Measuring amps on 9 volt battery

[George]

Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

 

[Jason Rosinski]

There is an internal source resistance to the battery. It cannot supply current faster than a
certain maximum rate. In this case it looks to be ~250Ohms. Think of it as a resister in series
with the battery internals.

George wrote:

Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

 

[George]

I'm not understanding this.
Are you saying there is a Resistor within the battery?

The "decreasing value" behaviour is not described in the MM manual.
It does say "The maximum input limit for DC/AC current measurement is 400 mA."
As shown below, I calculate with a 150 ohm R, the MM should read 60 mA.

I probably should question Radio Shack.



Jason Rosinski <Jason.DOT.Rosinski@zarlink.com> wrote in message news:<41816056.4080101@zarlink.com>...

There is an internal source resistance to the battery. It cannot supply current faster than a
certain maximum rate. In this case it looks to be ~250Ohms. Think of it as a resister in series
with the battery internals.

George wrote:
Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

 

[Bob Masta]

On 28 Oct 2004 13:06:30 -0700, tek1940@hotmail.com (George) wrote:

Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

Others have pointed out about internal battery resistance.
I'd like to point out that if you put the ammeter leads directly
across the battery, the current may easily exceed the limits of the
meter if the battery is *not* old and nearly dead. In that
case you will pop a little fuse which is typically hard to get
at in cheap meters, and is usually some bastard size (physical
size and electrical size) you don't have on hand.

So, as a good rule of thumb, *never* put ammeter leads across a
battery or other power source. Always insert the ammeter in series
with a branch of the circuit being tested.

Saves wear and tear on meters, and on your nerves!

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

 

[Tom Biasi]

"Bob Masta" <NoSpam@daqarta.com> wrote in message
news:4182387c.3151803@news.itd.umich.edu...

On 28 Oct 2004 13:06:30 -0700, tek1940@hotmail.com (George) wrote:

Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

Others have pointed out about internal battery resistance.
I'd like to point out that if you put the ammeter leads directly
across the battery, the current may easily exceed the limits of the
meter if the battery is *not* old and nearly dead. In that
case you will pop a little fuse which is typically hard to get
at in cheap meters, and is usually some bastard size (physical
size and electrical size) you don't have on hand.

So, as a good rule of thumb, *never* put ammeter leads across a
battery or other power source. Always insert the ammeter in series
with a branch of the circuit being tested.

Saves wear and tear on meters, and on your nerves!

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis

At the level the poster appears to be I should have mentioned that Bob, glad
that you did.


> www.daqarta.com

 

[George]

Thanks Tom and others for your info.
I guess this poster (me) is at a beginner level.
Learning OHM's law, relationship between each value and trying to
confirm with a 9V battery, resistors and MM.

Using letter R to represent ohms, showing Law and actual MM readings
at decreasing levels of magnitude:

Law: .06 mA = 8.2 V / 118,500 R
MM: .06 mA

Law: .556 mA = 8.2 V / 14/730 R
MM: .556 mA

Law: 3.79 mA = 8.2 V / 2162 R
MM: 3.75 mA (99% of Law)

Law: 55.40 mA = 8.2 V / 148 R
MM: 52.1 mA (94% of Law) ***

Law: 83.25 mA = 8.2 V / 98.5 R
MM: 76.5 mA (92% of Law) ***

Law: 138 mA = 8.2 V / 59.4 R
MM: 120 mA (87% of Law) ***

Law: 332 mA = 8.2 V / 24.7 R
MM: 253 mA (76% of Law) ***

*** At these levels, the MM showed initial 473 mA then the number
shown here then decreasing values. At the 24.7 R test, the rate of
decrease was in larger increments than at 148 R.

Did not proceed with lower R values because MM manual says it can read
max 400 mA.
In bottom two levels, I learned how to put resistors in Parallel and
calculate the net resistance. (98.7 and 148.7 for 59.4 R; 4 x 98.7
for 24.7 R)

Q1. Is the apparent error percentage due to my MM or would all meters
show these variances?

