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TPS82084: The input is 0V, the output is increased by 3.3V, will the chip be damaged?

Part Number: TPS82084

The TPS82084 input is 5V, the output is 3.3V, and the output is supplied to the MCU. When the MCU is programming, it directly adds 3.3V to the output. If the input is not added with 5V, will it damage the chip?

When  remove the input power, Output add the 3.3V power, the test input power is 2.7V, and the L22 current is 13mA.
After normal power-on, the TPS82084 works normally.

After communicating with TI technical support, 13mA is the 260Ω resistor consumption of the chip discharge. Calculated the 260Ω resistor power is 0.44W  Will the chip be burned after a long time of discharge?

If the normal input is 5V, but add 3.3V in output, What is the impact of the increased 3.3V on the chip? At this time, can the chip still work normally? Will it be damaged? 

  • Hi Rengui,

    I wrote this app note, found on the product page, to discuss exactly this question: www.ti.com/.../slyt689.pdf

    Based on your description, it sounds like there is very little or no current flow back to circuitry on the input. This is fine.
  • Input 5V, What is the impact of the increased 3.3V on the chip? At this time, can the chip still work normally? Will it be damaged?
  • Hi Rengai,

    I'm not sure that I understand your question. The app note should explain that your case is fine: input floating/open and output biased. There is no issue with current going into the discharge circuit.
  • Hi Chris Glaseri,

    Sorry, I have not stated clearly before.
    The reason for this problem is that 5V is open when programming the microcontroller, only 3.3V is added to the output. This has been confirmed to be no problem. However, the input 5V is controlled by the power diagnosis circuit of FPGA. When the FPGA is not programming or the power diagnosis circuit is abnormal, 5V may output normally. At this time, the added 3.3V and the 5V are present at the same time. At this time, the TPS82084 will Is it damaged?

    rengui sun
  • Hi Rengui,

    Thanks for re-explaining. So, you want to know about the case when the 5V is present at the input and the 3.3V is still present at Vout.

    This is ok as well. If the applied 3.3Vout is higher than the 3.3Vout given by the IC, then the IC will be in power save mode and not switch.

    If the applied 3.3Vout is lower then the 3.3Vout given by the IC, the IC will drive the output up to 3.3V.