Chapter 16, Problem 16.38P

(a)

To determine

The value of peak current in the inverter and the input voltage input voltage for the given specifications.

Answer to Problem 16.38P

The value of the current in the transistor is $234\text{μA}$ .

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 $3$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

  $\begin{array}{c}{K}_N=\frac{100\text{μA}/{\text{V}}^2}{2}\left(3\right)\\ =150\text{μA}/{\text{V}}^2\end{array}$

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 $7.5$ for ${\left(\frac{W}{L}\right)}_p$ in the above equation.

  $\begin{array}{c}{K}_P=\frac{40\text{μA}/{\text{V}}^2}{2}\left(7.5\right)\\ =150\text{μA}/{\text{V}}^2\end{array}$

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

  $V_{It}=\frac{V_{DD}}{2}$

Substitute $3.3\text{V}$ for $V_{DD}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}}{2}\\ =1.65\text{V}\end{array}$

The expression for the peak value of the current through the NMOS transistor is given by,

  $i_{D,peak}=K_N{\left(v_I-V_{TN}\right)}^2$

Substitute $150\text{μA}/{\text{V}}^2$ for $K_N$ , $0.4$ for $V_{TN}$ and $1.65\text{V}$ for $V_{It}$ in the above equation.

  $\begin{array}{c}{i}_{D,peak}=150\text{μA}/{\text{V}}^2{\left(1.65\text{V}-0.4\right)}^2\\ =234\text{μA}\end{array}$

Conclusion:

Therefore, the value of the current in the transistor is $234\text{μA}$ .

(b)

To determine

The value of peak current in the inverter and the input voltage input voltage for the given specifications.

Answer to Problem 16.38P

The maximum value of the current in the transistor is $188\text{μA}$ .

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.

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

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 $4$ for ${\left(\frac{W}{L}\right)}_p$ in the above equation.

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

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

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

Substitute $3.3\text{V}$ for $V_{DD}$ , $-0.4\text{V}$ for $V_{TP}$ , $80\text{μA}/{\text{V}}^2$ for $K_P$ and $200\text{μA}/{\text{V}}^2$ for $K_N$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{200\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}{1+\sqrt{\frac{200\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}}\\ =1.369\text{V}\end{array}$

The expression for the peak value of the current through the NMOS transistor is given by,

  $i_{D,peak}=K_N{\left(v_I-V_{TN}\right)}^2$

Substitute $200\text{μA}/{\text{V}}^2$ for $K_N$ , $0.4$ for $V_{TN}$ and $1.369\text{V}$ for $V_{It}$ in the above equation.

  $\begin{array}{c}{i}_{D,peak}=200\text{μA}/{\text{V}}^2{\left(1.369\text{V}-0.4\right)}^2\\ =188\text{μA}\end{array}$

Conclusion:

Therefore, the maximum value of the current in the transistor is $188\text{μA}$ .

(c)

To determine

The value of peak current in the inverter and the input voltage input voltage for the given specifications.

Answer to Problem 16.38P

The maximum value of the current in the transistor is $292\text{μA}$ .

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 $3$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

  $\begin{array}{c}{K}_N=\frac{100\text{μA}/{\text{V}}^2}{2}\left(3\right)\\ =150\text{μA}/{\text{V}}^2\end{array}$

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 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 $V_{TP}$ , $150\text{μA}/{\text{V}}^2$ for $K_P$ and $240\text{μA}/{\text{V}}^2$ for $K_N$ and $0.4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{3.3\text{V}+\left(-0.4\text{V}\right)+\sqrt{\frac{150\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}}\left(0.4\text{V}\right)}{1+\sqrt{\frac{150\text{μA}/{\text{V}}^2}{240\text{μA}/{\text{V}}^2}}}\\ =1.796\text{V}\end{array}$

The expression for the peak value of the current through the NMOS transistor is given by,

  $i_{D,peak}=K_N{\left(v_I-V_{TN}\right)}^2$

Substitute $150\text{μA}/{\text{V}}^2$ for $K_N$ , $0.4$ for $V_{TN}$ and $1.796\text{V}$ for $V_{It}$ in the above equation.

  $\begin{array}{c}{i}_{D,peak}=150\text{μA}/{\text{V}}^2{\left(1.796\text{V}-0.4\right)}^2\\ =292\text{μA}\end{array}$

Conclusion:

Therefore, the maximum value of the current in the transistor is $292\text{μA}$ .