Chapter 16, Problem 16.33P

(a)

To determine

The value of the input switching voltage and the input voltage for the given $v_O$

Answer to Problem 16.33P

The value of the input switching voltage is $1.7069\text{V}$ , input voltage for the output of $3.1\text{V}$ is $1.26\text{V}$ and the value of the input voltage for the output voltage of $0.2\text{V}$ is $2.157\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression to determine the trans-conductance parameter for NMOS is given by,

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

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

  $\begin{array}{c}{K}_n=\left(\frac{100\text{μA}/{\text{V}}^2}{2}\right)4\\ =200\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the trans-conductance parameter for PMOS is given by,

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

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

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

The expression to determine the transition points $V_{It}$ is given by,

  $V_{It}=\frac{V_{DD}+V_{TP}+\sqrt{\frac{k_n}{k_p}}{V}_{TN}}{1+\sqrt{\frac{k_n}{k_p}}{V}_{TN}}$

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

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{240\text{μA}/{\text{V}}^2}{200\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}{1+\sqrt{\frac{240\text{μA}/{\text{V}}^2}{200\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}\\ =1.7069\text{V}\end{array}$

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

  $K_n{\left(v_I-V_{TN}\right)}^2=K_p{\left[2\left(V_{DD}-v_I+V_{TP}\right){\left(V_{DD}-v_O\right)}^2-\left(V_{DD}-v_O\right)\right]}^2$

Substitute $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $3.1\text{V}$ for $v_O$ , $0.4\text{V}$ for $V_{TN}$ , $3.3\text{V}$ for $V_{DD}$ and $-0.4\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{l}\left[200\text{μA}/{\text{V}}^2{\left(v_I-0.4\text{V}\right)}^2=\left[240\text{μA}/{\text{V}}^2\left[\begin{array}{l}2\left(3.3\text{V}-v_I-0.4\text{V}\right)\\ \left(3.3\text{V}-3.1\text{V}\right)-{\left(3.3\text{V}-3.1\text{V}\right)}^2\end{array}\right]\right]\right]\\ 5v_I^2-1.6v_I-5.92=0\\ v_I=\frac{1.6\pm \sqrt{{\left(1.6\right)}^2+4\left(5.92\right)\left(5\right)}}{2\left(5\right)}\\ v_I=1.26\text{V}\end{array}$

The expression to determine the value of the input voltage when the output voltage is more than $1.25\text{V}$ is given by,

  $\frac{K_n}{K_p}\left[2\left(v_I-V_{TN}\right)v_O-v_O^2\right]={\left[V_{DD}-v_I+\left(V_{TP}\right)\right]}^2$

Substitute $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $0.2\text{V}$ for $v_O$ , $0.4\text{V}$ for $V_{TN}$ and $3.3\text{V}$ for $V_{DD}$ in equation (1).

  $\begin{array}{l}\frac{200\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}\left[2\left(v_i-0.4\text{V}\right)0.2\text{V}-{\left(0.2\text{V}\right)}^2\right]={\left[3.3\text{V}-v_I+\left(-0.4\text{V}\right)\right]}^2\\ 6v_I^2-36.8v_I+51.46=0\\ v_I=2.157\text{V}\end{array}$

Conclusion:

Therefore, the value of the input switching voltage is $1.7069\text{V}$ , input voltage for the output of $3.1\text{V}$ is $1.26\text{V}$ and the value of the input voltage for the output voltage of $0.2\text{V}$ is $2.157\text{V}$ .

(b)

To determine

The value of the input switching voltage and the input voltage for the given $v_O$

Answer to Problem 16.33P

The value of the input switching voltage is $1.2513\text{V}$ , input voltage for the output of $3.1\text{V}$ is $0.8554\text{V}$ and the value of the input voltage for the output voltage of $0.2\text{V}$ is $1.61\text{V}$ .

Explanation of Solution

Calculation:

The expression to determine the trans-conductance parameter for NMOS is given by,

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

Substitute $6$ for ${\left(\frac{W}{L}\right)}_n$ and $100\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ in the above equation.

  $\begin{array}{c}{K}_n=\left(\frac{100\text{μA}/{\text{V}}^2}{2}\right)6\\ =300\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the trans-conductance parameter for PMOS is given by,

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

Substitute $4$ for ${\left(\frac{W}{L}\right)}_p$ and $40\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_p$ in the above equation.

  $\begin{array}{c}{K}_p=\left(\frac{40\text{μA}/{\text{V}}^2}{2}\right)4\\ =80\text{μA}/{\text{V}}^2\end{array}$

The expression to determine the transition points $V_{It}$ is given by,

  $V_{It}=\frac{V_{DD}+V_{TP}+\sqrt{\frac{k_n}{k_p}}{V}_{TN}}{1+\sqrt{\frac{k_n}{k_p}}{V}_{TN}}$

Substitute $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $300\text{μA}/{\text{V}}^2$ for $k_n$ and $80\text{μA}/{\text{V}}^2$ for $k_p$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{300\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}{1+\sqrt{\frac{300\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}\\ =1.2513\text{V}\end{array}$

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

  $K_n{\left(v_I-V_{TN}\right)}^2=K_p{\left[2\left(V_{DD}-v_I+V_{TP}\right){\left(v_{DD}-v_O\right)}^2-\left(v_{DD}-v_O\right)\right]}^2$

Substitute $240\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $3.1\text{V}$ for $v_O$ , $0.4\text{V}$ for $V_{TN}$ , $3.3\text{V}$ for $V_{DD}$ and $-0.4\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{l}200\text{μA}/{\text{V}}^2{\left(v_I-0.4\text{V}\right)}^2=\left[240\text{μA}/{\text{V}}^2\left[\begin{array}{l}2\left(3.3\text{V}-v_I-0.4\text{V}\right)\\ \left(3.3-3.1\text{V}\right)-{\left(3.3-3.1\text{V}\right)}^2\end{array}\right]\right]\\ 15v_I^2-10.4v_I-2.08=0\\ v_I=0.8554\text{V}\end{array}$

The expression to determine the value of the input voltage when the output voltage is more than $1.25\text{V}$ is given by,

  $\frac{K_n}{K_p}\left[2\left(v_I-V_{TN}\right)v_O-v_O^2\right]={\left[V_{DD}-v_I+\left(V_{TP}\right)\right]}^2$

Substitute $240\text{μA}/{\text{V}}^2$ for $K_p$ , $300\text{μA}/{\text{V}}^2$ for $K_n$ , $0.2\text{V}$ for $v_O$ , $0.4\text{V}$ for $V_{TN}$ and $3.3\text{V}$ for $V_{DD}$ in equation (1).

  $\begin{array}{l}\frac{300\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}\left[2\left(v_i-0.4\text{V}\right)0.2\text{V}-{\left(0.2\text{V}\right)}^2\right]={\left[3.3\text{V}-v_I+\left(-0.4\text{V}\right)\right]}^2\\ 4v_I^2-29.2v_I+36.64=0\\ v_I=1.61\text{V}\end{array}$

Conclusion:

Therefore, the value of the input switching voltage is $1.2513\text{V}$ , input voltage for the output of $3.1\text{V}$ is $0.8554\text{V}$ and the value of the input voltage for the output voltage of $0.2\text{V}$ is $1.61\text{V}$ .