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Power dissipation and Junction temperature of MCIMX286CVM4B

Question asked by Rakshith Manjunath on Dec 9, 2019
Latest reply on Dec 10, 2019 by igorpadykov

Hi, I am working on calculating the power dissipation of MCIMX286CVM4B based on the current consumption on each of the power rail (See below) . All of the power source is from the internal DC/DC buck regulator of IMX286 without any battery connected. With the help of a software to measure the junction temperature, the software reads out 60°C at 24°C ambient. Doing the math with 36°C/W of thermal resistance considering 6-layer PCB, the total power dissipated is 1.0 Watt. The 5V power source to internal buck regulators measures 0.204 Amps giving 1.02 Watt of input power, this is what confuses me. How can power dissipation be almost equal to consumption. Am I doing anything incorrect? 

 

VDDIO(3.3V) = 0.0215 Amps ---> 0.0709 Watt 

VDDA (1.8V) = 0.0903 Amps ---> 0.1625 Watt

VDDD (1.2V) = 0.252 Amps ---> 0.3027 Watt                                            

Total -----------------------------------> 0.53625 Watt ( consumption ) 

 

Since I have the above consumption data, the 5V source measures 1.02 Watt including dissipation of lower voltage power rails. Therefore the dissipation should be 1.02 - 0.53625 = 0.4837 Watt of power dissipation resulting in 42.4°C of junction temperature, which is not the case. 

 

The processor is exposed to air at 24°C ambient and on a 6-layer PCB. 

 

Please advise. your inputs on this is much appreciated.

 

Thank you.

 

Best regards,

Rakshith Manjunath

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