Maximum Power Theorem

Maximum Power Transfer Theorem and Impedance Matching

Maximum Power Transfer Theorem and Impedance Matching 3.5

Why it is important to understand:[4, 5] This states that the power transfer occurs when the impedance of the source equals the impedance of the load or vice versa. It is also called Jacobi’s Law.[5]

A related concept is reflection-less impedance matching.[6] This is where the source impedance such as a transmitter matches the load impedance such as an antenna to avoid reflections in the transmission line.[6] In electronics, impedance matching is the practice of matching the input impedance of an electrical load or output impedance of its corresponding signal source to maximize power transfer or minimize reflections from the load.[7, 8, 9]

Maximum power transfer is extremely important in electronics, as well as all manner of low-frequency applications such as stereo sound systems, electrical generating plants, solar cells, and hybrid electrical cars.[10-15] Cell matching is very important where small signals are involved.[16, 18]

35.1 Maximum Power Transfer Theorems

A network that contains linear impedances and one or more voltage/current sources can be reduced to a Thevenin Equivalent circuit.[19, 20] When a load is connected to the terminals of the equivalent circuit, power is transferred from the source to the load.[21-24]

A Thevenin equivalent circuit with internal impedance Z = (r + jx)Ω and complex impedance Z = (R + jX) is shown below.[25-30] The idea is the internal impedance is fixed and you have to choose a matching load.[28, 36]

The maximum power transfer depends on the following four conditions:[31-33]

  • Condition 1: The load consists of a purely variable resistance.[34, 39, 43]
    R = √(r2 + x2) = |Z|[44]
    (Thus the load magnitude will have to be equal to the magnitude of the source impedance).[45-49]
  • Condition 2: Purely resistive load.[41, 46]
    R = r (which is the DC condition).[51]
  • Condition 3: The load has both variable resistance R and variable reactance X.[52, 53]
    In which case: X = -x and R = r for maximum power transfer.[54, 58, 59]
  • Condition 4: The load has variable resistance R and fixed reactance X.[60]
    The resistance R is given by: R = √(r2 + (x + X)2)[61, 62]

Worked Problems

Problem 1

Source: E = 120∠0° V, Z = (15 + j20)Ω.[66, 67]

The load is purely resistive. Match the load to the source and find the maximum power delivered to R.[68, 71, 73-75]

Solution:

R = √(r2 + x2) = √(152 + 202) = 25Ω[72]
Ztotal = (15 + j20) + 25 = 40 + j20[76, 78]
I = 120 / (40 + j20) = 2.683∠-26.56° A[79]
Pmax = I2 × R = 2.6832 × 25 = 179.96 W[81]

Problem 2

Load consists of a variable resistance R and variable reactance X (Condition 3).[83-86] Source impedance Z = (15 – j20)Ω.[88]

Solution:

X = -x ⇒ X = +20[89, 91]
R = r ⇒ R = 15Ω[87, 94]
Ztotal = (15 + j20) + (15 – j20) = 30Ω[95, 96]
I = 120V / 30Ω = 4A[97]
Pmax = I2 × R = 42 × 15 = 240 W[98]

Problem 3

Determine the value of the load resistor R required for maximum power transfer.[102-106]

Given E = 200∠0° V, f = 1kHz, C = 1μF.[108]

XC = 1 / (2πfC) = 1 / (2π × 1000 × 1 × 10-6) = 159.15Ω[109-113]

Calculating the equivalent impedance:

Zeq = (100 × -j159.15) / (100 – j159.15) = 71.69 – j45.05 Ω[115]
R = √(71.692 + (-45.05)2) = 84.67Ω[115]

Problem 6 (Thevenin Equivalent)

Find the Thevenin equivalent voltage E and impedance.[120, 243]

E = ((5 + j10) × 100∠30°) / (5 + 5 + j10) = 79.06∠48.43°[120, 124, 243, 244]
Zth = (5(5 + j10)) / (10 + j10) = 3.75 + j1.25 Ω[126, 245]

For maximum power transfer, conjugate matching is used:[128, 248]

ZL = 3.75 – j1.25 Ω[129, 249]

38.2 Impedance Matching

With transmission lines, the lines are ideally terminated with their characteristic impedance.[139-143] With dc sources, generators, or secondary cells, the internal impedance is usually very small. Attempting to match the load impedance as small as the internal impedance results in overloading the source.[142, 145, 147]

One method of achieving maximum power transfer between a source and a load is to adjust the value of the load impedance to match the source impedance. This can be achieved using a matching transformer.[146, 148-153]

For a transformer supplying a load impedance ZL:[154, 161]

V1 / V2 = N1 / N2 = I2 / I1[156]

The primary input impedance is given by:[167]

Z1 = V1 / I1 = (N1/N2 × V2) / (N2/N1 × I2) = (N1 / N2)2 × (V2 / I2)[172]

Since the load impedance ZL = V2 / I2,[173]

Z1 = (N1 / N2)2 × ZL

If the input and load impedance are purely resistive, the equation is:[176]

r = (N1 / N2)2 × RL[177, 191]

Problem 8

The output resistance of an amplifier is 448Ω. It is connected to a transformer with a turns ratio of 8:1. Determine the load resistance.[183-186]

Solution:

RL = r × (N2 / N1)2 = 448 × (1/8)2 = 7Ω[188, 191]

Problem 9

A generator has an output impedance of (450 + j60)Ω. Determine the turns ratio necessary to match a load of (40 + j19)Ω for the maximum transfer of power.[190, 192, 193]

Solution:

|Zsource| = √(4502 + 602) = 453.98Ω[196]
|Zload| = √(402 + 192) = 44.28Ω[197]
(N1 / N2) = √(|Zsource| / |Zload|) = √(453.98 / 44.28) = √10.25 = 3.20[201]