Chapter 9, Problem 9.1P

(a)

To determine

  $V_{OV}$ and $V_{GS}$ of each transistor of the NMOS differential pair.

Answer to Problem 9.1P

  $\begin{array}{l}{V}_{OV}=0.2\text{V}\\ V_{GS}=0.6\text{ V}\end{array}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{V}_{DD}=V_{SS}=1.0\text{ V}\\ V_{tn}=0.4\text{ V}\\ k_{{}_n}^{\text{'}}=0.4{\text{ mA/V}}^2\\ R_D=5\text{ k}\Omega \\ {\left(W/L\right)}_{1,2}=10\\ I=0.16\text{ mA}\end{array}$

Calculation:

The equation for current $I_D$ for each transistor in terms of bias current $I$ , and $k_{{}_n}^{\text{'}}$ is given as follows:

  $I_D=\frac{1}{2}{k}_{{}_n}^{\text{'}}\left(W/L\right)V_{OV}^2$

Where $I_D=\frac{I}{2}$

Plugging the values as follows:

  $\begin{array}{l}\frac{I}{2}=\frac{1}{2}{k}_{{}_n}^{\text{'}}\left(W/L\right)V_{OV}^2\\ I=k_{{}_n}^{\text{'}}\left(W/L\right)V_{OV}^2\end{array}$

Rewrite the equation of the overdrive voltage for each transistor, $V_{OV}$ as follows:

  $V_{OV}=\sqrt{\frac{I}{k_{{}_n}^{\text{'}}\left(W/L\right)}}$

Plugging the values as follows:

  $\begin{array}{l}{V}_{OV}=\sqrt{\frac{0.16\text{ mA}}{\text{0}{\text{.4 mA/V}}^2\times 10}}\\ V_{OV}=0.2\text{ V}\end{array}$

Write the equation for the gate to source voltage $V_{GS}$ for the transistor $Q_1$ as follows:

  $V_{GS}=V_{OV}+V_{tn}$

Plugging the values in the equation as follows:

  $\begin{array}{c}{V}_{GS}=0.2\text{ V}+0.4\text{ V}\\ \text{=}0.6\text{ V}\end{array}$

(b)

To determine

The $V_S$ , $I_{D1}$ , $I_{D2}$ , $V_{D1}$ , $V_{D2}$ , and $V_O$

For $V_{CM}=0$

Answer to Problem 9.1P

  $\begin{array}{l}{V}_S\text{=}-\text{0}\text{.6 V}\\ I_{D1}=I_{D2}=0.08\text{ mA}\\ V_{D1}=V_{D2}=0.6\text{ V}\\ V_O=0\end{array}$

Explanation of Solution

Given Information:

  $V_{CM}=0$

Calculation:

The equation for source voltage $V_S$ as follows:

  $V_S=V_{CM}-V_{GS}$

Plugging the values as follows:

  $\begin{array}{c}{V}_S=0-0.6\text{ V}\\ \text{=}-\text{0}\text{.6 V}\end{array}$

The drain current for each transistor is equal and is given as follows:

  $I_{D1}=I_{D2}=\frac{I}{2}$

Plugging the values as follows:

  $\begin{array}{l}{I}_{D1}=I_{D2}=\frac{0.16\text{ mA}}{2}\\ I_{D1}=I_{D2}=0.08\text{ mA}\end{array}$

Similarly, the drain voltage for each transistor is equal and is given as follows:

  $V_{D1}=V_{D2}=V_{DD}-\left(\frac{I}{2}{R}_D\right)$

Plugging the values as follows:

  $\begin{array}{l}{V}_{D1}=V_{D2}=1\text{ V}-\left(\frac{0.16\text{ mA}}{2}\times 5\text{ k}\Omega \right)\\ V_{D1}=V_{D2}=1\text{ V}-\left(0.08\text{ mA}\times 5\text{ k}\Omega \right)\\ V_{D1}=V_{D2}=1\text{ V}-0.4\text{ V}\\ V_{D1}=V_{D2}=0.6\text{ V}\end{array}$