Q2. Same question but regarding the "decreasing value" behavior.

Q3. If the meter could read larger amounts of current, and I used
lower amounts of R, the Law predicts higher current.
Law: 4.1 A = 8.2 V / 2 R
Can an 8.2 V battery produce 4.1 A?
Is there an additional math factor needed in the Law to filter out
"impossible" results?

Q4. Continuing to lower R to Almost zero (metal of the MM lead wires),
if the MM can read up to 400 mA, why isn't it safe to measure the
current of a 9 V battery? Is there another way to find available
current?

Thanks for your patience.




"Tom Biasi" <tombiasi@REMOVETHISoptonline.net> wrote in message news:<ezxgd.11344$UC4.6157193@news4.srv.hcvlny.cv.net>...

"Bob Masta" <NoSpam@daqarta.com> wrote in message
news:4182387c.3151803@news.itd.umich.edu...
On 28 Oct 2004 13:06:30 -0700, tek1940@hotmail.com (George) wrote:

Using a 9 volt battery, and Radio Shack Pen Style MM.
If I use one Resistor with a high value (eg. 15K), the MM shows .6mA,
which appears OK.
9V/15000R=.6mA

But when I use a R with small value (eg. 150), the MM shows about a
22mA value, then steadily decreasing values.
9V/150R=60mA

What should the MM show if the leads are put on the battery directly,
with very little R?
9V/.0001R=90,000A, which is impossible for the battery, correct?

On a used 9V battery the MM shows 37mA, then decreasing.

What causes the MM to show the apparently wrong and decreasing values?

Thanks

Others have pointed out about internal battery resistance.
I'd like to point out that if you put the ammeter leads directly
across the battery, the current may easily exceed the limits of the
meter if the battery is *not* old and nearly dead. In that
case you will pop a little fuse which is typically hard to get
at in cheap meters, and is usually some bastard size (physical
size and electrical size) you don't have on hand.

So, as a good rule of thumb, *never* put ammeter leads across a
battery or other power source. Always insert the ammeter in series
with a branch of the circuit being tested.

Saves wear and tear on meters, and on your nerves!

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis



At the level the poster appears to be I should have mentioned that Bob, glad
that you did.


www.daqarta.com

 

[George]

Thanks John,

John Fields <jfields@austininstruments.com> wrote in message news:<bk97o0tvmm7ra6497kuak2526fnil3t54s@4ax.com>...

snip
---
Since you haven't mentioned anything to the contrary, it seems you're
_assuming_ that the battery voltage will remain constant at 8.2V
regardless of the load on it. It won't.

This is the method I've been using:
1. Measure all the components separately (no load).
2. Using Ohm's Law, calculate the expected result.
3. Connect the circuit
4. Measure with a MM.
I did the first and last tests again and measured the battery under
load.
In the first test (highest R), the battery under load showed same 8.2
V.
In the last test (lowest R), it showed 6.15 V.
Question: Does the Law refer to unloaded or loaded voltage source?

You also seem to be having trouble with the concept of the internal
resistance of the battery. Consider it a variable resistance in
series with the battery, located in the same housing as the battery,
but which you can't physically get to.

But, you _can_ measure it.
snip

I understand there is internal resistance in the battery. But I
assumed it is internal and that whatever is produced by the battery is
*after* the fact. (Example, say my company makes a product. We buy a
component widget. The widget company may have various "internal
resistance" to make it, but when it arrives at our company, all we
care is that meets our specs.)

I did a search on "battery internal resistance", found this quote: "A
battery's internal impedance increases with decreasing capacity due to
various conditions such as age, ambient temperature, discharge history
etc."

Note it does not say it changes according to the load.

Instruments that measure battery internal resistance have an upper
range of 40 ohms. Other web pages had graphs showing an upper limit of
420 milli ohms, I don't know what size battery they were talking
about, possibly those used in a cell phone.

