Chapter 3, Problem 3.39P

a.

To determine

The design parameters for the given circuit.

To sketch: The load line along with Q -point.

Answer to Problem 3.39P

The design parameters are:

  $R_D=8\text{ k}\Omega$ , $R_S=4.38\text{ k}\Omega$ , $V_{DSQ}=3.81\text{ V}$

The load line laong with Q -point is shown in Figure 1.

Explanation of Solution

Given Information:

The given values are:

  $V_{TN}=1.4\text{ V, }{K}_n=0.25\text{ mA}/{\text{V}}^2,I_{DQ}=0.5\text{ mA, }{V}_D=1\text{ V}$

The given circuit is shown below.

  

Calculation:

Applying Ohm’s law to drain resistor,

  $R_D=\frac{V^{+}-V_D}{I_D}=\frac{5-1}{0.5}$

  $R_D=8\text{ k}\Omega$

Then,

  $\begin{array}{l}{I}_{DQ}=K_n{\left(V_{GS}-V_{TN}\right)}^2\\ 0.5=0.25\times {\left(V_{SGQ}-1.4\right)}^2\end{array}$

  $V_{GS}=2.81\text{ V}$

Applying Ohm’s law to drain resistor,

  $R_S=\frac{0-V_{GS}-V^{-}}{I_D}=\frac{-2.81+5}{0.5}$

  $R_S=4.38\text{ k}\Omega$

Then,

  $V_{DSQ}=V_D\text{+}{V}_{GS}\text{=1+2}\text{.81}=3.81\text{ V}$

  $V_{DSQ}=3.81\text{ V}$

The load line equation is,

  $\begin{array}{l}{V}_{DS}=(V^{+}-V^{-})-I_D(R_S+R_D)\\ V_{DS}=10-12.38I_D\end{array}$

When $V_{DS}=0\Rightarrow I_D=0.8078\text{ mA and }{I}_D=0\Rightarrow V_{SD}=10\text{ V}$

The load line and the Q -point shown below.

  

Figure 1

b.

To determine

The Q -point values for the given circuit with standard resistor values.

Answer to Problem 3.39P

The Q -point values are $I_{DQ}=0.506\text{ mA}$ , $V_{DSQ}=\text{3}\text{.675 V}$

Explanation of Solution

Given Information:

The given values are:

  $V_{TN}=1.4\text{ V, }{K}_n=0.25\text{ mA}/{\text{V}}^2,I_{DQ}=0.5\text{ mA, }{V}_D=1\text{ V}$

The given circuit is shown below.

  

Calculation:

Using the calculated resistor values in part (a), identify the standard resistor values close to them.

The values from part (a) are $R_D=8\text{ k}\Omega$ , $R_S=4.38\text{ k}\Omega$ .

So, choose the standard values as $R_D=8.2\text{ k}\Omega$ , $R_S=4.3\text{ k}\Omega$ .

Then,

  $\begin{array}{l}{I}_D=\frac{0-V_{GS}-V^{-}}{R_S}=K_p{\left(V_{GS}-V_{TN}\right)}^2\\ \frac{5-V_{GS}}{4.3}=0.25{\left(V_{GS}-1.4\right)}^2\\ 5-V_{GS}=1.075\left(V_{GS}^2-2.8V_{GS}+1.96\right)\\ 1.075V_{GS}^2-2.01V_{GS}-2.893=0\to (1)\end{array}$

Solving the above equation,

  $V_{GSQ}=\text{2}\text{.823 V}$

Then,

  $\begin{array}{l}{I}_D=K_p{\left(V_{GS}-V_{TN}\right)}^2\\ I_D=0.25{\left(2.823-1.4\right)}^2\end{array}$

  $I_{DQ}=0.506\text{ mA}$

Then,

  $\begin{array}{l}{V}_{SD}=(V^{+}-V^{-})-I_D(R_S+R_D)\\ V_{DS}=10-12.5\times 0.506\text{ V}\end{array}$

  $V_{DSQ}=\text{3}\text{.675 V}$

c.

To determine

The minimum and maximum value of $I_{DQ}$ .

Answer to Problem 3.39P

For $R_S=4.3-\text{10\% k}\Omega =3.87\text{ k}\Omega$ : $I_{DQ}=0.5476\text{ mA}$

For $R_S=4.3\text{ +10\% k}\Omega =4.73\text{ k}\Omega$ : $I_{DQ}=0.4706\text{ mA}$

Explanation of Solution

Given Information:

The given values are:

  $V_{TN}=1.4\text{ V, }{K}_n=0.25\text{ mA}/{\text{V}}^2,I_{DQ}=0.5\text{ mA, }{V}_D=1\text{ V}$

The given circuit is shown below.

  

Calculation:

From part (b), the drain current is

  $\begin{array}{l}{I}_D=\frac{0-V_{GS}-V^{-}}{R_S}=K_p{\left(V_{GS}-V_{TN}\right)}^2\\ 0.25R_S{\left(V_{GS}-1.4\right)}^2+V_{GS}-5=0\\ 0.25R_S{\left(V_{GS}^2-2.8V_{GS}+1.96\right)}^2+V_{GS}-5=0\end{array}$

For $R_S=4.3\text{ +10\% k}\Omega =4.73\text{ k}\Omega$

  $\begin{array}{l}0.25R_S{\left(V_{GS}^2-2.8V_{GS}+1.96\right)}^2+V_{GS}-5=0\\ 0.25\times 4.73\left(V_{GS}^2-2.8V_{GS}+1.96\right)+V_{GS}-5=0\\ 1.1825V_{GS}^2-2.311V_{GS}-2.6823=0\end{array}$

Solving the above equation,

  $V_{GS}=\text{2}\text{.772 V}$

Then,

  $\begin{array}{l}{I}_D=K_p{\left(V_{GS}-V_{TN}\right)}^2\\ I_D=0.25{\left(2.722-1.4\right)}^2\end{array}$

  $I_{DQ}=0.4706\text{ mA}$

For $R_S=4.3-\text{10\% k}\Omega =3.87\text{ k}\Omega$

  $\begin{array}{l}0.25R_S{\left(V_{GS}^2-2.8V_{GS}+1.96\right)}^2+V_{GS}-5=0\\ 0.25\times 3.87\left(V_{GS}^2-2.8V_{GS}+1.96\right)+V_{GS}-5=0\\ 0.9675V_{GS}^2-1.709V_{GS}-3.1037=0\end{array}$

Solving the above equation,

  $V_{GS}=\text{2}\text{.88 V}$

Then,

  $\begin{array}{l}{I}_D=K_p{\left(V_{GS}-V_{TN}\right)}^2\\ I_D=0.25{\left(\text{2}\text{.88}-1.4\right)}^2\end{array}$

  $I_{DQ}=0.5476\text{ mA}$