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DRV8412 thermal calculations/limitation

Daryoush Sahebjavaher
Intellectual 410 points
Other Parts Discussed in Thread: DRV8412

Hi,

I am confused about the Power ratings of your DRV8412 motor driver IC . I have used DRV 8412 samples on our new PCB design and would like to know what maximum continuous current it could theoretically withstand before overheating. The data sheet suggests a 6A continuous output current per chip and that in can drive motors up to 50V. This would suggest that using the parallel mode, this driver IC can drive a 300W motor which doesn't make sense. On page 3 of the data sheet under "PACKAGE HEAT DISSIPATION RATINGS" chart it is suggested that the resistance from junction to ambient is 26.1 degrees/watt. Since the DRV8412 is %97 efficient T_junction= 125 = (100-97)/100 x outputpower * 26.1 + 25 => outputpower = 3.83 watt. On the same page (page 3), the "PACKAGE POWER DERATINGS" suggest that the pakage power rating is 5watts at room temperature. So overall I am confused about the power rating of the DRV8412 device. If the recommended data sheet exposed power pad / heat slug area is met what is the power rating of the package? or in other words how many watts can it continuously output? My motors are 6-10 Volts and I'm trying to find out how much continuous current the DRV8412 can withstand.

Thank you for you help,

Darius

  • 0
    TI__Guru 55605 points

    Darius,

    You are on the right track.

    Not sure where the 26.1 C/W number came from as page 3 shows 25 C/W in the table you were referencing.  The package rating is calculated based on a maximum junction temperature of 150C (not 125C).  At 150C, the device will go into thermal protection and that is considered the maximum rated temperature.  

    So, the 5W comes from the following:

    Tj (max) = 150C = Pdissipated * 25C/W + 25C (ambient)  ---->  Substitute 5W into the equation and that is your package rating at 25C ambient.  For larger ambient temperatures, the allowed power to dissipate goes down as shown in the table.  

    When the outputs are parallel connected in the DRV8412, the effective Rds(on) is cut in half  So, 110mohms drops to 55mohms.  This gives you a calculated efficiency (taking into account other losses like quiescent current, GVDD/VDD current, switching losses @ 20kHz) of ~ 97.8 when operating at 50V.  

    Your equation above is correct, so you can calculate for 6A out at 50V as follows:

    Tjunction @ 25C ambient  =  (100-97.8)/100 * (50V*6) * 25C/W + 25C = 190C  ----->  which exceeds the package rating with Pdis = 6.6W.  

    Using the same equation, you can calculate the maximum current at that supply or back off the supply voltage in order to keep the power dissipation below the package rating. You will also have to consider your ambient which is probably more than 25C.