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Buck Converters: Why NFETs and not Diodes

  1. May 25, 2006 #1
    Hi,

    I am going through Franco's "Design with Operation Amplifiers and Analog Integrated Circuits" (an undergraduate textbook) and I am on the chapter about regulators and my question is specifically about the buck kind of switching regualtors.

    As I understand the basic operation of a buck regulator to be when the switch is closed the magnetic field of the coil is charged. When the switch is open, then energry stored in the coil is used to power the load. However this requires the circuit to be closed as circuit with an open would have no current. But the switch was opened, so they way to get around this is to put a diode with the positive terminal to ground on the inductor terminal that connects to the switch. This way the diode will conduct with the switch is closed allowing current to flow from the coil to the load.

    So everything seemed pretty straight forward until I googled for a buck switcher datasheet to check some out. Almost all the ones I found had NFETs instead of diodes to conduct current when the switch is open. Now I am trying to figure out why. Well, more like I want to understand the tradeoff, diode vs NFET.

    I think it should be possible to make a diode with a lower on resistance than a FET so my only guess is that is has something to do with Vds. But this seems weird to me because to saturate the transistor Vds > Vgs. This makes it seem like Vds should be bigger than the forward drop of a diode which means the duty cycle of the switch will have to be longer to maintain the same output voltage, lowering efficency.

    Sorry for the long question but I wanted to show that I put some thought into this and unfortunately it seems like way too much. I should be doing my homework!
     
  2. jcsd
  3. May 25, 2006 #2

    berkeman

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    Staff: Mentor

    That diode is often referred to as the "flywheel" diode, because it does what you say -- it provides a path for the current return when the highside switch transistor is open.

    The 0.7V forward voltage drop (or a little less with a Schottky diode) is a main source of efficiency loss in the buck regulator, so the alternative that you are looking at is to use a transistor in saturation (0.2V or 0.3V) to provide the flywheel function. It's more complicated, though, because the buck regulator has to synchronously turn on the flywheel element at just the right times, in order to keep the current flowing smoothly. The diode solution turns itself on and off, so it does not require a synchronous flywheel drive signal.

    The active circuit with the synchronous flywheel transistor is generally more expensive than the simple passive diode circuit, but the gain in efficiency (Pout/Pin) is sometimes worth the extra cost.
     
  4. May 25, 2006 #3
    Thanks Berkeman,

    I think I understand. So what parameters are most important in the Flywheel transistor? Obviously it must be able to handle the maximum amount of load current but I guess it should also have a saturation resistance low enough so that the Vds is small when it's conducting. Or I guess one could put two in parallel.

    Does the regulator try to turn off and on the high and low transistors at exactly the same time? This way they would both in linear mode. If they weren't it seems like there would be a short period of time where the path was closed, due to both being off, which would make Di/Dt very high and send the coil crazy. Or they would both be on which would essential short the source supply to ground through the transistors.

    I also saw some datasheets that have a zener diode in parallel with the NFET. Is this for protection in case the flywheel element is not turned on at precisely the right time? Perhaps those that have it can turn on the flywheel accurately, and those that don't can't.

    Thanks Again!
     
  5. May 25, 2006 #4

    berkeman

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    Staff: Mentor

    Probably a Schottky in parallel with the flywheel transistor, rather than a Zener. A well saturated transistor will have a lower Vds than the Schottky diode, but the diode will help ease/guarantee the switching transition that you mention.
     
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