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LM61460-Q1: Is there current limit under 6A in this device?

zhiyuan zhang
Intellectual 620 points
Part Number: LM61460-Q1
Other Parts Discussed in Thread: LM61440

Hello expert,

    Input 9~16V，output 7V 5A，power coil is 4.7uH,below is the schemitic

    I want to test the output over current limit function of this device. But when I increase the Eload CC current from 5A to 6A and then 7A,the BUCK is still working on 7V 7A. Is seems that no current limit function.Can you confirm? 

   And if no current limit on this device, how to ensure the safety when output load is over current?

  • 0
    TI__Guru 50120 points

    Hi,

    Can you provide waveform image of your inductor current at the different loadings?

    The current limit in this device is based on peak and valley current limit so if your inductor current does not exceed IL_HS and IL_LS during the upslope and downslope of inductor current, then the part will still be regulating and not hit soft current limit. 

    Regards,

    Jimmy

  • 0 in reply to Jimmy Hua
    Intellectual 620 points

    Hello expert,

        I have calculate the ripple of coil is 0.83A.So the Ip is 0.41A higher that the Io.For IL(HS)=11.5A, does it mean the output current limit is 11.5-0.41=11.09A for my design?

        I feel so surprised for a CV regulator to have a output capacity of 6A but integrate a output limit as 11A. 

  • 0 in reply to zhiyuan zhang
    TI__Guru 50120 points

    Hello,

    The typical IL(HS) is 10.3A and represents the most likely parametric norm at Tj = 25degC. 

    There are two current limits peak and valley, IL(HS) and IL(LS) that the inductor current must be below to not be in current limit. 

    The inductor high-side switch current should be below 10.3A, and the inductor low-side current should be below 7.1A. 

    The reason this device has a high current limit valley is to account for 6A DC current at max duty cycle. 

    If you need the design to have a precise CC (constant current) perhaps you can configure the device for CC/CV so that it never goes above 5A (CC/CV app note). Otherwise you can use the LM61440 (4A) device which has a IL(HS) of 7A and IL(LS) of 4.8A. 

    Regards,

    Jimmy 

  • 0 in reply to Jimmy Hua
    Intellectual 620 points

    Hello expert,

        How to understand IL(LS)  should be under 7.1A? Also in my design the the ripple of coil is 0.83A.So the IL(LS) is 0.41A lower that the Io. So I should have a output current limit as 7.1+0.41=7.51A,am I right?  And IL(HS) now is 7.51+0.41=7.92A not exceed the IL(HS) limit

  • 0 in reply to zhiyuan zhang
    Intellectual 620 points

    And when I short the output,I find a high peak current occur periodically，about 25A. Very higher than the IL(HS).Could you explain this? Could the device broken down?Vin is 16V in my test.

  • 0 in reply to zhiyuan zhang
    TI__Guru 50120 points

    Hello,

    Looking at the scope shot in short circuit condition, the output current rails to the typical high side current limit and is inline with the datasheet parameters. The waveform in green looks to be roughly 1 division 10A/div. The device is operating as expected under short circuit condition. 

    Regards,

    Jimmy

  • 0 in reply to Jimmy Hua
    Intellectual 620 points

    Hello expert,

          I confirmed that the 25A is just the output inrush due to cap discharge when the short occur,so no issue.

         And the CH4 is the output current,not the power coli current,so I don't think we can get the IL(HS) value from that.

        From the waveform the output current is 7.6A when the device try to build up the output.That is inline with my calculation(7.51A).So can I consider  IL_LS is the key point to define  the output current limit in my case?

  • 0 in reply to zhiyuan zhang
    TI__Guru 50120 points

    The key point that defines the output current limit is both IL_HS and IL_LS. If either one of those current limit is tripped then the part will be in current limit mode. Again refer back to the image of current limit. So depending on your inductor selection and average load current (IOUT), you can create a current limit at 6A by calculating what the necessary inductor value to have a small enough ripple such that the inductor valley current never reaches IL-LS.

    Note that the device would still be regulating however it's switching frequency may be scaled back as per description of the overcurrent protection Section 9.3.12 in the datasheet.

    Regards,

    Jimmy