Chapter 16, Problem 16.32P

(a)

To determine

The transition points for the p-channel and the n-channel MOSFET.

To sketch: The voltage transfer characteristics.

To find: The value of the input voltage for the different values of the input output voltages.

Answer to Problem 16.32P

The value of the transition voltages are $V_{It}$ is $1.65\text{V}$ , $V_{OPt}$ for $\text{2}\text{.05}\text{V}$ , $V_{ONt}$ is $1.25\text{V}$ . The plot for the voltage transfer characteristics is shown in Figure 2. The value of the input voltage for the output voltage of $0.25\text{V}$ is $2.032\text{V}$ and for the output voltage of $3.05\text{V}$ is $1.268\text{V}$ .

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

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

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

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

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 $V_{TP}$ , $100\text{μA}/{\text{V}}^2$ for $k_n$ and $100\text{μA}/{\text{V}}^2$ for $k_p$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{100\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}{1+\sqrt{\frac{100\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}\\ =1.65\text{V}\end{array}$

The expression to determine the value of the voltage $V_{OPt}$ is given by,

  $V_{OPt}=V_{It}-V_{TP}$

Substitute $1.65\text{V}$ for $V_{It}$ and $-0.4\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{c}{V}_{OPt}=1.65\text{V}-\left(-0.4\text{V}\right)\\ =\text{2}\text{.05}\text{V}\end{array}$

The expression to determine the value of the voltage $V_{ONt}$ is given by,

  $V_{ONt}=V_{It}-V_{TP}$

Substitute $1.65\text{V}$ for $V_{It}$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{ONt}=1.65\text{V}-0.4\text{V}\\ =1.25\text{V}\end{array}$

The voltage transfer characteristics is shown below.

The required diagram is shown in Figure 2

  

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

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

Substitute $100\text{μA}/{\text{V}}^2$ for $K_p$ , $100\text{μA}/{\text{V}}^2$ for $K_n$ , $0.25\text{V}$ 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.

  $\begin{array}{l}100\text{μA}/{\text{V}}^2\left[2\left(v_I-0.4\text{V}\right)\left(0.25\text{V}\right)-{\left(0.25\text{V}\right)}^2\right]=100\text{μA}/{\text{V}}^2{\left(3.3\text{V}-v_I-0.4\text{V}\right)}^2\\ v_I^2-6.3v_I+8.6725=0\\ v_I=\frac{2\pm \sqrt{2^2-4\left(1\right)\left(0.49\right)}}{2}\\ v_I=4.268\text{V,}2.032\text{V}\end{array}$

The value of the voltage must be less than the supply voltage, thus the input voltage is $2.032\text{V}$ .

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

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

Substitute $100\text{μA}/{\text{V}}^2$ for $K_p$ , $100\text{μA}/{\text{V}}^2$ for $K_n$ , $3.05\text{V}$ for $v_O$ , $0.4\text{V}$ for $V_{TN}$ , $3.3\text{V}$ for $V_{DD}$ and $3.3\text{V}$ for $V_{DD}$ in equation (1).

  $\begin{array}{l}100\text{μA}/{\text{V}}^2{\left(v_I-0.4\text{V}\right)}^2=100\text{μA}/{\text{V}}^2\left[2\left(3.3\text{V}-v_I-0.4\text{V}\right)\left(3.3\text{V}-3.05\text{V}\right)-{\left(3.3\text{V}-v_O\right)}^2\right]\\ v_I^2-0.3v_I-1.2275=0\\ v_I=\frac{0.3\pm \sqrt{{\left(0.3\right)}^2-4\left(-1.2275\right)\left(1\right)}}{2}\\ v_I=1.268\text{V}\end{array}$

Conclusion:

Therefore, the value of the transition voltages are $V_{It}$ is $1.65\text{V}$ , $V_{OPt}$ for $\text{2}\text{.05}\text{V}$ , $V_{ONt}$ is $1.25\text{V}$ . The plot for the voltage transfer characteristics is shown in Figure 2. The value of the input voltage for the output voltage of $0.25\text{V}$ is $2.032\text{V}$ and for the output voltage of $3.05\text{V}$ is $1.268\text{V}$ .

