VIH 5VDD + 3VTN + 5VTP VOL VDD - VTN + VrP and

Consider a simple CMOS inverter:

The pull-up network is equivalent to pMOS and the pull-down network is equivalent to nMOS of the inverter.

Considering nMOS=off and pMOS=on, then the inverter transfer characteristic is:

Here, input voltage

$0<V_I<V_{TN}$

The n channel is OFF and the p channel is ON and the output voltage, V_O = V_DD

When nMOS is saturated and pMOS is Linear:

Both transistors are ON, the input voltage is

$V_I>V_{TN}$

and the output voltage is such that the currents through both the transistors are equal.

The output voltage is given by:

$V_O =(V_I + V_{TP})+\sqrt{(V_{DD} - V_{TN} - V_{TP})(V_{DD} - 2V_I + V_{TN} - V_{TP})}$

for

$V_I \le \frac{V_{DD} + V_{TN} - V_{TP}}{2}$

Now, when the input voltage is

$V_I=\frac{ V_{DD} + \sqrt{\beta }{V}_{TN}-V_{TP}}{1+\sqrt{\beta }}$

then both the transistor are saturated.

The range of output voltage is given by:

$V_I - V_{TN} \le V_O \le V_I+ V_{TP}$

If the input voltage is further increased then nMOS becomes linear and pMOS is saturated. Then the range of input voltage is:

$\frac{ V_{DD} + \sqrt{\beta }{V}_{TN}-V_{TP}}{1+\sqrt{\beta }}< V_I< V_{DD}- V_{TP}$

The current in this condition is:

$\begin{array}{rcl}{I}_D & =& \frac{K_P}{2} {(V_{DD} - V_I- V_{TP})}^2 =K_n \left[\left(V_I- V_{TN}\right)V_O-\frac{1}{2}{V_O}^2\right]\end{array}$

The quadratic equation we get from this is:

$\begin{array}{rcl}& & \frac{1}{2}\end{array}\begin{array}{rcl}& & V_O\end{array}\begin{array}{rcl}& & {}^2\end{array}-\begin{array}{rcl}& & \left(V_I- V_{TN}\right)\end{array}\begin{array}{rcl}& & V\end{array}\begin{array}{rcl}& & {}_O\end{array}+\frac{{\left(V_{DD} - V_I - V_{TP}\right)}^2}{2\beta }=0$

The solution to this equation is:

$V_O=\begin{array}{rcl}& & \left(V_I- V_{TN}\right)\end{array}-\sqrt{{\left(\begin{array}{rcl}& & V\end{array}\begin{array}{rcl}& & \end{array}\right)}^{}}$_I− V_TN2-VDD − VI − VTP2β

When $\beta =1$

$V_O=\left(V_I-V_{TN}\right)-\sqrt{\left(V_{DD}-V_{TN}-V_{TP}\right)\left(2V_I-V_{DD}-V_{TN}+V_{TP}\right)}$