Chapter 16, Problem 16.46P

(a)

To determine

The noise margins of a CMOS inverter biased at given value.

Answer to Problem 16.46P

The value of $N{M}_H$ is $\text{1}\text{.025}\text{V}$ and $N{M}_L$ is $\text{1}\text{.025}\text{V}$ .

Explanation of Solution

Calculation:

The expression to determine the value of the $K_N$ is given by,

  $K_N=\frac{{k^{\prime }}_n}{2}{\left(\frac{W}{L}\right)}_n$

Substitute $100\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $2$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

  $\begin{array}{c}{K}_N=\frac{100\text{μA}/{\text{V}}^2}{2}\left(2\right)\\ =100\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the value of the $K_P$ is given by,

  $K_P=\frac{{k^{\prime }}_p}{2}{\left(\frac{W}{L}\right)}_p$

Substitute $40\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_p$ and $5$ for ${\left(\frac{W}{L}\right)}_p$ in the above equation.

  $\begin{array}{c}{K}_P=\frac{40\text{μA}/{\text{V}}^2}{2}\left(5\right)\\ =100\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the value of the low level of the input voltage is given by,

  $V_{IL}=V_{TN}+\frac{3}{8}\left(V_{DD}+V_{TP}-V_{TN}\right)$

Substitute $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{IL}=0.4\text{V}+\frac{3}{8}\left(3.3\text{V}+\left(-0.4\text{V}\right)-0.4\text{V}\right)\\ =1.3375\text{V}\end{array}$

The expression to determine the value of the high level of input voltage is given by,

  $V_{IH}=V_{TN}+\frac{5}{8}\left(V_{DD}+V_{TP}-V_{TN}\right)$

Substitute $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{IH}=0.4\text{V}+\frac{5}{8}\left(3.3\text{V}+\left(-0.4\text{V}\right)-0.4\text{V}\right)\\ =1.9625\text{V}\end{array}$

The expression to determine the value of the high level of output voltage is given by,

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

Substitute $1.3375\text{V}$ for $V_{IL}$ , $100\text{μA}/{\text{V}}^2$ for $K_p$ , $100\text{μA}/{\text{V}}^2$ for $K_n$ , $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{OHU}=\frac{1}{2}\left(\begin{array}{l}\left(1+\frac{100\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}\right)\left(1.2716\text{V}\right)+2.5\text{V}-\left(\frac{100\text{μA}/{\text{V}}^2}{50\text{μA}/{\text{V}}^2}\right)\left(1.3375\text{V}\right)\\ -\left(-0.4\text{V}\right)+0.4\left[\right]\text{V}\end{array}\right)\\ =2.9875\text{V}\end{array}$

The expression to determine the value of the high level of output voltage is given by,

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

Substitute $1.9625\text{V}$ for $V_{IH}$ , $100\text{μA}/{\text{V}}^2$ for $K_p$ , $100\text{μA}/{\text{V}}^2$ for $K_n$ , $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{OLU}=\frac{\left(1.9625\text{V}\right)\left(\frac{100\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}\right)-3.3\text{V}-\left(\frac{100\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}\right)04\text{V}-\left(-0.4\text{V}\right)}{2\left(\frac{100\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}\right)}\\ =0.3125\text{V}\end{array}$

The low level of the noise margin is given by,

  ${\text{NM}}_L=V_{IL}-V_{OLU}$

Substitute $1.3375\text{V}$ for $V_{IL}$ and $0.3125\text{V}$ for $V_{OLU}$ in the above equation.

  $\begin{array}{c}{\text{NM}}_L=1.3375\text{V}-0.3125\text{V}\\ =1.025\text{V}\end{array}$

The low level of the noise margin is given by,

  ${\text{NM}}_H=V_{OHU}-V_{IH}$

Substitute $1.9625\text{V}$ for $V_{IH}$ and $2.9875\text{V}$ for $V_{OHU}$ in the above equation.

