qq企鹅图标等级:Measuring SWR

  Measuring SWR   We're going to combine voltage measurements with current measurements in such a way that SWR pops out. Our meter will have an input connector, and an output connector, just like the current measuring peak-detector discussed above. From one of the connector center pins to ground, we can measure voltage too (with the voltage peak detector).
    Both our voltage detector and current detector give an output voltage in DC volts. We could combine these voltages, each weighted equally, to determine when the SWR is 1:1. When the load is exactly 50 ohms, our voltage probe gives exactly five times more output voltage than our current probe. The circuit shown could measure SWR, but only for resistive loads! Once again, the phase problem between voltage and current foils this circuit when the load is reactive. You'd also have to be prepared to reverse your multimeter probes when using this meter, because polarity can change on the DC output. Another problem: you can't easily measure power. This circuit has too many flaws.
    We can get around the reactive load problem, the polarity problem, and the power-measuring problem(sorta). The voltage and current must be added together before being detected by the diodes. In this way, phase will affect the amplitude of the combined sinewaves. Thus, reactive loads will throw off the detected 1:1 SWR, just as resistive loads (of the wrong value).
    Notice that we're going to add the RF waveforms of voltage and current - not multiply them. Multiplying would be wonderful, but requires some complex transistor circuits (or I.C's). Because v and i are added together, the SWR scale, (and the power scale) of our meter will be non-linear. That is, we won't be able to simply scale the meter reading to get SWR (or power) - we'll have to do a calculation. Or, if we use a dedicated meter to display the result, we'll have to make up our own non-linear scales. This is a fairly small price to pay for simplicity.
    Instead of detecting the RF voltage directly, this circuit uses a broadband step-down transformer to reduce the voltage amplitude. Remember, we need to weight the current-transformer's output, and the voltage equally when we combine them. Here's how the weighting is done...
    The current transformer will use a 0.5 ohm current sense resistor. Instead of 100 ohms, we'll use 50 ohms, along with a 1:10 step-up transformer.
    The voltage transformer will also use a 10:1 step-down winding ratio, so that its voltage output is one tenth of the RF voltage on the center conductor of the BNC's. We'll add a 50 ohm resistor to this (low impedance) winding too.
    Its important that the current waveform and voltage waveform each have the same impedance when we add them. This ensures that they're added with equal weight. It also ensures that the SWR output will be correct under all load conditions. With a 50 ohm dummy load connected to the output of this SWR meter, let's calculate how much voltage that the current-sense transformer supplies to its 50 ohm load, and how much voltage the voltage step-down transformer supplies to its 50 ohm load.
    With 5 watts coming into the meter, the line voltage will be 0.3162(rms) amps flowing through the 0.5 ohm (equivalent) current sense resistor. This gives 0.3162 x 0.5 = 0.1581 volts on the current transformer's one-turn primary winding. This voltage is stepped up by a factor of ten to give 1.581 volts(rms) across the 50 ohm resistor.
    With 5 watts dissipated into the 50 ohm dummy load, 15.811 volts(rms) appears at the BNC's center conductor. This voltage is stepped down by a factor of ten by the voltage stepdown transformer to 1.581 volts (rms).
    Notice that the current output (1.581v) and the voltage output (1.581v) are exactly the same amplitude. They're exactly the same phase too, since voltage and current are in-phase for resistive loads (remember, we've got a 50 ohm dummy load attached).
    Now here's the meat of how the SWR trick is done....the current sense resistor R1 has one end connected to ground. So the RF voltage at its hot end represents forward power, and its 1.581v (rms) can be detected with a peak-detector circuit to give a DC output voltage that a multimeter can read.
    Notice that the voltage-measuring transformer has its one-turn winding floating (neither end grounded), with its 50 ohm resistor in series (R2), connected to ground. The voltage from this winding (follow winding sense carefully in all the transformers) is actually subtracted from the current-sense output. Since the two are equal, there's no net RF voltage on R2. The voltage across R2 represents the reflected power of our SWR meter. All it takes is another diode peak detector to give a DC result that a multimeter can read. The only condition that gives a zero-volt result across this resistor is the one outlined here - with a 50 ohm resistive (dummy) load connected to the SWR meter's output port.

    To reinforce this result, let's examine the forward voltage and reverse voltage on R1 & R2 under extreme (infinte SWR) conditions. Consider the case where there's no load at all on the SWR meter's output port. RF current is zero, so there's nothing contributed by the current-sense transformer. The voltage sense transformer measures full RF voltage, and its one-turn winding dumps its voltage equally between R1 and R2. Thus the forward output is equal to the reflected output. This is indicative of an infinte SWR: the meter reading is full scale whether measuring forward or reverse.
    Now let's short the output port of the SWR to ground. This renders output voltage zero, but there's plenty current flowing. This time the current transformer will distribute its output equally between R1and R2, and the voltage transformer will contribute nothing. Again, since the forward and reverse voltages are equal, the SWR is infinite.
    With a reactive load, the forward and reverse voltages are equal once again...when you add (or subtract) two waves differing by 90 degrees, they don't even know each other are there. So once again, the current contribution distributes evenly between the two 50 ohm resistors, and the voltage contribution distributes evenly between the two resistors. Again, with equal forward and reverse contributions, SWR is infinte.

Accuracy of the SWR meter
    Does the forward and reverse output of this SWR meter accurately represent voltage standing wave ratio (or current standing wave ratio) under all load conditions? Yes, within about 1%, excluding diode detector errors. Error could be reduced if the turns ratio of the transformer were increased. But doing so will result in smaller DC output voltages - this'll increase error due to diode voltage drop. For QRP use, the 10:1 turns ratio suggested is a fair compromise between these two errors. Error due to diode drop can be significantly reduced with an op-amp compensator, a really good idea for QRPP work.
    Does this SWR meter eat up much transmitted power? A PSPICE run of the circuit was made, with the assumption that the transformers had complete coupling between their primary and secondary windings. The transformers were also assumed to have no resistive losses. With 50 ohm source and load, 99.5% power was delivered to the load. This meter could be kept in-line with very little loss.

Interpreting the output voltage
    Its too bad that this simple meter's DC output is non-linear (because we've added v and i instead of multiplying). To find the actual SWR, a calculation must be made. Or we've got to make up a non-linear scale for the output meter. We're interested mostly in the magnitude of the SWR. If we represent the load as a complex impedance (RL + jXL), then rho= sqrt[ (50 - RL)² + (XL)² / ((50 + RL)² + (XL)²)]
and SWR = (rho + 1)/(rho - 1). SWR is independent of output power.