Connect with us

Measuring amps on 9 volt battery

Discussion in 'Electronic Basics' started by George, Oct 28, 2004.

Scroll to continue with content
  1. George

    George Guest

    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
     
  2. 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.
     
  3. George

    George Guest

    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.
     
  4. Guest

    Heee hee hee hee!! :)

    Let us know what they say <g>

    John
     
  5. Tom Biasi

    Tom Biasi Guest

    There is a 'resistance' within the battery.

    The battery is not a perfect device, it can not supply voltage with
    unlimited current to a load.

    You seem to know this already. What does limit the battery's ability to
    supply current to a load?

    The main factor is the internal resistance. You are testing the battery at
    levels beyond its capability and are seeing the output drop as you heat up
    the battery and increase the internal resistance even more.

    Regards,

    Tom
     
  6. Graham Knott

    Graham Knott Guest

     
  7. Bob Masta

    Bob Masta Guest

    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
     
  8. Tom Biasi

    Tom Biasi Guest



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

    George Guest

    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.
     
  10. John Fields

    John Fields Guest

    ---
    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.

    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. Consider:

    +----------+
    | |
    +---|----+ |
    | | | |
    | [R1] | |
    | | | [R2]
    | |+ | |
    | [BAT] | |
    | | | |
    +---|----+ |
    | |
    +----------+

    Where R1 is the internal resistance of the battery, BAT is the stuff
    in the battery making the voltage, and the box around them indicates
    that they are inseparable. But, let's go ahead and separate them
    anyway, just for fun, like this:

    +-----+---->E1
    | |
    | [R1]
    |+ |
    [BAT] +---->E2
    | |
    | [R2]
    | |
    +-----+---->

    Now, let's assume that the stuff in the battery makes BAT a true
    voltage source which can supply an infinite current into a zero ohm
    load. If that's true, then the current BAT can supply into R1 and R2
    will only be limited by the sum of the resistances according to Ohm's
    Law:

    E
    I = ---
    R

    which, in this case, becomes


    E1
    I = ---------
    R1 + R2


    Using your data, we can say that in all cases E1 will be 8.2V, and if
    we retabulate the data for convenience we get:

    VOLTS mA OHMS
    E1 I R2
    -------|-------|--------

    8.2 0.06 118500

    8.2 0.556 14730

    8.2 3.75 2162

    8.2 52.1 148

    8.2 76.5 98.5

    8.2 120 59.4

    8.2 253 24.7


    If we rearrange I = E/R to solve for resistance we get:


    E
    R = ---
    I

    and if we plug in the first set of values you got we get:


    E 8.2V
    E = --- = ------ = 136667 ohms
    I 60µA


    Notice that the resistance you used to get that 60µA was 118500 ohms,
    some 18167 ohms short of what Ohm's law says it should be, and since
    Ohm's law doesn't lie, where did that resistance come from?

    From inside the battery, and it's the resistance we're calling R1

    Continuing in the same vein and filling out the rest of the table, we
    wind up with:

    VOLTS mA OHMS OHMS OHMS
    E1 I RT R2 R1
    -------|------|--------|--------|-------

    8.2 0.06 136667 118500 18167

    8.2 0.556 14748 14730 18

    8.2 3.75 2186 2162 24

    8.2 52.1 157 148 9

    8.2 76.5 107 98.5 8.5

    8.2 120 68.3 59.4 8.9

    8.2 253 32.4 24.7 7.7


    Notice that R1 is the internal resistance of the battery, and notice
    that it changes with load.
    ---
    ---
    A1. Different ammeters would show the same variances if they all had
    the same internal resistances.
    ---
    ---
    A2. Same answer
    ---
    ---
    A3A. If it has a low enough internal resistance and the resistance of
    the load is also low enough, yes.
    ---
    ---
    A3B. No, just proper reasoning.
    ---

    ---
    A4A. The reason you shouldn't connect an ammeter directly across a
    battery is because the internal resistance of the battery may be very
    low, which will allow a large current to flow through the meter. Then,
    if the meter isn't protected by a fuse, it could be damaged.
    ---
     
  11. George

    George Guest

    Thanks John,

    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?

    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.
     
  12. 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.

    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.
    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.
    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.
    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
     
  13. John Fields

    John Fields Guest

    ---
    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 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.
     
