• State Resolved
  • Locked Locked
  • Replies 7 replies
  • Subscribers 63 subscribers
  • Views 312 views
  • Users 0 members are here
Support feedback
Options
Options
Related

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.

LMZM23600: 5V supply only gives out .7V

Brian Kelley2
Prodigy 30 points
Part Number: LMZM23600
Other Parts Discussed in Thread: LMZM23601

Hello,

I have had a thread opened about this chip. Case number CS0191236. Please read this case.

Long story short:

Problem: I have this 5V chip creating 5V. I have only two chips on output, but only .7V output is present. I remove one chip, then 5V powers up fine. Then after power is up, I add the second chip and 5V is still good. Therefore, it is a powering up issue. I have 20uf on output.

Two solutions:

1. Add more resistance to 5V output (50 Ohm load minimum) then supply will power up to 5V while having the two chips on output of 5V.  This is not desirable as it eats power.

2. Increase the output capacitance to 86uF. Then 5V chip is able to power up while driving these two chip.

Question:

Why does this chip require more output capacitance than what the datasheet says? I was originally told that it might be a current limit upon power up due to a current inrush. Then not powering up correctly. But actually, I think adding more capacitance would actually cause more of a current in rush upon power up. If possible, I need a better explanation as to why more capacitance on the output fixes my issue. I would like to understand this more before using this chip in a manufacturing environment.

Thanks,

Brian

5V chip : LMZM23600V5silt

  • 0
    TI__Guru 50120 points

    HI Brian,

    Can you provide the schematic of your system? Do you have an existing E2E post that I can refer to since I was not able to find any instance of CS0191236 on the E2E forum. I want to do a quick schematic review to make sure you have the appropriate components for typical 5V operation following the datasheet recommendation. What are these chips and does it have a minimum capacitance requirement to get it operational? Are you switching the load (two chips) in and out of the system? 

    What type of condition are you expecting in your application? (VIN=? ; VOUT = 5V; IOUT = ?). You are correct in assuming that larger output capacitance will have larger inrush current because of the following relationship (App note reference ; I_inrush = Cload * dv/dt). Usually adding more output capacitance is for stability purposes. Since this part has an internally fixed soft-start at 3.5ms there is nothing soft-start related that you can configure.

    Please also provide waveform of VIN, VOUT, and SW for both instance of single chip vs dual chip power-up sequence. The switch node can be probed on the top-side of the exposed inductor. Please make sure not to short SW node to ground as this will permanently damage the part. 

    Regards,

    Jimmy 

  • 0 in reply to Jimmy Hua
    TI__Guru 50120 points

    Please provide updates and comments on my questions above.

    Regards,

    Jimmy 

  • 0 in reply to Jimmy Hua
    Prodigy 30 points

    Hello Jimmy,

    I have talked to TI in case number CS0191236 for several days. They helped me in adding more capacitance.

    It would take a lot of work to re-explain everything. Can you try again to access case number?

    Below is original schematic. I have an output of only .7V referenced to -5V. 1K Ohm load along with two LT6600IS8-20#PBF and 20uF.

    When the two chips are both installed, the supply only gives out .7V. Normally the two chips are always there. But while troubleshooting, I took out one chip and the 5V was produced.

    The while power was still on (5V good), then I added second chip back into circuit and 5V is still good. It will not however, power up 5V when these two chips are installed. No hot switching will be performed when circuit is working normally. Hot switching was just troubleshooting. And now circuit works when having a total of 86uF on output of chip (C57) instead of 20uF.

    Now to be clear and to state my circuit in another way, I am actually generating -5V with this chip. During Powerup: -5V is ~gnd, supply is 24V, output is Grounded.

    Currently the circuit is working fine due to the extra capacitance added. I would just like a better explanation as to why this happened and do I now have a robust design?

    Thanks,

    Brian

    PS. How do I send pictures?

     

     

     

     

  • 0 in reply to Brian Kelley2
    TI__Guru 50120 points

    Okay that makes more sense. I initially thought you were running a regular buck (step-down) conversion but that is not the case as you've clarified.You configured the device from regular buck into an inverting buck boost topology. Doing this will result in a different stability criteria than in regular buck topology. Conversions between buck to inverting buck boost will require more output capacitance to remain stable. 

    For reference you can read my app note about positive input to negative output application pertaining to the LMZM23601 device. 

    With that said having two chips will pull more current than one chip and may result in instability during startup. By having more output capacitance, this will help aid the device's operational stability during both startup and steady state. In your operating since you are not hitting hiccup mode during startup, the amount of capacitance you have tested for (86uF) should suffice for a stable design.

    Regards,

    Jimmy 

  • 0 in reply to Jimmy Hua
    TI__Guru 50120 points

    My previous answer has been rejected but no comment was mentioned, so I assume you have remaining questions on why you need more output capacitance. What specifically are you looking to get more information about that I can help provide insight on? 

    Regards,

    Jimmy 

  • 0 in reply to Brian Kelley2
    TI__Guru 50120 points

    Thank you for using the E2E forum.