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Power consumption of pulse coils

  1. Apr 3, 2008 #1
    I'm trying to optimize pulse coils and determine power consumption.

    According to Ohm's law:
    P = V2/R
    where P=power / V=voltage / R=resistance

    Why isn't Ohm's law providing the same results as actual meter testing??

    Testing coils with 1 "C" battery: 1.5 volts

    #1) 2.3 ohm coil: (1.5x1.5)/2.3= .98W
    Meter: 1.03V x .43A = .44W

    #2) .7 ohm coil: (1.5x1.5)/.7= 3.21W
    Meter: .6V x .9A = .54 watts

    #3) 2.8 ohm coil: (1.5x1.5)/2.8= .80W
    Meter: 1.12V x .42A = .47W

    #4) 30.7 ohm coil: (1.5x1.5)/30.7 = .073W
    Meter: 1.42V x .044A = .062W

    #4 is the only one even CLOSE! #1 and #3 are about half projected, and #2 is off by a MILE!

    I'm taking the digital mulimeter readings correctly: volts - across coil ends / amps - inserted into circuit

    Why is the actual metered consumption way less than the projected consumption of Ohm's law? :confused:

    Thanks!

    CH
     
  2. jcsd
  3. Apr 3, 2008 #2

    Integral

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    If in your first calculation ([itex] \frac {v^2} R [/itex]) You use the measured voltage drop your results are much better. The assumption that the battery can provide a full 1.5V to your low resistance loads is clearly bad. It looks like the missing voltage is dropped by the batteries internal resistance rather then the coil.
     
  4. Apr 3, 2008 #3
    Ahhh - thanks so much for the reply!

    I was banging my head! Thanks a ton! :cool:

    I guess Ohm's law should have an asterisk stating circuit voltage for guys like me! (rather than battery voltage measured before hooking it up!)

    And I guess I need to add battery resistance to my coil resistance....

    What is the internal battery resistance of a "C" 1.5V/4mA battery? (tiny it appears)
     
  5. Apr 3, 2008 #4

    Integral

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    A batteries internal resistance is variable, it depends on the age of the battery and the current draw. Note that the coil with the highest resistance had the best results. So lower current means lower losses. If you use a resistor in series with the coil to limit circiut current to something less then 40ma you should get very nice results.

    You may also want to look at [itex]P= I^2 R [/itex] using your measured I.
     
  6. Apr 3, 2008 #5
    Thanks for the reply Integral!

    Some further details/parameters:

    I'm trying to maximize Amp-Turns(AT) while limiting current to under 1A @ 1.5V and under .5W consumption. ~200 AT should suffice.

    Coil #2: #22 AWG, 160 turns x .9A = 144 AT (.54 watts)

    Coil #4: #30 AWG, 1200 turns x .044A = 52 AT (.062 watts)

    I was considering tripling the turns on coil #2 to get: 480T x .43A = 206 AT (.44 watts)

    This would give me ~50% more AT while reducing watts by ~20%.

    Looking at it now - and considering your suggestion - it appears I'll get better results per watt by upsizing the #30 AWG coil.
    Space isn't an issue - so bigger coils is fine. The concern would be saturation and at what point does increasing coil size not equal an increase in Q?

    The coil cores are 3/8" x 3" tap bolts... Not laminated electrical steel mind you - but easy to get! Suggestions/ideas on core material/sizing is welcome!!

    On the resistors - part of the idea of properly sizing my coils is to eliminate losses - like resistor "waste" heat.

    As such - I'm also using reed switches. The additional losses of a Hall IC or Phototransistor etc. might figure in ok later -
    but I'm sticking with lean and mean for now.

    Cool forum! :approve:

    CH
     
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