Chapter 16, Problem 16.2P

(a)

To determine

The transition point and value of $v_o$ for the given input.

Answer to Problem 16.2P

The transition point is,

  $\begin{array}{c}{v}_{It}=1.2185\text{V}\\ v_{Ot}=0.7185\text{V}\end{array}$

The value of $v_O$ for the given input is $v_O=0.1161\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The parameters are:

  $\begin{array}{l}{V}_{DD}=3.3\text{V}\\ K_n=50\mu \text{A}/{\text{V}}^2\\ R_D=100\text{k}\Omega \\ v_I=3.3\text{V}\\ V_{TN}=0.5\text{V}\end{array}$

Calculation:

For the transition point, the equation for finding the input voltage is,

  $K_n{R}_D{\left(v_{It}-V_{TN}\right)}^2+\left(v_{It}-V_{TN}\right)-V_{DD}=0$

Substituting the values,

  $\begin{array}{l}\left(\left(50\times {10}^{-6}\right)\left(100\times {10}^{-3}\right){\left(v_{It}-0.5\right)}^2\right)+\left(v_{It}-0.5\right)-3.3=0\\ 5\left(v_{It}{}^2+0.25-v_{It}\right)+\left(v_{It}-0.5\right)-3.3=0\\ 5v_{It}{}^2+1.25-5v_{It}+v_{It}-0.5-3.3=0\\ 5v_{It}{}^2-4v_{It}-2.55=0\end{array}$

Solving the above quadratic equation,

  $v_{It}=1.2185\text{V }\text{or }-0.4185\text{V}$

Ignoring the negative term, the input voltage of transition point is $v_{It}=1.2185\text{V}$

Now, the output voltage at the transition point is

  $\begin{array}{c}{v}_{Ot}=v_{It}-V_{TN}\\ =1.2185-0.5\\ =0.7185\text{V}\end{array}$

It is given that input voltage is $v_I=3.3\text{V}$ . It is equal to $V_{DD}$ means input is high. So, the output voltage in this case is given by equation,

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

Substituting the values in the above equation,

  $\begin{array}{l}{v}_O=3.3-\left(\left(50\times {10}^{-6}\right)\left(100\times {10}^3\right)\left(2\left(3.3-0.5\right)v_O-v_O{}^2\right)\right)\\ v_O=3.3-\left(5\left(5.6v_O-v_O{}^2\right)\right)\\ v_O=3.3-28v_O+5v_O{}^2\\ 5v_O{}^2-29v_O+3.3=0\end{array}$

Solving the above quadratic equation,

  $v_O=5.6838\text{V}\text{or}\text{0}\text{.1161}\text{V}$

Since, the output voltage cannot be greater than $V_{DD}$ , so only one value is considered.

The output voltage will be

  $v_O=0.1161\text{V}$

Conclusion:

Therefore, the transition points are

  $\begin{array}{c}{v}_{It}=1.2185\text{V}\\ v_{Ot}=0.7185\text{V}\end{array}$

Also, the value of $v_O$ for the given input is $v_O=0.1161\text{V}$

(b)

To determine

The transition point and value of $v_o$ for the given input.

Answer to Problem 16.2P

The transition point is,

  $\begin{array}{c}{v}_{It}=1.6869\text{V}\\ v_{Ot}=1.1869\text{V}\end{array}$

The value of $v_O$ for the given input is $v_O=0.3733\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The parameters are:

  $\begin{array}{l}{V}_{DD}=3.3\text{V}\\ K_n=50\mu \text{A}/{\text{V}}^2\\ R_D=30\text{k}\Omega \\ v_I=3.3\text{V}\\ V_{TN}=0.5\text{V}\end{array}$

Calculation:

For the transition point, the equation for finding the input voltage is,

  $K_n{R}_D{\left(v_{It}-V_{TN}\right)}^2+\left(v_{It}-V_{TN}\right)-V_{DD}=0$

Substituting the values,

  $\begin{array}{l}\left(\left(50\times {10}^{-6}\right)\left(30\times {10}^{-3}\right){\left(v_{It}-0.5\right)}^2\right)+\left(v_{It}-0.5\right)-3.3=0\\ 1.5\left(v_{It}{}^2+0.25-v_{It}\right)+\left(v_{It}-0.5\right)-3.3=0\\ 1.5v_{It}{}^2+0.375-1.5v_{It}+v_{It}-0.5-3.3=0\\ 1.5v_{It}{}^2-0.5v_{It}-3.425=0\end{array}$

