0
\$\begingroup\$

I've been trying to build a CMOS-level XOR gate. I've been told my implementation won't work because some of the transistors are in "source follower mode".

I've tried looking this up, but I can't find a simple description of the term.

In simple terms, what does source follower mode mean?

(Here's the XOR circuit for reference)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

2 Answers 2

3
\$\begingroup\$

This is a source follower.

schematic

simulate this circuit – Schematic created using CircuitLab

The output will be equal to the input minus the MOSFET's threshold voltage; the source 'follows' the gate. It's a configuration with no voltage gain (i.e. gain ≈ 1), but high current gain. The lack of voltage gain means it's to be avoided in logic circuitry; you want outputs that snap high or low quickly, even if the input goes a bit slowly.

In your schematic, transistors M8, M7, M3, and M4 are source followers.

\$\endgroup\$
1
\$\begingroup\$

A source follower configuration is a MOSFET configuration in which the output is taken from the source side of the MOSFET. When operated as a linear amplifier, the output potential on the source of the MOSFET is approximately \$V_{th}\$ less than the potential at the gate. So, the output/source signal "follows" (at a lower level) the input/gate voltage.

There are indeed MOSFETs in your schematic configured as source followers.

\$\endgroup\$
4
  • \$\begingroup\$ So it's considered a "source follower" because the: "Source Voltage -> Follows -> the Voltage of the Gate"? \$\endgroup\$
    – Connor
    Commented Mar 10, 2023 at 21:09
  • 1
    \$\begingroup\$ Yes, as long as the MOSFET is conducting. \$\endgroup\$ Commented Mar 10, 2023 at 21:10
  • \$\begingroup\$ What causes this to happen? \$\endgroup\$
    – Connor
    Commented Mar 10, 2023 at 21:10
  • 1
    \$\begingroup\$ If the gate-to-source voltage is high enough, the MOSFET will conduct and "pull-up" the source. \$\endgroup\$ Commented Mar 10, 2023 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.