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WEBENCH® Tools/LMZ13610: LMZ13610

Part Number: LMZ13610

Tool/software: WEBENCH® Design Tools


I have an input of 12V/24V to my system. Also there will be a 7.4V battery input to the system if main power is not there.

We are planning to use LMZ13610 to convert the input voltage (12V/24V/7.4) to 5V. The maximum requirement on the 5V rail will be 8.4A.

 have searched in TI website with my requirements and I found the regulator LMZ13610.

But when I try to design it with WEBNCH, It is showing that these design cannot be created.

I have tried with different Vin minimum. With 9V Vin min, It creates design.

But with less than 9V as Vin min, It is not creating any design.

I need clarification on below ponts.

  • Is these issue with the tool?
  • If not, the datsheet has a efficiency graph which shows that with 8V input . 5V output efficiency is more than 90%. Why the tool can't create that design?
  • Or LMZ13610 cannot support this  requirements?
  •  If LMZ13610 can't work with less voltage inputs, What will be the maximum possible output current at below condition?
    • Vin =6V. Vout=5V : Iout=?
    • Vin =7V. Vout=5V : Iout=?
    • Vin =8V. Vout=5V : Iout=?
    • Vin =8.4Vout=5V : Iout=?

Best Regards,


  • Hello Shafeeq,

    The datasheet specifies typical maximum duty cycle as 85%. 7.4V to 5V is 67.5% (ideal) duty cycle. It sounds like the tool may be trying to satisfy some maximum duty cycle. It is possible that the tool is more conservative, considering corner cases of temperature, current, off time requirements, Rdson, inductor DCR, etc. We will contact the WEBENCH team to see why the designs are not created. There could be an error in the tool.

    I would recommend getting an evaluation board and testing it in your specific system conditions (VIN, ambient temperature, loading). The user guide for the evaluation board design states that 7V minimum is needed for 5V output, so clearly it is possible.


  • Hi,

    Thanks for the reply.

    As per your reply, with 7V input I will be able to get 5V output.

    So I want to use the UVLO function of the chip.

    If the input voltage goes below 7V, I want5 to shutdown the IC.

    I have calculated the the resistor divider circuit values.

    RENT / RENB = (VIN UVLO / 1.274 V) – 1 

    With this equation 49K/10K=(7V/1.274)-1

    But with VEN(rising)  and VEN(falling) I am getting other voltage values.

    So can you please let me know the value of RENT, RENB and RENH for my requirement.

    I have to shutdown the system if input voltage goes below 7V.

    Thanks and Regards,


  • Hello Shafeeq,

    I assume you tried this on the evaluation board. You mentioned you are getting other voltage values for the EN rising and falling. What were these values?


  • Hi Denislav,

    I didn't try with evaluation board.

    I am little bit confused with the three equations.

    RENT / RENB = (VIN UVLO / 1.274 V) – 1

    VEN(rising) = 1.274 ( 1 + (RENT|| 2 meg)/ RENB)

    VEN(falling) = VEN(rising) – 13 µA ( RENT|| 2 meg || RENTB + RENH )

     Can you help me with these equations for my requirement.

    For my requirement i have to shut down the regulator if input goes below less than 7V.

    Best Regards,


  • Hello Shafeeq, 

    I took an evaluation board and changed the top EN resistor to 5.76kohm. The bottom was already at 1.27kohm. I had two supplies OR'd at the input to simulate your system. I turned off the higher voltage supply and slowly ramped down the other to simulate discharging battery. See waveform:

    I created a quick calculator in excel with the equations from the datasheet. See attached. This should help you get started. I would suggest playing with higher resistance values (e.g. 12.7k and 57.6k) to get more hysteresis. Also, I would suggest leaving a footprint/space for a hysteresis resistor so that you can adjust it for clean turn off without chatter when unloading from high load current. 


    I hope this helps.