LMZ14202H: thermal efficiency

Hi  NS:

the derating will make sure the internal hot point like IC's temperature does not over certain temperature. but the efficiency will still change by temperature, higher temperature will have lower efficiency, but when temperature go up to 80deg C, to keep the internal hot point not over its max limitation, we suggest customer to derating the current to keep the power loss flat.

Thanks

Hi Daniel.

Very thanks for your answer.

I think i need to check some more after checking yours.

I thought there would be no problem because the end-of-the-line load is using only a very low current (less than approximately 0.2A) compared to the load allowed by the IC (approximately 2A).  After I check your answer, I think I have to protect the IC through the derating even under this low load, is this right?

Also, is there any data that I can check the amount of efficiency change according to temperature? I think it would be good to have data under a very low load. Please let me know if you can't provide it.

I have one more question. If the efficiency decreases due to the temperature rise, is there an automatic shutdown function that protects the IC itself?

Regards.

NS You.

Hi Namisk:

I don't think you need to protect IC in light load condition, if the current is lower than derating current at any temperature, all the component temperature in module will be safe range.

I don't have efficiency data at light load, but you can get the simulation data from wenbench. for thermal, power loss is key instead of the efficiency. assume at 42Vin 5V 0.5A , the module is 90% efficiency compare to 5V 1A 92% efficiency, 5V 1A will have 0.43W power loss compare to 0.27W under 5V 0.5A. 

the IC will have OTP protection .

Thanks

Hi Daniel.

Thank you for your answer. It's really helpful.

Because of the sufficient capacity of the designed PSU, I thought that even if the power drop is increased, the sufficient power will be supplied to the IC without problems. When abnormal operation is occurred at high temperatures, slightly lowering the output voltage solves the problem. so I was looking at it as a problem of the temperature and efficiency. However, after checking the example of the relationship between output power and efficiency, it seems that there is little connection with the current problem. This is because if the output voltage of a problematic sample is lowered, it is operating at higher temperatures without any problems.

Only one sample has encountered the problem, and lowering the output voltage solves the symptom. Therefore, is it reasonable to judge this problem as follows?

'the temperature's power dissipation of the sample in question is slightly higher than that of other samples.'

If you have any other opinions, please give me advice.

Regards.

Hi ddaniel.

Your answer was very helpful. Thank you.

I hope you have a great day.

Thank you.

NS You.