Chapter 14, Problem 14.1TYU

Using the circuit and transistor parameters of Example 13.11, and assuming threshold voltages of $V_{TN}=0.5\text{V}$ and $V_{TP}=-0.5\text{V}$ , determine the maximumrange of common−mode input voltage for the all−CMOS folded cascode circuit ofFigure 14.1(b). (Ans. $V^{-}-0.184\le {\upsilon }_{CM}\le V^{+}-1.13\text{V}$ )

To determine

The maximum range of the common mode input voltage.

Answer to Problem 14.1TYU

  $V^{-}-0.184\text{V}\le v_{CM}\le V^{+}-1.13\text{V}$

Explanation of Solution

Given information:

Transistor parameters are:

  $\begin{array}{l}{I}_{REF}=100\text{μA}\\ K_n{}^{\text{'}}=80{\text{μA//V}}^{\text{2}}\\ K_p{}^{\text{'}}=40{\text{μA//V}}^{\text{2}}\\ W/L=25\end{array}$

Threshold voltage is

  $\begin{array}{l}{V}_{TN}=0.5\text{V}\\ V_{PN}=-0.5\text{V}\end{array}$

The given circuit is shown below.

  

Calculation:

The common mode input voltage range is given by

  $\begin{array}{l}{v}_{CM}\left(\max \right)=V^{+}-V_{SD1}\left(sat\right)-V_{SG1}\\ v_{CM}\left(\min \right)=V^{-}+V_{DS4}\left(sat\right)+V_{SD1}\left(sat\right)V_{SG1}\end{array}$

Let $R_2=140\text{k}\Omega$ and $R_1 =10\text{k}\Omega$

For M₁

  $I_D=\frac{k_p{}^{\text{'}}}{2}\left(\frac{W}{L}\right){\left(V_{SG1}+V_{TP}\right)}^2$

Substituting the values,

  $\begin{array}{l}50=\frac{40}{2}\left(25\right){\left(V_{SG1}-0.5\right)}^2\\ {\left(V_{SG1}-0.5\right)}^2=0.1\\ V_{SG1}=0.816\text{V}\\ \text{So,}\\ V_{SD}\left(sat\right)=0.816-0.5\\ V_{SD}\left(sat\right)=0.316\text{V}\\ \text{Then,}\\ {\text{v}}_{CM}\left(\max \right)=V^{+}-0.316-0.816\\ {\text{v}}_{CM}\left(\max \right)=V^{+}-1.13\text{V}\end{array}$

Similarly for M4

  $I_D=\frac{k_n{}^{\text{'}}}{2}\left(\frac{W}{L}\right){\left(V_{GS4}-V_{TN}\right)}^2$

Substituting the values,

  $\begin{array}{l}100=\frac{80}{2}\left(25\right){\left(V_{GS4}-5\right)}^2\\ V_{SG4}=0.816\text{V}\\ \text{So,}\\ V_{DS4}\left(sat\right)=0.816-0.5\\ V_{DS4}\left(sat\right)=0.316\text{V}\\ \text{Then}\\ {\text{v}}_{CM}\left(\min \right)=V^{-}+0.316+0.816\\ {\text{v}}_{CM}\left(\min \right)=V^{-}-0.184\text{V}\\ \text{So}\\ V^{-}-0.184\text{V}\le v_{CM}\le V^{+}-1.13\text{V}\end{array}$