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Re: Amperage pull of a computer?

Discussion in 'Electrical Engineering' started by [email protected], Jan 27, 2004.

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  1. Guest

    You might have missed the idea DarkMatter posted about
    using a resistor - it is buried in the posts. It is the
    best answer in the thread to give an easy way to determine
    amperage use on your own.

    It will let you determine the amperage used by any
    of your devices with pretty good accuracy and at low
    cost.
     
  2. Ross Mac

    Ross Mac Guest

    You are assuming he has a DMM to finish the "low cost" job....Not to mention
    the ability to do the conversions...
    An amp clamp is a simple and easy solution for an "approximation" that he
    was looking for...After all, this was not a current study on his PC but just
    a power problem he has in an old building with several pieces of hardware.
     
  3. For some reason, that made me think of "Green Acres" even though I
    haven't watched re-runs of it in years. <wink>
     
  4. Ross Mac

    Ross Mac Guest

    Brings back some old memories......
     
  5. DarkMatter

    DarkMatter Guest

    Extremely good accuracy. In fact. As accurate as the resistor, and
    meter are. :-]
     
  6. DarkMatter

    DarkMatter Guest

    With a 0.1 Ohm resistor, you'd have to be a total MacTard (I owe you
    several thousand of those... bwuahahahahaha) to have a problem with
    the "conversion". You can do it in your head. It's called "shift
    decimal, record reading".... Done! DIRECT conversion.

    1V = 10A

    0.1V = 1A

    Any monkey can do it. Move decimal ONE place.
     
  7. Ross Mac

    Ross Mac Guest

    I see the drugs are kicking in....enjoy until you pass out!
    And I am glad to see you are an expert at Ohms Law...now that's a tough
    one....I will sleep better to night, SnippetTroll, knowing that you can do
    simple arithmetic....
     
  8. Guest

    No, Ross, I am assuming nothing. A DMM with an
    amp clamp costs more than a DMM without an amp
    clamp. A 1% resistor would cost about 5 bucks.
    As to the conversion: just exactly how hard is
    it for him to multiply by 10?
     
  9. Guest

    Yes!! It's often a pain in the ass to see the
    error introduced when using an ammeter in series
    with the load - and sometimes worse than that,
    screwing up the circuit. I can't believe that some
    respondents can't see the value of the method.
     
  10. DarkMatter

    DarkMatter Guest


    For some reason, when those two respond to me, they don blinders,
    and not only fail to read what I post, but what others post as well.
    They would rather troll me than actually participate, then they
    manipulate what I posted into something I did NOT say, and claim *my*
    stupidity. It don't get much worse than that. I could outline the
    RossTard's omissions, and shirks from day one. He would rather sing
    little wussy troll boy Usenet troll songs though.

    He has his reward!

    Hell, I even found (and posted) a link the next day to a national
    semi site that explains the error!
     
  11. Ross Mac

    Ross Mac Guest

    I was speaking of a standalone amp clamp....take your pick of available
    devices...The resistor is not necesary and by the way....how do you plan to
    pick the wattage.....just a guess?....Until you have an approximation of
    current you cannot pick the wattage....by the time you have an
    approximation...you already made your measurement......As far as
    multilplying by ten....when did anyone in this thread ever bring that one
    up???
     
  12. Ross Mac

    Ross Mac Guest

    What type of ammeter do you use for such measurements? And what type of
    accuracy are you looking for? And how does that relate to the OP who wanted
    an approximation?.....Just a curiosity at this point.
     
  13. DarkMatter

    DarkMatter Guest

    You obviously know nothing about current shunts.

    For my 10 Amp power strip installed version, one would get a
    resistor that allows for 15 or 20 Amp passes and still be within half
    the wattage range of the resistor(s). There is nothing wrong with a
    pair of resistors at twice the calculated value in parallel to
    distribute the energy either. Just keep them thermally coupled.

    The resistor(s) used will actually be a 0.01 or a 0.001 ohm value.
    Somewhere in that range (not my original 0.1), so it take quite some
    current through it before it drops enough voltage to see its rated
    wattage. Sheesh, this is an easy engineering exercise to do. Try
    again, RossTard!
     
