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RMS or average current to determine resistoring heating?

Discussion in 'Electronic Design' started by mook Johnson, Mar 16, 2008.

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  1. mook Johnson

    mook Johnson Guest

    RMS has always been described to me as the AC current level to create the
    same amount or heat in a resistive load as a DC current. This as always
    been with sinusoidal currents.

    In this case I have narrow 50uS pulses of current at 5 amps with repetition
    rates of 500uS.

    If I calculate the RMS current vs. the average current there is a sizable
    difference. the average current is 0.5A the RMS current(from spice
    simulation) is 1.6A.

    Which if the two should be used when determining the heat generated by a
    power resistor.

    Assume the resistor chosen can handle the instantaneous power during the
    current spikes.

    I'm more concerned about the heatload this will add to the heatsink.

    There is a 10X difference in power between the two.


    thanks
     
  2. Mike Silva

    Mike Silva Guest

    RMS is correct. In your case (and assuming a 1 Ohm load for
    simplicity), you have 25W generated for 50us, and 0 for the remaining
    450us, giving an average over 500us of 2.5W. The current
    corresponding to 2.5W is 1.58A (your 1.6A figure).
     
  3. mook Johnson

    mook Johnson Guest

    That was simple enough.

    thanks.
     
  4. Think of it this way.

    Heat generated is proportional to power (I*I*R).
    The average is then the average of the I*I*R terms.
    Therefore you are looking at a RMS term.

    Obviously cooling figures in here so frequency matters but over a
    relatively broad range of frequencies RMS will give you a good place to
    start and will probably be sufficient.

    Also take a look at your resistor's power derating curves. They will
    sometimes provide some additional information that will help.

    If the resistor will dissipate the heat between pulses then your
    calculation may rely more on the pulse rating of the resistor than the
    power rating and of course there is some current for which the resistor
    is underated no matter how short the pulse but only a few parts have
    that value fully specified that I have seen.

    Robert
     
  5. Phil Allison

    Phil Allison Guest

    "mook Johnson"
    ** No matter what the wave shape - the rms value is the one for heat
    dissipation in a resistor.

    ** The rms one.
    ** Correct.

    BTW:

    The most dramatic example of this I have seen in a commercial product is
    with " Glow Plug " drivers for model size methanol engines ( 2 or 4 stroke).
    Model size glow plugs are made with platinum wire and draw up to 5 amps at
    1.2 volts - so the best way to power them is from a single Ni-Cd or NIMH
    cell.

    However, some modellers like to use their 12 volt lead-acid starter motor
    battery to drive the glow plug too.

    But as 5 amps at 12 volts is 60 watts, a 1.2 volt linear regulator is not
    practical - a switching regulator would be ideal but there is a much
    cheaper ( and nastier) solution that the Asian makers of cheap glow driver
    use. Simple PWM.

    Yep - the 12 volt battery is switched ( with a FET ) straight onto the 1.2
    volt glow plug at a high (but audible ) frequency with a very low duty
    cycle - about 1%.

    This means the current pulses are in the order of 50 amps !!!!

    The average current draw from the 12 volt battery is then 50/100 = 0.5
    mp - while the glow plug has 5 amps rms flowing through it.




    ........ Phil
     
  6. John Fields

    John Fields Guest

    ---
    If your pulse is rectangular, then the current won't be RMS, it'll
    be peak, and you should use average.
    ---
    ---
    Just for grins, let's say your pulses have an amplitude of 5 volts.

    That means that if they're pushing 5 amps through the resistor the
    resistor must have a resistance of one ohm, and the power being
    dissipated in the resistor, when the pulse is on, must be:


    P = IE = 5V * 5A = 25 watts.


    Now, let's further assume that your pulses are on for 50µs and off
    for 50µs.

    That means that the resistor will be dissipating 25 watts for 50µs
    and 0 watts for 50µs and, since the pulses have rectangular edges
    and current has the dimension of time, the resistor will be
    dissipating an average of 12.5 watts.

    Assuming a constant load impedance, then, your example of a 10% duty
    cycle will result in 1/10th of the peak power being dissipated by
    the load and transferred to the heat sink.
     
  7. Phil Allison

    Phil Allison Guest

    John Fields is LOSING IT BIG TIME



    ** Wow - now THAT is really crazy bullshit.

    Rectangular pulses have no RMS value ??

    The average value of a rectangular pulse stream gives the heating effect
    voltage ?



    The song for John:
    ----------------------

    They're coming to take me away, HA HA

    They're coming to take me away, HO HO HEE HEE HA HA

    To the funny farm

    Where life is beautiful all the time

    And I'll be happy to see

    Those nice, young men

    In their clean, white coats

    And they're coming to take me away, Ha-haaa!




    ....... Phil
     
  8. Greg Neill

    Greg Neill Guest

    Any waveform has an RMS value, including rectangular
    pulses. Of course, for rectangular pulses that are
    always greater than or equal to zero the average
    will work out to be the same as the RMS value. For
    waveforms that traverse zero, the average can be
    any value between the peaks including zero (consider a
    square wave with equal positive and negative peaks).

    [snip]
    The RMS value for your 505s/505s waveform is about 3.536 A.
    With a one ohm load the power dissipated is then
    (3.536 A)^2 * 1 Ohm = 12.5 W, same as using the average.
     
