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
Hi Rakshith
right, power dissipation is almost equal to consumption since
most current from these supplies is consumed by i.MX28 internally.
For example VDDD current is consumed by i.MX28 core.
Power supplies usage is described in AN4199
Using the i.MX28 Power Management Unit and Battery Charger
Typical i.MX28 VDDIO,VDDA,VDDD consumption can be found in Table 12
i.MX28 Applications Processors for Consumer Products
Also one can take into account temperature sensor accuracy https://community.nxp.com/thread/468500
and dcdc efficiency (about 85%).
Best regards
igor
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