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Regarding a MOSFET's reverse recovery power loss

  1. Jul 29, 2016 #1
    Hello everyone, so I'm currently trying to find out how much power will be lost during reverse recovery of a MOSFET's diode. I'm using this paper for reference, https://www.fairchildsemi.com/application-notes/AN/AN-6005.pdf

    the equation they end up with is
    P= Qrr*Vin*Freq_sw

    the problem is, for my application this gives an enormous number with the MOSFET I'm using. I have:
    Qrr = 1.9uC
    Vin = 380V
    Freq_sw = 100,000Hz

    which gives us a total power loss of 72.2W.

    My question is, does this seem right? I can't believe that almost 100 watts will be lost in just the reverse recovery process...

    Any help is greatly appreciated!
  2. jcsd
  3. Jul 29, 2016 #2


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    What is your design output voltage ?
  4. Aug 1, 2016 #3
    300V using PWM. I figured I MIGHT have to multiply by the duty cycle (reverse recovery time * (frequency)) and this gives a much more reasonable answer, but I'm not sure if this is the correct way to go about it.
  5. Aug 1, 2016 #4


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    You appear to be applying a technique that is appropriate below 5V output to an application with a 300V output. The use of a MOSFET as a synchronous diode is only an efficiency advantage when the output voltage becomes comparable with the diode voltage drop.

    If your input voltage is 380V and your output is 300V then you should certainly NOT be using a MOSFET as a synchronous diode.
    A fast-recovery Schottky diode will do the job much more efficiently.

    You should also consider replacing the main switch MOSFET with an IGBT or similar transistor.
    That will again lower switching losses in a high voltage buck converter.

    Look at the design of high voltage inverters and industrial three phase motor speed controllers. Those circuits are more applicable to your application.
  6. Aug 1, 2016 #5


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    Are you talking power efficiency, cost efficiency, or both? If power, why?

  7. Aug 2, 2016 #6


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    Both, but most definitely energy conversion efficiency.
    A MOSFET has a much higher capacitance than a Schottky diode or an IGBT. Each cycle of the ideal diode changes the voltage by 300V twice. That charge must be supplied and then removed quickly, which is the reason for the large power loss.

    The diode in the buck converter is effectively in series with the load. If the diode dropped 1 volt and the output voltage was 2.2V then you could not do much better than 2.2 / 3.2 = 70%. If you replaced the diode with a synchronous diode made with a MOSFET, the diode voltage could be 50mV which makes 2.2 / 2.25 = 98%.
    For 300V output with a 1 volt diode drop the efficiency is close to 300 / 301 = 99.7% which will not be improved by using a high capacitance MOSFET as a synchronous diode.
  8. Aug 4, 2016 #7
    Synchronous rectification using a FET is attractive at low voltage high current uses. Under 5 volts, & more than 1 or 2 amps will benefit from FET use as a rectifier. But at the switching frequencies used in power converters, the FET losses become excessive for high output voltages. Switching loss includes turn-on, turn-off, drain-source capacitance, gate charge, body diode conduction, & reverse recovery loss. These all increase as frequency increases. Turn-on, turn-off, & drain-source capacitance losses, increase with the voltage the device incurs while switching. So for 300 volts, an ultra-fast p-n diode, or Schottky will provide lower loss. The forward drop is very small compared to the output & has little impact on efficiency.

    Regarding the use of IGBT over FETs, that depends on the voltage, current, & frequency. FETs can incur lower loss at high frequencies compared to an IGBT. For high voltage low current, FETs are less lossy. As the current increases, IGBTs become less lossy. IGBTs are generally not offered below 300 volts because they are not competitive with FETs at low voltage. But for applications 300 volts & above they become viable. To determine which is better, the math must be done taking losses into account. For your app unless the current is very large, FET should be fine.

    Have I helped. BR.

  9. Aug 4, 2016 #8


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    The application note AN-6005, linked in the OP, shows how to convert a 12V input to 1.5V output at 15 amp using a synchronous converter. The high losses predicted in AN-6005 when running a 380V input to a 300V output are correct and what would be expected at high voltages.
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