  14. Guest

    Whichever one you pick, whether it is a clamp that
    you use with a DMM, or a stand-alone amp clamp that
    includes a readout, will cost more than a DMM and
    precision resistor to get the same accuracy. If I am
    wrong, please supply me with a make/model - I'll
    buy one!
    The method I referred to requires a resistor.
    Other methods do not.
    Of course you can. You should know how.
    P = I^2R. A 15W resistor will do fine.
    Even if his PC draws a whopping 10 amps
    (a grossly high number), the resistor would
    need to dissipate only 10 watts. Likewise
    for the other devices he wants to know about
    such as a synthesizer and audio mixer - we're
    not talking high power, here.
    Not true. See above. And an approximation is NOT a
    measurement.
    You did. You said "Not to mention the ability to do
    the conversions" . The conversion from volts measured
    across a .1 ohm resistor to amps requires multiplying
    by ten. Did you have something else in mind when you
    mentioned the ability to do the conversions?
     
  15. Guest

    I'll answer this question and others inline, as thoroughly
    as I can, below, to try to resolve this as best I can.
    That makes it long, of necessity.

    That depends on what I'm doing. A couple of examples
    with a project I'm currently working on:
    I've got a battery pulser that sends a short pulse
    of 3.2 amps into the battery. Running the pulse in
    series through an amp meter, or putting a clamp on
    the wire produces a pronounced affect on the waveform
    due to the change in inductance. Hell - I can't even
    move the short connecting wires without changing the
    waveform. So I have to use a resistor in the circuit,
    and measure the voltage across it with the scope to get
    a current measurement. When I discharge the battery
    (12V auto battery) into an auto headlamp, I am looking
    for a precision discharge to exactly (as close as my
    Fluke's accuracy) 11.89 volts at room temperature. I
    want to be able to see the exact current drawn at every
    10 mv step along the way below 12.2, again, within the
    accuracy of my test equipment. I can do that with a .01
    ohm 1% precision resistor in series with the headlamp,
    measuring the voltage across the resistor, and get a more
    accurate reading than the Fluke gives, because there is a
    voltage drop across the Fluke when I put it in series with
    the load to measure amps directly.

    In general, I've run into this problem over and over
    when I want "exact" (as close as my equipment will allow)
    measurements. There is a voltage drop across the
    ammeter when it is in series with the circuit being
    measured. Try this yourself: set up a circuit that
    draws (as measured with your ammeter in series with
    the load) current that is a little less than the
    maximum for the range the meter is set to. Now,
    turn the meter range selector up to the next higher
    range. You'll often (always?) get two different
    readings. Ask yourself which one is correct? Why
    are they different? (Answers below)
    Try the test I mentioned above. The accuracy
    needed would depend on the circuit - but I
    would like readings on 2 different scales
    to be within 2% of each other. That is unreasonable:
    say one range is 2000 mA and the other is 200 mA.
    A 1% error on 2000 mA is 20 mA. A 1% error on the
    200 mA scale is 2 mA. If I am measuring 100 mA,
    the meter could say 98 mA on the 200 mA scale,
    and it could say 120 mA on the 2000 mA scale.
    Now try the test, measuring voltage across a
    precision resistor. Ask yourself why that method
    yields a far more consistent current. With the
    precision resistor and different voltage scales,
    I get what I want. I know my measurement is within
    the accuracy of my equipment. The reason is that
    putting an amp meter in series with the load
    changes the circuit, by adding an unknown R
    (the resistance of the test leads and internal
    meter shunt) - and changing the meter setting
    changes the circuit again, by changing the shunt
    R. With the .1 ohm precision resistor measurement,
    there is no change to the circuit, other than the
    known R. On voltage scales, the internal impedance
    of the meter is negligible with respect to circuit,
    regardless of scale. An ugly thing with DMM's is
    that on the amps scales, the closer the scale is
    to the current being measured, the more the meter
    affects the current being drawn. The farther away
    it is over the actual current being drawn, the less
    it impacts the circuit - but the lower the resolution
    is. Going under the range (e.g. setting the meter to
    the 200 mA scale to measure 300 mA) is a non-starter:
    you can't get a reading and you may blow the fuse
    or cook the meter if it's not protected.

