MOSFET partially on question

Discussion in 'General Electronics Chat' started by 01cooperl, Jan 9, 2013.

  1. 01cooperl

    Thread Starter New Member

    Jan 9, 2013
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    I am using an N-channel MOSFET (500V, 100A, 0.055Ω) for a project where I wish to use the transistor to switch on a circuit containing a fully charged capacitor 330V, 80uF, therefore allowing it to discharge. I wanted to supply an analog voltage pulse to the transistor to that it can fully/partially switch the circuit depending on the pulse. The pulse lengths will be ~0.5ms and will varying in voltage within this pulse.

    My initial question is, will the MOSFET produce alot of heat when partially switched on even for the full 0.5ms? and will this heat energy come from the 330V capacitor supply on the drain and source pins, or the voltage pulse to the gate pin? As the focus of this project is energy efficiency from the capacitor.

    If this does create a significant heat energy losses from the capacitor, would it be better to can a series of short, sharp digital signals to a MOSFET IC Driver which can rapidly turn the MOSFET on and off to almost mimic the partially on sections of the pulse?

    Any help on this would be greatly appreciated. Thanks
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    3,820
    Heat (power) dissipation is calculated as (voltage difference from Drain-to-Source) x (amperage flowing from Drain-to-Source).

    As you can see, there is no current flow when the MOSFET is off, so no heat. There is also little voltage difference when the MOSFET is ON, so no heat.

    You will be operating in the linear range as the capacitor discharges so lots of heat will be built up. For this applications, you may be better off with a bipolar transistor (voltage drop BASE to EMITTER - check datasheet but a constant value in any given transistor - commonly 0.7 volts (unless it is a darlington).

    As you suggested, varying the duty cycle at some frequency (Pulse-width modulation) is a better way to vary the time-averaged power through a MOSFET but requires more complicated circuitry. Remember, unless you vary the duty cycle (change the aspect ratio of your rectangle wave) you do not vary the duty cycle. Swiching it on/off faster simply causes more heat in the MOSFET but no change in power to the load (see: Switching losses).
     
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  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    OK, let's take your numbers: 80uF charged to 330V, and discharging in 0.5 mS. That gives us:

    I = C dV/dT = 80uF * 330V / 0.5mS = 5.28 amps

    The energy in the cap is:

    E = 1/2 C V^2 = 1/2 80uF (330V)^2 = 4.356 Joules

    Now a watt is 1 joule in 1 second, so if you do this once a second you are only dissipating 4.356 watts in the FET.

    However, there is a big assumption here, that the current will be a constant 5.28 amps. You would need some further circuitry (such as the following image) to guarantee that is the current.

    [​IMG]

    Note this is a starting point for discussion... the fet and the R1 resistor would need different values. The pot to the fixed voltage would be replaced with your variable voltage pulse input.
     
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  4. crutschow

    Expert

    Mar 14, 2008
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    As ErnieM noted, the energy stored on the charged capacitor is 4.356 Joules. Neglecting any other series resistance, this energy will be dissipated in the transistor, independent of how fast you turn the transistor on or even if you rapidly turn it on and off. The power dissipation in the transistor then equals the number of discharges per second times 4.356J. Even doing it once per second will require a heat sink for the MOSFET.
     
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  5. 01cooperl

    Thread Starter New Member

    Jan 9, 2013
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    Thank you all for you posts. This has been helpful just to see where I can at least move forward. Apologies for this, but I will later be adding a copper wire coil of 520 turns over 30mm (turn density = 17.3 turns/mm) with an internal diameter of 4mm and a wire diameter of 0.5mm. This is place on the high side of this circuit. The aim is to avoid, where possible, any energy losses due to the switching of the transistor.

    I expect that with the coil in place, a fully charged capacitor would then discharge and cause minimal heating effects if the transistor is switched fully on straight away and from the duration of the discharge. However, it is when I switch the MOSFET partially, which will be in the linear region, that I am concerned at will cause heating effects as it is providing a resistance to the discharge. The main aim is to manipulate the discharge pulse from an exponential decay curve, to something that is more gradual or even some kind of pre defined curve that would eventually be sent to the gate pin via a computer or microcontroller.
     
  6. crutschow

    Expert

    Mar 14, 2008
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    An inductor in series with the capacitor will obviously limit the current based upon the resonant frequency of the LC circuit. In that case, limiting the current rise by switching the MOSFET rapidly on and off will definitely reduce its power dissipation over operating the MOSFET in the linear region.
     
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