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LMX2592: What is the function of pin 30 on this device? Can a lock problem be related to it?

Mikhail Matusov
Intellectual 340 points
Part Number: LMX2592
Other Parts Discussed in Thread: , LMX2582, USB2ANY, LMX2594

Hello gurus,

The LMX2592EVM schematic shows pin 30 connected to GND through a resistor. However, according to both LMX2582 (shown in the schematic) and LMX2592 datasheets this pin is NC. Does it have a purpose? 

The reason I am asking is because I am debugging a 3rd party board that uses LMX2592 as a clock synthesizer and I found that it never locks properly. The same configuration file loaded on LMX2592EVM works fine. There is a difference in the loop filters but I don't believe that to be a problem. Other than that and the pin 30 I can't see any obvious differences in the designs.

Thanks,
/Mikhail

  • 0
    TI__Genius 12705 points

    Hello,

    The typical application diagram in the datasheet is correct. For both LMX2592 and LMX2582 pin 30 should be left no connect.

    Regards,

  • 0 in reply to Liam Keese
    TI__Guru* 81580 points

    Hi Mikhail,

    Pin 30 is NC. However, there is no harm even if you connect it to ground via a resistor.

    Were you using USB2ANY and TICS Pro to make the EVM lock? Can you use them to program the device in the third party board? This will help isolate if this is a software or hardware issue.

  • 0 in reply to Noel Fung
    Intellectual 340 points

    Hi Noel,

    Yes, I am using USBA2ANY and TICS Pro to program the EVM. The chip on the 3rd party board is programmed using the same file generated by TICS Pro but with our own software and FPGA hardware. This hardware/software combination has been proven in our previous design, which used LMX2594. I am confident that the registers are being written correctly. I found that this 3rd party board has a ferrite bead in VCC path of the chip. So, it makes me wonder if that might be a source of the problem... I am probably going to try to replace it with a zero Ohm resistor. Please see below:

    Thanks,
    /Mikhail

  • 0 in reply to Mikhail Matusov
    Intellectual 340 points

    So, the ferrite bead was the source of the problem. Replaced it with a zero Ohm resistor and everything looks good now.

    /Mikhail

  • 0 in reply to Mikhail Matusov
    TI__Guru* 81580 points

    Hi Mikhail,

    Glad that you have resolved the problem. 

    I don't think the ferrite will introduce problem to the chip, could you also try replace the ferrite with a new one and see if it will lock or not? 

  • 0 in reply to Noel Fung
    Intellectual 340 points

    Hi Noel,

    It is well known amongst SI engineeres that ferrite beads placed in the power leads of a device can make things worse. I call them a well-intentioned evil. They can lead to undesired resonances or to a device simply not being able to draw current from the power plane fast enough. In the presence of a ferrite bead, placement of bulk capacitance becomes more critical. In this case there is very little capacitance at the device's side of the bead. 

    /Mikhail

  • 0 in reply to Mikhail Matusov
    TI__Genius 12705 points

    Hello Mikhail,

    Your comments are noted. Primarily these can be used for isolating the device from a noisy power supply. If you have a clean supply often the ferrite bead can be removed or replaced with a lower impedance ferrite bead depending on the system implementation.

    Regards,