Chapter 16, Problem 16.9EP

A CMOS inverter is biased at $V_{DD}=1.8\text{V}$ . The transistor parameters are $V_{TN}=0.4\text{V}$ , $V_{TP}=-0.4\text{V}$ , $K_n=200\mu {\text{A/V}}^{\text{2}}$ , and $K_p=80\mu {\text{A/V}}^{\text{2}}$ . (a) Determine the transition points. (b) Find the critical voltages $V_{IL}$ and $V_{IH}$ , and the corresponding output voltages. (c) Calculate the noise margins $N{M}_L$ and $N{M}_H$ . (Ans. (a) $V_{It}=0.7874\text{V}$ , $V_{OPt}=1.187\text{V}$ , $V_{ONt}=0.3874\text{V}$ ; (b) $V_{IL}=0.6323\text{V}$ , $V_{IH}=0.8767\text{V}$ , $V_{OHU}=1.7065\text{V}$ , $V_{OLU}=0.1337\text{V}$ ; (c) $N{M}_L=0.4986\text{V}$ , $N{M}_H=0.8298\text{V}$ )

To determine

The transition points.

Answer to Problem 16.9EP

The transition parameter is $V_{It}=0.7874\text{V}$ the transition parameter $V_{OPt}$ is $1.187\text{V}$ , the transition parameter $V_{ONt}$ is $0.3874\text{V}$ .

Explanation of Solution

Given:

  $V_{DD}=1.8\text{V}$

  $V_{TN}=0.4\text{V}$ , $V_{TP}=-0.4\text{V}$ , $K_n=200{\text{μA/V}}^{\text{2}}$ , and $K_p=80{\text{μA/V}}^{\text{2}}$ .

Calculation:

The transition parameter of a CMOS inverter $V_{lt}$ is,

  $V_{lt}=\frac{V_{DD}+V_{TP}+\sqrt{\frac{K_n}{K_p}}{V}_{Tn}}{1++\sqrt{\frac{K_n}{K_p}}}$

Here, $V_{DD}$ is the biased voltage, $V_{TP}$ is the PMOS threshold voltage, $V_{TN}$ is the threshold voltage of the NMOS, $K_n$ is the process parameter of NMOS, and $K_p$ is the process parameter of the PMOS.

Substitute the values,

  $\begin{array}{c}{V}_{lt}=\frac{V_{DD}+V_{TP}+\sqrt{\frac{K_n}{K_p}}{V}_{Tn}}{1++\sqrt{\frac{K_n}{K_p}}}\\ =\frac{\left(1.8\right)+\left(-0.4\right)+\sqrt{\frac{200\mu }{80\mu }}\left(0.4\right)}{1+\sqrt{\frac{200\mu }{80\mu }}}\\ =\frac{\left(1.8\right)+\left(-0.4\right)+\sqrt{\frac{200}{80}}\left(0.4\right)}{1+\sqrt{\frac{200\mu }{80\mu }}}\\ =0.7874\text{V}\end{array}$

The transition parameter of a CMOS inverter $V_{OPt}$ is,

  $V_{OPt}=V_{lt}-V_{TP}$

Substitute the values.

  $\begin{array}{c}{V}_{OPt}=V_{lt}-V_{TP}\\ =0.7874-\left(-0.4\right)\\ =1.187\text{V}\end{array}$

The transition parameter of a CMOS inverter $V_{ONt}$ is,

  $V_{ONt}=V_{lt}-V_{TN}$

Substitute $0.7874\text{V}$ for $V_{lt}$ , and $0.4\text{V}$ for $V_{TN}$ .

  $\begin{array}{c}{V}_{ONt}=V_{lt}-V_{TN}\\ =0.7874-\left(0.4\right)\\ =0.3874\text{V}\end{array}$

Conclusion:

Therefore, the transition parameter is $V_{It}=0.7874\text{V}$ , the transition parameter $V_{OPt}$ is $1.187\text{V}$ , the transition parameter $V_{ONt}$ is $0.3874\text{V}$ .

