# Switching DC with MOSFET: p-Channel or n-Channel; Low Side Load or High Side Load?

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

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.".)

• 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

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

Q: You mentioned that the circuits 2 & 3 are very common, why is that?
The switches can work in both quadrants, what makes one to choose P channel to N channel, high side to low side? –

A: This is largely covered in the original answer if you go through it carefully. But ...

ALL circuits operate only in 1st quadrant when on: Your question about 2 quadrant operation indicates a misunderstanding of the above 4 circuits. I mentioned 2 quadrant operation at the end (above) BUT it is not relevant in normal operation. All 4 of the circuits above are operating in their 1st quadrant - ie Vgs polarity = Vds polarity at all times when turned on.
2nd quadrant operation is possible ie
Vgs polarity = - Vds polarity at all times when turned on
BUT this usually causes complications due to the inbuilt "body diode" in the FET - see "Body Diode" section at end.

In circuits 2 & 3 the gate drive voltage always lies between the power supply rails, making it unnecessary to use "special" arrangements to derive the drive voltages.

In circuit 1 the gate drive must be above the 400V rail to get enough Vgs to turn on the MOSFET.

In circuit 4 the gate voltage must be below ground.

To achieve such voltages "bootstrap" circuits are often used which usually use a diode capacitor "pump" to give the extra voltage.

A common arrangement is to use 4 x N Channel in a bridge.
The 2 x low side FETs have usual gate drive - say 0/12 V, and the 2 high side FETS need (here) sav 412V to supply +12V to the high side FETS when the FET is turned on. This is not technically hard but is more to do, more to go wrong and must be designed. The bootstrap supply is often driven by the PWM switching signals so there is a lower frequency at which you still get upper gate drive. Turn off the AC and the bootstrap voltage starts to decay under leakage. Again, not hard, just nice to avoid.

Using 4 x N channel is "nice" as
all are matched,
Rdson is usually lower for same $than P channel. NOTE !!!: If packages are isolated tab or use insulated mounting all can go together on the same heatsink - BUT do take due CARE!!! In this case • The lower 2 have • switched 400V on the drains and • sources are grounded, • gates are at 0/12V say. while • the upper 2 have • permanent 400V on the drains and • switched 400V on the sources and • 400/412 V on the gates. Body diode: All FETS that are usually encountered* have an "intrinsic" or "parasitic" reverse biased body diode between drain and source. In normal operation this does not affect intended operation. If the FET is operated in the 2nd quadrant (eg for N Channel Vds = -ve, Vgs = +ve) [[pedantry: call that 3rd if you like :-) ]] then the body diode will conduct when the FET is turned off when Vds is -ve. There are situations where this is useful and desired but they are not what is commonly found in eg 4 FET bridges. *The body diode is formed due to the substrate that the device layers are formed on is conductive. Device with an insulating substrate (such as Silicon on Saphire), do not have this intrinsic body diode, but are usually very expensive and specialised). • you menioned that the circuits 2 & 3 are very common, why is that. The switches can work in both quadrants, what makes one to choose p channel to n channel, high side to low side? – seetharaman Mar 28 '16 at 7:19 • @seetharaman In circuits 2 & 3 the gate drive voltage lies always between the power supply raisl, making it unnecessary to use "special" arrangements to derive the drive voltages. In cct 1 the gate drive must be above the 400V rail to get enough Vgs to turn on the MOSFET. In cct 4 the gate voltage must be below ground. | To achieve such voltages "bootstrap" circuits are often used which usually use a diode capacitor "pump" to give the extra voltage. | A common arrangement is to use 4 x N Channel in a bridge. The 2 x low side FETs have usual gate drive - say 0/12 V, and the 2 high side .... – Russell McMahon Mar 29 '16 at 13:04 • .... FETS need (here) sav 412V to supply +12V to the high side FETS when the FET is turned on. This is not technically hard but is more to do, more to go wrong and must be designed. The bootstrap supply is often driven by the PWM switching signals so there is a lower frequency at which you still get upper gate drive. Turn off the AC and the bootstrap voltage starts to decay under leakage. Again, not hard, just nice to avoid. | Using 4 x N channel is "nice" as all are matched, Rdson is usually lower for same$ than P channel. If pkgs are isolated tab all can go together on heatsink - CARE!!! – Russell McMahon Mar 29 '16 at 13:08
• @seetharaman - (1) See addition to answer. (2) For best treatment of your posts (even comments/questions) you should always: Capitalise correctly (eg You not you). | Use correct punctuation (eg "Why is that?" needs the question mark. | Aim for correct spelling (don't copy me :-) ). (eh menioned -> mentioned). | You may think that I am being "picky". And I am. BUT many people care about such things and will treat your answers badly if they are not well presented. Also aim for sentences that do not "run on" (not always bad). eg " ... quadrants. What ...". What you wrote was OK but it .... – Russell McMahon Mar 29 '16 at 13:26
• .... can help the reader to break things up a little more. POR use eg "Given that the switches can work in both quadrants, what makes ...". | Note that the body diode direction usually makes 2 & 3 better - see answer addition. – Russell McMahon Mar 29 '16 at 13:28

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.

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.