Chapter 3, Problem 3.47P

a.

To determine

The width to length ratio of the transistors to meet the requirements.

Answer to Problem 3.47P

  ${\left(\frac{W}{L}\right)}_1=2.315$ , ${\left(\frac{W}{L}\right)}_2=1.585$ , ${\left(\frac{W}{L}\right)}_3=3.693$

Explanation of Solution

Given Information:

The given values are:

  $V_{TN}=0.6\text{ V, }{k}_n^{\text{'}}=120\mu \text{A}/{\text{V}}^2,I_{DQ}=0.8\text{ mA, }{V}_1=2.5\text{ V, }{V}_2=6\text{ V}$

The given circuit is shown below.

  

Calculation:

The voltage is,

  $V_{GS3}=V_1=2.5\text{ V}$

Also,

  $\begin{array}{c}{V}_{GS2}=V_2-V_1\\ =6-2.5\\ =\text{3}\text{.5 V}\end{array}$

And,

  $\begin{array}{c}{V}_{GS1}=V^{+}-V_2\\ =9-6\\ =\text{3 V}\end{array}$

Then the constant current is

  $\begin{array}{c}{I}_{DQ}=\frac{k_n^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_1{\left(V_{GS1}-V_{TN}\right)}^2\\ {\left(\frac{W}{L}\right)}_1=\frac{2I_{DQ}}{k_n^{\text{'}}{\left(V_{GS1}-V_{TN}\right)}^2}\\ =\frac{2\times 0.8}{0.12{\left(3-0.6\right)}^2}\\ =2.315\end{array}$

Similarly,

  $\begin{array}{c}{I}_{DQ}=\frac{k_n^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_2{\left(V_{GS2}-V_{TN}\right)}^2\\ {\left(\frac{W}{L}\right)}_2=\frac{2I_{DQ}}{k_n^{\text{'}}{\left(V_{GS2}-V_{TN}\right)}^2}\\ =\frac{2\times 0.8}{0.12{\left(3.5-0.6\right)}^2}\\ =1.585\end{array}$

And,

  $\begin{array}{c}{I}_{DQ}=\frac{k_n^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_3{\left(V_{GS3}-V_{TN}\right)}^2\\ {\left(\frac{W}{L}\right)}_3=\frac{2I_{DQ}}{k_n^{\text{'}}{\left(V_{GS3}-V_{TN}\right)}^2}\\ =\frac{2\times 0.8}{0.12{\left(2.5-0.6\right)}^2}\\ =3.693\end{array}$

b.

To determine

The change in the output voltages for the $k_n^{\text{'}}$ parameter change.

Answer to Problem 3.47P

When $k_n^{\text{'}}=120\times 1.05=126\mu \text{A}/{\text{V}}^2$

  $V_1=2.5\text{ V}$ , $V_2=6\text{ V}$

When $k_n^{\text{'}}=120\times 0.95=114\mu \text{A}/{\text{V}}^2$

  $V_1=2.5\text{ V}$ , $V_2=6\text{ V}$

Explanation of Solution

Given Information:

The given values are:

  $V_{TN}=0.6\text{ V, }{k}_n^{\text{'}}=120\mu \text{A}/{\text{V}}^2,I_{DQ}=0.8\text{ mA, }{V}_1=2.5\text{ V, }{V}_2=6\text{ V}$

The given circuit is shown below.

  

Parameter $k_n^{\text{'}}$ changes by $+5\%$ and $-5\%$

Calculation:

When there is a change in $k_n^{\text{'}}$ for all transistors with the same percentage, then it will change current through each transistor which is equal in all transistors. According to the below equation, there is no change in $V_{GSQ}$ . So, there is no effect on output voltages $V_1\text{ and }{V}_2$ .

  $I_{DQ}=\frac{k_n^{\text{'}}}{2}\left(\frac{W}{L}\right){\left(V_{GSQ}-V_{TN}\right)}^2$

From part (a), ${\left(\frac{W}{L}\right)}_1=2.315$ , ${\left(\frac{W}{L}\right)}_2=1.585$ , ${\left(\frac{W}{L}\right)}_3=3.693$

For each case, $k_n^{\text{'}}=k_{n1}^{\text{'}}=k_{n2}^{\text{'}}=k_{n3}^{\text{'}}$

Then,

  $\begin{array}{l}{I}_{D2}=I_{D3}\\ \frac{k_{n2}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_2{\left(V_{GS2}-V_{TN}\right)}^2=\frac{k_{n3}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_3{\left(V_{GS3}-V_{TN}\right)}^2\\ 1.585{\left(V_2-V_1-0.6\right)}^2=3.693{\left(V_1-0.6\right)}^2\\ V_2=2.526V_1-0.3156\to (1)\end{array}$

