written 6.6 years ago by | • modified 6.6 years ago |
2-Input NOR Gate
CMOS voltage transfer characteristics
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:
$V_{IH}(1+\frac{K_p}{K_n}) =2V_{out}+ V_{Tn}+\frac{K_p}{K_n}(V_{DD}-\begin{vmatrix}V_{Tp}\end{vmatrix})$