Chapter 8, Problem 8.5P

(a)

To determine

To sketch:The safe operating area for the transistor and the load line on the same graph.

Answer to Problem 8.5P

Thearea for the safe operation of the transistor and the load line for the transistor is shown in Figure 1.

Explanation of Solution

Calculation:

The sketch for the safe operating area of the transistor is shown below.

The required diagram is shown in Figure 1

  

The above figure shows the safe region under which the transistor must be operated. The graph is plotted under the linear current and voltage scale. The sketch for the load line is dotted and is given by $\frac{-1}{R_D}$ .

Conclusion:

Therefore, the area for the safe operation of the transistor and the load line for the transistor is shown in Figure 1

(b)

To determine

The value of the drain current.

Answer to Problem 8.5P

Thepower at $V_{GG}$ equal to $5\text{V}$ is $9.375\text{W}$ , at $V_{GG}$ equal to $6\text{V}$ is $30\text{W}$ , at $7\text{V}$ is $39.375\text{W}$ , at $8\text{V}$ is $10.82\text{W}$ and at $9\text{V}$ is $7.106\text{W}$ .

Explanation of Solution

Calculation:

The expression for the voltage $V_{GSn}$ is given by,

  $V_{GSn}=V_{GG}$

The expression for the drain current is given by,

  $i_D=K_n{\left(V_{GSn}-V_{Th}\right)}^2$ ……. (1)

Substitute $0.25$ for $K_n$ , $5\text{V}$ for $V_{GSn}$ and $4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{i}_D=\left(0.25\right){\left(\left(5\text{V}\right)-4\text{V}\right)}^2\\ =0.25\text{A}\end{array}$

The expression for the drain to source voltage is given by,

  $v_{DS}=V_{DD}-i_D{R}_D$ …… (2)

Substitute $40\text{V}$ for $V_{DD}$ , $0.25\text{A}$ for $i_D$ and $10\Omega$ for $R_D$ in the above equation.

  $\begin{array}{c}{v}_{DS}=40\text{V}-\left(0.25\text{A}\right)\left(10\Omega \right)\\ =37.5\text{V}\end{array}$

The expression for the power dissipation is given by,

  $P_D=v_{DS}{I}_D$ …… (3)

Substitute $0.25\text{A}$ for $I_D$ and $37.5\text{V}$ for $v_{DS}$ in the above equation.

  $\begin{array}{c}{P}_D=\left(37.5\text{V}\right)\left(0.25\text{A}\right)\\ =9.375\text{W}\end{array}$

Substitute $0.25$ for $K_n$ , $6\text{V}$ for $V_{GSn}$ and $4\text{V}$ for $V_{TN}$ in equation (1).

  $\begin{array}{c}{i}_D=\left(0.25\right){\left(6\text{V}-4\text{V}\right)}^2\\ =1\text{A}\end{array}$

Substitute $40\text{V}$ for $V_{DD}$ , $1\text{A}$ for $i_D$ and $10\Omega$ for $R_D$ in equation (2).

  $\begin{array}{c}{v}_{DS}=40\text{V}-\left(1\text{A}\right)\left(10\Omega \right)\\ =30\text{V}\end{array}$

Substitute $1\text{A}$ for $I_D$ and $30\text{V}$ for $v_{DS}$ in equation (3).

  $\begin{array}{c}{P}_D=\left(30\text{V}\right)\left(1\text{A}\right)\\ =30\text{W}\end{array}$

Substitute $0.25$ for $K_n$ , $7\text{V}$ for $V_{GSn}$ and $4\text{V}$ for $V_{TN}$ in equation (1).

  $\begin{array}{c}{i}_D=\left(0.25\right){\left(7\text{V}-4\text{V}\right)}^2\\ =2.25\text{A}\end{array}$

Substitute $40\text{V}$ for $V_{DD}$ , $2.25\text{A}$ for $i_D$ and $10\Omega$ for $R_D$ in equation (2).

  $\begin{array}{c}{v}_{DS}=40\text{V}-\left(2.25\text{A}\right)\left(10\Omega \right)\\ =17.5\text{V}\end{array}$

Substitute $2.25\text{A}$ for $I_D$ and $17.5\text{V}$ for $v_{DS}$ in equation (3).