Also note that my MM is an inexpensive model. In the first test I
reported that it showed .06 mA. But in my second test I noticed it
also showed .05 and .07. My point is that some of my data may not be
accurate enough.

This is a great learning experience.

 

[Peter Bennett]

On 31 Oct 2004 06:40:28 -0800, tek1940@hotmail.com (George) wrote:

Thanks John,

John Fields <jfields@austininstruments.com> wrote in message news:<bk97o0tvmm7ra6497kuak2526fnil3t54s@4ax.com>...
snip
---
Since you haven't mentioned anything to the contrary, it seems you're
_assuming_ that the battery voltage will remain constant at 8.2V
regardless of the load on it. It won't.


This is the method I've been using:
1. Measure all the components separately (no load).
2. Using Ohm's Law, calculate the expected result.
3. Connect the circuit
4. Measure with a MM.
I did the first and last tests again and measured the battery under
load.
In the first test (highest R), the battery under load showed same 8.2
V.
In the last test (lowest R), it showed 6.15 V.
Question: Does the Law refer to unloaded or loaded voltage source?

When you use Ohm's Law to calculate the current through a resistor,
you must use the voltage across the resistor at the time the current
is flowing - that is the only voltage the resistor knows about.

I understand there is internal resistance in the battery. But I
assumed it is internal and that whatever is produced by the battery is
*after* the fact. (Example, say my company makes a product. We buy a
component widget. The widget company may have various "internal
resistance" to make it, but when it arrives at our company, all we
care is that meets our specs.)

The internal resistance of the battery is indeed inside the battery,
but it is still in series with the battery terminals, so it will cause
the battery terminal voltage to drop when current is drawn from the
battery.

I did a search on "battery internal resistance", found this quote: "A
battery's internal impedance increases with decreasing capacity due to
various conditions such as age, ambient temperature, discharge history
etc."

Note it does not say it changes according to the load.

Yes it does: "discharge history, etc.".

The battery operation depends on chemical reactions. When you draw
current from the battery, "used" chemicals may accumulate on the
electrodes, and obstruct further reactions. At low currents, these
used chemicals can diffuse back through the electrolyte, so they have
little effect on the battery operation. At high currents, the used
chemicals are created faster than they can diffuse, so they have a
greater detrimental effect on the battery operation.

Instruments that measure battery internal resistance have an upper
range of 40 ohms. Other web pages had graphs showing an upper limit of
420 milli ohms, I don't know what size battery they were talking
about, possibly those used in a cell phone.

A lead-acid battery (car battery) will have a very low internal
resistance. A small battery, like the tiny cells in a 9 volt battery,
will have a much greater internal resistance. Both the cell size and
chemistry affect the internal resistance.

Also note that my MM is an inexpensive model. In the first test I
reported that it showed .06 mA. But in my second test I noticed it
also showed .05 and .07. My point is that some of my data may not be
accurate enough.

The specs for most digital meters say "+/- 1 count". If my
meter,which will read to .0001 mA,reads .0600 mA, yours, reading to
only .01 mA, can legitimately read .05, .06 or .07 mA.



--
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

 

[George]

snip

Question: Does the Law refer to unloaded or loaded voltage source?

When you use Ohm's Law to calculate the current through a resistor,
you must use the voltage across the resistor at the time the current
is flowing - that is the only voltage the resistor knows about.

If I measure the battery at 8.2 V, can you not assume that is what is
realized in the circuit?
If you're taking an exam and you calculate an answer based on the
unloaded battery (before connecting to the circuit), will the answer
be correct?

I understand there is internal resistance in the battery. But I
assumed it is internal and that whatever is produced by the battery is
*after* the fact. (Example, say my company makes a product. We buy a
component widget. The widget company may have various "internal
resistance" to make it, but when it arrives at our company, all we
care is that meets our specs.)

