According to all the simple diagrams on the internet, there should be no difference between the source and drain terminals of a MOSFET. So can the two terminals be interchanged in a circuit?
I read something about there being a 'body diode' between the source and the drain, but I didn't fully understand that..

 

According to all the simple diagrams on the internet, there should be no difference between the source and drain terminals of a MOSFET. So can the two terminals be interchanged in a circuit?

One very important point you're missing is that a standard N-ch MOSFET is turned OFF when Vgs=0v, and ON when Vgs=10v; Vgs means voltage on the gate using the source terminal as the reference point.

Let's say that you are using a standard N-ch MOSFET normally; the source terminal is connected to ground, the drain sinks current from the load (which is connected to a 12v supply), and the gate is controlled by a voltage that is switched between 0v and 10v. The MOSFET turns on and off normally.

Now you flip the MOSFET around so that the source and drain connections are swapped. The drain is now at ground potential, but what about the source terminal? Well, the gate is still being controlled by a voltage that is switched between 0v and 10v, but the source terminal is at 12v! So, Vgs will be -2v or -12v, and the MOSFET will never be able to get turned on.

The voltage on the gate needs to always be referenced to the source terminal. If you forget to do that, you will wind up having problems.

I read something about there being a 'body diode' between the source and the drain, but I didn't fully understand that..

That's the other reason you can't just swap the drain and source. An N-ch MOSFET has a diode with anode on the source, cathode on the drain. If the source terminal is about 0.6v more positive than the drain, current will start flowing from the source to the drain whether the MOSFET is turned on or off.

 

But structurally, there is no difference between the drain and the source.. So why wouldn't the MOSFET be ON if ,say, the gate was positive relative to the drain?

 

The body diode is typically in power MOSFets and that will prevent you from swapping the drain and source.

Theoretically, the drain and source can be swapped, and when you do this, the source becomes the drain and the drain becomes the source. For an N-MOSFET, the source is the lower potential, and the drain is the higher potential. The Vgs then still functions as a proper Vgs in this arrangement. But, the body diode is going to be a problem.

Bipolar transistors similarly allow swapping of the emitter and collector, although the transistor properties will be significantly different for most transistor designs.

In principle, one could make a symmetrical BJT or a symmetrical MOSFET that allowed swapping of the terminals with identical performance, but such devices would not be optimized for specific purposes, so I expect it's pretty rare to see this.

 

You said only power mosfets have the body diode.. does that mean that in theory, other types of MOSFETS could be used in either polarity?

I also have one more question:
Why does the gate have to be positive relative to the drain or the source? It attracts electrons from the the p type semi-conductor of the substrate. Wouldn't that mean that the gate should be positive relative to the substrate?

 

You said only power mosfets have the body diode.. does that mean that in theory, other types of MOSFETS could be used in either polarity?

Yes, I think so. I believe that I tried this experimentally over 25 years ago, but even if I did, I can't remember which part it was.

Why does the gate have to be positive relative to the drain or the source? It attracts electrons from the the p type semi-conductor of the substrate. Wouldn't that mean that the gate should be positive relative to the substrate?

The field is what controls the flow in a MOSfet, so it is the voltage relative to the most negative (in N-channel device) terminal. The substrate connection might complicate the device symmetry. For example, the substrate might be tied to the source internally. My device physics knowledge is too rusty to determine all the consequences of asymmetric designs in relation to this question. The simple answer is that the terminals can be swapped in principle, but specific design features may make the performance asymmetric. So, unless you have a good reason and a good understanding of the particular device design, you shouldn't actually do the swapping in practice.

 

While I have never personally tried this, my understanding is a MOSFET is truly a 4 quadrant device, so the following could be used to control an AC signal:

Note the drive is referenced to the midpoint of the 2 devices.

It's on my list of things to try for AC load switching.

I don't remember exactly the type of FET involved, but we do have one product using a bare chip device and the geometry drawing shows the D and S terminals marked as "D/S" and "S/D" and they are 100% interchangeable.

 

Thank you very much.. You're posts have been very informative.

 

I believe that SD210/SD214 are truly symmetrical, with interchangeable source and drain. The body (substrate) is brought out on a separate pin, and is not connected to the source.
These are small signal parts, and are not suitable for high current switching.