Substituting PMOS for NMOS?

[carbonzit]

Maybe a dumb question, perhaps already answered elsewhere:

Can one substitute a P-channel MOS transistor for a N-channel one in a working circuit, by swapping D and S? (Assuming a PMOS complementary to the NMOS.)

If not, why not?

 

[Sceadwian]

Generally no, because then if you switch drain/source relative to the load the load will be on the wrong side of the FET. It MAY work under some circumstances, mind you that pfets given the same physical junction size will always have a higher on resistance than an nfet, so that will further alter under what conditions simply swapping things might work with as it will effect power dissipation drastically if the load changing position relative to the source/drain will in many cases cause it to be in it's linear region much smoke will come from this.

 

[alec_t]

Simply swapping as you describe may also mean that the gate drive voltage will need to be inverted and/or level-shifted.

 

[crutschow]

If you swap an P-Channel for an N-Channel the source and drain are interchanged. So if the circuit was originally a common-source circuit it will become a source-follower, and if it was originally a source-follower it will become a common-source circuit. Generally the operation of those two types of circuits is quite different with the common-source circuit having voltage gain >1 and the source-follower having a gain of slightly <1.

 

[simonbramble]

it is (roughly) the same as asking if you can switch an npn for a pnp and change over the emitter and collector terminals. Generally the answer is No. Also with PMOS and NMOS FETs you have the body diode (that opposes conventional current flow) to think of

 

[carbonzit]

If you swap an P-Channel for an N-Channel the source and drain are interchanged. So if the circuit was originally a common-source circuit it will become a source-follower, and if it was originally a source-follower it will become a common-source circuit. Generally the operation of those two types of circuits is quite different with the common-source circuit having voltage gain >1 and the source-follower having a gain of slightly <1.

That makes perfect sense. However, I should have pointed out that the application I have in mind uses the MOSFET as a switch (in a boost converter). So does this situation still apply? Seems to me it should make little difference when the "load" is simply the current passing through the transistor.

 

[ericgibbs]

That makes perfect sense. However, I should have pointed out that the application I have in mind uses the MOSFET as a switch (in a boost converter). So does this situation still apply? Seems to me it should make little difference when the "load" is simply the current passing through the transistor.

hi cz,
In the majority of cases it would not be reasonable just to swap the two types.

If you have a specific application, a circuit diagram would help us to explain what problems swapping types may cause.

 

[crutschow]

That makes perfect sense. However, I should have pointed out that the application I have in mind uses the MOSFET as a switch (in a boost converter). So does this situation still apply? Seems to me it should make little difference when the "load" is simply the current passing through the transistor.

It makes no difference to the load but it makes a big difference for the bias polarity the transistor gate requires to turn on the transistor. Remember that an N-MOSFET must have a plus gate-source voltage and a P-MOSFET must have a minus gate-source voltage to turn on.

Thus, for example, if you had a typical boost positive-output switching regulator with the switch going to ground and you use a P-MOSFET with a grounded drain (for correct transistor polarity), you will need a negative voltage to fully turn it on as a switch. That means you need a negative as well as a positive supply voltage.

 

[carbonzit]

Thus, for example, if you had a typical boost positive-output switching regulator with the switch going to ground and you use a P-MOSFET with a grounded drain (for correct transistor polarity), you will need a negative voltage to fully turn it on as a switch. That means you need a negative as well as a positive supply voltage.

Ah, so (this was also mentioned earlier by alec_t). So if this is the normal configuration, using an N-MOSFET:

**broken link removed**

Then this scheme, using a P-channel device:

**broken link removed**

would only work if the gate drive polarity was as shown (negative with respect to ground), is that correct?

I'm a little confused (polarities always do that to me). Isn't the gate polarity and voltage reckoned with respect to the source (the datasheet parameter is Vgs)? So does a P-MOSFET need its gate driven negative (with respect to the source) in order to turn it on?

All this is pretty academic, as it's pretty obvious I'm not going to be able to do a P-for-N device swap in this application, but I'd like to get this concept a little clearer in my mind.

 

[crutschow]

Ah, so (this was also mentioned earlier by alec_t).
Then this scheme, using a P-channel device:...
would only work if the gate drive polarity was as shown (negative with respect to ground), is that correct?

I'm a little confused (polarities always do that to me). Isn't the gate polarity and voltage reckoned with respect to the source (the datasheet parameter is Vgs)? So does a P-MOSFET need its gate driven negative (with respect to the source) in order to turn it on?

Yes the P-MOSFET gate would need to be driven negative with respect to the source to turn on. Since, when the transistor is turned fully on as a switch in your circuit, the source and drain are essentially at the same voltage (common or 0V), this means the gate voltage must be negative with respect to ground.

The gate must also must be set near the positive supply voltage to turn it off.