The equation for output voltage $V_O$ as follows:

  $V_O=V_{D2}-V_{D1}$

Plugging the values as follows:

  $\begin{array}{c}{V}_O=0.6\text{ V}-0.6\text{ V}\\ \text{=0}\end{array}$

(c)

To determine

The $V_S$ , $I_{D1}$ , $I_{D2}$ , $V_{D1}$ , $V_{D2}$ , and $V_O$

For $V_{CM}=0.4\text{ V}$

Answer to Problem 9.1P

  $\begin{array}{l}\text{For}{V}_{CM}=0.4\text{ V:}\\ V_S\text{=}-\text{0}\text{.2 V}\\ I_{D1}=I_{D2}=0.08\text{ mA}\\ V_{D1}=V_{D2}=0.6\text{ V}\\ V_O=0\end{array}$ :

Explanation of Solution

Given Information:

  $V_{CM}=0.4\text{ V}$

Calculation:

Write the equation for source voltage $V_S$ as follows:

  $V_S=V_{CM}-V_{GS}$

Plugging the values as follows:

  $\begin{array}{c}{V}_S=0.4\text{ V}-0.6\text{ V}\\ \text{=}-\text{0}\text{.2 V}\end{array}$

The drain current for each transistor is equal and is given as follows:

  $I_{D1}=I_{D2}=\frac{I}{2}$

Plugging the values as follows:

  $\begin{array}{l}{I}_{D1}=I_{D2}=\frac{0.16\text{ mA}}{2}\\ I_{D1}=I_{D2}=0.08\text{ mA}\end{array}$

Similarly, the drain voltage for each transistor is equal and is given as follows:

  $V_{D1}=V_{D2}=V_{DD}-\left(\frac{I}{2}{R}_D\right)$

Plugging the values as follows:

  $V_{D1}=V_{D2}=1\text{ V}-\left(\frac{0.16\text{ mA}}{2}\times 5\text{ k}\Omega \right)$

  $\begin{array}{l}{V}_{D1}=V_{D2}=1\text{ V}-\left(0.08\text{ mA}\times 5\text{ k}\Omega \right)\\ V_{D1}=V_{D2}=1\text{ V}-0.4\text{ V}\\ V_{D1}=V_{D2}=0.6\text{ V}\end{array}$

The equation for output voltage $V_O$ as follows:

  $V_O=V_{D2}-V_{D1}$

Plugging the values as follows:

  $\begin{array}{c}{V}_O=0.6\text{ V}-0.6\text{ V}\\ \text{=0}\end{array}$

(d)

To determine

The $V_S$ , $I_{D1}$ , $I_{D2}$ , $V_{D1}$ , $V_{D2}$ , and $V_O$

For $V_{CM}=-0.1\text{ V}$

Answer to Problem 9.1P

For $V_{CM}=-0.1\text{ V}$ ,

  $\begin{array}{l}{V}_S\text{=}-\text{0}\text{.7 V}\\ I_{D1}=I_{D2}=0.08\text{ mA}\\ V_{D1}=V_{D2}=0.6\text{ V}\\ V_O=0\end{array}$

Explanation of Solution

Given Information:

  $V_{CM}=-0.1\text{ V}$

Calculation:

The equation for source voltage $V_S$ as follows:

  $V_S=V_{CM}-V_{GS}$

Plugging the values as follows:

  $\begin{array}{c}{V}_S=-0.1\text{ V}-0.6\text{ V}\\ \text{=}-\text{0}\text{.7 V}\end{array}$

The drain current for each transistor is equal and is given as follows:

  $I_{D1}=I_{D2}=\frac{I}{2}$

Plugging the values as follows:

  $\begin{array}{l}{I}_{D1}=I_{D2}=\frac{0.16\text{ mA}}{2}\\ I_{D1}=I_{D2}=0.08\text{ mA}\end{array}$

