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Original question:

UCC12040: Best isolated device at VIN=5V, VOUT=5V, IOUT=7mA

UCC12041-Q1: UCC12041-Q1 OUTPUT CAP

Hangjie Wang
Intellectual 880 points
Part Number: UCC12041-Q1

Hi expert,

It was found that there was no output capacitor, the UCC12041-Q1 chip was very hot. Could you help analyze what the reason for this phenomenon? thanks.

BR,

Hangjie

  • 0
    TI__Expert 7366 points

    Hi Hangjie,

    -Output capacitance: for any application using UCC12041, it is recommended to use a 100nF HF decoupling cap (0402) next to the IC and a 10uF filter cap (0603 or 0805). Please see picture below.

    -About the temperature of the IC: It depends how much power it is running and the Input voltage applied.

    Thank you.

  • 0
    TI__Intellectual 880 points

    Hi Manuel,

    Thanks for your quick reply, when I didn’t put the output cap, the device is very hot, and if I have add a cap for circuit, which will work normal. I want to know why the device be hot while no cap.

  • +1 in reply to Hangjie Wang
    TI__Expert 7366 points

    Hi Wang.

    If you do not connect Cout, Vout ripple will significantly increase and switching and conduction losses increases, this leads to more power dissipation, lower efficiency and therefore the device will get hot.

    Thank you

  • 0 in reply to Manuel Alva Hernandez
    TI__Intellectual 880 points

    Hi Manuel,

    Thanks for your reply, we have do two test as below, could you please share the detailed working of the internal circuit to explain why a large VPP causes such large losses? thanks a lot.

    1. Co=10uF, Input=5V/45mA, Output=5V/8mA

    2.Co=100nF, Input=5V/160mA, Output=4.8V/8mA

  • 0 in reply to Hangjie Wang
    TI__Expert 7366 points

    Hi Hangjie.

    TI is in a long holiday and this answer might be delayed. 

    Thank you

  • 0 in reply to Manuel Alva Hernandez
    TI__Expert 7366 points

    Hi Hangjie,

    -For Cout=100nF, the RMS current increases (higher peaks for harmonics that are not well filtered due to Cout low). This large/extra RMS current needs to be provided by the input current. This high RMS current at the input leads to higher conduction losses at the primary switched of the full bridge configuration (Pcond_loss=Irms^2*Ron_switch).

    -This is reflected in the higher input current for Cout=100nF (160mA). The input supply needs to deliver more current to deliver almost the same power than the first case (Cout=10uF). This leads to a lower efficiency and more power dissipation.

    Thank you.