The internal resistance of the battery is indeed inside the battery,
but it is still in series with the battery terminals, so it will cause
the battery terminal voltage to drop when current is drawn from the
battery.

Not understanding this yet. If you measure the V at the battery's
external posts, does this not include the internal resistance? Does
the internal resistance only get activated when a load is connected?
I'd like to see a circuit diagram of the innards, showing the path
from one post to the other, thru the resistance.
Evidently I missed "Batteries 101".

I did a search on "battery internal resistance", found this quote: "A
battery's internal impedance increases with decreasing capacity due to
various conditions such as age, ambient temperature, discharge history
etc."

Note it does not say it changes according to the load.

Yes it does: "discharge history, etc.".

Several web sites I've read indicate the internal resistance only
increases with age; a higher discharge history would increase the
resistance faster.
I didn't see indications that it's a variable resistance changing with
the load, as John stated.

The battery operation depends on chemical reactions. When you draw
current from the battery, "used" chemicals may accumulate on the
electrodes, and obstruct further reactions. At low currents, these
used chemicals can diffuse back through the electrolyte, so they have
little effect on the battery operation. At high currents, the used
chemicals are created faster than they can diffuse, so they have a
greater detrimental effect on the battery operation.

Instruments that measure battery internal resistance have an upper
range of 40 ohms. Other web pages had graphs showing an upper limit of
420 milli ohms, I don't know what size battery they were talking
about, possibly those used in a cell phone.

A lead-acid battery (car battery) will have a very low internal
resistance. A small battery, like the tiny cells in a 9 volt battery,
will have a much greater internal resistance. Both the cell size and
chemistry affect the internal resistance.

Do you know how much internal resistance there is on a new 9 V
battery?
Why would it be higher than a 12 V car battery?
I've sent an inquiry to a 9 V manufacturer; it's not stated on their
data sheets.
I'm not convinced a 9 V battery could have 18167 ohms.

 

[George]

Thanks John,
I was just reporting more info, no offense intended.
With your info and suggestions, I'll be learning more.

I'm sure all of this will sink in eventually; I'm one of those "slow"
learners, needing to know more details. (another example - I feel the
need to know what materials are utilized in the battery to provide 1.5
volts and how all those amps are stuffed in there, and how they get
out. I force myself to accept it as a "black box" and proceed with my
projects.)

Perhaps my basic question about measuring amps will prove to be
unimportant;
it's just that I saw unusual patterns and behavior; did not know if it
was due to my MM or something else.
The question has brought up others, like how to measure the battery's
amp capacity (how many amps are left?). I guess this cannot be done
with a cheap MM; just use the related Voltage capacity, since it
degrades at about the same rate.

It does seem important to know the basics before moving on to more
complicated circuits.

My career has been in computers, but at 64 I'm finally getting to
learn the electronics side. One of my first electronic projects is a
simple robot, to follow a line. "Robot Building for Beginners" by
David Cook.

Think of it this way. 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?
Pretend the robot is in an assembly line. It can measure each
component with an expensive/precise MM before making the simple
circuit.




John Fields <jfields@austininstruments.com> wrote in message news:<4jrao05ek34vgh443222q9e5m49051jpdn@4ax.com>...

On 31 Oct 2004 06:40:28 -0800, tek1940@hotmail.com (George) wrote:

Thanks John,

John Fields <jfields@austininstruments.com> wrote in message news:<bk97o0tvmm7ra6497kuak2526fnil3t54s@4ax.com>...
snip
---
Since you haven't mentioned anything to the contrary, it seems you're
_assuming_ that the battery voltage will remain constant at 8.2V
regardless of the load on it. It won't.


This is the method I've been using:
1. Measure all the components separately (no load).
2. Using Ohm's Law, calculate the expected result.
3. Connect the circuit
4. Measure with a MM.
I did the first and last tests again and measured the battery under
load.
In the first test (highest R), the battery under load showed same 8.2
V.
In the last test (lowest R), it showed 6.15 V.
Question: Does the Law refer to unloaded or loaded voltage source?

