Tool/software:
Hi Team,
Use TPS51397A, with an application range of 20V to 5V and a current output of 9.3A. There are the following issues:
1. If the efficiency is 90%, the heat consumption is 5.1W. How to allocate 5.1W of heat consumption to ICs, inductors, capacitors, etc;
2. Do I need to add a 470uF electrolytic capacitor to the output? Or do you have any good suggestions?
Hi Reed,
For your questions:
1. The loss mainly includes the core loss of the inductor, the copper loss of the inductor and the IC loss. You can use Webench Model to get the power loss of this device under your test condition.
https://webench.ti.com/power-designer/switching-regulator/customize/130?VinMin=20&VinMax=20&O1V=5&O1I=9.3&base_pn=TPS51397A&AppType=None&Flavor=None&op_TA=30&origin=pf_panel&lang_chosen=en-US&optfactor=3&Topology=Buck&flavor=None&VoltageOption=None
2. Please make sure that the output capacitance does not exceed 330uF. Our chip adopts DCAP3 control mode, and the internal compensation circuit has been added, so there is no need to use a large ESR output capacitor externally.
Thanks.
Aurora
