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  • Basic resistor wattage question

    Just looking for some clarification on something reference resistors and heat.

    I have heard two theories on resistor wattage that contradict. In speaker crossovers, I have heard that low resistance resistors (say 1 ohm) will get hotter than higher (say 20 ohm) because the one ohm will pass more current than the 20 ohm.

    I have also heard that in the application of an electronic circuit, like in an amplifier, that a higher resistance (20K ohm) will get hotter than lower (5K ohm) because the 20k will drop more voltage across it than the 5k and have to dump the voltage as heat.

    Which is right? or are they both right in their own application? The second theory makes more sense to me based on ohms law.

    Thanks again guys

  • #2
    The temperature to which a resistor will rise is based on the current flowing through it, and that current is determined by the voltage across the resistor divided by its resistance. A 1-ohm resistor does not automatically get hotter than a 20-ohm resistor unless it has it has enough current flowing to dissipate more power than the 20-ohm resistor. When any circuit is designed, whether in a speaker crossover or in an amplifier, one determines how much power will be dissipated by a resistor when operating as intended, then a resistor with an appropriate power rating is chosen.
    Paul

    Originally posted by dynamo View Post
    Just looking for some clarification on something reference resistors and heat.

    I have heard two theories on resistor wattage that contradict. In speaker crossovers, I have heard that low resistance resistors (say 1 ohm) will get hotter than higher (say 20 ohm) because the one ohm will pass more current than the 20 ohm.

    I have also heard that in the application of an electronic circuit, like in an amplifier, that a higher resistance (20K ohm) will get hotter than lower (5K ohm) because the 20k will drop more voltage across it than the 5k and have to dump the voltage as heat.

    Which is right? or are they both right in their own application? The second theory makes more sense to me based on ohms law.

    Thanks again guys

    Comment


    • #3
      The temp is directly correlated to the power being dissipated by the resistor. In a resistor ...

      Power = V x A

      Since V = R x A, Power can be written several ways for convenience of calculation:

      Power = V2 / R; or

      Power = A2 x R

      For any known resistor, you know what R is. Then you need to have either the voltage drop across the resistor or the amps flowing thru it to derive power. Most of us size the power dissipation by watts as resistors are spec.'d with a wattage rating (e.g., 5 ohm 10 watt resistor, or 10K ohm 1/2 watt resistor as two examples). Temp really isn't a factor unless you have overall heat dissipation issues or heat sensitive materials in an electronic system.

      Comment


      • #4
        Mike nailed it... you have to have voltage and amperage for power. At that point, watts is watts. A 5W resistor will get the exact same temperature with 5W dissipated, regardless of the resistance value. A 1K resistor with 70.7V across it for 10.7mA of current will be the same temperature as a 1 ohm resistor with 2.23V / 2.23A given the same resistor size and environment.
        Electronics engineer, woofer enthusiast, and musician.
        Wogg Music
        Published projects: PPA100 Bass Guitar Amp, ISO El-Cheapo Sub, Indy 8 2.1 powered sub, MicroSat, SuperNova Minimus

        Comment


        • #5
          It's the Law! Click image for larger version

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          "Not a Speaker Designer - Not even on the Internet"
          “Pride is your greatest enemy, humility is your greatest friend.”
          "If the freedom of speech is taken away, then dumb and silent we may be led, like sheep to the slaughter."

          Comment


          • #6
            Thanks guys, I for the most part have a general understanding of Ohm's law, and was able to correctly calculate resistor rating on some small electronic circuits I have played with. I know that E x I is wattage, but don't know either E or I for a given amplifier output power, just the total of the two in watts. If I knew either the current or voltage of my total amplifier wattage, then I would have enough data for the equation. I have always just guessed and probably gone unnecessarily high on the wattage ratings for crossover resistors, but it would be nice to be able to more accurately calculate them. Any guidance as to where to find or what to start out for either voltage or current when trying to do the equation for a crossover?

            Comment


            • #7
              • Series Circuits:
              • Voltage drops add to equal total voltage.
              • All components share the same (equal) current.
              • Resistances add to equal total resistance.
              • Parallel Circuits:
              • All components share the same (equal) voltage.
              • Branch currents add to equal total current.
              • Resistances diminish to equal total resistance.


              https://www.swtc.edu/ag_power/electr...l_circuits.htm
              "Not a Speaker Designer - Not even on the Internet"
              “Pride is your greatest enemy, humility is your greatest friend.”
              "If the freedom of speech is taken away, then dumb and silent we may be led, like sheep to the slaughter."

              Comment


              • #8
                Originally posted by dynamo View Post
                Just looking for some clarification on something reference resistors and heat.

                I have heard two theories on resistor wattage that contradict. In speaker crossovers, I have heard that low resistance resistors (say 1 ohm) will get hotter than higher (say 20 ohm) because the one ohm will pass more current than the 20 ohm.