To determine

The critical voltages $V_{IL}$ and $V_{IH}$ , and the corresponding output voltages.

Answer to Problem 16.9EP

The low critical voltage $V_{IL}$ is $0.6323V$ , the high critical voltage $V_{IH}$ is $0.8767\text{V}$ , the voltage $V_{OHU}$ at the point $V_{IL}$ is $1.7065V$ ,and the voltage $V_{OLU}$ at the point $V_{IL}$ is $0.1337\text{V}$ .

Explanation of Solution

Given:

  $V_{DD}=1.8\text{V}$

  $V_{TN}=0.4\text{V}$ , $V_{TP}=-0.4\text{V}$ , $K_n=200{\text{μA/V}}^{\text{2}}$ , and $K_p=80{\text{μA/V}}^{\text{2}}$ .

Calculation:

The low level critical voltage $V_{lL}$ is,

  $V_{lL}=V_{TN}+\frac{\left(V_{DD}+V_{TP}-V_{TN}\right)}{\left(\frac{K_n}{K_p}-1\right)}\left[2\sqrt{\frac{\frac{K_n}{K_p}}{\frac{K_n}{K_p}+3}}-1\right]$

Substitute the values,

  $\begin{array}{c}{V}_{lL}=V_{TN}+\frac{\left(V_{DD}+V_{TP}-V_{TN}\right)}{\left(\frac{K_n}{K_p}-1\right)}\left[2\sqrt{\frac{\frac{K_n}{K_p}}{\frac{K_n}{K_p}+3}}-1\right]\\ =\left(0.4\right)+\frac{\left(\left(1.8\right)+\left(-0.4\right)-\left(0.4\right)\right)}{\left(\frac{200\mu }{80\mu }-1\right)}\left[2\sqrt{\frac{\frac{\left(200\mu \right)}{\left(80\mu \right)}}{\frac{200\mu }{80\mu }+3}}-1\right]\\ =\left(0.4\right)+\frac{\left(\left(1.8\right)+\left(-0.4\right)-\left(0.4\right)\right)}{\left(\frac{200}{80}-1\right)}\left[2\sqrt{\frac{\frac{\left(200\right)}{\left(80\right)}}{\frac{200}{80}+3}}-1\right]\\ =0.6323\text{V}\end{array}$

The high level critical voltage $V_{lH}$ is,

  $V_{lH}=V_{TN}+\frac{\left(V_{DD}+V_{TP}-V_{TN}\right)}{\left(\frac{K_n}{K_p}-1\right)}\left[\frac{2\frac{K_n}{K_p}}{\sqrt{3\frac{K_n}{K_p}+1}}-1\right]$

Substitute the values,

  $\begin{array}{c}{V}_{lH}=V_{TN}+\frac{\left(V_{DD}+V_{TP}-V_{TN}\right)}{\left(\frac{K_n}{K_p}-1\right)}\left[\frac{2\frac{K_n}{K_p}}{\sqrt{3\frac{K_n}{K_p}+1}}-1\right]\\ =\left(0.4\right)+\frac{\left(\left(1.8\right)+\left(-0.4\right)-\left(0.4\right)\right)}{\left(\frac{200\mu }{80\mu }-1\right)}\left[\frac{2\frac{200\mu }{80\mu }}{\sqrt{3\frac{200\mu }{80\mu }+1}}-1\right]\\ =\left(0.4\right)+\frac{\left(\left(1.8\right)+\left(-0.4\right)-\left(0.4\right)\right)}{\left(\frac{200}{80}-1\right)}\left[\frac{2\frac{200}{80}}{\sqrt{3\frac{200}{80}+1}}-1\right]\\ =0.8767\text{V}\end{array}$

The output voltage $V_{OHU}$ at the point $V_{IL}$ is,

  $V_{OHU}=\frac{1}{2}\left[\left(1+\frac{K_n}{K_p}\right)V_{IL}+V_{DD}-\left(\frac{K_n}{K_p}\right)V_{TN}-V_{TP}\right]$

