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Why does a MOSFET heat up with slow rise/fall times at gate?

  1. Dec 3, 2009 #1
    I have a pulse width modulator thats composed of a triangular wave generator and a comparator. This outputs a pulse width modulated signal, whose duty cycle I set by changing the comparator's V- voltage.

    I have measured the rise and fall times of my output waveform and they are 27uS (for both). I have been told by someone that such a slow waveform will cause power losses in the MOSFET I am using to drive a simple, small, DC Motor. I was told that this because with with a slow rise/fall time, the current and voltage waveforms overlap by a small bit and since P=VI, this causes power loss.

    Is this true? I'm skeptical because I don't understand why current and voltage are out of phase in the first - is this because the motor is an inductive load?
     
  2. jcsd
  3. Dec 3, 2009 #2
    With a slow rise and fall, there is an increased time during which the MOSFET is carrying some current AND has some voltage drop across it. Result. It gets hot.
    Bob S
     
  4. Dec 3, 2009 #3
    Thank you!

    Does this phenomena occur in BJTs too? If so, why are inverter BJTs circuits used to improve the rise/fall times of the waveform?

    In my case, I was recommended that I should insert a BJT inverter between the output of the comparator and the input of the MOSFET. There is a resistor connected between the collector and power supply. This resistor ends up making a RC circuit with the input capacitance of the MOSFET and due to a time constant of 1uS, the rise/fall time is much faster.

    Does this make sense?

    Also, do slow rise/fall times affect the switch speed of a transistor?
     
  5. Dec 3, 2009 #4
    Exactly, an inverter circuit would be able to source/sink more current than the comperator, allowing the mosfet to switch on and off more quickly, meaning that it will be in the linear range for a shorter amount of time, thus reducing switching losses.
     
  6. Dec 3, 2009 #5

    vk6kro

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    Is this true? I'm skeptical because I don't understand why current and voltage are out of phase in the first - is this because the motor is an inductive load?

    In the following circuit,

    FET driver.PNG

    Can you see that if the FET is ON, there will be very little voltage across it, so this multiplied by the current results in very little power.
    If the FET is OFF, there will be maximum voltage across the FET but no current, so the power will be zero or very low.

    It is only when the FET is turning ON and turning OFF that there can be a large product of voltage and current. If this takes a long time, it can result in heating of the FET.
     
  7. Dec 3, 2009 #6
    OSFETs come in four different types. They may be enhancement or depletion mode, and they may be n-channel or p-channel. We are only interested in n-channel enhancement mode MOSFETs, and these will be the only ones talked about from now on. There are also logic-level MOSFETs and normal MOSFETs. We can use either type.

    The source terminal is normally the negative one, and the drain is the positive one (the names refer to the source and drain of electrons). The diagram above shows a diode connected across the MOSFET. This diode is called the "intrinsic diode", because it is built into the silicon structure of the MOSFET. It is a consequence of the way power MOSFETs are created in the layers of silicon, and can be very useful. In most MOSFET architectures, it is rated at the same current as the MOSFET itself.
     
  8. Dec 3, 2009 #7
    Please don't blindly copy and paste stuff from other sites. If you have a link that you think might provide good information, just post it.

    http://robots.freehostia.com/SpeedControl/MosfetBody.html

    Absolutely. A faster transient time allows the resulting waveform to maintain its shape longer as frequency increases until it turns into mush.
     
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