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12306 Views 3 Replies Latest reply: Mar 3, 2012 1:28 PM by DontFeedTheTrolls RSS
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Jul 25, 2011 5:05 PM

FAN73711 High Side Driver for Nfet

Trying to run a FDD2572 n channel mosfet on the high end.  The supply voltage is 48 V and the current draw on the fet is 4 amps.  The 12 supply to FAN73711 gives gate voltage of 60 to 48 V supple.  Just need help setting it up .  The all high end fet drivers require a diode from the supply to the Vb and cap to ramp up the gate voltage.  The data sheet doesn't really help on the diode and the cap as far as parameters. 



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  • designspecialist-power 62 posts since
    Mar 22, 2011
    Currently Being Moderated
    Jul 26, 2011 11:45 AM (in response to TinFoil)
    Re: FAN73711 High Side Driver for Nfet

    TinFoil,

     

    I couldn't find any information in the datasheet about the bootstrap circuit either, but the circuit should be simple enough in order to figure out what components are appropriate.

     

     

    Cboot will need to charge up the input capacitance of the FDD2572 without dropping its own voltage too much.  This means that you should just make it a much larger value, say a couple orders of magnitude larger, than the FDD2572 input capacitance.  It looks like a conservative estimate of the input capacitance of the FDD2572 is 2nF, so make Cboot something like 0.2uF.  Naturally, you would need to be sure it can handle the voltage across it, VDD, without any problems as well.

     

    When Cboot is being charged it is basically part of a series RC circuit with Rboot.  That means Rboot Will limit the amount of current through Dboot when it is charging, and subsequently slow the charge time of Cboot.  If you don't need to switch quickly or at a high frequency you could just make the resistor something relatively large, say 1k Ohm, since at 12V input that would limit the current to less than 12mA.  If you did make Cboot 0.2uF and Rboot 1k Ohm, then the five time constant rule of thumb (5*RC) would make the charge time of Cboot around 1ms.

     

    If you limit the current through Dboot with a 1k Rboot such that 12mA is the max current, then you really just need to choose a Schottky diode for Dboot that can handle 12mA (easy to do ) and that has a reverse voltage high enough above your input voltage (48V in this case) to prevent reverse conduction.

     

    I hope that covers everything you were asking.

     

    Message was edited by: Alec I corrected a mistake I made about the reverse voltage the Schottky diode will see, differentially it would be whatever your input voltage is.

    • MauPham Novice 7 posts since
      Feb 28, 2012
      Currently Being Moderated
      Mar 3, 2012 9:49 AM (in response to Alec)
      FAN73711 High Side Driver for Nfet

      Alec:

       

      Rboot limits the peak charging current through Dboot and Cboot. The average current is much lower (just enough to replenish the average current used to drive the High Side FET FDD2572).

       

      If the time constant of Rboot and Cboot is too high compared to the shortest Off phase of a PWM cycle, the boostrap action will fail at the highest PWM duty cycle.

       

      For example, if the Buck Converter is designed to accomodate the highest duty cycle of 95% at a PWM frequency of 100 kHz, the shortest time available to replenish Cboot is only 0.5 us. The time constant of Rboot and Cboot shouldn't be highest than that by a factor of 5, i.e. 2.5 us.  In this case, Rboot shouldn't be larger than just 12.5 Ohms.

      • DontFeedTheTrolls Novice 3 posts since
        Mar 3, 2012
        Currently Being Moderated
        Mar 3, 2012 1:28 PM (in response to MauPham)
        FAN73711 High Side Driver for Nfet

        Once Cboot is fully charged during start up shouldn't it stay close to fully charged from that point on?  This is assuming it is a large enough value that the amount of charge it sources to the input capacitance of the FET during each switch cycle is a small enough percentage of it's total available charge.  The previous answer said make it something like two orders of magnitude larger than the input capacitance of the FET, so that should meet that requirement.  Therefore, since during a switch Cboot would only charge/discharge a very small amount of its full capacity at the given voltage across it, doesn't the five time constant rule cease to apply in the traditional sense after start up?

         

        A 1k Ohm resistor does seem awfully large, I think usually they are around the 10 Ohm range, but 1k Ohm might be fine if the switching frequency and duty cycle allow.  Here it appears the original poster is just using it as a high-side switch, not in any kind of DC/DC converter, so switching frequency and duty cycle might be irrelevant in this case.

         

        There is a Fairchild Semiconductor application note on bootstrap circuits http://www.fairchildsemi.com/an/AN/AN-6076.pdf that is pretty good in my opinion.  Sections 3 and the first parts of section 4 talk about choosing values and other considerations for the components being discussed.  They also talk about the time constant, but I'm still not convinced that it can be directly correlated to the amount of time afforded for charging Cboot as a function of switching frequency and duty cycle, because as I said before I think an appropriately sized Cboot should only charge/discharge a very small amount of its capacity during each switch.

         

        There's a good chance I'm wrong, but since I'm an Engineer and not a complete Tool I'm obligated to assume my own understanding is correct until someone successfully explains to me why my analysis is in error or otherwise provides me solid proof to the contrary.

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