This is the circuit I'm trying to understand:

Schmittt Trigger from Rabeay's book

What I understand:

Clearly, whenever \$V_{out}\$ is low \$M_{3}\$ is off and the strength of \$M_{2}\$ and \$M_{4}\$ surpasses that of \$M_{1}\$ so the trip point shifts to the right, conversely, when \$V_{out}\$ is high, the trip point shift to the left. So we have that when \$V_{out}\$ is high we have a switching threshold of \$V_{dd}/2-\Delta_{1}\$ and when \$V_{out}\$ is low we have a switching threshold of \$V_{dd}/2+\Delta_{2}\$

However, why does this even make it work? It's enough to know about whether \$V_{out}\$ is high or low, the Schmitt trigger should choose the switching threshold according to whether \$V_{in}\$ (which is eventually \$V_{out}\$) is increasing or not, so what am I missing?

  • \$\begingroup\$ So, you didn't notice the positive feedback? See this cdn.intechopen.com/pdfs/35073/… page 14. \$\endgroup\$
    – G36
    Jun 16, 2021 at 19:11
  • \$\begingroup\$ To be honest, I'm unaware of how does it help. I'll check out the link, thank you. \$\endgroup\$
    – Essam
    Jun 16, 2021 at 19:19

1 Answer 1


It was all cleared up after reading what @G36 sent, the \$V_{DD}/2+\Delta_{2}\$ threshold is set due to feedback whenever we're at \$V_{in}=0\$ then, the \$VDD/2+\Delta_{1}\$ is set once after that threshold (or for example if we're at \$V_{in}=V_{DD}\$.

The notion of "\$V_{in}\$ is eventually \$V_{out}\$ is incorrect" the buffer has one threshold at a time and would provide a clean output that trips provided an input that say changes from \$0\$ to \$V_{DD}\$ linearly.


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.