This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TPS3852: The voltage on pin ~MR

Part Number: TPS3852
Other Parts Discussed in Thread: TPS3851

The datasheet of TPS3852 says that MR is pulled up to VDD (we use 3.3V). But when I measure the voltage on MR with an oscilloscope voltmeter with 10MΩ input resistance, it is only 1.2V. Why the weak pull up resistance is so large? From 500nA pull up current, I suppose the resistance could not be so large.

  • User,

    As long as /MR is above 0.8V, the logic status is "high". /MR is considered logic low when under 0.25V. Does the device cause a reset when you measure /MR at 1.2V?

    -Michael

  • Dear Michael,
    It was not reset at 1.2V. But I am affraid it will be reset in high and low temperature(-40℃-85℃)。And I don't think 1.2V on MR is normal state.
  • User,

    You are correct that /MR should be pulled up to VDD. We have not seen this before. Can you provide a schematic of your circuit? Is anything connected to the /MR or VDD pins?

    -Michael

  • Michael,

    There is noting connected on MR. See the figure below. R803 is depopulated. please check your IC and measure voltage on MR on your site. Thanks.

  • User,

    I tested the EVM for TPS3851 which has the same /MR internal structure as TPS3852 and I measured /MR to be 1.2V regardless of VDD. After speaking to our systems team, it seems like this device family pulls /MR up to an internal reference voltage making the VIH and VIL specs independent of VDD and also keeping the /MR current independent from Iq. It is correct for /MR to be 1.2V when not connected to anything as it pulls up internally to a reference voltage. As long as the /MR pin is above 0.8V the device will not reset. We gaurantee the correct functionality with /MR floating which means the /MR pin will never drop below 0.8V across all devices and all temperature when /MR is floating.

    I hope this answers your question and clears up any confusion. Please let me know if you have any other questions. Thanks!

    -Michael