trivial to some question from newbie (re: Ohm's law)

Discussion in 'The Projects Forum' started by dogfuel, Apr 13, 2009.

  1. dogfuel

    Thread Starter Member

    Apr 13, 2009
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    Ok ... don't laugh

    I have a 24v 7.5mA source ...

    If I insert a 2.2k ohm resistor into the circuit, after this do I see about 7.5 v 7.5mA (24 - 2200 * .0075).:confused:

    I need to drop the voltage to connect to a device that can take 10v max.

    Thanks!
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    It always helps to draw a schematic, even of the simplest circuit. It clarifies your thoughts, and helps go to the next step.

    What is happening is your source can provide 7.5ma max. It's current is regulated, just as the voltage is. So if you try to pull more than 7.5ma, it will drop the voltage to limit the current.

    Work Ohm's Law from reverse, figure a resistor that drops 7.5ma at 10V.

    E = IR, or R = E/I, which is R = 10V/0.0075A = 1.33333 KΩ ≈ 1.3KΩ

    You're measured voltage is off. Going through Ohm's Law, it should be; E = IR = 0.0075A 2.2KΩ = 16.5V

    So, if your numbers for voltage and current are correct, R = E/R = 7.5V / 0.0075A = 1KΩ

    The math clearly shows one of your numbers is off, either the resistor, current, or voltage. You need to remeasure all three. Have you actually measured the 7.5ma, or are you assuming it is that number because of the power supply rating? A cheap off the shelf DVM will do all three measurements, with excellent accuracy.
     
    Last edited: Apr 13, 2009
  3. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    You can use the "LED Equation", just swap in 24V for Vsupply, 10V for Vled, and 0.0075 for Iled

    R=\fra{V_{\small{Supply}} - V_{\small{LED} }}{I_{\small{LED}}}
     
  4. Wendy

    Moderator

    Mar 24, 2008
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  5. dogfuel

    Thread Starter Member

    Apr 13, 2009
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    Thank you both ...

    This is not a LED problem ... I think thatoneguy was just trying to put it in a practical context saying it was like an LED.

    It is an output from a welder power supply intended to be treated like a process loop using the current to measure. As the power supply sensing is basically binary, it is either 24v 7.5mA or nothing.

    Rather than read it as an analog input, I'm trying to read it as digital looking for either voltage or not. I'm trying to get the voltage down to something that will not trash my data acqusition device which can only take 5-10 volts and use a pull down resistor to make sure it is either this reduced voltage or 0.

    Can you tell I'm new to this??? Thanks again for humoring the ignorant.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The 7.5mA current, is that drawn by another sensor, the minimum draw needed, or the maximum allowed draw from the 24V indicator supply?

    Are there other devices connected to this 24V line that would cause a drain? If not, 2 resistors as a voltage divider to draw 1mA or so would allow the 5V signal between the two resistors to provide a logic level output. 5V when 24V is present, pulled to ground when no voltage is present.

    ASCII Circuit:
    24V - 18kΩ -+- 4.7kΩ - GND would give ≈5v at the "+" point, drawing just over 1mA.
     
  7. dogfuel

    Thread Starter Member

    Apr 13, 2009
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    thatoneguy -

    this looks to be the answer in terms even I can understand - thanks!!!
     
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