PSIM-JMAG Interface

Yong Jun Lee_21635
Yong Jun Lee_21635 Altair Community Member
edited April 2 in Community Q&A

I have two questions.

1. When using the Thermal model, the thermal resistance unit is Kelvin or Celsius per Watts.
but PSIM only has a model for electrical resistance. For example, if the thermal resistance is 0.3K/W, does PSIM just enter 0.3 ohms?
Do I also need a conversion to the temperature unit?

2. Can a motor model with above 3 phases of input in JMAG be imported into PSIM and used?
For example, assuming that I modelled a 6-phase input motor in JMAG, after extracting the jcf file and xml file, which are the data files of this motor, can I use the MagCoupler (DL) model in this case?
If you have a sample file for this, can I get some?

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Best Answer

  • Rhonda_20369
    Rhonda_20369 New Altair Community Member
    edited March 28 Answer ✓

    1. This depends on what type of electrical signal is used in your circuit to represent the power loss.

    For example, the thermal resistance is 0.3K/W. 

    Here, "W" represents the power loss in Watt.

    If your power loss is 100W, and it is represented by a current signal of 100A, you can use a resistor to represent this thermal resistance. Therefore,

    The temperature drop on this thermal resistance = I * R = power_loss * 0.3 = 100W * 0.3K/W = 30 K

    Please note, this is the temperature drop, not the absolute temperature. It is the drop between the two terminals of the resistor. Therefore, 30K is the same as 30C.

    In PSIM, usually, Celsius is used.

    If one end of your thermal resistance is connected to a 25V, this end is set at 25degC. Then, at the other end, the temperature would be:

    30 + 25 = 55 (degC)

     

    However, if your power loss is not represented as a current signal, you might need to use a different way to represent your thermal resistance in order to calculate the temperature drop on it.

     

    2. Using JMAG models.

    Please click "Help >> Index", and then, type "JMAG", you will find there are many ways you can use JMAG models in PSIM. You will also find the pdf tutorials as well as video tutorials.

     

    Sincerely.

    Rhonda

Answers

  • Rhonda_20369
    Rhonda_20369 New Altair Community Member
    edited March 28 Answer ✓

    1. This depends on what type of electrical signal is used in your circuit to represent the power loss.

    For example, the thermal resistance is 0.3K/W. 

    Here, "W" represents the power loss in Watt.

    If your power loss is 100W, and it is represented by a current signal of 100A, you can use a resistor to represent this thermal resistance. Therefore,

    The temperature drop on this thermal resistance = I * R = power_loss * 0.3 = 100W * 0.3K/W = 30 K

    Please note, this is the temperature drop, not the absolute temperature. It is the drop between the two terminals of the resistor. Therefore, 30K is the same as 30C.

    In PSIM, usually, Celsius is used.

    If one end of your thermal resistance is connected to a 25V, this end is set at 25degC. Then, at the other end, the temperature would be:

    30 + 25 = 55 (degC)

     

    However, if your power loss is not represented as a current signal, you might need to use a different way to represent your thermal resistance in order to calculate the temperature drop on it.

     

    2. Using JMAG models.

    Please click "Help >> Index", and then, type "JMAG", you will find there are many ways you can use JMAG models in PSIM. You will also find the pdf tutorials as well as video tutorials.

     

    Sincerely.

    Rhonda

  • Yong Jun Lee_21635
    Yong Jun Lee_21635 Altair Community Member
    edited March 28
    Rhonda_20369 said:

    1. This depends on what type of electrical signal is used in your circuit to represent the power loss.

    For example, the thermal resistance is 0.3K/W. 

    Here, "W" represents the power loss in Watt.

    If your power loss is 100W, and it is represented by a current signal of 100A, you can use a resistor to represent this thermal resistance. Therefore,

    The temperature drop on this thermal resistance = I * R = power_loss * 0.3 = 100W * 0.3K/W = 30 K

    Please note, this is the temperature drop, not the absolute temperature. It is the drop between the two terminals of the resistor. Therefore, 30K is the same as 30C.

    In PSIM, usually, Celsius is used.

    If one end of your thermal resistance is connected to a 25V, this end is set at 25degC. Then, at the other end, the temperature would be:

    30 + 25 = 55 (degC)

     

    However, if your power loss is not represented as a current signal, you might need to use a different way to represent your thermal resistance in order to calculate the temperature drop on it.

     

    2. Using JMAG models.

    Please click "Help >> Index", and then, type "JMAG", you will find there are many ways you can use JMAG models in PSIM. You will also find the pdf tutorials as well as video tutorials.

     

    Sincerely.

    Rhonda

    Thanks for reply.

    Your answer was helpful.
    Then, the values of these thermal resistors, junction to case, and case to sink are displayed in K/W on the data sheet of a typical MOSFET or IGBT.
    Is it okay to enter and use this value in PSIM as it is?

     

  • Rhonda_20369
    Rhonda_20369 New Altair Community Member
    edited April 2
    Yong Jun Lee_21635 said:

    Thanks for reply.

    Your answer was helpful.
    Then, the values of these thermal resistors, junction to case, and case to sink are displayed in K/W on the data sheet of a typical MOSFET or IGBT.
    Is it okay to enter and use this value in PSIM as it is?

     

    Yes.

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