peak or average? homework problem

Discussion in 'Homework Help' started by electricmaniac, Sep 26, 2012.

  1. electricmaniac

    Thread Starter New Member

    Jul 25, 2011
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    Hello,

    When you select a resistor, among the various details, you should consider also the wattage rating.

    Concerning this, I stumbled on a basic problem not so basic to me. Simply stated, the exercise asks what should be the wattage rating of each resistor. In order to solve this problem I've simulated the behaviour of three resistors in series acting as a voltage divider. Looking at the wattage profile of the first resistor in series - R1 - (see attached plots) should one consider the peak value of 90W, therefore choosing a 100W resistor (or, better, 180W) or it is the time average of 22.6W the value to consider, choosing a 40/45W resistor?

    As far as the other two resistor is concerned, the things seem to be clearer: Choose a 10W resistor for R2 (average of 6.2W) and a 3 or 5W resistor for R3 (average of 2W). Am I wrong?

    Thank you.
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,750
    4,797
    The answer really depends on the time scale involved and thermal characteristics of the resistors and environment they are in.

    In general, the power rating needed is based on the maximum operating temperature that is permitted for the component. For most resistors, this rating is for when it is being operated at constant power. But when the power is changing, all bets are off. For instance, you would never put your hand on a stove burner after the heat has been turned on for several minutes, but if you put your hand on it when it is cool (has been off for a long time), you can keep your hand on there for several seconds after you turn it on (talking electric elements, here). The reason is that there is a thermal time constant associated with the element that says it will take so much time for a change in power to achieve a given change in temperature. If you are applying power for a sufficiently short interval of time such that the device's thermal time constant is much larger than the time interval, then you can apply extremely high instantaneous power to the device (your limit will usually be determined by some other factor, such as some kind of breakdown voltage). So in this case, your power rating (based on thermal concerns) will be dictated primarily by your average power dissipation. But if your thermal time constant is on the order of or less than the time interval involved, then your power rating needs to be based on the peak power.
     
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  3. electricmaniac

    Thread Starter New Member

    Jul 25, 2011
    13
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    The problem here is clearly with R1: Does an impulse between 90W and 20W lasting for about 20 seconds imply the resistor's instantaneous breakdown? For R2 and R3 I read a clear asymptotic value, therefore I should take into account the average power dissipated. I understand the reasoning about the thermal time constant, but:

    Can I find the thermal time constant of a resistor on the datasheet?

    Could someone give me an example of the calculations involving a (even fictitious) thermal time constant and concerning the power dissipation of a given resistor?

    BTW the resistance values are as follows: R1 10kOhm, R2 60kOhm, R3 30kOhm.

    Thank you folks.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,750
    4,797
    Hard to say. If it is a physically small resistor, then anything longer than a couple seconds is probably longer than the thermal time constant. But if it is a large resistor with good heat sinking, then it's thermal time constant could enough greater than 20s so that you can use a larger time window.

    Sometimes, but only for components that are intended to see this kind of stress. Many components will give the thermal conductivity between the part os the component that is most sensitive to thermal breakdown and ambient air. This is given in terms of W/C (watts per degree celsius of temperature difference). The other factor you need, which is less commonly found, is the thermal mass, or J/C (or its recipricol) which is the amount of energy (joules) needed to raise the temperature by one degree celsuis. You can actually then set up a first-order circuit analogy between the thermal circuit and an RC electrical circuit to get the time constant -- both are governed by the same differential equation.

    Perhaps later. I don't have time to construct an example right now. But you can ponder what I've said before and perhaps do it yourself or at least be better prepared to follow what I, or someone else, throws out later.
     
  5. electricmaniac

    Thread Starter New Member

    Jul 25, 2011
    13
    0
    Hi WBahn,

    The things are getting harder here :).
    I'm an enthusiastic user of ltspice and I found particularly interesting your idea of setting up a circuit thermal analogy. Could you suggest any reading, book etc. for an absolute beginner?

    Thank you for your time, anyway.
     
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