---
It refers to the current which will be forced to flow through a
resistance with a known voltage across it. _Measure_ the voltage
across the resistance, then divide that voltage by the resistance of
the resistor and you'll have the current flowing through the resistor.

If you don't, and you make the assumption that a 9V battery will stay
at 9V regardless of whether 1 milliamp or half an amp is being taken
from it you'll be wrong.
---



You also seem to be having trouble with the concept of the internal
resistance of the battery. Consider it a variable resistance in
series with the battery, located in the same housing as the battery,
but which you can't physically get to.

But, you _can_ measure it.
snip

I understand there is internal resistance in the battery. But I
assumed it is internal and that whatever is produced by the battery is
*after* the fact.

---
You still don't understand. It's not after the fact, it's part of the
deal.

The internal resistance is the same as a resistor that you can't get
rid of that's in series with the battery, and since the current in
series resistances is the same in all the resistances, the internal
resistance will drop some voltage when the load is drawing current,
and that voltage won't be available for the load to get the current it
needs.


Let's say you have a 9 volt battery with a 1 ohm internal resistance
hooked up to some load. It'll look like this:


+------------------------+
+-|-[9V BATTERY+]--[1 OHM]-|-----+
| +------------------------+ |
| [LOAD]
| |
+--------------------------------+


Now let's say the load is 9 ohms. That'll make the circuit look like
this:

+------------------------+
+-|-[9V BATTERY+]--[1 OHM]-|-----+
| +------------------------+ |
| [9R]
| |
+--------------------------------+

Since resistors in series add, and the current in series resistances
is the same in all the resistances, we have 9 volts and 10 ohms, so
the current flowing in the circuit will be

E 9V
I = --- = ---- = 0.9A
R 10R


Now, if your 9 ohm load is a 9 watt light bulb and you expect to get a
certain amount of light from it when you connect it across the 9 volt
battery, you're in for a surprise, since you'll only be able to get
0.9A out of the battery because of its internal resistance.
---

I did a search on "battery internal resistance", found this quote: "A
battery's internal impedance increases with decreasing capacity due to
various conditions such as age, ambient temperature, discharge history
etc."

Note it does not say it changes according to the load.

---
So you don't know what the hell you're talking about, but I'm wrong
because you looked up something that didn't mention it?
Get a goddam clue.
---

Instruments that measure battery internal resistance have an upper
range of 40 ohms. Other web pages had graphs showing an upper limit of
420 milli ohms, I don't know what size battery they were talking
about, possibly those used in a cell phone.

Also note that my MM is an inexpensive model. In the first test I
reported that it showed .06 mA. But in my second test I noticed it
also showed .05 and .07. My point is that some of my data may not be
accurate enough.

---
The problem is that you didn't take _enough_ data. At every point
where you measured the current you should also have measured the
voltage across the resistor.
---

This is a great learning experience.

---
You're welcome.

 

[John Fields]

On 1 Nov 2004 06:03:06 -0800, tek1940@hotmail.com (George) wrote:

Thanks John,
I was just reporting more info, no offense intended.
With your info and suggestions, I'll be learning more.

I'm sure all of this will sink in eventually; I'm one of those "slow"
learners, needing to know more details. (another example - I feel the
need to know what materials are utilized in the battery to provide 1.5
volts and how all those amps are stuffed in there, and how they get
out. I force myself to accept it as a "black box" and proceed with my
projects.)

---
For general info:
http://www.duracell.com/oem/default.asp

For 9V alkaline:
http://www.duracell.com/oem/Pdf/MX1604.pdf

Perhaps my basic question about measuring amps will prove to be
unimportant;
it's just that I saw unusual patterns and behavior; did not know if it
was due to my MM or something else.
The question has brought up others, like how to measure the battery's
amp capacity (how many amps are left?). I guess this cannot be done
with a cheap MM; just use the related Voltage capacity, since it
degrades at about the same rate.

