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LM61495-Q1:Adjustable IC(10A) Short

MIN JE BANG
Prodigy 10 points
Part Number: LM61495-Q1


Tool/software:

I am applying 29.4V (battery power applied), giving feedback through a variable resistor, and creating an output between 22V and 24V. However, during testing, VIN (2, 14PIN), SW (16PIN), and GND (1, 15PIN) became shorted, and I cannot find the cause even after looking at the data sheet. Please tell me the reason for the failure. However, during testing, VIN (2, 14PIN), SW (16PIN), and GND (1, 15PIN) became shorted, and I cannot find the cause even after looking at the data sheet. Please tell me the reason for the failure.

  • 0
    Prodigy 10 points

    I am in charge of HW development and I am contacting you because I encountered a problem during the PCB function evaluation using TI components.

    1 Parts used: LM61495QRPHRQ1 (DC/DC switching regulator)

        1.1 Desired performance: DC24~29V input power, 22V, 10A output.

        1.2 Problems

          a. I confirmed that there is no problem with the output current performance of 1A when the input power is about 10V.

          b. However, when it exceeds about 22V, the IC is internally destroyed even in the no-load state, and the VIN and SW pins are short-circuited.

            b.1 I tested several parts, but they are the same.

    2 There is a possibility that it is a circuit and PCB design error due to a development mistake, but I am contacting you because that possibility is slim after debugging.

    3 I also applied the application provided by TI in the same way, but the problem is the same.

    4 If I am missing something, please advise.




    5 I also tested the above circuit provided by WEBENCH® POWER DESIGNER.

  • 0
    TI__Genius 11995 points

    Hi Min Jae,

    The biggest issue I see here is the output capacitance. 1mF is a very high output capacitance - my recommendation would be to lower the output capacitance and check the behavior. It is recommended to stick to the recommended values in the datasheet since the soft start and the compensation are internal to the IC. 

    You can simulate this circuit on WEBENCH to gauge the performance. 

    Thanks,

    Best Regards,

    Taru

  • 0 in reply to Taru Kapoor
    Prodigy 10 points

    Thank you for your reply.

    1. Is there any problem with the PIN setting?

    2. I thought the RESET pin should be supplied with less than <20V, so could you explain 1, 2, and 3 of the error conditions of this pin in more detail?

  • 0 in reply to MIN JE BANG
    TI__Genius 11995 points

    Hi Min Je,

    Apologies for the delay - I was out of office for a few days.

    1. Is there any problem with the PIN setting?

    I guess you mean the schematic - as I mentioned - please test with a lower output capacitance. Also depopulate the output filters (beyond VCC_VOUT) when testing.

    2. I thought the RESET pin should be supplied with less than <20V, so could you explain 1, 2, and 3 of the error conditions of this pin in more detail?

    RESET is an output pin. It gives the status of the output - It is an open drain output which needs to be pulled up to your desired voltage level. I see you are not using the pin out any where. If you are not using the RESET functionality - you can leave it NC.

    Considering your output is 21-24V - the voltage at the RESET pin with the 100k + 100k resistor ladder will be ~11 - 12V so I guess that should not damage the device. In typical use cases - the pin is pulled up to some standard logic levels like 3.3V/5V which are read by an external MCU. Hence you can tie it to Vcc/BIAS as seen in the WEBENCH design. 

    Section 8.3.6 in the datasheet covers the operation of this pin in detail. You may want to refer to that. 

    Thanks,

    Best Regards,

    Taru