# Tri-state inverting buffer analog signal analysis

I'm using a tri-state inverting buffer with a self-biasing resistor (represented in the second image) as an amplifier.

I'm trying to do a theoretical analysis, but I'm not sure how to get the gain expression. EN is a clock signal (0 is 0 V and 1 is 1.2 V), which means that during the 1 phase transistors M1 and M4 are basically resistors.

My question is:

Is the gain expression the following?

$$A_{DC}=-(gm_{M_{2B}}+gm_{M_{3B}})*((rds_{M_{1}}+rds_{M_{2B}})\parallel (rds_{M_{4}}+rds_{M_{3B}}))$$

Do gmM1 and gmM4 also enter in the expression in some form?

EDIT:

Considering the self-bising resistor and that the current going through it goes from the output to the input and using Kirchhoff's law on the input and output junction, every fraction containing the resistor is eliminated. I might be doing something wrong.

The W/L is 1um/120nm for the NMOS and 3um/120nm for the PMOS.

After simulating, I can confirm that M1 and M4 are in the ohmic region.

• Your last sentence - the actual question - got a bit garbled between your brain and your keyboard. What is the question and where is the '?'? – Transistor Jul 20 '17 at 17:27
• No gm1, gm4 are not relevant for gain from Vin to Vout, M1 & M4 are either in triode or cut-off. So to be more precise you can replace gm2,gm3 with the degenerated gm from Rdson of M1&M2 – sstobbe Jul 21 '17 at 6:20