---
You can't even really do that, since remaining capacity varies with
load and if you don't have a valid Voltage Decay VS Load profile it's
all pretty much guesswork. Educated guesswork, but guesswork
nonetheless.
---

It does seem important to know the basics before moving on to more
complicated circuits.

My career has been in computers, but at 64 I'm finally getting to
learn the electronics side. One of my first electronic projects is a
simple robot, to follow a line. "Robot Building for Beginners" by
David Cook.

Think of it this way. 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?
Pretend the robot is in an assembly line. It can measure each
component with an expensive/precise MM before making the simple
circuit.

---

start: measure r

120k: bne 120kR 12k
r = 1.2E5
bra calci

12k: bne 12kR 1200
r = 1.2E4
bra calci

1200: bne 1200R 120
r = 1.2E3
bra calci

120: bne 120R 12
r = 1.2E2
bra calci

12: bne 12R 1r2
r = 1.2
bra calci

1r2: bne 1.2R 0r12
r = 0.12
bra calci

0r12: bne err1
r = 0.12

calci: connect resistor to 9V battery
measure battery voltage
divide battery voltage by resistance
quotient = i
write i to RAM for later use
bra end

err1: resistor out of range
write error flag to RAM

end: bra end
---



--
John Fields

 

[John Fields]

On Mon, 01 Nov 2004 08:47:10 -0700, uvcceet@juno.com wrote:

+--------------------------+
| |_
| +-[9V SOURCE+]--[Rint]--->_|+9V
| | |
| | |
| | |_
| +------------------------>_|-9V
| |
+--------------------------+


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
than it appears?

---
Seems most of us use a program I always forget the name of and the URL
to, but I prefer to do it "by hand" since I think it makes for a more
compact drawing.
---

They are handy, and given the right font do a great job. Just seems rather
labor intensive to me.

---
After you get used to it it's not bad.
---

Hey, if its by hand, and takes as long as I think, kudos to all who do it as
its a big help.

---
Thanks! 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

 

[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%27s-ascii-circuit&meta=group%3Dsci.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)?

---

start: measure r

120k: bne 120kR 12k
r = 1.2E5
bra calci

12k: bne 12kR 1200
r = 1.2E4
bra calci

1200: bne 1200R 120
r = 1.2E3
bra calci

120: bne 120R 12
r = 1.2E2
bra calci

12: bne 12R 1r2
r = 1.2
bra calci

1r2: bne 1.2R 0r12
r = 0.12
bra calci

0r12: bne err1
r = 0.12

calci: connect resistor to 9V battery
measure battery voltage
divide battery voltage by resistance
quotient = i
write i to RAM for later use
bra end

err1: resistor out of range
write error flag to RAM

end: bra end
---

 

[John Fields]

On 2 Nov 2004 21:35:53 -0800, tek1940@hotmail.com (George) wrote:

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.

start: measure r

120k: bne 120kR 12k
r = 1.2E5
bra calci

12k: bne 12kR 1200
r = 1.2E4
bra calci

1200: bne 1200R 120
r = 1.2E3
bra calci

120: bne 120R 12
r = 1.2E2
bra calci

12: bne 12R 1r2
r = 1.2
bra calci

1r2: bne 1.2R 0r12
r = 0.12
bra calci

0r12: bne err1
r = 0.12

calci: connect resistor to 9V battery
measure battery voltage
divide battery voltage by resistance
quotient = i
write i to RAM for later use
bra end

err1: resistor out of range
write error flag to RAM

end: bra end
---

--
John Fields

 

[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]

On 4 Nov 2004 14:47:29 -0800, tek1940@hotmail.com (George) wrote:

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"?

---
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?
---

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?

---
Dunno...


--
John Fields