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Stupid Question on Xsim and Resistors

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  • Stupid Question on Xsim and Resistors

    Hello from Oz

    My basic understanding is that when you parallel resistors, the value is the larger minus the smaller. I have a design which needs a 4.7 ohm resistor as part of an L-Pad; I don't have a 4.7 ohm, but I have a 5.6 and a 1 ohm. So, if I'm right in my understanding, wiring the resistors in parallel should give me 4.6, close enough. But according to Xsim, it doesn't, at least in the FR graph, so can anyone see where I'm wrong please?

    Of course, I'll buy the 4.7 ohm resistors, but I'd like to know where I'm wrong.

    Thank you

    Geoff

    PS sorry about the image size
    Attached Files

  • #2
    Nah that will give you something like 0.85ohm. I don't know the exact formula but here's a website I found with a calculator.
    https://www.allaboutcircuits.com/too...ce-calculator/
    Constructions: Dayton+SB 2-Way v1 | Dayton+SB 2-Way v2 | Fabios (SB Monitors)
    Refurbs: KLH 2 | Rega Ela Mk1

    Comment


    • Geoff Millar
      Geoff Millar commented
      Editing a comment
      Thanks DeZZar: I've seen some crossovers where two resistors are paralleled to increase power handling, eg two 100 ohms were used to give 50 ohms in a PETT project in the woofer shunt circuit. I think that's where I got the idea from.

      My MTMs use a 10 ohm and 8.2 ohm wired in parallel in the tweeter circuit, before the tweeter cap, so what value would that make, please?

      Thank you

      Geoff

    • DeZZar
      DeZZar commented
      Editing a comment
      10 & 8.2 = 4.51

      I've used resistors doubled up for power handling as well - usually just double or triple the value though not a mix-n-match of different values.

  • #3
    Product over the sum.

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    • DeZZar
      DeZZar commented
      Editing a comment
      That'll only work if there's just two of them.

      Otherwise its essentially 1/sum(1/r, 1/r .....for as many as there are)

  • #4
    In any case the final value will be slightly less than the smallest resistance value, yes?

    Comment


    • DeZZar
      DeZZar commented
      Editing a comment
      well....

      if I had 12.2, 13.3, 2, 15, 25, 5.6 = 1.06

      if I had 1.9, 5.6, 2.1, 5.6, 3.3, 2.1 = 0.46

      so I don't know if a generalized rule of thumb really works...probably best to stick with the formula

  • #5
    I get your point as each additional resistor adds some conductivity that ads-up but the OP suggested that he was dealing with two resistors if I am not mistaken?



    Help a guy out?

    Comment


    • DeZZar
      DeZZar commented
      Editing a comment
      lol....hey man I'm just googling this stuff! :P

    • Steve Lee
      Steve Lee commented
      Editing a comment
      Thanks - even an old dog like me can learn new tricks/concepts if he isn't afraid to ask questions here without fear of being ridiculed, right - mate?.
      My recollection of 70's math is failing me of late . . .


  • #6
    Add together the reciprocals of the values, then take the reciprocal of that.

    IF you've got "Calculator" on your Windows machine, you can use the "1/x" function.
    1/5.6 + 1/1(same as "1") =
    0.179 + 1 = 1.179,
    the recip. of which is 1/1.179 = 0.85 ohms.

    Case 2: (10n(ohm) & 8.2n in parallel) ...
    = 1/10 + 1/8.2
    = 0.1 + 0.122 = 0.222, which inverted = 1/0.222 = 4.5ohms.

    Comment


    • fpitas
      fpitas commented
      Editing a comment
      Yes, mathematically speaking you add conductances. To get back to resistance you take the reciprocal.

    • Steve Lee
      Steve Lee commented
      Editing a comment
      Thanks for the refresher.

  • #7
    Thank you all for the quick answers, this now makes sense!

    Geoff

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