TPS7A47-Q1: About temperature

Hello,

Thank you for your question.

What load current is being used in your system? This will help me to determine if the temperature increase may be a concern.

Best Regards,

Hannah

Hi，

I used it to power a clock distribution chip and a microcontroller，and the current was about 0.1A

Hello,

It is normal for the temperature to be increased to 100 degrees at an input of 20V. When calculating using the junction to ambient thermal resistance, we can see that the temperature will increase by 31.1 degrees C above the ambient with every Watt dissipated. 

1. Using the first case 5V to 3.3V:

(0.1A)*(5V-3.3V)=0.17W-> 0.17W*31.1=5.287+25(ambient temperature)=30 C

2. Using the second case 20V to 3.3V:

(0.1A)*(20V-3.3V)=1.67W-> 1.67W*31.1=51.937+25=77 C

The 77 C here is a simulated value, not a tested value, which means that it is likely that the 100-degree value you found is reasonable for this case. 

3. Using the third case 35V to 3.3V:

(0.1A)*(35V-3.3V)=3.17W-> 3.17W*31.1=124 C

Again, this is a simulated value, not a tested value, so the real value would be larger. This value could be around the maximum operating temperature of 145C. 

Heat dissipation is a concern based on this case. To increase the heat dissipation around the part, we recommend expanding copper around the device and integrating vias in your design. 

Please see this note that explains thermal performance: Digital Designer's Guide to Linear Voltage Regulators & Thermal Mgmt (LDO) (Rev. A) (ti.com)

Please see this note for updated thermal metrics: Semiconductor and IC Package Thermal Metrics (Rev. C) (ti.com)

Please let me know if you have any additional questions. 

Thank you,

Hannah