Chapter 16, Problem 16.37P

(a)

To determine

The peak current in the inverter during the switching cycle.

Answer to Problem 16.37P

The value of the current in the transistor is $24.2\text{μA}$ .

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 $80\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $2$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

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

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 $1.8\text{V}$ for $V_{DD}$ , $-0.35\text{V}$ for $V_{TP}$ , $80\text{μA}/{\text{V}}^2$ for $K_P$ and $80\text{μA}/{\text{V}}^2$ for $K_N$ and $0.35\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{1.8\text{V}+\left(-0.35\text{V}\right)+\sqrt{\frac{80\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}\left(0.35\text{V}\right)}{1+\sqrt{\frac{80\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}}\\ =0.9\text{V}\end{array}$

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

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

Substitute $80\text{μA}/{\text{V}}^2$ for $K_N$ , $0.35$ for $V_{TN}$ and $V_{It}$ for $V_I$ in the above equation.

  $i_{D,\max }=80\text{μA}/{\text{V}}^2{\left(V_{It}-0.35\text{V}\right)}^2$

Substitute $0.9\text{V}$ for $V_{It}$ in the above equation.

  $\begin{array}{c}{i}_{D,\max }=80\text{μA}/{\text{V}}^2{\left(0.9\text{V}-0.35\text{V}\right)}^2\\ =24.2\text{μA}\end{array}$

Conclusion:

Therefore, the value of the current in the transistor is $24.2\text{μA}$ .

(b)

To determine

The peak current in the inverter during the switching cycle.

Answer to Problem 16.37P

The maximum value of the current in the transistor is $29.\text{34}\text{μA}$ .

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 $80\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $2$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

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

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 $6$ for ${\left(\frac{W}{L}\right)}_p$ in the above equation.

  $\begin{array}{c}{K}_P=\frac{40\text{μA}/{\text{V}}^2}{2}\left(6\right)\\ =120\text{μA}/{\text{V}}^2\end{array}$

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 $1.8\text{V}$ for $V_{DD}$ , $-0.35\text{V}$ for $V_{TP}$ , $120\text{μA}/{\text{V}}^2$ for $K_P$ and $80\text{μA}/{\text{V}}^2$ for $K_N$ and $0.35\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{1.8\text{V}+\left(-0.35\text{V}\right)+\sqrt{\frac{80\text{μA}/{\text{V}}^2}{120\text{μA}/{\text{V}}^2}}\left(0.35\text{V}\right)}{1+\sqrt{\frac{80\text{μA}/{\text{V}}^2}{120\text{μA}/{\text{V}}^2}}}\\ =0.95561\text{V}\end{array}$

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

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

Substitute $80\text{μA}/{\text{V}}^2$ for $K_N$ , $0.35$ for $V_{TN}$ and $V_{It}$ for $V_I$ in the above equation.

  $i_{D,\max }=80\text{μA}/{\text{V}}^2{\left(V_{It}-0.35\text{V}\right)}^2$

Substitute $0.95561\text{V}$ for $V_{It}$ in the above equation.

  $\begin{array}{c}{i}_{D,\max }=80\text{μA}/{\text{V}}^2{\left(0.95561\text{V}-0.35\text{V}\right)}^2\\ =29.\text{34}\text{μA}\end{array}$

Conclusion:

Therefore, the maximum value of the current in the transistor is $29.\text{34}\text{μA}$ .

(c)

To determine

The peak current in the inverter during the switching cycle.

Answer to Problem 16.37P

The maximum value of the current in the transistor is $33.22\text{μA}$ .

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 $80\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $4$ for ${\left(\frac{W}{L}\right)}_n$ in the above equation.

  $\begin{array}{c}{K}_N=\frac{80\text{μA}/{\text{V}}^2}{2}\left(4\right)\\ =160\text{μA}/{\text{V}}^2\end{array}$

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,

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

Substitute $1.8\text{V}$ for $V_{DD}$ , $-0.35\text{V}$ for $V_{TP}$ , $80\text{μA}/{\text{V}}^2$ for $K_P$ and $160\text{μA}/{\text{V}}^2$ for $K_N$ and $0.35\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{It}=\frac{1.8\text{V}+\left(-0.35\text{V}\right)+\sqrt{\frac{160\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}\left(0.35\text{V}\right)}{1+\sqrt{\frac{160\text{μA}/{\text{V}}^2}{80\text{μA}/{\text{V}}^2}}}\\ =0.8056\text{V}\end{array}$

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

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

Substitute $80\text{μA}/{\text{V}}^2$ for $K_N$ , $0.35$ for $V_{TN}$ and $V_{It}$ for $V_I$ in the above equation.

  $i_{D,\max }=80\text{μA}/{\text{V}}^2{\left(V_{It}-0.35\text{V}\right)}^2$

Substitute $0.8056\text{V}$ for $V_{It}$ in the above equation.

  $\begin{array}{c}{i}_{D,\max }=80\text{μA}/{\text{V}}^2{\left(0.8056\text{V}-0.35\text{V}\right)}^2\\ =33.22\text{μA}\end{array}$

Conclusion:

Therefore, the maximum value of the current in the transistor is $33.22\text{μA}$ .