A 13.56-MHz, ISM (industrial, scientific, and medical)-band RF-measurement device had the test requirement of a 50-hour, 1-kW burn-in. The device under test needed to be simultaneously stressed at an RF potential and RF current equivalent to 1 kW, but the only spare RF source at hand was a 100W RF generator. Besides, saving energy seemed important. The circuit in Figure 1 develops 1 kW from the 100W RF source by storing energy in a transmission line. The circuit comprises two 43° sections (approximately 6 feet) of RG-213 coaxial cable with UHF connectors at each end. The device under test, which has an electrical length of 4°, connects between the two lines, T₁ and T₂, making the total line length 90°. An off-the-shelf amateur-radio antenna tuner matches the 50Ω RF generator to the line's input impedance.

Circuit operation is simple. The RF energy connects to the line's input through the antenna tuner. The energy travels to the shorted end of the line, where it is reflected and travels back to the input. The reflected energy then reflects off the conjugate match, which the antenna tuner provides, and combines with the next half-cycle of input energy to flow toward the short again. This process continues with the stored energy continuing to build until the circuit losses equal the generator power. Considering the circuit operation in terms of impedances reminds you that the input impedance of a shorted, lossless, 90° transmission line is infinite. At the shorted end of the line, V/I is zero, and, at the input of the line, V/I is infinite. At the center, where the device under test is located, the magnitude of V/I is equal to the characteristic impedance of the transmission line—50Ω in this case. The RF voltage and current are 90° out of phase, but that fact does not affect the burn-in of the device.

Consider how much line input power is required to develop 1 kW at the device under test. The loss of each 6-foot section of RG-213 is 0.025 dB, and the device-under-test loss is 0.05 dB. The loss for a wave traveling down the line is therefore 0.1 dB. The return loss, R_L, is twice this amount, or 0.2 dB, because the wave must travel down the line to the short circuit and return to the source. Now, you can calculate the reflected power, P_R, for an incident power, P_IN, of 1000W using the following formula: P_R=(P_IN)10^(RL/10)=(1000)10^(–0.2/10)=955W.

So, when 1000W flows down the line, 955W returns to the input. The line input power required is equal to the incident power minus the reflected power, which is 1000–955, or 45W. Because the line loss and device-under-test loss are both 0.05 dB, half of the 45W loss is dissipated in the coax, and half is dissipated in the device under test. The measured antenna-tuner loss is 40W, which makes the total circuit loss 85W. You can determine the line's input impedance by calculating the line-input complex reflection coefficient (Γ) and solving for the input impedance using Γ=10^(RL/20)=10^(–0.2/20)=0.9772, and

The antenna tuner must match the 50Ω generator's output impedance to the 4.3-kΩ line-input impedance. You can confirm the circuit's operation by measuring the amplitude and phase of the device's voltage and current. The measurement uses an oscilloscope with voltage and current probes. A power meter at the device under test measures forward and reflected power of 1 kW. Because of the high circuit Q, you'll find that adjusting the RF-source frequency to obtain circuit resonance is easier than trimming cable lengths to obtain circuit resonance. The primary limiting factors with this circuit are the temperature rise of the coaxial cable and the losses incurred in the impedance-matching circuit. Coax having lower loss allows you to achieve a higher "power multiplication" and higher device-under-test power.

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