4.8 Maximum Power Transfer

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4.8 Maximum Power Transfer

In many practical situations, a circuit is designed to provide power to a load. There are applications in areas such as communications where it is desirable to maximize the power delivered to a load. We now address the problem of deli vering the maximum po wer to a load when gi ven a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load.

The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance RL. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig. 4.48, the power delivered to the load is

$$p = i^2 R_L = \left(\frac{V_{\text{Th}}}{R_{\text{Th}} + R_L}\right)^2 R_L$$

(4.21)

For a given circuit, VTh and RTh are fixed. By varying the load resistance RL, the power delivered to the load varies as sketched in Fig. 4.49. We notice from Fig. 4.49 that the power is small for small or large values of RL but maximum for some value of RL between 0 and ∞. We now want to show that this maximum power occurs when RL is equal to RTh. This is known as the maximum power theorem.

Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load (^RL = ^RTh).

To prove the maximum power transfer theorem, we differentiate p in Eq. (4.21) with respect to RL and set the result equal to zero. We obtain

dp ____ dRL = V² Th [ (RTh + RL) 2 ⁻2RL(RTh + RL) ________________________ (^RTh + RL) 4 ] = V² Th [ (RTh + RL − 2RL) _______________ (*^R*Th + RL) 3 ] ⁼⁰

Figure 4.48

The circuit used for maximum power transfer.

Figure 4.49 Power delivered to the load as a function of RL.

This implies that

$$0 = (R_{\text{Th}} + R_L - 2R_L) = (R_{\text{Th}} - R_L)$$

\n(4.22)

which yields

$$R_L = R_{\text{Th}} \tag{4.23}$$

showing that the maximum power transfer takes place when the load resistance RL equals the Thevenin resistance RTh. We can readily confirm that Eq. (4.23) gives the maximum power by showing that d² p∕dR^{2 L}< 0.

The maximum po wer transferred is obtained by substituting Eq. (4.23) into Eq. (4.21), for

pmax = V2 ____Th 4RTh (4.24)

Equation (4.24) applies only when RL = RTh. When RL ≠ RTh, we compute the power delivered to the load using Eq. (4.21).

Find the v alue of RL for maximum po wer transfer in the circuit of Example 4.13 Fig. 4.50. Find the maximum power.

The source and load are said to be

matched when ^RL = ^RTh.

Solution:

We need to find the Thevenin resistance RTh and the Thevenin voltage VTh across the terminals a-b. To get RTh, we use the circuit in Fig. 4.51(a) and obtain

$$R_{\text{Th}} = 2 + 3 + 6 \parallel 12 = 5 + \frac{6 \times 12}{18} = 9 \ \Omega$$

For Example 4.13: (a) finding RTh, (b) finding VTh.

To get VTh, we consider the circuit in Fig. 4.51(b). Applying mesh analysis gives

$$-12 + 18i_1 - 12i_2 = 0, \qquad i_2 = -2 \text{ A}$$

Solving for i1, we get i¹ = −2∕3. Applying KVL around the outer loop to get VTh across terminals a-b, we obtain

−12 + 6i1 + 3i2 + 2(0) + VTh = 0 ⇒ VTh = 22 V

For maximum power transfer,

$$R_L = R_{\text{Th}} = 9 \ \Omega$$

and the maximum power is

$$p_{\text{max}} = \frac{V_{\text{Th}}^2}{4R_L} = \frac{22^2}{4 \times 9} = 13.44 \text{ W}$$

Determine the value of RL that will draw the maximum power from the rest of the circuit in Fig. 4.52. Calculate the maximum power.

Answer: 126.67 Ω, 96.71 mW.