  9. Phil Allison

    Phil Allison Guest

    "Greg Neill"
    "John Fields"
    Any waveform has an RMS value, including rectangular
    pulses. Of course, for rectangular pulses that are
    always greater than or equal to zero the average
    will work out to be the same as the RMS value.


    ** WHAAAATTTT ???????????????

    For single polarity, rectangular pulses, the average and rms values are
    NOT the same !!!!!!


    I av = I pk times duty cycle.

    I rms = Ipk times sq.rt duty cycle.


    Get real, dickhead.




    ....... Phil
     
  10. John Fields

    John Fields Guest

     
  11. Mook Johnson

    Mook Johnson Guest

    BTW:

    Thanks

    I too run RC boats (IMPBA) and have one of that hobbico power panels that
    make the glow plugs queel. :) I never considered the heating affects of the
    50Amp pulses through the batteries internal impedance. Good thing we only
    need them for a few seconds during startup.
     
  12. Phil Allison

    Phil Allison Guest

    "Mook Johnson"

    ** How did you know I was ( once) a model boat racer ?

    It's the AMPBA here.

    I ran a couple of monos and a tunnel hull (aka catamaran), 3.5cc and 7.5cc.


    ** Those "power panels" must be the WORST possible way to light up a glow
    plug.

    Some of the evil problems include:

    1. The 12 volt battery's voltage sags and hence plug wire temp DROPS when
    you use the electric starter, often so much so that the engine will not
    start. Particularly true if the engine is a little flooded.

    2. Accidental contact between the starter motor frame and the engine sends
    12 volts straight to the plug = instant plug death.

    3. When the engine starts and you throttle it up briefly, the plug wire
    becomes white hot and may fail. This is because the power delivered to the
    plug by the panel increases with resistance, which is suddenly raised when
    the engine runs.

    None of this happens with a Ni-Cd cell - even a 1400mAH sub C size is
    fine.

    Waaaaaay better than any stupid MOSFET power panel.



    ....... Phil
     
  13. Greg Neill

    Greg Neill Guest

    Your method works fine for a postive-going
    waveform (as in this particular case), but would
    not be correct for a general waveform. If the
    OP's pulses were rectangular but not anchored
    on zero volts, the average and RMS currents would
    be different, and the RMS value would have to be
    used to find the power dissipated in a resistive
    load.
     
  14. Mook Johnson

    Mook Johnson Guest

    7.5 and 15CC outriggers here.

    Yup had the same problems you discribe. Was using 4AH gell cells and the 90
    boat would start relaibly. Switched to low impedance 7.5Ah gell cells (low
    internal resistiance) and that went away but the field pack got heavy.

    We run 50% - 60% nitro and the plugs are COLD and we run rich. So we need
    the 4 - 5A get them glowing. I had a setup that used a Alkaline D cell for
    lighter. Worked well but then I "graduated" to a power panel. :(

    Might be time to flunk back a grade. That D cell seemed to last FOREVER.
    (my boats start pretty easy)
     
  15. BobW

    BobW Guest

    Phil,

    Everyone's allowed one mistake in their life. Take your mother, for example.

    Bob
     
  16. Fred Bloggs

    Fred Bloggs Guest

    You shouldn't be concerned with either one. Power dissipation is the
    rate at which the resistor is converting electrical to thermal energy in
    joules per unit of time. When the periodicity is small compared to the
    thermal time constant of the heat sink, then you can you can use the
    average value of joule dissipation per cycle, because much like a low
    pass filter, the heat sink temperature will not respond to the high
    frequency component of the thermal dissipation profile, it will only
    respond to its DC value, and another way of saying 'DC value' is
    'average'. You might notice that computing average power dissipation per
    cycle is the exact same thing as computing RMS current and multiplying
    by R. Forget about the formulas and consider what it is you really want
    to know.
     
  17. John Fields

    John Fields Guest

     
  18. Phil Allison

    Phil Allison Guest

    " John Fields is LOSING IT BIG TIME "

    ** Completely insane.


    They're coming to take him away, HA HA

    They're coming to take him away, HO HO HEE HEE HA HA

    To the funny farm

    Where life is beautiful all the time

    And he'll be happy to see

    Those nice, young men

    In their clean, white coats

    And they're coming to take him away, Ha-haaa!



    ....... Phil
     
  19. Phil Allison

    Phil Allison Guest

    "Greg Neill"
    "John Fields"
    Your method works fine for a postive-going
    waveform (as in this particular case), but would
    not be correct for a general waveform. If the
    OP's pulses were rectangular but not anchored
    on zero volts, the average and RMS currents would
    be different, and the RMS value would have to be
    used to find the power dissipated in a resistive
    load.

    ** WHAAAATTTT ???????????????

    For single polarity, rectangular pulses, the average and rms values are
    NOT the same !!!!!!


    I av = I pk times duty cycle.

    I rms = Ipk times sq.rt duty cycle.


    Get real, dickhead.




    ....... Phil
     

  20. She made more than one:

    1. She let his dad get her pregnant.
    2. She didn't have an abortion.
    3. She didn't flush Phil down he toilet they day he clawed his way out
    of her body.
    4. She didn't have him committed, for life.


    --
    aioe.org is home to cowards and terrorists

    Add this line to your news proxy nfilter.dat file
    * drop Path:*aioe.org!not-for-mail to drop all aioe.org traffic.

    http://improve-usenet.org/index.html
     
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