  $\begin{array}{c}{\text{NM}}_H=2.9875\text{V}-1.9625\text{V}\\ =\text{1}\text{.025}\text{V}\end{array}$

Conclusion:

Therefore, the value of $N{M}_H$ is $\text{1}\text{.025}\text{V}$ and $N{M}_L$ is $\text{1}\text{.025}\text{V}$ .

(b)

To determine

The noise margins of a CMOS inverter biased at $V_{DD}$

Answer to Problem 16.46P

The value of s is $0.9413\text{V}$ and $N{M}_L$ is $1.1112\text{V}$ .

Explanation of Solution

Calculation:

The expression to determine the value of the $K_N$ is given by,

  $K_N=\frac{{k^{\prime }}_n}{2}{\left(\frac{W}{L}\right)}_n$

Substitute $100\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $4$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

  $\begin{array}{c}{K}_N=\frac{100\text{μA}/{\text{V}}^2}{2}\left(2\right)\\ =100\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the value of the $K_P$ is given by,

  $K_P=\frac{{k^{\prime }}_p}{2}{\left(\frac{W}{L}\right)}_p$

Substitute $40\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_p$ and $12$ for ${\left(\frac{W}{L}\right)}_p$ in the above equation.

  $\begin{array}{c}{K}_P=\frac{40\text{μA}/{\text{V}}^2}{2}\left(12\right)\\ =240\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the value of the low level of the input voltage is given by,

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

Substitute $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{IL}=0.4\text{V}+\frac{3.3\text{V}+\left(-0.4\text{V}\right)-0.4\text{V}}{\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}-1\right)}\left(2\sqrt{\frac{\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}}{\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}+3}}-1\right)\\ =1.4127\text{V}\end{array}$

The expression to determine the value of the high level of input voltage is given by,

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

Substitute $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{IH}=0.4\text{V}+\frac{3.3\text{V}+\left(-0.4\text{V}\right)-0.4\text{V}}{\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}-1\right)}\left(\frac{2\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)}{\sqrt{3\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)+1}}-1\right)\\ =2.0370\text{V}\end{array}$

The expression to determine the value of the high level of output voltage is given by,

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

Substitute $1.4127\text{V}$ for $V_{IL}$ , $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{OHU}=\frac{1}{2}\left(\left(1+\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)\left(1.4127\text{V}\right)+3.3\text{V}-\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)\left(0.4\text{V}\right)-\left(-0.4\text{V}\right)\right)\\ =2.9783\text{V}\end{array}$

The expression to determine the value of the high level of output voltage is given by,

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

Substitute $2.0370\text{V}$ for $V_{IH}$ , $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $0.4\text{V}$ for $V_{TN}$ , $-0.4\text{V}$ for $V_{TP}$ and $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{OLU}=\frac{2.0370\text{V}\left(1+\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)-3.3\text{V}-\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)\left(0.4\text{V}\right)-\left(-0.4\text{V}\right)}{2\left(\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\right)}\\ =0.3007\text{V}\end{array}$

The low level of the noise margin is given by,

  ${\text{NM}}_L=V_{IL}-V_{OLU}$

Substitute $1.4127\text{V}$ for $V_{IL}$ and $0.3007\text{V}$ for $V_{OLU}$ in the above equation.

  $\begin{array}{c}{\text{NM}}_L=1.4127\text{V}-0.3007\text{V}\\ =1.1112\text{V}\end{array}$

The low level of the noise margin is given by,

  ${\text{NM}}_H=V_{OHU}-V_{IH}$

Substitute $2.0370\text{V}$ for $V_{IH}$ and $2.9783\text{V}$ for $V_{OHU}$ in the above equation.

  $\begin{array}{c}{\text{NM}}_H=2.9783\text{V}-2.0370\text{V}\\ =0.9413\text{V}\end{array}$

Conclusion:

Therefore, the value of $N{M}_H$ is $0.9413\text{V}$ and $N{M}_L$ is $1.1112\text{V}$ .