Chapter 16, Problem 16.8P

(a)

To determine

The value of the voltage $v_{OH}$ and $v_{OL}$ for the given voltage $V_B$ .

Answer to Problem 16.8P

The maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $0.364\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

The expression for the voltage $V_{OH}$ is given by,

  $V_{OH}=5\text{V}-V_{TN}$

Substitute $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{OH}=5\text{V}-1\text{V}\\ =\text{4}\text{V}\end{array}$

The expression for the enhancement mode NMOS inverter equation is given by,

  $M_D\left(2\left(v_{IH}-V_{TN}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=M_L{\left(V_B-V_{OL}-V_{TN}\right)}^2$

Substitute $1\text{mA}/{\text{V}}^2$ for $M_D$ , $0.4\text{mA}/{\text{V}}^2$ for $M_L$ , $1\text{V}$ for $V_{TN}$ , $5\text{V}$ for $v_{IH}$ and $4\text{V}$ for $V_B$ in the above equation.

  $\begin{array}{l}\left(1\text{mA}/{\text{V}}^2\right)\left(2\left(5\text{V}-1\text{V}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=0.4\text{mA}/{\text{V}}^2{\left(4\text{V}-V_{OL}-1\text{V}\right)}^2\\ V_{OL}=0.364\text{V}\end{array}$

Conclusion:

Therefore, the maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $0.364\text{V}$ .

(b)

To determine

The value of the voltage $v_{OH}$ and $v_{OL}$ for the given voltage $V_B$ .

Answer to Problem 16.8P

The maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $0.62\text{V}$ .

Explanation of Solution

Calculation:

The expression for the voltage $V_{OH}$ is given by,

  $V_{OH}=5\text{V}-V_{TN}$

Substitute $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{OH}=5\text{V}-1\text{V}\\ =\text{4}\text{V}\end{array}$

Consider the input voltage $v_{IH}$ for the logic 1 is $5\text{V}$ which is equal to $V_{DD}$ .

The expression for the enhancement mode NMOS inverter equation is given by,

  $M_D\left(2\left(v_{IH}-V_{TN}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=M_L{\left(V_B-V_{OL}-V_{TN}\right)}^2$

Substitute $1\text{mA}/{\text{V}}^2$ for $M_D$ , $0.4\text{mA}/{\text{V}}^2$ for $M_L$ , $1\text{V}$ for $V_{TN}$ , $5\text{V}$ for $v_{IH}$ and $5\text{V}$ for $V_B$ in the above equation.

  $\begin{array}{l}\left(1\text{mA}/{\text{V}}^2\right)\left(2\left(5\text{V}-1\text{V}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=0.4\text{mA}/{\text{V}}^2{\left(5\text{V}-V_{OL}-1\text{V}\right)}^2\\ V_{OL}=0.62\text{V}\end{array}$

Conclusion:

Therefore, the maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $0.62\text{V}$ .

(c)

To determine

The value of the voltage $v_{OH}$ and $v_{OL}$ for the given voltage $V_B$ .

Answer to Problem 16.8P

The maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $0.935\text{V}$ .

Explanation of Solution

Calculation:

The expression for the voltage $V_{OH}$ is given by,

  $V_{OH}=5\text{V}-V_{TN}$

Substitute $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{OH}=5\text{V}-1\text{V}\\ =\text{4}\text{V}\end{array}$

Consider the input voltage $v_{IH}$ for the logic 1 is $5\text{V}$ which is equal to $V_{DD}$ .

The expression for the enhancement mode NMOS inverter equation is given by,

  $M_D\left(2\left(v_{IH}-V_{TN}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=M_L{\left(V_B-V_{OL}-V_{TN}\right)}^2$

Substitute $1\text{mA}/{\text{V}}^2$ for $M_D$ , $0.4\text{mA}/{\text{V}}^2$ for $M_L$ , $1\text{V}$ for $V_{TN}$ , $5\text{V}$ for $v_{IH}$ and $6\text{V}$ for $V_B$ in the above equation.

  $\begin{array}{l}\left(1\text{mA}/{\text{V}}^2\right)\left(2\left(5\text{V}-1\text{V}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=0.4\text{mA}/{\text{V}}^2{\left(6\text{V}-V_{OL}-1\text{V}\right)}^2\\ V_{OL}=0.935\text{V}\end{array}$

Conclusion:

Therefore, the maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $0.935\text{V}$ .

(d)

To determine

The value of the voltage $v_{OH}$ and $v_{OL}$ for the given voltage $V_B$ .

Answer to Problem 16.8P

The maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $1.3138\text{V}$ .

Explanation of Solution

Calculation:

The expression for the voltage $V_{OH}$ is given by,

  $V_{OH}=5\text{V}-V_{TN}$

Substitute $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{V}_{OH}=5\text{V}-1\text{V}\\ =\text{4}\text{V}\end{array}$

Consider the input voltage $v_{IH}$ for the logic 1 is $5\text{V}$ which is equal to $V_{DD}$ .

The expression for the enhancement mode NMOS inverter equation is given by,

  $M_D\left(2\left(v_{IH}-V_{TN}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=M_L{\left(V_B-V_{OL}-V_{TN}\right)}^2$

Substitute $1\text{mA}/{\text{V}}^2$ for $M_D$ , $0.4\text{mA}/{\text{V}}^2$ for $M_L$ , $1\text{V}$ for $V_{TN}$ , $5\text{V}$ for $v_{IH}$ and $7\text{V}$ for $V_B$ in the above equation.

  $\begin{array}{l}\left(1\text{mA}/{\text{V}}^2\right)\left(2\left(5\text{V}-1\text{V}\right)V_{OL}-{\left(V_{OL}\right)}^2\right)=0.4\text{mA}/{\text{V}}^2{\left(7\text{V}-V_{OL}-1\text{V}\right)}^2\\ V_{OL}=1.3138\text{V}\end{array}$

Conclusion:

Therefore, the maximum value of the voltage $V_{OH}$ is $4\text{V}$ and the value of the minimum low output voltage $V_{OL}$ is $1.3138\text{V}$ .