    The op wanted 1) an approximation of the current his
    PC would use OR 2) an easy way to determine it
    for himself. In addition, 3) he wanted to know the
    current other devices would draw. Here's the
    part of his post that addresses what he wants
    beyond the approximation of what his PC uses:
    "or if there is an easy way for me to determine
    this on my own. (in U.S. with 120 V circuit).
    I have other devices that I'd like to know also, such
    as electronic music synthesizers and an audio mixer. "

    It is the cheapest way he can measure the current
    for himself of the PC and the other devices. It also
    happens to be far more accurate than an approximation.
    It avoids the safety problem of opening the AC supply
    line and having 120 volts AC through the meter leads
    & connections. DM described putting the resistor in
    the neutral side inside a power strip, so the voltage
    at the meter leads would be low. As far as I know, no
    one in the thread described a safe way to use an inline
    ammeter. A clamp on is more expensive (I'm hoping
    you'll provide a cheaper source than I have found for
    them) and the use of a clamp requires opening things
    up to clamp on a single lead - again a safety exposure.
    It is cheaper, safer and more accurate way to provide
    answers to parts 2 and 3 of what he wanted - and
    in answering part 2, he gets the answer to part 1,
    only it is far better than an approximation.
     
  16. Ross Mac

    Ross Mac Guest

    The comment was not about MY ability to do the conversions. They are
    rudimentary electronics. It was of the OP. Also you can use P=IE or I^2R or
    whatever permutation you would like, but the OP would still need to know the
    current. All the OP had, assuming he even knew the power formula was P=?E
    .......so now, how do you figure the wattage for the resistor? Each device
    would require a different wattage....even more guess work. An amp clamp is
    cheap...and the easiest way to approximate the current.
    The question was obviously not from an engineer or a technician so why would
    anyone want to send him on such a wild goose chase...... While your resistor
    is cheap, did you think of the other supporting hardware he might not also
    have? To do this safely, you would probably want to build it up in a box
    with an outlet and cord. I think by the time you spent the time and expense
    with all this, you are back to a cheap amp clamp.
    Enjoy the Superbowl if you are so inclined...Ross : )
     
  17. Ross Mac

    Ross Mac Guest

    I meant to also mention, he was not JUST measuring his PC, so a guess at 10
    amps might not be a good one and if he had the current listed on the
    devices, he would not need to measure it in the first place...
     
  18. Ross Mac

    Ross Mac Guest

    Take an extension cord (lamp cord style), separate one of the power leads
    and amp clamp it....No safety issue, no big expense and most likely a pretty
    good approximation. How do you figure? You are going to buy a resistor with
    a "guessed" wattage, then have to build some sort of box with a cord and
    outlet. You make it sound like "all" you need is a resistor. Does your
    resistor just plug into the wall?...You still need additional hardware. And
    how do you figure the OP can handle all this since he asked a very simple
    question to begin with? A 2% error, if that's what you got, certainly would
    not be an issue by reading the OP. Are you even sure he can perform the
    simple arithmetic to figure the current based on voltage measurement across
    the resistor? Based on the OP, I would say no......
     
  19. DarkMatter

    DarkMatter Guest

    When you read the voltage across the resistor, and convert, you get
    the exact current flowing in the entire circuit.

    Back to school for you!
     
  20. DarkMatter

    DarkMatter Guest

    They are not cheap. Opening the line up to expose a single line is
    still required as well. I have already told you this once.
    People have been measuring current this way for well over a hundred
    years. You are the wild goose, and your brain is what you are
    chasing.
    A simple meter. Doh!
    A power strip with a line cord attached (they all have them -DOH-)

    Open the line cord! D O H !!! A real toughy!
    They are not cheap. DOH!
    Grow up and admit you are clueless if you are so inclined.
     
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