Chapter 3, Problem 3.8EP

(a) Consider the circuit shown in Figure 3.33. The transistor parameters are $V_{TP}=-0.40\text{V}$ and $K_p=30\mu {\text{A/V}}^{\text{2}}$ . Design the circuit such that $I_{DQ}=60\mu \text{A}$ and $V_{SDQ}=2.5\text{V}$ . (b) Determine the variation in Q−point values if the $V_{TP}$ and $K_p$ parameters vary by ±5 percent. (Ans. (a) $R_S=19.77\text{k}\Omega$ , $R_D=38.57\text{k}\Omega$ ; (b) $58.2\le I_{DQ}\le 61.08\mu \text{A},2.437\le V_{SDQ}\le 2.605\text{V}$ )

To determine

The design parameters of the circuit.

Answer to Problem 3.8EP

  $\begin{array}{l}{R}_S=19.77\text{kΩ}\\ R_D=38.57\text{kΩ}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{TP}=-0.4\text{V},K_P=30\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ I_{DQ}=60\text{μA},V_{SDQ}=2.5\text{V}\end{array}$

Calculation:

Assuming the transistor operates in saturation region.

  $\begin{array}{l}{I}_{DQ}=K_P{\left(V_{SG}+V_{TP}\right)}^2\\ 60=30{\left(V_{SG}-0.4\right)}^2\\ {\left(V_{SG}-0.4\right)}^2=2\\ V_{SG}=1.414+0.4\\ V_{SG}=1.814\text{V}\end{array}$

From above calculation:

  $\begin{array}{l}{V}_{SD}>V_{SG}+V_{TP}\\ 2.5>1.814-0.4\\ 2.5>1.414\end{array}$

Hence, the assumption is correct and the transistor operates in saturation region.

The voltage of source terminal is:

  $\begin{array}{l}{V}_{SG}=V_S-V_G\\ 1.814=V_S-0\\ V_S=1.814\text{V}\end{array}$

The voltage of drain terminal is:

  $\begin{array}{l}{V}_{SD}=2.5\\ V_S-V_D=2.5\\ 1.814-V_D=2.5\\ V_D=-0.686\text{V}\end{array}$

The value of $R_S$ is:

  $\begin{array}{l}{I}_D=\frac{3-V_S}{R_S}\\ 0.06=\frac{3-1.814}{R_S}\\ R_S=\frac{1.186}{0.06}\\ R_S=19.77\text{kΩ}\end{array}$

The value of $R_D$ is:

  $\begin{array}{l}{I}_D=\frac{V_D-\left(-3\right)}{R_D}\\ 0.06=\frac{-0.686+3}{R_D}\\ R_D=\frac{2.314}{0.06}\\ R_D=38.57\text{kΩ}\end{array}$

To determine

The range of values of $I_D,V_{SD}$ .

Answer to Problem 3.8EP

  $\begin{array}{l}58.6\text{μA}\le I_{DQ}\le 61.2\text{μA}\\ 2.429\text{V}\le V_{SDQ}\le 2.58\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{TP}=-0.4\text{V},K_P=30\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ I_{DQ}=60\text{μA},V_{SDQ}=2.5\text{V}\end{array}$

Calculation:

The value of $V_{SG}$ is:

  $V_{SG}=1.814\text{V}$

Values of $K_P,V_{TP}$ :

  $\begin{array}{l}{K}_P=30\pm \left(5\raisebox{1ex}{$o$}\!\left/ \!\raisebox{-1ex}{$o$}\right.\text{of}30\right)\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ K_P=30\pm \left(\frac{5}{100}\times 30\right)\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ K_P=30\pm \left(1.5\right)\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ {\left(K_P\right)}_{\max }=\left(30+1.5\right)\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ {\left(K_P\right)}_{\max }=31.5\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ {\left(K_P\right)}_{\min }=\left(30-1.5\right)\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ {\left(K_P\right)}_{\min }=28.5\raisebox{1ex}{$\text{μA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\end{array}$

  $\begin{array}{l}{V}_{TP}=-0.4\pm \left(5\raisebox{1ex}{$o$}\!\left/ \!\raisebox{-1ex}{$o$}\right.\text{of}-0.4\right)\text{V}\\ V_{TP}=-0.4\pm \left(\frac{5}{100}\times \left(-0.4\right)\right)\text{V}\\ V_{TP}=-0.4\pm \left(-0.02\right)\\ V_{TP}{}_{\max }=-0.4+\left(-0.02\right)\\ =-0.42\text{V}\\ V_{TP}{}_{\min }=-0.4-\left(-0.02\right)\\ =-0.38\text{V}\end{array}$

The value of $I_D$ is maximum for ${\left(K_p\right)}_{\max },V_{TP}{}_{\max }$ .

  $\begin{array}{l}{\left(I_D\right)}_{\max }={\left(K_P\right)}_{\max }{\left(V_{SG}+V_{TP}{}_{\max }\right)}^2\\ {\left(I_D\right)}_{\max }=\left(31.5\times {10}^{-6}\right){\left(1.814-0.42\right)}^2\\ {\left(I_D\right)}_{\max }=61.2\text{μA}\end{array}$

Applying Kirchhoff’s voltage law in drain-source terminals:

  $\begin{array}{l}{\left(V_{SD}\right)}_{\min }=3-{\left(I_D\right)}_{\max }\left(R_D+R_S\right)-\left(-3\right)\\ {\left(V_{SD}\right)}_{\min }=6-\left(0.0612\right)\left(19.77+38.57\right)\\ {\left(V_{SD}\right)}_{\min }=2.429\text{V}\end{array}$

The value of $I_D$ is minimum for ${\left(K_p\right)}_{\min },V_{TP}{}_{\min }$ .

  $\begin{array}{l}{\left(I_D\right)}_{\min }={\left(K_P\right)}_{\min }{\left(V_{SG}+V_{TP\min }\right)}^2\\ {\left(I_D\right)}_{\min }=\left(28.5\times {10}^{-6}\right){\left(1.814-0.38\right)}^2\\ {\left(I_D\right)}_{\min }=58.6\text{μA}\end{array}$

Applying Kirchhoff’s voltage law in drain-source terminals:

  $\begin{array}{l}{\left(V_{SD}\right)}_{\max }=3-{\left(I_D\right)}_{\min }\left(R_D+R_S\right)-\left(-3\right)\\ {\left(V_{SD}\right)}_{\max }=6-\left(0.0586\right)\left(19.77+38.57\right)\\ {\left(V_{SD}\right)}_{\max }=2.58\text{V}\end{array}$

The values of $I_D,V_{SD}$ is:

  $\begin{array}{l}58.6\text{μA}\le I_D\le 61.2\text{μA}\\ 2.429\text{V}\le V_{SD}\le 2.58\text{V}\end{array}$