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

tri-state inverting buffer

tri-state with resistor

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?


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.

  • \$\begingroup\$ Your last sentence - the actual question - got a bit garbled between your brain and your keyboard. What is the question and where is the '?'? \$\endgroup\$ – Transistor Jul 20 '17 at 17:27
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    \$\begingroup\$ 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 \$\endgroup\$ – sstobbe Jul 21 '17 at 6:20

EN is an ENable signal. Depending on its signal the Mosfets M2A, M1A, M3A, M4 are blocked or active, enabling the output Vout when M2A, M1 and M4 are active together.

When working as digital inverter the M1 and M4 are used to activate/ deactivate the circuit, but in your circuit there is a self-biasing resistor that will alter the transistors biasing and operation mode, thus the mosfets aren't in the cut-off/ohmic regions. Thus the bias resistor should be considered in the gain equation. In this configuration the system will work like a push-pull(Totem Pole) amplifier, so it will be subject to the cross-over effect, and I think you cannot compensate that. I suggest you to take a look on this links, maybe them help:

  • \$\begingroup\$ The corssover effect isn't present due to the fact that there is always current flowing through the transistors. \$\endgroup\$ – Syphirint Jul 21 '17 at 11:48

If M2b and M3B are long channel, their Rout will be HIGH and the gain will be high.

Thus the schematic, without W/L, the uA/volt^2, and the Lambda Effect, does not provide enough information to evaluate the gain.

  • \$\begingroup\$ I agree, but I'm using short channel transistors and for theoretical expression, those values aren't necessary unless one want to make assumptions about the circuit. \$\endgroup\$ – Syphirint Jul 21 '17 at 11:07

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