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LM66100: How the RCB circuit works ?

Stanley Xu
Intellectual 1620 points
Part Number: LM66100

Hi TI engineer，

I have a question about RCB application, datasheet show that  the max turn on threshold is VCE- VIN=80mV

 when the power on so the switch on and  VOUT = VCE=VIN, why switch can be maintained ON state? 

by the way, because of R-ON, there are some voltage  drop in the R-ON, but in light load condition or VOUT pin is float, V-drop is too low , I don't think this V-drop can exceed on state threshold.

 

  • +1
    TI__Genius 13300 points

    Hello Stanley,

    It can stay on due to the ON buffer and previous state. If the device is on and Vce begins to increase it will turn the FET off 35mV above, but will only turn back on ~150mV below. In order to activate the always-ON Reverse current blocking feature you have to connect the CE' pin to Vout. However, the channel will not switch off to stop reverse current until Vout is forced above Vin by the constant VOFF. In the case of the LM66100, this means the channel will turn off when Vout - Vin >= 35mV (typ)
    With a Ron of 91 mohm (typ for Vin = 3.6V), this voltage difference could mean a reverse current of 0.38A or more (calculated using Ohm's law) I = 35mV / 91mohm
    A potential workaround is adding a series resistance to your output and connecting the CE' pin to the end of the output series resistor. This will force the comparator to calculate the difference between Vout minus the Voltage across the resistor, and Vin. (Vout - VR) - Vin >= 35mV typ
    Therefore, the current needed to trigger this condition can be calculated as I = 35mV / (Ron + Rout)
    However, this will make your device more sensitive to voltage differences and will increase the resistance of your circuit. 

    Regards,

    Kalin Burnside

  • 0 in reply to Kalin Burnside
    TI__Intellectual 1620 points

    Hi Kalin,

    I have completed a simple circuit simulation by Pspice, circuit show as blow, it‘s look like works well.

    VCE-VIN=0, now is ON state and due to hysteresis only when VCE-VIN>= 35mV (typ) can the  MOSFET transfer from ON state to OFF state，right？

    I have saw that VCE-VIN turn off threshold(min) in datasheet is 0V, I don't why it can be 0V, in my opinion it can be work well with 0V turn off threshold, at this RCB case when the MOSFET turn on, VOUT=VIN=VCE, so the turn off condition met(0V threshold), and MOSFET turn off, then turn on condition met

    and MOSFET turn on.........   So the device output oscillation happens.

     

  • 0 in reply to Stanley Xu
    TI__Intellectual 1620 points

     it can’t be work well with 0V turn off threshold

  • 0 in reply to Stanley Xu
    TI__Intellectual 1620 points

     it can’t be work well with 0V turn off threshold

  • +1 in reply to Stanley Xu
    TI__Genius 13300 points

    Hello Stanley,

    If you would like to test this on an EVM they are available. In the case of Vin ≈ Vce = Vout (very low voltage drop) the state will depend on the previous state. When the device initially turns on Vin -> 5V and Vout follows with Vout < Vin so the FET turns on. When Vout ≈ Vin the device will remain ON due to the previous state being ON. When Vout > Vin by 35mV(typ) the FET will shut off and pass Vin-Vfwd. The device will then turn back on when Vin > Vout+150mV. This means that if a reverse current event occurs and the output is higher than the input the FET will shut off and block reverse current and then turn back on once the output has fallen below Vin again (since Vin - Vfwd will cause the device to turn on once the event ends). See the below figure to get a visual understanding of what I mean. 

    Regards,

    Kalin Burnside

  • 0 in reply to Kalin Burnside
    TI__Intellectual 1620 points

    thank you Kalin, your answer very clear!