Chapter 16, Problem 16.79P

To determine

To plot: The graphof $v_{O1}$ and $v_O$ with respect to $v_1$ .

To find: The value of $v_{O1}$ and $v_O$ at different values of the input voltage.

Answer to Problem 16.79P

The required plots are shown in Figure 2 and Figure 3. The value of the voltage $v_{O1}$ at input voltage of $1.5\text{V}$ is $2.88\text{V}$ , at $1.6\text{V}$ is $2.6928\text{V}$ , at $1.8\text{V}$ is $0.607\text{V}$ and at $1.7\text{V}$ is $0.607\text{V}<v_{O1}<2.693\text{V}$ . The value of the voltage $v_O$ at input voltage of $1.5\text{V}$ is $0\text{V}$ , at $1.6\text{V}$ is $0.00979\text{V}$ , at $1.8\text{V}$ is $3.298\text{V}$ and at $1.7\text{V}$ is $0.00979\text{V}<v_{O1}<3.298\text{V}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1.

  

Calculation:

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

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

Substitute $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{0.2\text{mA}/{\text{V}}^2}{0.2\text{mA}/{\text{V}}^2}}\left(0.5\text{V}\right)}{1+\sqrt{\frac{0.2\text{mA}/{\text{V}}^2}{0.2\text{mA}/{\text{V}}^2}}}\\ =1.6\text{V}\end{array}$

Case(i)

The expression to determine the output voltage of the first CMOS inverter is given by,

  $K_n{\left(v_I-V_{TN}\right)}^2=K_p\left[2\left(V_{DD}-v_I+V_{TP}\right)-{\left(V_{DD}-v_{O1}\right)}^2\right]$

Substitute $1.5\text{V}$ for $v_I$ , $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}0.2\text{mA}/{\text{V}}^2{\left(1.5\text{V}-0.5\text{V}\right)}^2=0.2\text{mA}/{\text{V}}^2\left[2\left(\left(3.3\text{V}\right)-v_I-0.4\text{V}\right)-{\left(3.3\text{V}-v_{O1}\right)}^2\right]\\ {\left(v_{O1}\right)}^2-3.8v_{O1}+2.65=0\\ v_{O1}=\frac{-\left(-3.8\right)\pm \sqrt{{\left(-3.8\right)}^2-4\left(1\right)\left(2.65\right)}}{2}\\ v_{O1}=2.88\text{V},0.9202\text{V}\end{array}$

The expression to determine the output voltage of the second CMOS inverter is given by,

  $K_p{\left(V_{DD}-v_{O1}+V_{TP}\right)}^2=K_n\left[\left(v_{O1}-V_{TN}\right)v_O-v_O^2\right]$

Substitute $2.88\text{V}$ for $v_{O1}$ , $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}0.2\text{mA}/{\text{V}}^2{\left(3.3\text{V}-2.88\text{V}-0.4\text{V}\right)}^2=0.2\text{mA}/{\text{V}}^2\left[\left(2.88\text{V}-0.5\text{V}\right)v_O-v_O^2\right]\\ {\left(v_O\right)}^2-4.7v_O+0.0004=0\\ v_O=4.7599\text{V,}0.0001\text{V}\end{array}$

Thus the value of the of appropriate voltage is $0\text{V}$ .

Case (ii)

The expression to determine the output voltage of the first CMOS inverter is given by,

  $K_n{\left(v_I-V_{TN}\right)}^2=K_p\left[2\left(V_{DD}-v_I+V_{TP}\right)-{\left(V_{DD}-v_{O1}\right)}^2\right]$

Substitute $1.6\text{V}$ for $v_I$ , $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}0.2\text{mA}/{\text{V}}^2{\left(1.6\text{V}-0.5\text{V}\right)}^2=0.2\text{mA}/{\text{V}}^2\left[2\left(\left(3.3\text{V}\right)-1.6\text{V}-0.4\text{V}\right)-{\left(3.3\text{V}-v_{O1}\right)}^2\right]\\ {\left(v_{O1}\right)}^2-4v_{O1}+3.52=0\\ v_{O1}=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(3.52\right)}}{2}\\ v_{O1}=2.6928\text{V},1.3072\text{V}\end{array}$

The bias voltage is $3.3\text{V}$ , thus the appropriate value of the output voltage is $2.693\text{V}$

