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Mosfets, NPN tranistors and mosfet drivers

Discussion in 'General Electronics Discussion' started by Lord_grezington, May 3, 2013.

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  1. Lord_grezington


    May 3, 2013
    Hello all

    I want to switch this Mosfet (FDP61N20) with a gate voltage of 10V.

    Instead of using a Mosfet driver (which can be expensive) is it not possible to just switch it using a fast switching PNP transistor or fast switching little mosfet as mosfets are only voltage driven?

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    The problem is that whilst mosfets are voltage driven, the gate has a certain capacitance.

    To switch the mosfet on or off you need to charge or discharge this capacitance.

    The speed at which the mosfet switches is determined almost completely by how much current you can sink or source from your gate driver.

    As the mosfet switches it goes from a zero dissipation (off) state through to a low dissipation (on) state. In the on state, the losses are i^2R, with R being Rds(on) (presuming you have it turned hard on).

    However, the mosfet (as we've seen previously) doesn't switch on or off instantaneously. It changes over a period of time from an essentially open circuit to a low resistance (often between a few ohms and a few thousandths of an ohm). Whilst it is switching, the instantaneous power dissipation rises (perhaps to a very high value) before dropping to the low I^2R losses where R is Rds(on).

    Some fairly simple calculus can tell you what these switching losses are.

    Thee losses will heat up your device. If the switching is slow, the device will heat up a lot (and may fail). If the switching is fast, the device will heat up far less.

    The losses in a single switching event may be fairly low, even though the peak power might be very high. However, if instead of just switching the device on or off once, you switch it on and off many times per second, you can have bigger problems.

    Let's say our mosfet dissipates 0.1W while turned on (and 0W when turned off). An example might be a 1A load and a 0.1 ohm Rds(on).

    For this particular device, 0.1W is enough for it to feel noticably warm, but not hot (perhaps it's in a TO-92 case)

    Let's also say that a single switching event causes a loss of 1mJ in 10us. If we switch that infrequently, that's 1mW averaged over a second. Not enough to even notice above the 0.1W it normally dissipates when on.

    However that's an average dissipation of 0.001/0.0001 = 100W during that millionth of a second the device is switching. Presumably the device is quite capable of handling this single switching event.

    Now ask it to switch the load faster. Let's ask it to switch the load 1000 times per second. Maybe this is a PWM control of a resistive heating element or a light globe (an essentially resistive load).

    Assume 2000 switching events and a 50% duty cycle.

    So now the losses while the device is on amount to 0.1/2 = 0.05W. And the losses whitest switching are (2000*0.001)/1 = 2W.

    The total power dissipated in now 2.05W. If this device is in a TO-92 case, it's going to fry. So perhaps we now need a device in a TO-220 case and a small heatsink.

    But what else can we do? we don't want to redesign for a larger mosfet?

    Let's increase the current available at the gate by a factor of 10. This will reduce the switching time to (say) 1us. Roughly speaking, our switching losses will also be reduced by a factor of 10.

    So now the switching losses are (2000*0.0001)/1 = 0.2W. The total dissipation is now 0.25W, within the ability of the TO-92 package to dissipate.

    Now let's say we want to use the same transistor in a switchmode power supply. This will be switching at 100kHz, and for ease of calculation we will also assume the other factors remain the same (50% duty cycle, the new faster 1us switching time, etc)

    The dissipation whilst on remains the same: 0.05W.

    The switching losses are now (200000*0.0001)/1 = 20W.

    These losses are *huge*. We need a fairly well heatsinked TO-220 package.

    Again, we need to increase the available gate current, perhaps by another factor of 10, which will drop the dissipation down to 2W.

    So now we are supplying 100 times more current to the base of the mosfet than we were initially.

    If you look at the specifications for power mosfets, you can easily determine the gate currents they are specifying their devices at. As an example here is a datasheet for a relatively low current (3A) surface mount mosfet.

    On the second page you can see specification of maximum Rds(on) of 0.1 ohms (the same as my example)

    You will also see a gate charge of 22nC. if your gate driver was able to supply 2.2mA to the gate, it could turn the device on or off in 10uS (for a 10V supply that's a resistance of about 2k2)

    Once we get to the SMPS example, the gate current rises to 220mA. Clearly we need something able to supply significant current.

    Digging a little deeper, we find that the device is capable of a rise time of 30ns. The rough calculations tell me you need about 750mA of gate current to do that.

    Going back to our example, if you did that, the switching losses would fall to about 6W (which is still beyond what the example device is capable of).

    In this case we need to choose another device. Rds(on) is less important than how fast we can switch it. And that's important, because as we chase lower and lower Rds(on) the gate capacitance rises and we need to employ larger and larger gate currents to switch the device.

    If out load is very high current (so that Rds(on) needs to be low) we are likely to need many amps of gate current just to stop the mosfet going into meltdown.

    And these are the reasons why gate drivers are required.

    However, yes, you can build a gate driver with a pair of transistors. Remember that you need one transistor to turn the gate on, and the other one to turn it off -- a resistor to turn off the mosfet will often not be sufficient.

    Also, since the transistors are driving a capacitive load, you need to place a resistor in series with the gare to ensure that you do not exceed the maximum current for the transistors or the mosfet's gate.

    Your attention to the gate driver is generally needed in proportion to the voltage and current of the load and the frequency you are switching it.
  3. Lord_grezington


    May 3, 2013
    Hi Steve

    Thanks, useful explanation (I need to pull the gate of the Mosfet up and down). Using NPN's or small Mosfet (with a low gate capacitance) will complicate things and not save component cost as I need two of them.

    Will this ( be ok to drive the FDP61N20?

    Another problem which having issue with is the gate resistor, How do I spec this? The Data sheet mentions 25 Ohms but this feels very low. I have tried some calculations and come up with 130 Ohms. Can you help with this?

  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    That seems to be an interesting device. And for the price it may not be worth using discrete components.

    It looks like you'd use a Vcc of 10V in your application, and I would recommend decoupling the input power to this chip with both a 0.1uF capacitor and several uF of low ESR capacitor.

    A 25 ohm gate resistor seems fine. Go with the recommendation unless you have a reason to want slower switching.

    The datasheet for this device is not very revealing about how the input to this device must be driven. It seems that you need to provide 10mA of input current, and that, in turn, suggests some form of current rather than voltage drive.
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