11.3 Maximum Average Power Transfer

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11.3 Maximum Average Power Transfer

In Section 4.8 we solv ed the problem of maximizing the po wer delivered by a power-supplying resistive network to a load RL. Representing the circuit by its Thevenin equi valent, we pro ved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance RL = RTh. We now extend that result to ac circuits.

Consider the circuit in Fig. 11.7, where an ac circuit is connected to a load ZL and is represented by its Thevenin equivalent. The load is usually represented by an impedance, which may model an electric motor, an antenna, a TV, and so forth. In rectangular form, the Thevenin impedance ZTh and the load impedance ZL are

$$\mathbf{Z}_{\mathrm{Th}} = R_{\mathrm{Th}} + jX_{\mathrm{Th}} \tag{11.13a}$$ $$\mathbf{Z}_L = R_L + jX_L \tag{11.13b}$$

The current through the load is

ough the load is
\n

$$I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + Z_L} = \frac{V_{\text{Th}}}{(R_{\text{Th}} + jX_{\text{Th}}) + (R_L + jX_L)}$$

\n(11.14)

From Eq. (11.11), the average power delivered to the load is

1), the average power delivered to the load is
\n

$$P = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{|\mathbf{V}_{\text{Th}}|^2 R_L/2}{(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2}$$

\n(11.15)

Our objective is to adjust the load parameters RL and XL so that P is maximum. To do this we set ∂P∕∂RL and ∂P∕∂XL equal to zero. From Eq. (11.15), we obtain

num. To do this we set

$$\partial P/\partial R_L$$

and $\partial P/\partial X_L$ equal to zero. From
1.15), we obtain

$$\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}|^2 R_L (X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}$$

(11.16a)

$$\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}| \mathbf{r}_L(X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}$$

(11.16a)

$$\frac{\partial P}{\partial R_L} = \frac{|\mathbf{V}_{\text{Th}}|^2 [(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2 - 2R_L(R_{\text{Th}} + R_L)]}{2[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2}$$

(11.16b)

Setting ∂P∕∂XL to zero gives

$$X_L = -X_{\text{Th}} \tag{11.17}$$

and setting ∂P∕∂RL to zero results in

gives us the maximum average power as

ero results in
\n

$$R_L = \sqrt{R_{\text{Th}}^2 + (X_{\text{Th}} + X_L)^2}$$

\n(11.18)

Combining Eqs. (11.17) and (11.18) leads to the conclusion that for maximum average power transfer, ZL must be selected so that XL = −XTh and RL = RTh, i.e.,

$$Z_L = R_L + jX_L = R_{Th} - jX_{Th} = Z_{Th}^*$$

(11.19)

For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance ZTh.

This result is known as the maximum average power transfer theorem for the sinusoidal steady state. Setting RL = RTh and XL = −XTh in Eq. (11.15) When ZL = Z*Th, we say that the load is matched to the source.

In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (11.18) by setting XL = 0; that is,

*^P*max = ∣VTh^∣

2 _____ 8RTh

$$R_L = \sqrt{R_{\text{Th}}^2 + X_{\text{Th}}^2} = |\mathbf{Z}_{\text{Th}}|$$

(11.21)

(11.20)

Figure 11.7 Finding the maximum average power

transfer: (a) circuit with a load, (b) the Thevenin equivalent.

This means that for maximum a verage power transfer to a purely resis tive load, the load impedance (or resistance) is equal to the magnitude of the Thevenin impedance.

Figure 11.8 For Example 11.5.

Example 11.5 Determine the load impedance ZL that maximizes the a verage po wer drawn from the circuit of Fig. 11.8. What is the maximum a verage power?

Solution:

First we obtain the Thevenin equivalent at the load terminals. To get ZTh, consider the circuit shown in Fig. 11.9(a). We find

$$\mathbf{Z}_{\text{Th}} = j5 + 4 || (8 - j6) = j5 + \frac{4(8 - j6)}{4 + 8 - j6} = 2.933 + j4.467 \ \Omega$$

Figure 11.9

Finding the Thevenin equivalent of the circuit in Fig. 11.8.

To find VTh, consider the circuit in Fig. 11.8(b). By voltage division,

$$\mathbf{V}_{\text{Th}} = \frac{8 - j6}{4 + 8 - j6} (10) = 7.454 \underline{\text{/}} - 10.3^{\circ} \text{ V}$$

The load impedance draws the maximum power from the circuit when

$$\mathbf{Z}_L = \mathbf{Z}_{\text{Th}}^* = 2.933 - j4.467 \ \Omega$$

According to Eq. (11.20), the maximum average power is

$$P_{\text{max}} = \frac{|\mathbf{V}_{\text{Th}}|^2}{8R_{\text{Th}}} = \frac{(7.454)^2}{8(2.933)} = 2.368 \text{ W}$$

For the circuit shown in Fig. 11.10, find the load impedance ZL that absorbs the maximum average power. Calculate that maximum average power.

8 Ω 5 Ω ‒j4 Ω j10 Ω Z^L 12 A Figure 11.10

Practice Problem 11.5

For Practice Prob. 11.5.

Answer: 3.415 − j0.7317 Ω, 51.47 W.

In the circuit in Fig. 11.11, find the value of RL that will absorb the Example 11.6 maximum average power. Calculate that power.

Solution:

We first find the Thevenin equivalent at the terminals of RL.

rst find the Thevenin equivalent at the terminals of R_L.
Z_Th = (40 − j30)

$$||j20 = \frac{j20(40 - j30)}{j20 + 40 - j30} = 9.412 + j22.35 Ω$$

By voltage division,

division,
\n

$$\mathbf{V}_{\text{Th}} = \frac{j20}{j20 + 40 - j30} (150/30^{\circ}) = 72.76/134^{\circ} \text{ V}$$

The value of RL that will absorb the maximum average power is

$$V_L \text{ that will absorb the maximum average po}$$

\n

$$R_L = |\mathbf{Z}_{\text{Th}}| = \sqrt{9.412^2 + 22.35^2} = 24.25 \ \Omega$$

The current through the load is

t through the load is
\n

$$I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + R_L} = \frac{72.76/134^{\circ}}{33.66 + j22.35} = 1.8/100.42^{\circ} \text{ A}$$

The maximum average power absorbed by RL is

$$P_{\text{max}} = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{1}{2} (1.8)^2 (24.25) = 39.29 \text{ W}$$

In Fig. 11.12, the resistor RL is adjusted until it absorbs the maximum average power. Calculate RL and the maximum average power absorbed by it.

Answer: 30 Ω, 23.06 W.