  14. George

    George Guest

    snip

    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?


    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".

    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.

    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.
     
  15. No - the actual battery voltage once you connect a load will depend on
    the current drawn by the load, and the age of the battery.
    Exams are not real life - on exams you can probably assume that you
    are given an ideal voltage source, which will provide the specified
    voltage, regardless of how much current you try to draw.
    It would definitely be much higher than a car battery, but I don't
    know by how much - and I don't want to risk damaging my meter (or
    killing a battery) trying to find out.

    A car battery has very large plates, so that it can deliver the 200
    amps or so that are required to operate the starter without the
    voltage dropping too much.

    That sounds very much too high for a fresh battery.

    I'd guess you might be able to get 250 mA or so through a short
    circuit - that would be about 36 ohms. However, if you attempt to
    draw 250 mA from a 9 volt battery, the chemicals near the plates will
    be rapidly depleted, so the internal resistance will rise, and the
    current and voltage will drop. If you let the battery rest for a
    while, it will recover somewhat, so you will be able to draw a fairly
    high (but probably not so high) current for another short time.


    --
    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
     
  16. George

    George Guest

    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.




     
  17. John Fields

    John Fields Guest

    ---
    Only if you measure the battery voltage when the battery is under
    load.
    ---
    ---
    No, but if the the purpose of the exam is just to see whether or not
    you know how to use Ohm's law, it won't matter. However, if the
    purpose of the exam is to see whether you know what internal
    resistance is about, you'll flunk! ;)
    ---
    ---
    AHA!!! There it is...

    Yes. it does, but you don't see it because the resistance of the meter
    is so much higher than the internal resistance of the battery.

    Taking a look at this circuit,

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

    the 9v source and Rint represent the battery and its internal
    resistance, and the box around them represents the case; just like a
    regular 9V battery is put together.

    Now, let's say that Rint is one ohm and that you've got your voltmeter
    hooked from +9V to -9V, like this:

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

    Let's also say that the internal resistance of your voltmeter is 10
    megohms.

    Then, for convenience, we can redraw the circuit to look like this:


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

    Where R1 is the internal resistance of the battery, R2 is the internal
    resistance of the meter, E1 is the voltage from the 9V source, and E2
    is the voltage across the meter.


    Now, if we calculate the current being drawn from the battery, we can
    say:

    E 9V
    I = ------- = ------------- = 0.00000089999991A = 899.9991nA
    R1+R2 10,000,001R

    and if R2 is equal to 10 megohms, the voltage the meter will see will
    not be 9.0V, it'll be 8.9999991V, because the other 0.0000009V will
    have been dropped across R1, the battery's internal resistance.

    Considering that R1R2 is a voltage divider, there's a much more
    convenient way to find out what the voltage across the meter will be,
    and that's to use


    E1R2
    E2 = -------
    R1+R2


    For the 10 Megohm case you'll get:


    E1R2 9V * 10000000R
    E2 = ------- = ----------------- = 8.9999991V
    R1+R2 1R + 10000000R

    which is just what we got before.


    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.

    If we use a voltmeter with a 1000 ohm iron vane movement to measure
    the battery voltage, it'll draw 9mA from the battery when it's
    connected and it'll read:


    E1R2 9V * 1000R
    E2 = ------- = ------------ = 8.991001V
    R1+R2 1R + 1000R


    The reading dropped again because of the higher current flowing
    through the battery's internal resistance, and the higher the load
    current goes, the higher the drop across the internal resistance will
    be.

    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
    ---
    No, it's always there but it doesn't matter until you start taking
    current out of the battery. Kind of like even though the electricity
    from the power company is always there, on the other side of the
    switch, you don't get charged for it until you turn on the switch and
    start using it.
    ---
    ---
    Read what follows "AHA!!! There it is...", previous.
    ---
    ---
    Never mind what you read, find out for yourself what the deal is.
    Make your measurements again measuring the battery voltage and the
    current through the load for each measurement you make, then calculate
    the various internal resistances you get for the battery with
    different loads and post what you find.
     
  18. Guest


    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?

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

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

    John
     
  19. John Fields

    John Fields Guest

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

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

    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
    ---
     
  20. John Fields

    John Fields Guest

    ---
    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.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-