It’s not always needed, but here’s how to achieve it when it is

The maximum transfer of power from a source to its load requires that the load impedance be matched to the source impedance. This applies to both DC and AC circuits. In DC circuits, this match is purely resistive. In AC circuits (RF is AC), the source or load (or both) can be a complex impedance. A complex impedance consists of a real part (resistance) and an imaginary part (reactance). In an RF circuit, the load impedance must be the complex conjugate of the source impedance in order for optimum power transfer to occur. If the source impedance is 30-j20 ohms, then the load impedance must be 30+j20 ohms. The figure 30+j20 is said to be the complex conjugate of 30-j20 and vice versa.

Figure 1 shows a transmitter that is connected to a transmission line through a pi network consisting of three impedance branches: C_{1}, L_{1} and C_{2}. The transmission line is an odd multiple of one-quarter wavelength, and the antenna impedance is 35+j15 ohms. The pi network is used to provide a conjugate match between the transmission line (at the antenna side) and the antenna. Notice that the pi-matching network is located at the transmitter and that by adjusting the values of the pi network components we can produce a conjugate match at the far end of the transmission line where it connects to the antenna.

An important point to understand is that a conjugate match is achieved at every junction (points A, B and C) in Figure 1. At point A, the impedance presented to the transmitter by the pi network is 50±j0 ohms, or a pure resistance of 50 ohms. Figure 2 and Figure 3 describe the conjugate match at point B, looking in opposite directions. Figure 4 shows a conjugate match at point C. The antenna impedance is 35+j15 ohms and the impedance looking back toward the transmitter from point C is 35.4-j15.1 ohms — a conjugate match to the antenna. This example was worked out using winSMITH^{®} software, so the actual figures are not exact but very close to the ideal. The transmission line is considered to be lossless.

Frequently, amateur radio operators get into arguments over the use of an antenna tuner located near the transmitter. Some argue that the antenna is not actually tuned by the tuner but that the transmitter is simply tricked into seeing a 50-ohm impedance at the transmitter output. They argue that the tuner must be placed at the antenna in order to actually tune the antenna. Figures 1 through 4 prove that this is not the case. Some argue that the antenna is not tuned unless it has a pure resistance of 50 ohms. Perhaps the terms resonant and tuned are confused. Although the antenna has reactance, the reactance is tuned out by the conjugate match at point C. For further reading, I recommend *Reflections II* by Walter Maxwell.

Real-world transmission lines and components are not ideal — they exhibit loss or attenuation. This causes some deviation from the true conjugate-match theorem described above. However, transmission line loss should be kept at a minimum within practical limits. The lower the line loss, the closer to the conjugate-match theorem the system operates. Lower line loss also equates to less additional attenuation with standing waves on the line.

Sometimes, it is possible to change the length of coaxial cable between a transmitter, or source, and the load. Changing the length of line only works when the load is not a pure resistance of 50 ohms. Changing the line length can change the impedance seen by the transmitter. Different line lengths can transform the complex load impedance from capacitive to inductive, and even to purely resistive. Often, a line length can be found that will allow better tuning and loading of the transmitter output stage — even though a perfect match isn't achieved. Sometimes, this is the case when connecting a transmitter to a duplexer.

Maximum power transfer (MPT) is not always required or even desired. In low-noise RF circuits, trade-offs between optimum power transfer and noise figure are required. For example, the point of optimum power transfer and optimum noise figure may not be the same, so a deliberate impedance mismatch may be required to reach a compromise between the two criteria. In another example, suppose that the load that your home presented to the power line was operated at maximum power transfer. Who could afford the electric bill? Who could afford the required wiring?

Another case in point is the automotive lead-acid storage battery, as the reserve power capability of this type of battery is much greater than the normal operating requirements. In other words, the power level required to start the engine is much greater than that required to keep it running and to operate the electrical accessories.

Figure 5 shows a typical DC circuit using a storage battery. The resistor R_{S} represents the internal resistance, or source resistance, of the battery and the resistor R_{L} represents the load resistance. If the internal resistance of the battery is 0.05 ohms and the open-circuit battery voltage is 12 volts, the maximum current that the battery could deliver (under short-circuit conditions) would be 12/0.05 = 240 amperes. All of the power would be dissipated within the internal resistance of the battery and the dissipated power would be equal to the voltage multiplied by the current, or 12 × 240 = 2880 watts.

Because maximum power transfer occurs under matched-load conditions, the load resistance would have to be equal to the internal resistance of the battery, or 0.05 ohms. In this case, half the open-circuit battery voltage would be dropped across the internal resistance of the battery and half the voltage would appear across the 0.05-ohm load resistor. Thus, with 6 volts across the load resistor, the power dissipated within the load resistor would be E^{2}/R = 6^{2}/0.05 = 36/0.05 = 720 watts. Because the internal resistance of the battery is equal to the load resistance, 720 watts would be dissipated within the internal resistance of the battery. And because the total power dissipated is 2 × 720 = 1440 watts, the efficiency of the circuit is at a maximum of 50%. Battery life would be very short if operated under MPT conditions except for short bursts. The maximum power transfer under matched-load conditions is defined by the formula in Equation 1, while power transfer under mismatched-load conditions is defined by the formula in Equation 2.

The internal resistance of a battery might vary with charge level and the load demand. Different battery types will exhibit different internal resistances. Typical battery loads are nowhere near the point of maximum power transfer. The object is to use a battery that will deliver the necessary power over a reasonable period of time between recharges. For non-rechargeable batteries, the power demand should be such that maximum battery life is obtained.

A “cheap and dirty” way to check a simple 1.5 V dry cell is to place an ammeter directly across the battery terminals and measure the resultant current flow. Because this procedure creates, in practical terms, a dead short, it must be done quickly to avoid draining the battery. This isn't recommended for batteries that have the ability to deliver a very high current. A better way is to place a moderate load across the battery and measure the voltage drop between open-circuit and loaded conditions.

Until next time — *stay tuned!*

*Harold Kinley is the author of three books, including* Standard Radio Communications Manual *and* The Radioman's Manual of RF Devices. *A certified electronics technician, he holds an FCC First Class Radiotelephone License and an amateur radio operator Extra class license.*