Chapter 3, Problem 3.27P

The transistor in the circuit in Figure P3.27 has parameters $V_{TN}=0.8\text{V}$ and $K_n=0.25{\text{mA/V}}^{\text{2}}$ . Sketch the load line and plot the Q−point for (a) $V_{DD}=4\text{V,}{R}_D=1\text{k}\Omega$ and (b) $V_{DD}=5\text{V,}{R}_D=3\text{k}\Omega$ . What is the operating bias region for each condition?

Figure P3.27

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.27P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{TN}=0.8\text{V,}{K}_n=0.25\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ V_{DD}=4\text{V},R_D=1\text{kΩ}\end{array}$

Calculation:

The value of $V_{GS}$ is:

  $\begin{array}{l}{V}_{GS}=V_G-V_S\\ V_{GS}=V_{DD}-0\\ V_{GS}=4\text{V}\end{array}$

The value of $V_{DS}\left(\text{sat}\right)$ is:

  $\begin{array}{l}{V}_{DS}\left(\text{sat}\right)=V_{GS}-V_{TN}\\ V_{DS}\left(\text{sat}\right)=4-0.8\\ V_{DS}\left(\text{sat}\right)=3.2\text{V}\end{array}$

Applying Kirchhoff’s voltage in drain-source terminal:

  $\begin{array}{l}{V}_{DS}=V_{DD}-I_D{R}_D\\ V_{DS}=4-1000I_D\mathrm{...}\left(1\right)\end{array}$

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

  $\begin{array}{c}{I}_D=K_n\left[2\left(V_{GS}-V_{TN}\right)V_{DS}-V_{DS}{}^2\right]\\ I_D=0.25\times {10}^{-3}\left[2\left(4-0.8\right)\left(4-1000I_D\right)-{\left(4-1000I_D\right)}^2\right]\\ I_D=0.25\times {10}^{-3}\left[6.4\left(4-1000I_D\right)-{\left(4-1000I_D\right)}^2\right]\\ I_D=0.25\times {10}^{-3}\left[\left(4-1000I_D\right)\left(6.4-4+1000I_D\right)\right]\\ I_D=0.25\times {10}^{-3}\left[\left(4-1000I_D\right)\left(2.4+1000I_D\right)\right]\\ I_D=0.25\times {10}^{-3}\left[9.6+4000I_D-2400I_D-{10}^6{I}_D{}^2\right]\\ I_D=2.4\times {10}^{-3}+0.4I_D-250I_D{}^2\\ 250I_D{}^2+0.6I_D-2.4\times {10}^{-3}=0\\ I_D=2.12\text{mA}\end{array}$

From equation (1):

  $\begin{array}{l}{V}_{DS}=4-1000I_D\\ V_{DS}=4-1000\left(2.12\times {10}^{-3}\right)\\ V_{DS}=1.88\text{V}\end{array}$

From above calculations:

  $V_{DS}<V_{DS}\left(\text{sat}\right)$

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

  $\left(V_{DSQ},I_{DQ}\right)=\left(1.88\text{V, }2.12\text{mA}\right)$

The diagram of load line is sketched as follows:

From equation (1):

  $\begin{array}{l}{V}_{DS}=4-1000I_D\\ I_D=-1\times {10}^{-3}{V}_{DS}+4\times {10}^{-3}\end{array}$

It is equation of straight line with negative slope.

  $y=-mx+c$

At $I_D=0$ ,

  $\begin{array}{l}{I}_D=-1\times {10}^{-3}{V}_{DS}+4\times {10}^{-3}\\ 0=-1\times {10}^{-3}{V}_{DS}+4\times {10}^{-3}\\ V_{DS}=4\text{V}\end{array}$

At $V_{DS}=0$ ,

  $\begin{array}{l}{I}_D=-1\times {10}^{-3}\times 0+4\times {10}^{-3}\\ I_D=4\text{mA}\end{array}$

  

Figure 1

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.27P

A load-line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{V}_{TN}=0.8\text{V,}{K}_n=0.25\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{${\text{V}}^{\text{2}}$}\right.\\ V_{DD}=5\text{V},R_D=3\text{kΩ}\end{array}$

Calculation:

The value of $V_{GS}$ is:

  $\begin{array}{l}{V}_{GS}=V_G-V_S\\ V_{GS}=V_{DD}-0\\ V_{GS}=5\text{V}\end{array}$

The value of $V_{DS}\left(\text{sat}\right)$ is:

  $\begin{array}{l}{V}_{DS}\left(\text{sat}\right)=V_{GS}-V_{TN}\\ V_{DS}\left(\text{sat}\right)=5-0.8\\ V_{DS}\left(\text{sat}\right)=4.2\text{V}\end{array}$

Applying Kirchhoff’s voltage in drain-source terminal:

  $\begin{array}{l}{V}_{DS}=V_{DD}-I_D{R}_D\\ V_{DS}=5-3000I_D\mathrm{...}\left(2\right)\end{array}$

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

  $\begin{array}{c}{I}_D=K_n\left[2\left(V_{GS}-V_{TN}\right)V_{DS}-V_{DS}{}^2\right]\\ I_D=0.25\times {10}^{-3}\left[2\left(5-0.8\right)\left(5-3000I_D\right)-{\left(5-3000I_D\right)}^2\right]\\ I_D=0.25\times {10}^{-3}\left[8.4\left(5-3000I_D\right)-{\left(5-3000I_D\right)}^2\right]\\ I_D=0.25\times {10}^{-3}\left[\left(5-3000I_D\right)\left(8.4-5+3000I_D\right)\right]\\ I_D=0.25\times {10}^{-3}\left[\left(5-3000I_D\right)\left(3.4+3000I_D\right)\right]\\ I_D=0.25\times {10}^{-3}\left[17+15000I_D-10,200I_D-9\times {10}^6{I}_D{}^2\right]\\ I_D=4.25\times {10}^{-3}+1.2I_D-2250I_D{}^2\\ 2250I_D{}^2-0.2I_D-4.25\times {10}^{-3}=0\\ I_D=1.42\text{mA}\end{array}$

From equation (2):

  $\begin{array}{l}{V}_{DS}=5-3000I_D\\ V_{DS}=5-3000\left(1.42\times {10}^{-3}\right)\\ V_{DS}=0.74\text{V}\end{array}$

From above calculations:

  $V_{DS}<V_{DS}\left(\text{sat}\right)$

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q-point is:

  $\left(V_{DSQ},I_{DQ}\right)=\left(0.74\text{V, 1}\text{.42}\text{mA}\right)$

The diagram of load line is sketched as follows:

From equation (1):

  $\begin{array}{l}{V}_{DS}=5-3000I_D\\ I_D=-\left(\frac{1}{3}\right)\times {10}^{-3}{V}_{DS}+\left(\frac{5}{3}\right)\times {10}^{-3}\end{array}$

It is equation of straight line with negative slope.

  $y=-mx+c$

At $I_D=0$ ,

  $V_{DS}=5\text{V}$

At $V_{DS}=0$ ,

  $I_D=1.67\text{mA}$

  

Figure 2