Chapter 8, Problem 8.1EP

(a)

To determine

The value of minimum value of $R_L$ such that the Q point of the transistor stays within the safe operating area for given value of $V_{CC}$ .

Answer to Problem 8.1EP

The value of resistance for which the Q point in within the safe area is $5.76\Omega$ , the maximum collector current is $4.17\text{A}$ and the value of maximum transistor power dissipation is $25\text{W}$ .

Explanation of Solution

Calculation:

The expression for the collector to the emitter load line is given by,

  $V_{CE}=V_{CC}-I_C{R}_C$

The expression for the power transistor is given by,

  $P_T=V_{CC}{I}_C-I_C^2{R}_L$

The expression for the value of current at which the maximum power is obtained is obtained by differentiating the above equation with respect to zero and is given by,

  $\begin{array}{l}\frac{d\left(V_{CC}{I}_C-I_C^2{R}_L\right)}{dt}=0\\ I_C=\frac{V_{CC}}{2R_L}\end{array}$

The expression for the voltage at the maximum power point is,

  $V_{CE}=V_{CC}-I_C{R}_L$

Substitute $\frac{V_{CC}}{2R_L}$ for $I_C$ in the above equation.

  $\begin{array}{l}{V}_{CE}=V_{CC}-\left(\frac{V_{CC}}{2R_L}\right)R_L\\ V_{CE}=\frac{V_{CC}}{2}\end{array}$

The expression for the maximum power dissipation in the circuit is given by,

  $P_T=V_{CE}{I}_C$

Substitute $\frac{V_{CC}}{2}$ for $V_{CE}$ and $\frac{V_{CC}}{2R_L}$ for $I_C$ in the above equation.

  $\begin{array}{c}{P}_T=\frac{V_{CC}}{2}\left(\frac{V_{CC}}{2R_L}\right)\\ =\frac{{\left(V_{CC}\right)}^2}{4R_L}\end{array}$

It is given that $V_{CC}=24\text{V}$ .

Substitute $24\text{V}$ for $V_{CC}$ and $25\text{W}$ for $P_T$ in the above equation.

  $\begin{array}{l}25\text{W}=\frac{{\left(24\text{V}\right)}^2}{4R_L}\\ R_L=5.76\Omega \end{array}$

The expression for the maximum collector current is given by,

  $I_{C\left(\max \right)}=\frac{V_{CC}}{R_L}$

Substitute $5.76\Omega$ for $R_L$ and $24\text{V}$ for $V_{CC}$ in the above equation.

  $\begin{array}{c}{I}_{C\left(\max \right)}=\frac{24\text{V}}{5.76\Omega }\\ =4.17\text{A}\end{array}$

The expression for the maximum power dissipation is given by,

  $P_{Q,\max }=P_T$

Subsume $25\text{W}$ for $P_T$ in the above equation.

  $P_{Q,\max }=25\text{W}$

Conclusion:

Therefore, the value of resistance for which the Q point in within the safe area is $5.76\Omega$ , the maximum collector current is $4.17\text{A}$ and the value of maximum transistor power dissipation is $25\text{W}$ .

(b)

To determine

The value of minimum value of $R_L$ such that the Q point of the transistor stays within the safe operating area for given value of $V_{CC}$ .

Answer to Problem 8.1EP

The value of resistance for which the Q point in within the safe area is $2.4\Omega$ , the maximum collector current is $5\text{A}$ and the value of maximum transistor power dissipation is $15\text{W}$ .

Explanation of Solution

Calculation:

The expression for the maximum collector current is given by,

  $I_{c,\max }=\frac{V_{CC}}{R_L}$

It is given that $V_{CC}=12\text{V}$

Substitute $5\text{A}$ for $I_{c,\max }$ and $12\text{V}$ for $V_{CC}$ in the above equation.

  $\begin{array}{c}5\text{A}=\frac{12\text{V}}{R_L}\\ R_L=2.4\Omega \end{array}$

The value of the maximum collector current is given by,

  $I_{C,\max }=\frac{V_{CC}}{R_L}$

Substitute $12\text{V}$ for $V_{CC}$ and $2.4\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{I}_{C,\max }=\frac{12\text{V}}{2.4\Omega }\\ =5\text{A}\end{array}$

The expression for the maximum power dissipation in the transistor is given by,

  $P_T=\frac{{\left(V_{CC}\right)}^2}{4R_L}$

Substitute $12\text{V}$ for $V_{CC}$ and $2.4\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{P}_T=\frac{{\left(12\text{V}\right)}^2}{4\left(2.4\Omega \right)}\\ =15\text{W}\end{array}$

Conclusion:

Therefore, the value of resistance for which the Q point in within the safe area is $2.4\Omega$ , the maximum collector current is $5\text{A}$ and the value of maximum transistor power dissipation is $15\text{W}$ .