What would the drain voltage be if NMOS source voltage is not ground

nmossourcetransistors

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Hi I'm a beginner trying to learn CMOS cuicuit.

Let's say we have two NMOS connected as shown.
Given: A, B, C are connected to Vdd = 5V;
VtM1 = 0.6V is the threshold voltage of M1;
VtM2 = 0.5V is the threshold voltage of M2.

Could someone please explain why the voltage of node "y" is Vy = Vdd – VtM1 – VtM2 ?
(which means if the input of M2, Vx = 5V – 0.6V = 4.4V, the output of M2 is Vy = 4.4V – 0.5V = 3.9V ?)

Best Answer

The voltage at node Y is not what you say it is:

$$V_Y\ne V_{DD}-V_{THM1}-V_{THM2}$$

Rather,

$$V_Y=V_{DD}-max\left(V_{THM1},V_{THM2}\right)$$

So, for \$V_{DD}=5\rm{V}\$, \$V_{THM1}=0.6\rm{V}\$, and \$V_{THM2}=0.5\rm{V}\$, then \$V_Y=4.4\rm{V}\$. M1 is in saturation with \$V_{DSM1}=0.6\rm{V}\$, while M2 is in triode with \$V_{DSM2}=0\rm{V}\$.

If you flipped them around, so that \$V_{THM1}=0.5\rm{V}\$ and \$V_{THM2}=0.6\rm{V}\$, then they would both be in saturation. M1 would have \$V_{DSM1}=0.5\rm{V}\$, and M2 would have \$V_{DSM2}=0.1\rm{V}\$. \$V_Y\$ would still be \$4.4\rm{V}\$.

I should note that, since the body is disconnected from the S/D terminals, these MOSFETs will experience body effect. Therefore their threshold voltages will be substantially higher than we have considered in this analysis.

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