Chapter 6, Problem 6.83P

For the circuit shown in Figure 6.57, the transistor parameters are $\beta =100$ and $V_A=100\text{V}$ , and the source resistor is $R_S=0$ . Determine the maximum undistorted signal power that can be delivered to $R_L$ if: (a) $R_L=1\text{kΩ}$ , and (b) $R_L=10\text{kΩ}$ .

To determine

The maximum undistorted signal power delivered to R_L.

Answer to Problem 6.83P

  $P_{R_L}=0.29\times {10}^{-3}\text{W}$

Explanation of Solution

Given:

  $\begin{array}{l}\beta =100\\ V_A=100\text{V}\\ R_s=0\\ R_L=1\text{k}\Omega \end{array}$

The given circuit is shown below.

  

Calculation:

Draw DC model of the given circuit

  

Applying KVL,

  $\begin{array}{l}{I}_{BQ}{R}_B+V_{BE\left(on\right)}+I_{EQ}{R}_E+V^{-}=0\\ I_{BQ}{R}_B+V_{BE\left(on\right)}+\left(1+\beta \right)I_{BQ}{R}_E+V^{-}=0\\ I_{BQ}\left(R_B+\left(1+\beta \right)R_E\right)+V_{BE\left(on\right)}+V^{-}=0\\ I_{BQ}=\frac{-V_{BE\left(on\right)}-V^{-}}{R_B+\left(1+\beta \right)R_E}\end{array}$

Substituting the values,

  $\begin{array}{l}{I}_{BQ}=\frac{-0.7-\left(-10\right)}{\left[100+\left(1+100\right)10\right]\times {10}^3}\\ I_{BQ}=\frac{9.3}{\left[100+1010\right]\times {10}^3}\\ I_{BQ}=8.738\times {10}^{-6}\\ I_{BQ}=8.738\text{μA}\end{array}$

Now draw the small signal model of given circuit.

  

From above small signal model calculate output voltage,

  $\begin{array}{l}{V}_0=I_0\left(r_0\parallel R_E\parallel R_L\right)\\ \because I_0=I_b+\beta I_b\text{and}{\text{r}}_0=\frac{V_A}{I_{CQ}}\\ \therefore V_0=I_b+\beta I_b\left(\frac{V_A}{I_{CQ}}\parallel R_E\parallel R_L\right)\\ V_0=\left(1+\beta \right)I_b\left(\frac{V_A}{I_{CQ}}\parallel R_E\parallel R_L\right)\end{array}$

Substituting the values,

  $\begin{array}{l}{V}_0=\left(1+100\right)\times \left(8.378\times {10}^{-6}\right)\left(\frac{100}{100\times 8.378\times {10}^{-6}}\parallel 10\times {10}^3\parallel R_L\right)\\ V_0=\left(101\right)\times \left(8.378\times {10}^{-6}\right)\left(\frac{1}{8.378\times {10}^{-6}}\parallel 10\times {10}^3\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(119.36\times {10}^3\parallel 10\times {10}^3\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(\frac{119.36\times {10}^7}{129.360\times {10}^3}\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(0.922\times {10}^4\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(0.922\times {10}^4\parallel 1\times {10}^3\right)\\ V_0=846.17\times {10}^{-6}\left(\frac{0.922\times {10}^4\times 1\times {10}^3}{0.922\times {10}^4+1\times {10}^3}\right)\\ V_0=846.17\times {10}^{-6}\left(\frac{9220000}{10220}\right)\\ V_0=846.17\times {10}^{-6}\left(902.15\right)\\ V_0=0.763\text{V}\end{array}$

Hence,

  $\begin{array}{l}{P}_{R_L}=\frac{V_{rms}{}^2}{R_L}\\ P_{R_L}=\frac{{\left(\frac{V_0}{\sqrt{2}}\right)}^2}{R_L}\\ P_{R_L}=\frac{V_0{}^2}{2R_L}\\ P_{R_L}=\frac{{\left(0.763\right)}^2}{2\times {10}^3}\\ P_{R_L}=0.29\times {10}^{-3}\text{W}\end{array}$

To determine

The maximum undistorted signal power delivered to R_L.