                I have also heard that in the application of an electronic circuit, like in an amplifier, that a higher resistance (20K ohm) will get hotter than lower (5K ohm) because the 20k will drop more voltage across it than the 5k and have to dump the voltage as heat.

                Which is right? or are they both right in their own application? The second theory makes more sense to me based on ohms law.

                Thanks again guys
                This isn't the answer to your question. . . .
                In most of the projects or circuits that I've built, I always do a "burn in" and allow anything, everything to warm up and or get hot before wrapping it all up. If anything does get too hot to touch then there is a thermal problem that needs to be addressed. I've seen plenty of commercial products with burned and/or scorched traces because of too small of wattage resistors.

                Comment


                • #9
                  Thanks guys. AE, that is a good idea, thanks. I know that if I have a circuit that draws 500mA and need a resistor to drop 15v to 10v, that there will be a 5v drop across the resistor, and that the 5v drop x the 0.5A current is 2.5w of heat that must be dissipated, so a 5w resistor would be more than enough. I just struggle to apply that logic to crossover resistors since I feel I don't have enough data to crunch the numbers.

                  I will look more at the data and resources you guys have provided. What I have been doing, guessing, and checking if I'm actually concerned, has been working fine. I don't listen long or loud enough to ever have had any get warm.

                  It just bothers me that if I wanted to actually calculate it I couldn't because I feel I still need to know either the voltage or current of the source power at expected peak power to determine it.

                  Sorry if I'm completely missing what you're telling me and giving you that banging head on wall feeling

                  I appreciate the help as always

                  Comment


                  • #10
                    And if you increase the voltage...
                    Audio amplifiers are sometimes evaluated on the basis of db gain.
                    if we know the gain of an amplifier and the magnitude of the input signal, we can calculate the magnitude of the output.
                    http://www.allaboutcircuits.com/text...mplifier-gain/
                    "Not a Speaker Designer - Not even on the Internet"
                    “Pride is your greatest enemy, humility is your greatest friend.”
                    "If the freedom of speech is taken away, then dumb and silent we may be led, like sheep to the slaughter."

                    Comment


                    • #11
                      Thanks Sydney. The voltage gain and explanation in the link is starting to make sense. So if I don't know the voltage gain of the amplifier, but know that my source is 2v rms and the amplifier input impedance is 20k ohms, and power output of the amplifier is 100w into 8 ohms, how do I know how many volts are gained to reach the 100w into 8 ohms from my 2v input (of .0001A I assume based on the 20k input impedance)?

                      Comment


                      • #12
                        If P=V(V) / r, is it as simple as for a 100w into 8r that 28.3(28.3) = 800 / 8r = 100, so there is 28.3v in a 100w signal? And therefore 28.3 / 8r = 3.5A of current in that same 100w amp output?

                        Comment


                        • #13
                          And then if I know I am working with up to 28.3v and 3.5A input, and plan to insert a 4 ohm series series resistor, 3.5A x 4r = 14v drop? This seems off since that would be 49w of power to dissipate as heat?

                          Comment


                          • #14
                            Originally posted by dynamo View Post
                            ...how do I know how many volts are gained to reach the 100w into 8 ohms ...
                            Oft times amps ( the better ones anyway ) will list input sensitivity; like a value of .775v being a zero reference.
                            If for instance an amp raised (amplified) the Vin from .775 to 7.75 ( 10x ) then the amplifier provides 20db of gain.
                            With 1V in the Vout would be 10V
                            With 2V in the Vout would be 20V...
                            https://support.biamp.com/General/Au..._output_levels

                            1 watt into an 8 ohm load means 2.83 Volts .35 Amps
                            100 watts into an 8 ohm load means 28.28 volts 3.5 Amps

                            http://www.rapidtables.com/calc/elec...calculator.htm
                            http://www.crownaudio.com/en/tools/calculators
                            "Not a Speaker Designer - Not even on the Internet"
                            “Pride is your greatest enemy, humility is your greatest friend.”
                            "If the freedom of speech is taken away, then dumb and silent we may be led, like sheep to the slaughter."

                            Comment


                            • #15
                              Originally posted by dynamo View Post
                              And then if I know I am working with up to 28.3v and 3.5A input, and plan to insert a 4 ohm series series resistor, 3.5A x 4r = 14v drop? This seems off since that would be 49w of power to dissipate as heat?
                              More like 9.43V, 2.36 Amps, 22.23 watts - Remember the Total load would now be 12 ohms ( the speaker * AND the series resistor ).
                              The other 18.86V is dropped across the 8 ohm speaker.
                              * based on nominal 8 ohm load.
                              "Not a Speaker Designer - Not even on the Internet"
                              “Pride is your greatest enemy, humility is your greatest friend.”
                              "If the freedom of speech is taken away, then dumb and silent we may be led, like sheep to the slaughter."

                              Comment

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