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CSD18542KTT: Power Dissipation and Thermal

Jorge Vega
Intellectual 300 points
Part Number: CSD18542KTT

Hello, I have a basic question concerning junction temperature on my fet and also concerning duty cycle.

I plan to use CSD18542 on the power block for a 3Phase motor.  The motor draws current in the order of 1.4A for about 50ms.  The Period I (hopefully correctly) calculated from the image shown below at ~ 80ms.  Which makes the duty cycle 0.62.

The image was obtained by using a clamp current probe on phase A.  Can you let me know if this is how u would have calculated the duty cycle.... that is the time any one of the 6 fets are conducting?

also, can I calculate Tj like this:  Tj = Tc + Id x Vds x D.  Where D is the duty cycle.

  • 0
    Intellectual 300 points

    I wanted to clarify that I am not accounting for any switching or conduction losses in the calculation of the junction temperature.

  • 0
    TI__Mastermind 32335 points

    Hi Jorge,

    Thanks for the inquiry. Please see the attachment. I would calculate the period of the signal shown as 160ms. The starting point and ending point should be the same to capture one full period. You can use the transient thermal impedance curves as shown in Figure 1 of the datasheet to calculate the temperature rise of the FET. At 50ms, you're nearly approaching the steady state value where the duty cycle does not make a big difference. I would calculate the conduction loss as Irms^^2 x rds(on).

    1541.Waveform.pdf

  • 0 in reply to John Wallace1
    Intellectual 300 points


    Thank you John.  I want to make sure I understood well.  Does the image below reflect back what you stated in your reply (apologizing for the messy capture)?

    OK now on temperature rise:

    Tj = Tc + 0.9C/W x I^2 x Rdson

  • 0 in reply to Jorge Vega
    TI__Mastermind 32335 points

    Hi Jorge,

    Yes, you have selected the correct curve for 30% duty cycle and 50ms pulse width. However, please note that ZthethaJC is a normalizing factor which you multiply times RthetaJC to get an equivalent thermal impedance. As you move further to the right on the graph, all curves approach a value of 1. In this case, you would multiply 0.9 x 0.6C/W = 0.54C/W equivalent thermal impedance for your operating conditions.