(b)

To determine

The transition points for the p-channel and the n-channel MOSFET.

To sketch: The voltage transfer characteristics.

To find: The value of the input voltage for the different values of the input output voltages.

Answer to Problem 16.32P

The value of the transition voltages are $V_{It}$ is $1.436\text{V}$ , $V_{OPt}$ for $\text{1}\text{.836V}$ , $V_{ONt}$ is $1.036\text{V}$ . The plot for the voltage transfer characteristics is shown in Figure 3. The value of the input voltage for the output voltage of $0.25\text{V}$ is $1.78\text{V}$ and for the output voltage of $3.05\text{V}$ is $1.056\text{V}$ .

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 $4$ 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 $5$ for ${\left(\frac{W}{L}\right)}_p$ and $40\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_p$ in the above equation.

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

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 $100\text{μA}/{\text{V}}^2$ for $k_p$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{200\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}{1+\sqrt{\frac{200\text{μA}/{\text{V}}^2}{100\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}\\ =1.436\text{V}\end{array}$

The expression to determine the value of the voltage $V_{OPt}$ is given by,

  $V_{OPt}=V_{It}-V_{TP}$

Substitute $1.436\text{V}$ for $V_{It}$ and $-0.4\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{c}{V}_{OPt}=1.436\text{V}-\left(-0.4\text{V}\right)\\ =\text{1}\text{.836V}\end{array}$

The expression to determine the value of the voltage $V_{ONt}$ is given by,

  $V_{ONt}=V_{It}-V_{TP}$

Substitute $1.436\text{V}$ for $V_{It}$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{ONt}=1.436\text{V}-0.4\text{V}\\ =1.036\text{V}\end{array}$

The voltage transfer characteristics is shown below.

The required diagram is shown in Figure 3

  

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

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

Substitute $100\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $0.25\text{V}$ 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.

  $\begin{array}{l}200\text{μA}/{\text{V}}^2\left[2\left(v_I-0.4\text{V}\right)\left(0.25\text{V}\right)-{\left(0.25\text{V}\right)}^2\right]=100\text{μA}/{\text{V}}^2{\left(3.3\text{V}-v_I-0.4\text{V}\right)}^2\\ v_I^2-6.8v_I+8.935=0\\ v_I=\frac{6.8\pm \sqrt{\left(6.8\right)-4\left(8.935\right)\left(1\right)}}{2}\\ v_I=5.02\text{V,}1.78\text{V}\end{array}$

The value of the voltage must be less than the supply voltage, thus the input voltage is $1.78\text{V}$ .

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

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

Substitute $100\text{μA}/{\text{V}}^2$ for $K_p$ , $200\text{μA}/{\text{V}}^2$ for $K_n$ , $3.05\text{V}$ for $v_O$ , $0.4\text{V}$ for $V_{TN}$ , $3.3\text{V}$ for $V_{DD}$ and $3.3\text{V}$ for $V_{DD}$ in equation (1).

  $\begin{array}{l}200\text{μA}/{\text{V}}^2{\left(v_I-0.4\text{V}\right)}^2=100\text{μA}/{\text{V}}^2\left[2\left(3.3\text{V}-v_I-0.4\text{V}\right)\left(3.3\text{V}-3.05\text{V}\right)-{\left(3.3\text{V}-3.05\right)}^2\right]\\ 2v_I^2-1.1v_I-1.0675=0\\ v_I=\frac{1.1\pm \sqrt{{\left(1.1\right)}^2-4\left(-1.0675\right)\left(2\right)}}{2\left(2\right)}\\ v_I=1.056\text{V}\end{array}$

Conclusion:

Therefore, the value of the transition voltages are $V_{It}$ is $1.436\text{V}$ , $V_{OPt}$ for $\text{1}\text{.836V}$ , $V_{ONt}$ is $1.036\text{V}$ . The plot for the voltage transfer characteristics is shown in Figure 3. The value of the input voltage for the output voltage of $0.25\text{V}$ is $1.78\text{V}$ and for the output voltage of $3.05\text{V}$ is $1.056\text{V}$ .