Also,

  $\begin{array}{l}{I}_{D1}=I_{D3}\\ \frac{k_{n1}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_1{\left(V_{GS1}-V_{TN}\right)}^2=\frac{k_{n3}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_3{\left(V_{GS3}-V_{TN}\right)}^2\\ 2.315{\left(9-V_2-0.6\right)}^2=3.693{\left(V_1-0.6\right)}^2\end{array}$

Substituting equation (1)

  $\begin{array}{l}2.315{\left(9.3156-2.526V_1-0.6\right)}^2=3.693{\left(V_1-0.6\right)}^2\\ 2.315{\left(9.3156-2.526V_1-0.6\right)}^2=3.693{\left(V_1-0.6\right)}^2\\ \left(9.3156-2.526V_1-0.6\right)=1.263\left(V_1-0.6\right)\\ V_1=\frac{9.4734}{3.789}\text{ V}\end{array}$

  $V_1=2.5\text{ V}$

From equation (1),

  $V_2=2.526\times 2.5-0.3156$

  $V_2=6\text{ V}$

So,

When $k_n^{\text{'}}=k_{n1}^{\text{'}}=k_{n2}^{\text{'}}=k_{n3}^{\text{'}}120\times 1.05=126\mu \text{A}/{\text{V}}^2$

  $V_1=2.5\text{ V}$ , $V_2=6\text{ V}$

When $k_n^{\text{'}}=k_{n1}^{\text{'}}=k_{n2}^{\text{'}}=k_{n3}^{\text{'}}=120\times 0.95=114\mu \text{A}/{\text{V}}^2$

  $V_1=2.5\text{ V}$ , $V_2=6\text{ V}$

c.

To determine

The output voltages for a specified change in each transistor.

Answer to Problem 3.47P

  $V_1=2.468\text{ V}$ , $V_2=5.919\text{ V}$

Explanation of Solution

Given Information:

The given values are:

  $V_{TN}=0.6\text{ V, }{k}_n^{\text{'}}=120\mu \text{A}/{\text{V}}^2,I_{DQ}=0.8\text{ mA, }{V}_1=2.5\text{ V, }{V}_2=6\text{ V}\text{.}$

The given circuit is shown below.

  

Also, $k_n^{\text{'}}$ parameter of M₁ decreases by 5% while $k_n^{\text{'}}$ parameter of M₂ and M₃ increases by 5%.

Calculation:

When there is a change in $k_n^{\text{'}}$ for all transistors with the different percentages for transistors, the current through transistors must be equal so that according to the below equation, there may be a change in $V_{GSQ}$ for each transistor. So that output voltages $V_1\text{ and }{V}_2$ will change.

  $I_{DQ}=\frac{k_n^{\text{'}}}{2}\left(\frac{W}{L}\right){\left(V_{GSQ}-V_{TN}\right)}^2$

For M₁ :

  $k_{n1}^{\text{'}}=120\times 0.95=114\mu \text{A}/{\text{V}}^2$

Now for M₂ and M₃ :

  $k_{n2}^{\text{'}}=k_{n3}^{\text{'}}=120\times 1.05=126\mu \text{A}/{\text{V}}^2$

From part(a), ${\left(\frac{W}{L}\right)}_1=2.315$ , ${\left(\frac{W}{L}\right)}_2=1.585$ , ${\left(\frac{W}{L}\right)}_3=3.693$

Then,

  $\begin{array}{l}{I}_{D2}=I_{D3}\\ \frac{k_{n2}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_2{\left(V_{GS2}-V_{TN}\right)}^2=\frac{k_{n3}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_3{\left(V_{GS3}-V_{TN}\right)}^2\end{array}$

  $\begin{array}{l}1.585{\left(V_2-V_1-0.6\right)}^2=3.693{\left(V_1-0.6\right)}^2\\ V_2=2.526V_1-0.3156\to (1)\end{array}$

Also,

  $\begin{array}{l}{I}_{D1}=I_{D3}\\ \frac{k_{n1}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_1{\left(V_{GS1}-V_{TN}\right)}^2=\frac{k_{n3}^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_3{\left(V_{GS3}-V_{TN}\right)}^2\\ \frac{114}{2}\times 2.315{\left(9-V_2-0.6\right)}^2=\frac{126}{2}\times 3.693{\left(V_1-0.6\right)}^2\end{array}$

Substituting equation (1)

  $\begin{array}{l}\frac{114}{2}\times 2.315{\left(9.3156-2.526V_1-0.6\right)}^2=\frac{126}{2}\times 3.693{\left(V_1-0.6\right)}^2\\ \frac{114}{2}\times 2.315{\left(9.3156-2.526V_1-0.6\right)}^2=\frac{126}{2}\times 3.693{\left(V_1-0.6\right)}^2\\ \left(9.3156-2.526V_1-0.6\right)=1.32784\left(V_1-0.6\right)\\ V_1=\frac{9.512304}{3.85384}\text{ V}\end{array}$

  $V_1=2.468\text{ V}$

From equation (1),

  $V_2=2.526\times 2.468-0.3156$

  $V_2=5.919\text{ V}$