Simple DC/DC Synchronous Buck Question

Discussion in 'General Electronics Chat' started by newbie217, Mar 12, 2010.

  1. newbie217

    Thread Starter Active Member

    Apr 12, 2009
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    I have a simple question concerning the operation of a synchronous buck DC/DC circuit. In the standard model most commonly used, a mechanical switch is used to connect the input to the output during the ON state, and a diode is used during the OFF state. When the input is disconnected, the diode forward conducts when the SW voltage falls to ~ -700 mV (diode drop). In this case, the SW node fluctuates from VIN (ON) to ~700 mV (OFF).

    For the case of a synchronous rectifier, an NMOS device is used instead of the free-wheeling diode. If my understanding is correct, the drain and source of a MOSFET are symmetrical and can be used interchangeably. That is, current can conduct in both directions when the device is ON (assuming VGS is above Vt).

    For the high side NMOS, the current from VIN flows from drain to souce. The drain is held at a higher potential than the source (Vsource = rds(on) x Iin)

    For the low side NMOS, however, the current flows from source to drain (if we denote the source as the terminal tied to GND). In this scenario, what happens to the voltage at SW node? Is SW node still negative like when using a conventional diode? Or, can we instead, interchange the source and drain terminals and say that the drain is tied to GND. If we do this, then the source will again be at the lower potential (- rds(on) x IL)? Does it make sense to interpret it this way? I guess it's strange for me to look at the FET and say that the drain would be at a lower potential than the source and still be conducting.

    Thanks for the help
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Yeah, it is kind of strange.

    Try to think of the MOSFET purely as a switch. If Vgs exceeds the threshold voltage sufficiently, it will basically act as a somewhat resistive switch (specified by Rds(on) in datasheets)
    If Vgs=0, it will be off - no conductivity.

    Synchronous rectification is really interesting. I'm pretty new to the concept myself. I was wondering where all the high-power diodes disappeared to when I stumbled on it.
     
  3. JDT

    Well-Known Member

    Feb 12, 2009
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    Yes, MOSFETs do conduct in the reverse direction.

    It is worth noting that all power MOSFETs have a reversed diode connected across the device. This is called the "body diode" and is a consequence of the way the device is constructed.

    This means that when the MOSFET is conducting in the reverse direction, if the voltage drop through the device is greater than about 0.7V, this diode will start to conduct current. Also, in your second circuit, if the rectifier MOSFET is not switched on the circuit will still work, but at lower efficiency, due to this diode.

    Also note that the rectifier device will not necessarily be switched ON all through the power switch OFF period. It should be switched OFF by the controller when the inductor current falls to zero (when the energy in the inductor is fully transferred to the load). A diode does this automatically of course.

    The use of a synchronous rectifier is very useful in circuits where the output voltage is low (example: computer CPU power supplies).
     
  4. newbie217

    Thread Starter Active Member

    Apr 12, 2009
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    So when the low side MOSFET is ON and conducting, say for a load of 120 A and a rds(on) of 6 mohms, then the voltage drop across the low side is -0.7 V. This should be enough to turn on the body-diode. If the low side MOSFEt is ON, do we look at the two resistances as being in parallel? (rds(on) and the diode resistance) Thanks.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    Well, you'll run into package power dissipation problems before that, depending on the package in use. 120A over 6mOhms is 86.4W of power. A TO220 package is limited to around 69A current maximum.

    Body diodes aren't all exactly the same. You'll have to look at the MOSFET in question to get the specifications. The Vf of the body diode will be a function of current through it.
     
  6. newbie217

    Thread Starter Active Member

    Apr 12, 2009
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    That brings up a good point. The NTMFS4833N power MOSFET is an SO-8 FL package rated for 191 A and rds(on) of 3 mohms. This is a power dissipation of 110 W. The part is rated for 125 W. How does so much power get dissipated in such a small package? Pardon my lack of experience, but does this not sound extreme?
     
  7. SgtWookie

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    Jul 17, 2007
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    You need a way to dissipate that heat. Look at layout considerations at the bottom of the datasheet. If you ignore that aspect, you will quickly blow the lid right off the IC with a loud bang, and the magic smoke will make a hasty exit.
     
  8. JDT

    Well-Known Member

    Feb 12, 2009
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    Correct. This is not necessarily a bad thing, provided the device can dissipate the heat (as SgtWookie says).

    But, it the voltage drop is high enough the bring the body diode into conduction then the ON resistance is too high and you are not getting any benefit from using a synchronous rectifier anyway. You need a larger device, or a number in parallel.

    In my experience, the specifications of MOSFET devices are very optimistic. The values of max current and power dissipation are instantanuous values. I have seen TO220 devices rated at more than 100A and in reality the copper pins will melt if this current was sustained for any length of time! Let alone the semiconductor chip!
     
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  9. SgtWookie

    Expert

    Jul 17, 2007
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    You might see MOSFETs with Id ratings of hundreds of Amperes, but that is merely theoretical. A TO220 package is limited to 69A maximum.

    The advantage of the higher Id rating is that the MOSFET power dissipation will be lower due to the low Rds(on). The trade-off is a significantly higher Qg (gate charge), as the physical size of the junction is higher. The higher gate charge can be quite challenging if you want to use even moderately high switch frequencies.
     
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