn't it be represented as R3?

---
It could be, since once we've figured out the total resistance we can
call that value anything we want. But, for the moment, just to keep
us from forgetting that there are really two resistors there let's
keep on calling it Rt.

OK, now if there are no meters in there we'll have:


+-----+---->E1
| |
| [R1]
+| |
9V +---->E2
| |
| [R2]
| |
+-----+---->0V

and if we start off with E1 = 9V, R = 1 ohm and R2 = 20 ohms,

we know from last time that we can write:


E1 * R2 9 * 20
E2 = --------- = -------- = 8.571429V
R1 + R2 1 * 20



Now, if we substitute Rt for R2 we'll have:

+-----+---->E1
| |
| [R1]
+| |
9V +---->E2
| |
| [Rt]
| |
+-----+---->0V


and, if we plug in the value we got for Rt with a 10Mohm DMM in there
we'll have:


E1 * Rt 9 * 19.999960
E2 = --------- = ------------- = 8.571428V
R1 + Rt 1 + 19.999960


If we plug in Rt for the analog voltmeter we'll have


E1 * Rt 9 * 19.998
E2 = --------- = ------------- = 8.57138
R1 + Rt 1 + 19.998



So you can see that things only start to get significant about four or
five places to the right of the decimal because the resistances of the
voltmeters are so large compared to the internal resistance of the
battery.

 

[John Fields]

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?

---
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]

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?

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

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

---
In the sense that it puts it in an endless loop, yes.
---

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

---
No.

Here's what you asked for:

"If I needed to program a robot to calculate
exactly how many amps will flow from a 9 v batttery in seven simple
circuits of only resistors at successive levels (120,000, 12,000,
1200, 120, 12, 1.2, and .12 ohms) what would be the steps?"

Notice that in calci the battery voltage is measured with the load
connected to it. Then, by Ohm's law, all you need to do to get the
current is divide the voltage across the load by the load resistance,
which is what calci does.

 

[Peter Bennett]

R2 and R3 are in parallel, and to determine the total resistance of
the pair we can write:



R2 * R3
Rt = ---------
R2 + R3


Where Rt is the total resistance of the pair.


For your digital multimeter, with a 10M input resistance we'll get:



R2 * R3 20 * 10E6
Rt = --------- = ---------- = 19 999 960 ohms
R2 + R3 20 + 10e6

Watch those pesky periods! That should be 19.999 960 ohms

and for the 20,000 ohms-per-volt analog multimeter we'll get


R2 * R3 20 * 200000
Rt = --------- = ------------- = 19 998 000 ohms
R2 + R3 20 + 200000

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

 

[John Fields]

Watch those pesky periods! That should be 19.999 960 ohms

 

[George]

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 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 112-10) / 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?

 

[John Fields]

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"?
---
Yup...
---

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?

---
They're more accurate.
---

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.

---
Here's what I get for A at 0.06mA:

0.06Ma-->
+---------+---->9.0V
| |
| [R1]
+| |
9V +---->8.18V
| |
| [118800R]
| |
+---------+---->0V

9V
R1 = -------- - 118800R = 31200R
0.06mA



And for B at 0.06mA:


0.06Ma-->
+---------+---->9.0V
| |
| [R1]
+| |
9V +---->8.18V
| |
| [118800R]
| |
+---------+---->0V

9.0V - 8.18V
R1 = --------------- = 13666R
0.06mA


Both widely different from what you got.

They're also different from each other by a factor of greater than
two, so my assumption that they'd be the same was incorrect.
---

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

---
For the very small currents being drawn, perhaps. You may want to try
to see what you get when you load the battery more heavily.
---

Why is Method A very different than B?

---
The only reason I can think of is because the internal battery voltage
isn't really 9V. If that's true, then it must be something that
satisfies both methods A and B in that the internal resistance comes
out the same in both cases. Care to have a crack at finding out what
it is?
---

 

[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)?

 

[Rusty Wright]

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

http://www.ibiblio.org/obp/electricCircuits/

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 someone asks a question demonstrating that they don't know or
understand electronics, I would point them there.

 

[John Fields]

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.
---

 

[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