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Heat dissipation of a configuration of resistors

Discussion in 'General Electronics Chat' started by valknut, Aug 24, 2017.

  1. valknut

    valknut New Member

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    Suppose we have a hypothetical one-ohm resistor that becomes damaged if one-hundred watts or greater passes through it. In other words, a voltage of ten volts would generate a current of ten amps and thus the resistor would be destroyed.

    If two of these resistors are wired in parallel then the resulting component W would have a total resistance of (1 x 1) / (1 + 1) = 1/2 ohms. Wiring two W components together in series would therefore result in a component X having a total resistance of one ohm.

    What happens if we now apply ten volts across X? Would every bit of one-hundred watts pass through each individual resistor, or would it be less? I suspect it would be something like fifty watts per resistor (and also fifty watts per W component). Is that correct, or am I way off?
     
  2. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    1) 1 ohm resistor, 10 volts, 100 watts
    2) I think you want to use 4 resistors in a series parallel mode.
    ----R1-----R2------
    ............|
    ----R3-----R4-----
    I can not draw well like that. LOL
    The 4 resistor will be 1 ohm.
    The power will divide up nicely. 25 watts each.
    Think of R1, R2. It is a 2 ohm resistor with 10 volts across the total. 5 amps 50 watts. 5 volts across R1 and 5 volts across R2.
     
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    As Ron says, the power will distribute evenly over the four resistors so each one sees 25 watts.

    However, if 100 watts destroys the single resistor then it should really be run more like at 50 watts. Thus two 2 ohm resistors in parallel should do it. With four in series parallel as you connected them, you will in fact get cooler operation though. You might find this information on the data sheet as some power resistor manufacturers show this information on the data sheet.
    Very roughly a resistor run at 1/2 power gets about up to a 57 percent rise of the maximum temperature rise. Again roughly, a resistor run at 1/4 power gets about up to a 33 percent rise of it's maximum temperature rise at full power.
    For example if a resistor gets up to 110 degrees C at 20 degrees C ambient at full power then when run at 1/4 power it should only get up to around 50 degrees C. This is a rough estimate though.
    Some power resistors are rated for very high temperature though, for example if one gets up to 200 degrees C at full power with an ambient of 20 C again then at 1/4 power it should get up to around 80 degrees C, which still might be considered too hot for the application.
    Then there is change of ambient to consider because if that goes up then the temperature of the resistor goes up too naturally.
     
    Last edited: Aug 24, 2017
  4. dave

    Dave New Member

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  5. valknut

    valknut New Member

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    I thought R1 and R2 (component W in my example) would be equal to 1/2 an ohm, the total resistance of two resistors in parallel being 1/(1/R1 + 1/R2) = (R1 * R2) / (R1 + R2). Is that incorrect?

    I was just curious how the energy would be distributed in a hypothetical sense. But yes, I will definitely read the data sheet and be sure to keep the dissipation as low as possible in a real circuit!

    It's surprising how the energy is distributed equally regardless of whether the resistors are arranged in series or parallel - not what I was expecting!
     
  6. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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  7. valknut

    valknut New Member

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    I see, the ASCII art you originally posted threw me off. The W component I was describing corresponds the second diagram (R1 and R3). It all makes sense now. Thanks for the help!
     
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    This phenomenon turns up now and then in various ways in Nature and is collectively known as symmetry.
    Whenever we have symmetry in a problem the solution is simplified simply because we have repetition. Break the symmetry however and then it has to be calculated more carefully noting how each part works in the solution.
    When we do have symmetry though the whole idea is to figure out how to exploit that symmetry in order to gain that simplification in the analysis. Resistor grids can be studied to find out more about how resistors work when connected together in various ways.

    This particular "connect four to get the same resistance" idea can also be found to be somewhat similar to the idea of sheet resistance. When measuring sheet resistance a square of the sheet always measures the same resistance, so they call it "resistance per square". Note this is not the same as "resistance per square meter" or "resistance per square inch", it's just resistance per square anything.
    You may or may not want to get into this just yet, but in case you do there is a lot on the web about sheet resistance.

