In a p-channel MOSFET, if I connect the bulk to the drain instead of connecting the bulk to the source, I will have my bulk-source pn junction forward biased. Since the source potential is by definition lower than the drain voltage (which is the same as the bulk voltage because bulk and drain will be connected), the voltage of the source will be lower than the voltage of the bulk. Will current flow from the bulk to the source? Is it to prevent this that we connect the bulk to the source?

  • \$\begingroup\$ A good discussion which may help a little can be found here: electronics.stackexchange.com/questions/72582/… \$\endgroup\$
    – jonk
    Jun 3, 2017 at 14:55
  • \$\begingroup\$ if I connect the bulk to the drain instead of connecting the bulk to the source, I will have my bulk-source pn junction forward biased. That is simply not true. In a PMOS the Bulk is N and drain and source are both P. So the Source-Bulk is P-N. To use a PMOS properly, the Bulk (N) must be at the highest voltage, the Source (P) must be at a lower voltage. If you forward bias the Source-Bulk PN junction, you're not using the PMOS properly ! \$\endgroup\$ Jun 3, 2017 at 15:28

1 Answer 1


In IC design, there isn't a real difference between the source and drain. Take a look at this image, showing a N-channel MOSFET in a P-Bulk process:

Crossection of N-channel MOSFET

There are two body diodes here - one form source to bulk and one from drain to bulk.

What happens in discrete mosfets is that we choose what the source terminal is by connecting the bulk to that terminal, instead of bringing out all 4 connections (in truth, usually we couldn't bring out all 4 even if we wanted to, due to the nature of how most discrete devices are manufacured - one of the two terminals is always shorted out to the bulk). Because the source and bulk are then at the same potential, as they are shorted, the source-bulk diode is of no intrest. It is still there, but no potential can be put accros it since it is shorted out.

In short: The source is not connected to the bulk because we decided that we should have the bulk and source connected - The source is the source precisely because it is connected to the bulk. If we connected the other terminal to the bulk, we would just call that the source instead.

This is also why sometimes, in IC design, no arrow is drawn on the symbol to show what terminal is the drain or the source. This happens mostly in digital, analog designers don't tend to use these symbols that often. Since there is no physical difference between the two terminals, why pretend there is? enter image description here

  • \$\begingroup\$ Nice. And the build-in source to drain flyback diode is because of the source->bulk->(pn junction between source and drain)->drain path? \$\endgroup\$
    – edamondo
    Jun 3, 2017 at 19:03

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