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I think, it is time I understand working principle of MOSFET transistors...

Illustration of Four Different Possibilities

Suppose that;

  • I want to switch voltage on a resistive load by a MOSFET transistor.
  • Any control signal between -500V and +500V can be easily generated.
  • The transistor models in the picture are not important, they can be of any other appropriate model as well.

Question #1
Which of the driving techniques are feasible? I mean, which of these four circuits would work with correctly applied control signals?

Question #2
What is the range of the voltage level of the control signals (CS1, CS2, CS3, CS4) that loads and unloads the resistor? (I understand that exact boundaries of on and off states must be calculated individually. But I'm asking for approximate values to understand the working principle. Please give statements like "In circuit (2), the transistor turns on when CS2 is below 397V and turns off when above 397V.".)

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Suppose that ... Any control signal between -500V and +500V can be easily generated - Some supposition! I guess we work in different industries. –  Kevin Vermeer Sep 1 '11 at 14:33
    
@Kevin Vermeer: I'm only trying to learn the theory for now. –  hkBattousai Sep 1 '11 at 14:45
    
Ah, that makes more sense. Are you learning with a goal of entering a high-voltage field? You may be adding complications that make learning harder by trying to work on 400V signals. –  Kevin Vermeer Sep 1 '11 at 15:36
    
@Kevin Vermeer: I wanted people to answer my question in a more general way, considering all. So that they can give me more information. –  hkBattousai Sep 1 '11 at 16:12
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3 Answers

up vote 11 down vote accepted

All the circuits are feasible when correctly driven, but 2 & 3 are far more common, far easier to drive well and far safer wrt not doing things wrong.

Rather than give you a set of voltage based answers I'll give you some general rules which are much more useful once you understand them.

  • MOSFETs have a safe maximum Vgs or Vsg beyond which they may be destroyed, This is usually about the same in either direction and is more a result of construction and oxide layer thicknesses.

  • MOSFET will be "on" when Vg is between Vth and Vgsm

    • In a positive direction for N Channel FETs.
    • In negative direction for P Channel FETs.

This makes sense of controlling the FETs in the above circuits.

Define a voltage Vgsm as the maximum voltage that gate may be more +ve than source safely.
Define -Vgsm as the most that Vg may be negative relative to s.

Define Vth as the voltage that a gate must be wrt source to just turn the FET on. Vth is +ve for N channel FETs and negative for P channel FETs.


SO

Circuit 3
MOSFET is safe for Vgs in range +/- Vgsm.
MOSFET is on for Vgs> +Vth

Circuit 2
MOSFET is safe for Vgs in range +/- Vgsm.
MOSFET is on for - Vgs > -Vth (ie gate is more negative than drain by magnitude of Vth.

Circuit 1 Exactly the same as circuit 3
ie the voltages relative to the FET are identical. No surprise when you think about it. BUT Vg will now be ~= 400V at all timed.

Circuit 4 Exactly the same as circuit 2
ie the voltages relative to the FET are identical. Again, no surprise when you think about it. BUT Vg will now be ~= 400V below the 400V rail at all times.

ie the difference in the circuits is related to the voltage of Vg wrt ground for an N Channel FET and +400V for a P channel FET. The FET does not "know" the absolute voltage its gate is at - it only "cares" about voltages wrt source.


Related - will arise along the way after the above discussion:

  • MOSFETS are '2 quadrant' switches. That is, for an N channel switch where the polarity of gate and drain relative to the source in "4 quadrants" can be + +, + -, - - , and - +, the MOSFET will turn on with

    • Vds = +ve and Vgs +ve

    OR

    • Vds negative and Vgs positive
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You don't specify if the control voltage is with respect to ground, or if it can float.

Circuit 3 is the most practical N-channel scheme. The source is at a fixed voltage with respect to ground, which means you can provide a fixed gate-source voltage to control it. The MOSFET will be 'on' anywhere from +2.5 to +12V above ground, depending on the device.

Circuit 1 is tricky. When the MOSFET is off, the source is somewhat of a floating node (imagine a resistor divider with the top resistor enormous) sitting somewhere close to zero. When the MOSFET is on, the source will be very close to 400V assuming saturation. A moving source means that the gate-to-ground control voltage would have to move as well to keep the MOSFET on.

Circuit 1 is better if you reference the control voltage to the source of the MOSFET and not to ground. This is trivial if you're intending to drive the MOSFET with a PWM signal with sufficiently small on-time to allow use of a pulse transformer or charge-pump driver. Fixing the control voltage to the source of the MOSFET means the MOSFET can float up and down as it wants to, without impacting the drive.

Circuit 2 is straightforward like circuit 3. If the control voltage is referenced to ground, proving 397.5V to 388V from gate to ground (-2.5 to -12V from gate to source) will turn the MOSFET on. The source is fixed (always at +400V) so controlling the gate means a fixed voltage is all you need. (Unless your 400V bus collapses, but that's another issue).

Circuit 4, like circuit 2, is tricky. When the MOSFET is off, the source sits near 400V. When it is on, it will fall to near zero. A variable source means a variable gate supply with respect to ground, which is again a messy proposition.

In general, keep your sources fixed where possible, or if they have to float, use a floating supply to control them.

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This is a good question! There are some nuances that the other answers have missed, so I thought I'd chime in.

The short answer is the following:

  • Topology #3 (low-side N-channel switch) is the most commonly used. Since the MOSFET source terminal is connected to ground, the gate drive for this is simple. Connect gate to ground to turn off. Connect gate to a voltage of 5-10V above ground to turn on. Read your MOSFET datasheet and it will tell you what gate voltage you need to provide.

When would you ever not use this topology? The only major reason to do so is if you have a load that needs to have one terminal tied to circuit ground, for electrical safety or to minimize electromagnetic radiation/susceptibility. Some motors/fans/pumps/heaters/etc must do this, in which case you are forced to use the high-side topology #1 or #2.

  • An N-channel high-side switch (Topology #1) has better performance than a comparably sized/priced P-channel high-side switch, but the gate drive is more complicated, and has to be relative to the N-channel MOSFET source terminal, which varies as the circuit switches, but there are specialized gate drive ICs which are meant to drive high-side N-channel MOSFETS. High-voltage or high-power applications generally use this topology.

  • A P-channel high-side switch (Topology #2) has worse performance than a comparably sized/priced N-channel high-side switch, but the gate drive is simple: connect gate to the positive rail ("+400V" in your drawing) to turn it off, and connect gate to a voltage that is 5-10V below the positive rail to turn it on. Well, mostly simple. At low supply voltages (5-15V), you can essentially just connect the gate to ground to turn the MOSFET on. At higher voltages (15-50V), you can often create a bias supply with a resistor and a zener diode. Above 50V, or if the switch has to switch on fast, this gets impractical and this topology is less often used.

  • The last topology #4 (low-side P-channel switch) has the worst of all worlds (worse device performance, complex gate drive circuit) and is essentially never used.

I've written a more detailed discussion in a blog post.

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