Substitute the values,

  $\begin{array}{c}{V}_{OHU}=\frac{1}{2}\left[\left(1+\frac{K_n}{K_p}\right)V_{IL}+V_{DD}-\left(\frac{K_n}{K_p}\right)V_{TN}-V_{TP}\right]\\ =\frac{1}{2}\left[\left(1+\frac{200\mu }{80\mu }\right)\left(0.6323\right)+\left(1.8\right)-\left(\frac{200\mu }{80\mu }\right)\left(0.4\right)-\left(-0.4\right)\right]\\ =\frac{1}{2}\left[\left(1+\frac{200}{80}\right)\left(0.6323\right)+\left(1.8\right)-\left(\frac{200}{80}\right)\left(0.4\right)-\left(-0.4\right)\right]\\ =1.7065\text{V}\end{array}$

The output voltage $V_{OLU}$ at the point $V_{IH}$ is,

  $V_{OLU}=\frac{\left[\left(1+\frac{K_n}{K_p}\right)V_{IH}+V_{DD}-\left(\frac{K_n}{K_p}\right)V_{TN}-V_{TP}\right]}{2\left(\frac{K_n}{K_p}\right)}$

Substitute the values,

  $\begin{array}{c}{V}_{OLU}=\frac{\left[\left(1+\frac{K_n}{K_p}\right)V_{IH}+V_{DD}-\left(\frac{K_n}{K_p}\right)V_{TN}-V_{TP}\right]}{2\left(\frac{K_n}{K_p}\right)}\\ =\frac{\left[\left(1+\frac{200\mu }{80\mu }\right)\left(0.8767\right)-\left(1.8\right)-\left(0.4\right)\left(\frac{200\mu }{80\mu }\right)-\left(-0.4\right)\right]}{2\left(\frac{200\mu }{80\mu }\right)}\\ =\frac{\left[\left(1+\frac{200}{80}\right)\left(0.8767\right)-\left(1.8\right)-\left(0.4\right)\left(\frac{200}{80}\right)-\left(-0.4\right)\right]}{2\left(\frac{200}{80}\right)}\\ =0.1337\text{V}\end{array}$

Conclusion:

Therefore, the low critical voltage $V_{IL}$ is $0.6323V$ , the high critical voltage $V_{IH}$ is

  $0.8767\text{V}$ , the voltage $V_{OHU}$ at the point $V_{IL}$ is $1.7065V$ ,and the voltage $V_{OLU}$ at the point $V_{IL}$ is $0.1337\text{V}$ .

To determine

The noise margins $N{M}_L$ and $N{M}_H$ .

Answer to Problem 16.9EP

The low noise margin $N{M}_L$ is $0.4895\text{V}$ , and the high noise margin $N{M}_H$ is $0.8298\text{V}$ .

Explanation of Solution

Given:

  $V_{DD}=1.8\text{V}$

  $V_{TN}=0.4\text{V}$ , $V_{TP}=-0.4\text{V}$ , $K_n=200{\text{μA/V}}^{\text{2}}$ , and $K_p=80{\text{μA/V}}^{\text{2}}$ .

Calculation:

The low noise margin $N{M}_L$ is,

  $N{M}_L=V_{IL}-V_{OLU}$

Substitute $0.6323V$ for $V_{IL}$ and $0.1337\text{V}$ for $V_{OLU}$

  $\begin{array}{c}N{M}_L=V_{IL}-V_{OLU}\\ =\left(0.6232\right)-\left(0.1337\right)\\ =0.4895\text{V}\end{array}$

The high noise margin $N{M}_H$ is,

  $N{M}_H=V_{OHU}-V_{IH}$

Substitute $1.7065V$ for $V_{OHU}$ and $0.8767\text{V}$ for $V_{lH}$

  $\begin{array}{c}N{M}_H=V_{OHU}-V_{IH}\\ =\left(1.7065\right)-\left(0.8767\right)\\ =0.829\text{8V}\end{array}$

Conclusion:

Therefore, the low noise margin $N{M}_L$ is $0.4895\text{V}$ , and the high noise margin $N{M}_H$ is $0.8298\text{V}$ .