Similarly, the drain voltage for each transistor is equal and is given as follows:

  $V_{D1}=V_{D2}=V_{DD}-\left(\frac{I}{2}{R}_D\right)$

Plugging the values as follows:

  $V_{D1}=V_{D2}=1\text{ V}-\left(\frac{0.16\text{ mA}}{2}\times 5\text{ k}\Omega \right)$

  $\begin{array}{l}{V}_{D1}=V_{D2}=1\text{ V}-\left(0.08\text{ mA}\times 5\text{ k}\Omega \right)\\ V_{D1}=V_{D2}=1\text{ V}-0.4\text{ V}\\ V_{D1}=V_{D2}=0.6\text{ V}\end{array}$

Write the equation for output voltage $V_O$ as follows:

  $V_O=V_{D2}-V_{D1}$

Plugging the values as follows:

  $\begin{array}{l}{V}_O=0.6\text{ V}-0.6\text{ V}\\ \text{=0}\end{array}$

(e).

To determine

The highest value of $V_{CM}$ for the transistor $Q_1$ and $Q_2$ to be in saturation mode.

Answer to Problem 9.1P

  $V_{CM\max }\text{=1 V}$

Explanation of Solution

Given Information:

Transistor $Q_1$ and $Q_2$ to be in saturation mode.

Calculation:

In saturation mode, the drain voltages and drain currents are maximum.

Write the equation for a maximum value of $V_{CM}$ for transistor $Q_1$ and $Q_2$ to be in saturation mode as follows:

  $V_{CM\max }=V_{D2}+V_{tn}$

Plugging the values as follows:

  $\begin{array}{c}{V}_{CM\max }=0.6\text{ V+}0.4\text{ V}\\ \text{=1 V}\end{array}$

(f).

To determine

The lowest value allowed for $V_S$ and $V_{CM}$ , if the current source $I$ requires a minimum of $0.2\text{ V}$ .

Answer to Problem 9.1P

  $V_{S\min }\text{=}-0.8\text{ V}$

  $V_{CM\min }\text{=}-0.2\text{ V}$

Explanation of Solution

Given Information:

The current source $I$ requires a minimum of $0.2\text{ V}$ voltage supply.

Calculation:

The equation for the minimum value of $V_S$ as follows:

  $V_{S\min }=-V_{SS}+V_{GS}$

Plugging the values as follows:

  $\begin{array}{c}{V}_{S\min }=-1\text{ V + }0.2\text{ V}\\ \text{=}-0.8\text{ V}\end{array}$

The minimum value of $V_{CM}$ is given as follows:

  $V_{CM\min }=V_{S\min }+V_{GS}$

Plugging the values as follows:

  $\begin{array}{c}{V}_{CM\min }=-0.8\text{ V + }0.6\text{ V}\\ \text{=}-0.2\text{ V}\end{array}$

(g).

To determine

Input common mode range.

Answer to Problem 9.1P

  $-0.2\text{ V}\le V_{ICM}\le 1\text{V}$

Explanation of Solution

Calculation:

As the input common mode range is defined over the saturation of transistors within the circuitry. If the transistors are in saturation, the input common mode range became equivalent to the common mode voltage range.

Write the equation for a maximum value of $V_{CM}$ for transistor $Q_1$ and $Q_2$ to be in saturation mode as follows:

  $V_{CM\max }=V_{D2}+V_{tn}$

Plugging the values as follows:

  $\begin{array}{c}{V}_{CM\max }=0.6\text{ V+}0.4\text{ V}\\ \text{=1 V}\end{array}$

The minimum value of $V_{CM}$ is given as follows:

  $V_{CM\min }=V_{S\min }+V_{GS}$

Plugging the values as follows:

  $\begin{array}{c}{V}_{CM\min }=-0.8\text{ V + }0.6\text{ V}\\ \text{=}-0.2\text{ V}\end{array}$