The expression to determine the output voltage of the second CMOS inverter is given by,

  $K_p{\left(V_{DD}-v_{O1}+V_{TP}\right)}^2=K_n\left[\left(v_{O1}-V_{TN}\right)v_O-v_O^2\right]$

Substitute $2.693\text{V}$ for $v_{O1}$ , $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}0.2\text{mA}/{\text{V}}^2{\left(3.3\text{V}-2.693\text{V}-0.4\text{V}\right)}^2=0.2\text{mA}/{\text{V}}^2\left[\left(2.693\text{V}-0.5\text{V}\right)v_O-v_O^2\right]\\ {\left(v_O\right)}^2-4.386v_O+0.0424829=0\\ v_O=4.3762\text{V,}0.00979\text{V}\end{array}$

Thus the value of the of appropriate voltage is $0.00979\text{V}$ .

Case (iii)

The expression to determine the output voltage of the first CMOS inverter is given by,

  $K_p{\left(V_{DD}-v_I+V_{TP}\right)}^2=K_n\left[2\left(v_I-V_{TN}\right)v_{O1}-v_{O1}{}^2\right]$

Substitute $1.7\text{V}$ for $v_I$ , $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}.2\text{mA}/{\text{V}}^2{\left(3.3\text{V}-1.7\text{V}-0.4\text{V}\right)}^2=.2\text{mA}/{\text{V}}^2\left[2\left(1.7\text{V}-0.5\text{V}\right)v_{O1}-v_{O1}{}^2\right]\\ {\left(v_{O1}\right)}^2-2.6v_{O1}+1.21=0\\ v_{O1}=\mathrm{1.99280.5}\text{V},0.6072\text{V}\end{array}$

The bias voltage is $3.3\text{V}$ , thus the appropriate value of the output voltage is $0.607\text{V}$

The expression to determine the output voltage of the second CMOS inverter is given by,

  $K_p{\left(V_{DD}-v_{O1}+V_{TP}\right)}^2=K_n\left[\left(v_{O1}-V_{TN}\right)v_O-v_O^2\right]$

Substitute $0.607\text{V}$ for $v_{O1}$ , $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $0.2\text{mA}/{\text{V}}^2$ for $K_P$ and $0.2\text{mA}/{\text{V}}^2$ for $K_N$ and $0.5\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}0.2\text{mA}/{\text{V}}^2{\left(3.3\text{V}-0.607\text{V}-0.4\text{V}\right)}^2=0.2\text{mA}/{\text{V}}^2\left[\left(0.607\text{V}-0.5\text{V}\right)v_O-v_O^2\right]\\ {\left(v_O\right)}^2-2.014v_O-4.2326=0\\ v_O=3.298\text{V}\end{array}$

Thus the value of the of appropriate voltage is $0.00979\text{V}$ .

Case (iv)

The output voltage of the first and the second CMOS inverter is switching point when the input voltage is the $V_{it}$ . When the transition voltage is between $1.7\text{V}$ the output voltage of the first CMOS inverter $v_{O1}$ at $v_I$ is $1.7\text{V}$ a variable voltage and is in the range of $0.607\text{V}\text{to}2.693\text{V}$ . In the same manner the voltage at the output of the second CMOS inverter is $1.7\text{V}$ and it ranger from $0.00979\text{V}\text{to}3.398\text{V}$ .

The plot for the voltage between $v_{O1}$ and $v_I$ is shown below.

  

The plot between the second CMOS inviter output $v_O$ and the input voltage is shown below.

The required plot is shown in Figure 3

  

Conclusion:

Therefore, the required plots are shown in Figure 2 and Figure 3. The value of the voltage $v_{O1}$ at input voltage of $1.5\text{V}$ is $2.88\text{V}$ , at $1.6\text{V}$ is $2.6928\text{V}$ , at $1.8\text{V}$ is $0.607\text{V}$ and at $1.7\text{V}$ is $0.607\text{V}<v_{O1}<2.693\text{V}$ . The value of the voltage $v_O$ at input voltage of $1.5\text{V}$ is $0\text{V}$ , at $1.6\text{V}$ is $0.00979\text{V}$ , at $1.8\text{V}$ is $3.298\text{V}$ and at $1.7\text{V}$ is $0.00979\text{V}<v_{O1}<3.298\text{V}$ .