    Good luck with your studies.
     
  9. valknut

    valknut New Member

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    Sheet resistance seems to me rather counter-intuitive, but I suppose it's analogous to the way that the pressure of a fluid is only dependent on depth? Grid resistance...looks pretty complicated. Found some material such as this and this, but that's all just way over my head at this point. But thanks for the thoughts, anyway, very interesting stuff.
     
  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well it's interesting that when we connect four resistors in a certain way we get the same resistance as one resistor.
    However, it does not stop there. If we call that new resistor (made from the 4 resistors) Rn1, then when we connect four of THEM in the same manner, we again get the same resistance which was the original resistance for one resistor.
    Now continuing in this manner, we can build a network of many resistors that still end up being the same value as just one but the construction gets bigger and bigger.

    Yes the sheet resistance is two dimensional so it's a little different. What might be interesting is to build a grid using the above method and then start to measure resistances inside the construction not just at the two ends that form that one big resistor.

    We can look at the heat dissipation vs surface area too which forms a simple relationship.
    When we have a resistor R1 with surface area S1 and power dissipation P1 that results in a temperature T1 in an environment at ambient temperature, when we increase the network to four resistors all the same value as before, the surface area goes up to 4*S1 and in this case the surface area went up in a linear fashion so the heat distribution is linear so we get power dissipation P1/4, or stated another way we can get power dissipation 4*P1 now.
    That's not how things usually work though, because the volume vs surface area vs power comes into play where when we increase the volume we dont increase the surface area enough, when it is all in the same package. So the individual packages allow a more even spread of the heat which is nice.

    I find that in real life applications i dont like to run the resistors at too high a temperature because they might melt the plastic case usually used with todays' stuff. I did a battery charger one time that had to charge twice a day automatically and used a 10 ohm single power resistor. I placed the resistor outside the case so that it could get free air flow and kept the dissipation down low for the size. I cant remember the numbers now but it was something like a 10 watt resistor dissipating just 1 watt.

    Also you may find interesting is there are more strict limits when we confine the resistor where it can not get free air flow.
    In theory, if we place a resistor in a confined perfectly sealed container where no heat can get out and fun that resistor at 1 microwatt, the temperature will eventually reach an infinite value which would melt any substance we know of. This basically says that all resistors must have some means of cooling from the outside world. This usually means a vent in the case if there is significant power dissipation.
     
  11. valknut

    valknut New Member

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    Got it, so be conservative in selecting power ratings and mindful of heat transfer. Very good advice, thank you!
     
    Last edited: Aug 30, 2017
  12. Ratchit

    Ratchit Well-Known Member

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    First of all, power does not pass through a resistor, charge does. Power is dissipated in any component if charge passes through it and resistance is encountered. If no current exists in a component, then no power is dissipated. For instance, no power is dissipated in a perfect capacitor no matter whatever voltage below its rated maximum is applied. This is because a perfect capacitor does not allow charge to pass through its dielectric.

    Yes, that is correct.

    You are way off. The circuit of one ohm and ten volts would dissipate 100 watts. Each of the four resistors would dissipate 25 watts each.

    Ratch
     
  13. valknut

    valknut New Member

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    Excellent point and yes I am aware of that fact. It seems to me perfectly reasonable to say that power "passes through it" however for the mere fact that we are talking about something that has resistance. Incidentally this was a major source of confusion for me at one point and not always well-covered in text books, so I'm glad that you pointed it out. Perhaps someone will stumble upon this thread at some point in the future and actually learn that crucial concept!
     
  14. Ratchit

    Ratchit Well-Known Member

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    If power passed through a device like charge does, it would have to pass through the wires in series with the device. Then the wires would dissipate the same energy as the device, which of course, it does not do. Only current is constant in a series circuit, no matter what the voltages or resistances are.

    Another example if a perfect coil. Any amount of charge can pass through the coil, but the lack of resistance means that no energy is dissipated.

    Ratch
     
  15. valknut

    valknut New Member

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    Aha, I see what you mean now! Thanks for the correction.
     

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