Solving the above quadratic equation,

  $v_{It}=1.6869\text{V}\text{or }-1.3535\text{V}$

Ignoring the negative term, the input voltage of transition point is $v_{It}=1.6869\text{V}$

Now, the output voltage at the transition point is

  $\begin{array}{c}{v}_{Ot}=v_{It}-V_{TN}\\ =1.6869-0.5\\ =1.1869\text{V}\end{array}$

It is given that input voltage is $v_I=3.3\text{V}$ . It is equal to $V_{DD}$ means input is high. So, the output voltage in this case is given by equation,

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

Substituting the values in the above equation,

  $\begin{array}{l}{v}_O=3.3-\left(\left(50\times {10}^{-6}\right)\left(30\times {10}^3\right)\left(2\left(3.3-0.5\right)v_O-v_O{}^2\right)\right)\\ v_O=3.3-\left(1.5\left(5.6v_O-v_O{}^2\right)\right)\\ v_O=3.3-8.4v_O+1.5v_O{}^2\\ 1.5v_O{}^2-9.4v_O+3.3=0\end{array}$

Solving the above quadratic equation,

  $v_O=5.8933\text{V}\text{or}\text{0}\text{.3733}\text{V}$

Since, the output voltage cannot be greater than $V_{DD}$ , so only one value is considered.

The output voltage will be

  $v_O=0.3733\text{V}$

Conclusion:

Therefore, the transition points are

  $\begin{array}{c}{v}_{It}=1.6869\text{V}\\ v_{Ot}=1.1869\text{V}\end{array}$

Also, the value of $v_O$ for the given input is $v_O=0.3733\text{V}$

(c)

To determine

The transition point and value of $v_o$ for the given input.

Answer to Problem 16.2P

The transition point is,

  $\begin{array}{c}{v}_{It}=2.6472\text{V}\\ v_{Ot}=2.1472\text{V}\end{array}$

The value of $v_O$ for the given input is $v_O=1.6631\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The parameters are:

  $\begin{array}{l}{V}_{DD}=3.3\text{V}\\ K_n=50\mu \text{A}/{\text{V}}^2\\ R_D=5\text{}\text{k}\Omega \\ v_I=3.3\text{V}\\ V_{TN}=0.5\text{V}\end{array}$

Calculation:

For the transition point, the equation for finding the input voltage is,

  $K_n{R}_D{\left(v_{It}-V_{TN}\right)}^2+\left(v_{It}-V_{TN}\right)-V_{DD}=0$

Substituting the values,

  $\begin{array}{l}\left(\left(50\times {10}^{-6}\right)\left(5\times {10}^{-3}\right){\left(v_{It}-0.5\right)}^2\right)+\left(v_{It}-0.5\right)-3.3=0\\ 0.25\left(v_{It}{}^2+0.25-v_{It}\right)+\left(v_{It}-0.5\right)-3.3=0\\ 0.25v_{It}{}^2+0.0625-0.25v_{It}+v_{It}-0.5-3.3=0\\ 0.25v_{It}{}^2+0.75v_{It}-3.7375=0\end{array}$

Solving the above quadratic equation,

  $v_{It}=2.6472\text{V }\text{or }-5.6472\text{V}$

Ignoring the negative term, the input voltage of transition point is $v_{It}=2.6472\text{V}$

Now, the output voltage at the transition point is

  $\begin{array}{c}{v}_{Ot}=v_{It}-V_{TN}\\ =2.6472-0.5\\ =2.1472\text{V}\end{array}$

It is given that input voltage is $v_I=3.3\text{V}$ . It is equal to $V_{DD}$ means input is high. So, the output voltage in this case is given by equation,

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

Substituting the values in the above equation,

  $\begin{array}{l}{v}_O=3.3-\left(\left(50\times {10}^{-6}\right)\left(5\times {10}^3\right)\left(2\left(3.3-0.5\right)v_O-v_O{}^2\right)\right)\\ v_O=3.3-\left(0.25\left(5.6v_O-v_O{}^2\right)\right)\\ v_O=3.3-1.4v_O+0.25v_O{}^2\\ 0.25v_O{}^2-2.4v_O+3.3=0\end{array}$

Solving the above quadratic equation,

  $v_O=7.9368\text{V}\text{or}1.6631\text{V}$

Since, the output voltage cannot be greater than $V_{DD}$ , so only one value is considered.

The output voltage will be

  $v_O=1.6631\text{V}$

Conclusion:

Therefore, the transition points are

  $\begin{array}{c}{v}_{It}=2.6472\text{V}\\ v_{Ot}=2.1472\text{V}\end{array}$

Also, the value of $v_O$ for the given input is $v_O=1.6631\text{V}$