Answer to Problem 6.83P

  $P_{R_L}=0.82\times {10}^{-3}W$

Explanation of Solution

Given:

  $\begin{array}{l}\beta =100\\ V_A=100\text{V}\\ R_s=0\\ R_L=10\text{k}\Omega \end{array}$

The given circuit is shown below:

  

Calculation:

Draw DC model of the given circuit.

  

Apply KVL,

  $\begin{array}{l}{I}_{BQ}{R}_B+V_{BE\left(on\right)}+I_{EQ}{R}_E+V^{-}=0\\ I_{BQ}{R}_B+V_{BE\left(on\right)}+\left(1+\beta \right)I_{BQ}{R}_E+V^{-}=0\\ I_{BQ}\left(R_B+\left(1+\beta \right)R_E\right)+V_{BE\left(on\right)}+V^{-}=0\\ I_{BQ}=\frac{-V_{BE\left(on\right)}-V^{-}}{R_B+\left(1+\beta \right)R_E}\\ \text{Plugging}\text{values}\\ I_{BQ}=\frac{-0.7-\left(-10\right)}{\left[100+\left(1+100\right)10\right]\times {10}^3}\\ I_{BQ}=\frac{9.3}{\left[100+1010\right]\times {10}^3}\\ I_{BQ}=8.738\times {10}^{-6}\\ I_{BQ}=8.738\text{μA}\end{array}$

Now, draw the small signal model of given circuit

  

From above small signal model calculate output voltage,

  $\begin{array}{l}{V}_0=I_0\left(r_0\parallel R_E\parallel R_L\right)\\ \because I_0=I_b+\beta I_b\text{and}{\text{r}}_0=\frac{V_A}{I_{CQ}}\\ \therefore V_0=I_b+\beta I_b\left(\frac{V_A}{I_{CQ}}\parallel R_E\parallel R_L\right)\\ V_0=\left(1+\beta \right)I_b\left(\frac{V_A}{I_{CQ}}\parallel R_E\parallel R_L\right)\end{array}$

Substituting the values,

  $\begin{array}{l}{V}_0=\left(1+100\right)\times \left(8.378\times {10}^{-6}\right)\left(\frac{100}{100\times 8.378\times {10}^{-6}}\parallel 10\times {10}^3\parallel R_L\right)\\ V_0=\left(101\right)\times \left(8.378\times {10}^{-6}\right)\left(\frac{1}{8.378\times {10}^{-6}}\parallel 10\times {10}^3\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(119.36\times {10}^3\parallel 10\times {10}^3\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(\frac{119.36\times {10}^7}{129.360\times {10}^3}\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(0.922\times {10}^4\parallel R_L\right)\\ V_0=846.17\times {10}^{-6}\left(0.922\times {10}^4\parallel 10\times {10}^3\right)\\ V_0=846.17\times {10}^{-6}\left(\frac{0.922\times {10}^4\times 10\times {10}^3}{0.922\times {10}^4+10\times {10}^3}\right)\\ V_0=846.17\times {10}^{-6}\left(\frac{92200000}{19220}\right)\\ V_0=846.17\times {10}^{-6}\left(4797.08\right)\\ V_0=4.059\text{V}\end{array}$

So,

  $\begin{array}{l}{P}_{R_L}=\frac{V_{rms}{}^2}{R_L}\\ P_{R_L}=\frac{{\left(\frac{V_0}{\sqrt{2}}\right)}^2}{R_L}\\ P_{R_L}=\frac{V_0{}^2}{2R_L}\\ P_{R_L}=\frac{{\left(4.059\right)}^2}{2\times 10\times {10}^3}\\ P_{R_L}=0.82\times {10}^{-3}\text{W}\end{array}$