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Draw 2 input CMOS NOR gate and using equivalent inverter approach and derive expression for $V_{IL},V_{IH}, V_{OH} ,V_{OL}$
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2-Input NOR Gate
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CMOS voltage transfer characteristics enter image description here

Region of Opertion of Transistors in a Symmetrical Inverter

Region Input Voltage $ V_i $ Output Voltage $ V_o $ NMOS Transistor PMOS Transistor
1 $ V_i \leq V_{TN}$ $ V_{OH} = V_{DD} $ Cutoff Linear
2 $ V_{TN} \lt V_i \leq V_o + V_{TP}$ High Saturation Linear
3 $ V_i \approx V_{DD} /2 $ $ V_{DD}/2 $ Saturation Saturation
4 $ V_o + V_{TN} \lt V_i \leq (V_{DD} + V_{TP})$ Low Linear Saturation
5 $ V_i \geq (V_{DD} + V_{TP})$ $ V_{OL}=0 $ Linear Cutoff

Calculation of $V_{IL}$

Equating currents for saturated nMOS and nonsatrated pMOS device (Region 2):

$\frac{K_n}{2}(V_{in}-V_{Tn})^2$

=$\frac{K_p}{2}(2(V_{DD}-V_{in}-\begin{vmatrix}V_{Tp}\end{vmatrix})(V_{DD}-V_{out})-(V_{DD}-V_{out})$

The derivation condition $(dV_{out}/dV_{in})=-1$ has to be evaluated for
$I_{Dn}(V_{in})=I_{Dp}(V_{in},V_{out}):$

$\frac{dV_{out}}{dV_{in}}=\frac{(dI_{Dn}/dV_{in})-(\partial I_{Dp}/\partial V_{in})}{(\partial I_{Dp}/\partial V_{out})}=-1$

Evaluating the derivative gives:
$V_{IL}(1+\frac{K_n}{K_p}) =2V_{out}+ \frac{K_n}{K_p}V_{Tn}-V_{DD}-\begin{vmatrix}V_{Tp}\end{vmatrix}$
This equation has to be solved together with the first equation=>$V_{IL}$


Calculation of $V_{IH}$
At the point $V_{IH}$ the NMOS device is nonsaturated and the PMOS transistor is saturated (region 4):
$\frac{K_n}{2}[2(V_{IH}-V_{Tn})V_{out}-V_{out}^2]=\frac{K_p}{2}(V_{DD}-V_{IH}-\begin{vmatrix}V_{Tp}\end{vmatrix})^2$

The derivation condition $(dV_{out}/dV_{in})=-1$ has to be evaluated for

$I_{Dn}(V_{in},V_{out})=I_{Dp}(V_{in}):$
$\frac{dV_{out}}{dV_{in}}=\frac{(dI_{Dn}/dV_{in})-(\partial I_{Dn}/\partial V_{in})}{(\partial I_{Dn}/\partial V_{out})}=-1$

which gives:
$V_{IH}(1+\frac{K_p}{K_n}) =2V_{out}+ V_{Tn}+\frac{K_p}{K_n}(V_{DD}-\begin{vmatrix}V_{Tp}\end{vmatrix})$

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