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LMR36006-Q1: How does he stop inductor current while Hiccup mode?

Part Number: LMR36006-Q1
Other Parts Discussed in Thread: LMR36006

Hi team, 

Could you tell me how LMR36006-Q1 stops inductor current while Hiccup mode?
Figure 12 shows short circuit condition and LMR36006-Q1 is in hiccup mode. After 94ms hiccup mode starts, inductor current flows again and it is over its current limit. LMR36006-Q1 starts to turn on high side MOSFET to flow supply inductor current.

In this condition, does high side MOSFET turn on over 20ms? Or does low side MOSFET turns on after high side switch turns on and it leads to hits high side over current? I guess low side valley current limit operates and skip turning on high side MOSFET, is it correct?

If yes, how do you stop inductor current through low side MOSFET? If hiccup is continue to operate, LMR36006-Q1 should stop inductor current. However, this current is able to flow through body diode of low side MOSFET. Is there additional B2B shutdown MOSFET on low side?

Regards,
Ochi

  • Hi Ochi,

    LMR36006 high-side current limit is 1A, the low-side current limit is 0.8A, so during short circuit condition, the inductor current peak is 1A and valley is 0.8A, the average value is 900mA, On Fig12, the IL should be output current which is same as inductor average value 900mA.

    After running in short circuit for 20ms, both high-side and low-side is turn-off, inductor current will flow on low-side FET body diode which Vf is 0.8V typical.

    Inductor current will decrease to zero within:

    T=Ipeak*L/0.8V=1A*10uH/0.8V=12us.

    B R

    Andy  

      

  • Hi Andy-san,

    Thank you for your supporting.

    Could you tell me what happen if inductor current reaches high side current limit 1A? Also what happen if inductor current reaches low side current limit 0.8A?

    In the case high side and low side MOSFET suddenly turns off if inductor current reaches their current limit, inductor current does not keep 1A or 0.8A for 20ms; inductor current goes zero in 12us instead of 20ms. 

    Or if inductor current reaches their current limit, high side and low side MOSFET operate such like constant current operation in their current limit; 1A and 0.8A for 20ms?

    Regards,
    Ochi

  • Hi Ochi,

    what happen if inductor current reaches high side current limit 1A? ------High-side switch turn-off immediately, then low-side turn-off after dead time

    what happen if inductor current reaches low side current reaches low side current limit 0.8A-------When the low-side switch turns on, the inductor current begins to ramp down. If the current does not fall below ILIMIT before the next high-side turn on cycle, then that cycle is skipped, and the low-side MOSFET keeps turn-on till the current falls below ILIMIT 0.8A.

    In the case high side and low side MOSFET suddenly turns off if inductor current reaches their current limit, inductor current does not keep 1A or 0.8A for 20ms; inductor current goes zero in 12us instead of 20ms------Yes. 

    Or if inductor current reaches their current limit, high side and low side MOSFET operate such like constant current operation in their current limit; 1A and 0.8A for 20ms?---------No, High-side switch turn-off immediately once current trig the 1A limit, in worst case if the output is hard short to 0V, low side FET RDS(on) is 0.15ohm, the L/R time constant is T=10uH/0.15ohm=66.7us, which mean the inductor current will drop below 0.8A within 0.2T=13.3us. so the converter will skip about 5 cycle 12.5us for Fsw=400KHz operation, then high-side will turn-on at 6th cycle. the inductor current is regulated between peak 1A and valley 0.8A by this way, you will see the average peak is 0.9A at output.

    B R

    Andy 

  • Hi Andy-san, 

    It is really clear for me to understand current limit and hiccup operation. Thank you so much.

    Regards,
    Ochi