Chapter 7, Problem 7.1P

To determine

The Co-ordinates of active region segment $\left(AB\right)$ of voltage transfer characteristics.

Answer to Problem 7.1P

Coordinates of Point A:

  $A\equiv \left(V_t,V_{DD}\right)=\left(0.5\text{ V},3\text{ V}\right)$

Coordinates of Point B:

  $B\equiv \left(V_{GS\left|B\right.},V_{DS\left|B\right.}\right)=\left(0.69\text{ V,}0.19\text{ V}\right)$

Explanation of Solution

Given:

  

Fig: NMOS amplifier

  $\begin{array}{l}{V}_{DD}=3\text{ V}\\ V_t=0.5\text{ V}\\ k_n=10\frac{\text{mA}}{{\text{V}}^2}\\ R_D=15\text{ k}\Omega \end{array}$

Calculation:

From the given circuit diagram, it is clear that the $x$ coordinate of the point $A$ is $V_t$ and $y$ coordinate of the point $A$ is $V_{DD}$ .

  $\begin{array}{l}\text{Coordinates of Point }A:\\ \text{X}-\text{Coordinate}:V_t\\ \text{Y}-\text{Coordinate}:V_{DD}\\ \text{Where,}\\ V_t=\text{Threshold volatge of transistor}\text{.}\\ V_{DD}=\text{Input DC supply}\end{array}$

Plugging the values of $V_t$ and $V_{DD}$

  $\begin{array}{l}\text{Coordinates of Point }A:\\ \text{X}-\text{Coordinate}:V_t=0.5\text{ V}\\ \text{Y}-\text{Coordinate}:V_{DD}=3\text{ Volt}\\ \text{Point }A:\left(0.5\text{ V,3 V}\right)\end{array}$

The expression of $X$ coordinate of the point $B$ that is $\left(V_{GS\left|B\right.}\right)$ given by

  $V_{GS\left|B\right.}=V_t+\frac{\sqrt{2k_n{R}_D{V}_{DD}+1}-1}{k_n{R}_D}$

Plugging the values and simplifying as follows:

  $\begin{array}{c}{V}_{GS\left|B\right.}=V_t+\frac{\sqrt{2k_n{R}_D{V}_{DD}+1}-1}{k_n{R}_D}\\ V_{GS\left|B\right.}=0.5\text{ V}+\frac{\sqrt{2\times 10\frac{\text{mA}}{{\text{V}}^2}\times 15\text{ k}\Omega \times \text{3 V+1}}-1}{10\frac{\text{mA}}{{\text{V}}^2}\times 15\text{ k}\Omega }\\ =\left(0.5\text{+}\frac{\sqrt{900+1}-1}{150}\right)\text{ V}\\ =0.6934\text{ V}\end{array}$

The expression of $Y$ coordinate of the point $B$

  $\left(V_{DS\left|B\right.}\right)$ is given as follows:

  $\left(V_{DS\left|B\right.}\right)=V_{GS\left|B\right.}-V_t$

Plugging the resultant obtained values of $V_{GS\left|B\right.}$ and given value of $V_t$

to simplify further as follows:

  $\begin{array}{c}\left(V_{DS\left|B\right.}\right)=V_{GS\left|B\right.}-V_t\\ =0.6934\text{ V}-0.5\text{ V}\\ =\text{0}\text{.1935 V}\\ \approx \text{0}\text{.19 V}\end{array}$

Conclusion:

The coordinate of point A is $\left(0.5\text{ V,3 V}\right)$ and coordinate of point B is $\left(0.69\text{ V,}0.19\text{ V}\right)$ .