  $\begin{array}{c}{P}_D=\left(17.5\text{V}\right)\left(2.25\text{A}\right)\\ =39.375\text{W}\end{array}$

The expression for the drain current in the non saturated region is given by,

  $i_D=k_n\left[2\left(v_{GS}-V_{TN}\right)v_{DS}-{\left(v_{DS}\right)}^2\right]$ …… (4)

Substitute $0.25$ for $k_n$ , $8\text{V}$ for $v_{GS}$ and $4\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{i}_D=0.25\left[2\left(8\text{V}-4\text{V}\right)v_{DS}-{\left(v_{DS}\right)}^2\right]\\ =2v_{DS}-0.25v_{DS}^2\end{array}$ …… (5)

The expression for the drain current by ohm’s law is given by,

  $i_D=\frac{V_{DD}-v_{DS}}{R_D}$

Substitute $40\text{V}$ for $V_{DD}$ and $R_D$ for $R_L$ in the above equation.

  $i_D=\frac{40\text{V}-v_{DS}}{10\Omega }$

Substitute $\frac{40\text{V}-v_{DS}}{10\Omega }$ for $i_D$ in equation (5).

  $\begin{array}{l}2v_{DS}-0.25v_{DS}^2=\frac{40\text{V}-v_{DS}}{10\Omega }\\ 2.5v_{DS}^2-21v_{DS}+40\text{V=0}\\ v_{DS}=2.919\text{V}\end{array}$

Substitute $2.919\text{V}$ for $v_{DS}$ in equation (4).

  $\begin{array}{c}{i}_D=2\left(2.919\text{V}\right)-0.25\left(2.919\right)\\ =3.71\text{A}\end{array}$

Substitute $2.919\text{V}$ for $v_{DS}$ and $3.71\text{A}$ for $i_D$ and $2.919\text{V}$ for $v_{DS}$ in equation (3).

  $\begin{array}{c}{P}_D=\left(2.919\text{V}\right)\left(3.71\right)\\ =10.82\text{W}\end{array}$

Substitute $0.25$ for $k_n$ , $9\text{V}$ for $v_{GS}$ and $4\text{V}$ for $V_{TN}$ in the equation (4).

  $\begin{array}{c}{i}_D=0.25\left[2\left(9\text{V}-4\text{V}\right)v_{DS}-{\left(v_{DS}\right)}^2\right]\\ =2.5v_{DS}-0.25v_{DS}^2\end{array}$ …… (6)

Substitute $\frac{40\text{V}-v_{DS}}{10\Omega }$ for $i_D$ in the above equation.

  $\begin{array}{l}2.5v_{DS}-0.25v_{DS}^2=\frac{40\text{V}-v_{DS}}{10\Omega }\\ 2.5v_{DS}^2-26v_{DS}+40\text{V}=\text{0}\\ v_{DS}=1.87\text{V}\end{array}$

Substitute $1.87\text{V}$ for $v_{DS}$ in equation (6).

  $\begin{array}{c}{i}_D=2.5\left(1.87\text{V}\right)-0.25{\left(1.87\text{V}\right)}^2\\ =3.80\text{A}\end{array}$

Substitute $1.87\text{V}$ for $v_{DS}$ and $3.80\text{A}$ for $i_D$ and $2.919\text{V}$ for $v_{DS}$ in equation (3).

  $\begin{array}{c}{P}_D=\left(1.87\text{V}\right)\left(3.80\text{A}\right)\\ =7.106\text{W}\end{array}$

Conclusion:

Therefore, the power at $V_{GG}$ equal to $5\text{V}$ is $9.375\text{W}$ , at $V_{GG}$ equal to $6\text{V}$ is $30\text{W}$ , at $7\text{V}$ is $39.375\text{W}$ , at $8\text{V}$ is $10.82\text{W}$ and at $9\text{V}$ is $7.106\text{W}$ .

(c)

To determine

Whether there is a possibility of transistor getting damage.

Answer to Problem 8.5P

Yes, thetransistor will get damaged at $V_{GG}=7\text{V}$ .

Explanation of Solution

The power of the transistor at $V_{GG}$ equal $7\text{V}$ is given by,

  $P=39.375\text{W}$

The above power is greater than the rated power of the transistor this means that the transistor gets damaged at $V_{GG}=7\text{V}$ .