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Temperature measurement in LTspice

Discussion in 'General Electronics Chat' started by alok1982, Feb 6, 2014.

  1. alok1982

    alok1982 Member

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    How to measure temperature of a component in LTspice during simulation?
     
  2. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi,
    Press the 'S' key on keyboard.
    Enter this line into text box .step temp 0 50 10

    This will step temp from 0C thru 50C in 10 C steps

    OK.?
    E

    To measure the wattage dissipation of a component , place the cursor on the components and press and hold down the ALT key, a thermometer cursor should appear
    Read the Watts on the bottom screen screen status text
     
  3. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Carefully read what Eric wrote. There is no measure the temperature.
    There is a set the temp. There is measure the wattage.
     
  4. dave miyares

    Dave New Member

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  5. mading2018

    mading2018 Member

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    So if I want to test my component and see the power losses (w losses) at 100 degree only, can I simply write: .step temp 100 ?
     
  6. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Write .temp 100
     
  7. mading2018

    mading2018 Member

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    Hmm, I forgot to ask. After I have write the .temp 100 as a SPICE directive, I can just simply use the thermometer cursor to see the watt losses, right?
    I just want to make sure that I understand this correctly.
     
  8. dave miyares

    Dave New Member

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  9. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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  10. mading2018

    mading2018 Member

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    Okay, that's good. :)
    Maybe I doing something wrong, but the losses during the same period of time, 10 ms, seems not to be any large difference between
    to have .temp 100 and not having .temp 100.
     

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  11. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Why would you expect a large difference?
     
  12. mading2018

    mading2018 Member

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    I thought that if the component is having a high temperature (100 degree C), then I expected that would lead to higher W losses (heat losses).
     
  13. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Depends what the component is and how it is used. An NTC thermistor, for example, might dissipate less power when hot. Semiconductor junctions drop less voltage when hot.
     
  14. mading2018

    mading2018 Member

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    Oh, then it sounds it is hard to measure the losses at a certain temperature :(..
     
  15. kubeek

    kubeek Well-Known Member

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    You still don´t say why do you expect the losses to change? For example mosfets will have higher Rdson, but diodes will have lower forward drop. So it really depends on the circuit in question.
     
  16. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    A resistor will have the same loss.
    -----------------------
    In the real world we put the power supply in a hot room and see if it survives.
    In SPICE we can not have a hot room. We can only force the inside of all parts to be at 25C or 100C or .......
     
  17. crutschow

    crutschow Well-Known Member Most Helpful Member

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    I think y0u are confusing heat loss with a device's power dissipation.
    The power dissipation is determined by the power loss in the device and does not vary with ambient temperature.
    The temperature of the device increases until the power dissipation to its surroundings equals the internal power dissipation of the device, as determined by the thermal resistance to ambient.

    Spice does not determine that temperature rise.
    You can determine the rise by measuring the device wattage (dissipation) in Spice, and then calculating the rise from the device's thermal resistance to ambient, which is determined by the device's package, how its mounted on a circuit board, and whether it has a heatsink.
     
  18. mading2018

    mading2018 Member

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    Yes, thats is excatly what I want to test. If I force all parts to be 100 C, then I just want to see power dissipation (w losses).
     
  19. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Probably the difference in power loss from 25C to 100C is small.

    ----------------------------------------------------------
    I don't have time to look into this but you are not having 3k watts in D2. We are doing something wrong.
    [​IMG]
     
  20. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Spice cannot tell you that.
    It can only tell you the dissipation due to the circuit operation.

    What do you mean (w losses)?
    What losses?

    To determine what you want, you would need to know the thermal resistance of the device, junction to ambient.
     
  21. mading2018

    mading2018 Member

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    I mean the power dissipation/losses.
     
  22. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Okay, I'm confused about what you are trying to do.
    You cannot learn anything about power dissipation by changing the temperature of the device in simulation.
    Power dissipation is a function of the power dissipated in the device by the circuit voltages and currents.
    That's what causes the junction temperature to change.
    And the only way you can find the actual junction temperature is to know the devices power dissipation and it's thermal resistance to ambient.
    Is that clear?

    Okay, lets work backwards as you seem intent on doing.
    Let's say the junction temperature of a device is 100°C at a room ambient of 25°C, with a junction to ambient thermal resistance of 10°C/W.
    That means the device is dissipating (100-25)/10 = 7.5W.
    Note you can only make that calculation if you know the device's thermal resistance. and that's not part of the Spice models.
     

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