One of the most misused and misunderstood pieces of test equipment in the radio amateur's shack must surely be the good old SWR bridge.   So let us examine what it does, how it does it and perhaps dispel a few myths into the bargain.

    Let us start with forward and reverse (reflected) power.   We should have no problem coping with the concept of forward power.   The generator (transmitter) supplies a R.F. voltage to the transmission line which behaves very much like a resistor whose value is equal to the characteristic impedance of the line.   So it is no surprise to learn that a current of  Vf/Zo  flows down the line and that this current is in phase with the applied R.F. voltage.

    Reflected power is perhaps not so easy to imagine.   So let us picture the forward wave traveling down the transmission line at a speed nearly that of light.   Very soon the wave is going to reach the end of the transmission line,    what then ?  Well if at that end  there is a resistor equal in value to the characteristic impedance of the line,  another length of line of the same impedance  or a load such as an antenna that behaves like a resistor at the frequency in use,    then the wave will be totally absorbed and no power will be reflected.

    However if the line is not terminated with a resistance equal to the characteristic impedance,  then not all of the power in the wave will be absorbed and some power will be reflected back as a wave flowing in the reverse direction.

    What do we mean by the 'direction' of the wave?   It should help if we consider the case of a simple rechargeable cell,   when we charge the cell the current flows from the charger into the cell,  we could say the current is flowing 'in phase' with the applied voltage.   When we come to use the charged cell the voltage across the cell retains the same polarity as when the cell was being charged,  but now the current is flowing out of the cell i.e. the reverse direction.   we could now say that the current is flowing 180degrees out of phase to the voltage.

    So going back to our transmission line we can now see (I hope) that the current of the reflected wave is 180degrees out of phase to the voltage of the reflected wave.

    As the reflected wave travels back towards the generator (transmitter) it passes the forward wave traveling in the other direction.   At each point the total voltage and current will be the sum of the voltages and currents of the forward and reflected waves.

    At some points on the line the voltages of the forward and reflected waves will be in phase and therefore will add together.   At other points the voltages will be out of phase and the reflected voltage will be subtracted from the forward voltage,     (the reflected voltage will always be less than or equal to the forward voltage).

    The ratio of maximum voltage on the line (Vf+Vr ) to the minimum voltage on the line (Vf-Vr) is called the VSWR (Voltage Standing Wave Ratio) usually shortened to SWR.

    In the early days of amateur radio  transmission lines were of the open wire parallel line type,   and to measure the VSWR (SWR),   an RF voltmeter could be connected across the lines and slid along,   the maximum and minimum values recorded and used to calculate VSWR

i.e. VSWR = Vmax/Vmin.

    With coaxial lines the inner conductor is totally enclosed by the outer sheath and measuring the voltage at more than one point on the line is not a practical proposition,    there has to be a better way!.

    The modern SWR bridge in one of its many forms is the answer!  to find out how it does it first we must take look at the Ohm's law that most of us wrestled with before taking the amateur radio exam.

    The very definition of resistance (and impedance) is the ratio of voltage to current in a component (or circuit), so take the equation (sorry about the maths) for resistance R = V/I,

    That means, the value of a resistor is simply the voltage across it divided by the current through it.

So what has that got to do with the SWR bridge?

    The answer is everything!,   the SWR bridge compares the voltage across a transmission line with the current flowing in it,   and that sounds suspiciously like measuring resistance (or impedance).

    The typical commercial SWR bridge is a bridge circuit which is balanced for 50 ohms,   that is when the line 'sees' 50 ohms the bridge is balanced and the "reflected" power indication is zero.

    To achieve balance,   the voltage across the line (in volts) has to be fifty times the value of the current in the line (in amps) and the voltage and current must be in phase,   that is the positive peaks of the voltage across the line and the current in the line must coincide precisely.

    If there is reflected power flowing in the line,    the voltage measured across the line will be due to the voltage of the forward wave plus the voltage of the reflected wave.

    However the current measured in the line will be the forward voltage divided by 50 (for Zo = 50) minus the reflected voltage divided by 50 (since the current flowing in the reflected wave is flowing 180degrees out of phase with the voltage of the reflected wave).

    Resorting to a little mathematics:-

 
               Vm = Vf + Vr                           (voltage measured)
 
                Im = If - Ir = Vf - Vr               (current measured)
                                      50   50

 If we divide the voltage measurement by fifty then we can do something very interesting with the two measurements :-
 

               Vm = Vf + Vr                           (voltage measured divided by fifty)
                50      50   50

               Im = Vf - Vr
                        50   50

          Vm + Im = 2Vf                                (reflected voltages cancel
           50             50                                      forward voltage only)

          Vm - Im = 2Vr                                 (forward voltages cancel
           50            50                                    reflected voltage only)

    If the maths have left you cold,   it is easy enough to sum up the operation in words,    the SWR bridge produces two voltages internally,    one proportional to the voltage across the line and one proportional to the current in the line at a particular point.

    These two voltages are adjusted so that they are equal when the SWR is 1:1,  if they are equal,  when one is subtracted from the other there will be no indication on the meter.

    The presence of reflected power on the line will always cause the bridge to become unbalanced,    although the reflected voltage is always fifty times the reflected current the same as the forward wave,   since the current in the reflected wave is 180degrees out of phase with the reflected voltage,   when the voltages are in phase and add,   the currents are out of phase and subtract,  and vice versa.

    The normal method of using an SWR bridge,  is to first apply a small amount of RF energy to the line through the bridge,  with the switch in the forward position where both voltages are added together,   then the SET control is adjusted so that the needle on the meter aligns with the SET position on the SWR meter dial.

    If the line is correctly terminated the voltage sensed from the voltage across the line will be equal to (and in phase with) the voltage generated by the current sensor,    in the forward switch position these are added together producing a voltage 2V,    when the switch is now set to the SWR position one of these two voltages is subtracted from the other and the difference is indicated on the meter,     if they are equal and in phase then there is no reading,   indicating that the line is correctly terminated.

    However if power is being reflected from the load end of the transmission line there will be a difference between the two voltages,   they will not cancel out completely,  reflected power and hence a mismatched load is indicated.
 
    So simple,  what can go wrong?  Well its more a case of misunderstanding the readings than the readings being erroneous,  now we know how it works,    lets put a few myths to rest.

    Firstly can you try altering the length of the transmission line to improve the SWR reading?  Well it sometimes seems to work but the reading is usually all that changes,   not the SWR,  why?,   if the antenna is balanced and fed with unbalanced coaxial cable there is a good chance that some of that RF will return to the shack on the outside of the coax,    this can cause the SWR bridge to give false readings which will change depending on the length of the coax.
 
    Any genuine change in the SWR is by the RF on the outside of the coax being reflected back to the antenna and changing the feed impedance,    changing the length of the coax will alter the phase of this reflected voltage when it returns to the feed point.

    Even if the antenna is for unbalanced feed or a balanced antenna is fed with a balun,   it is still possible for the outside of the coax to pick up some RF and carry it back to the shack,  if the coax is not carefully sited with reference to the antenna.

    Is reflected power is lost power?   no!   the SWR bridge does give a reasonably accurate indication of forward and reflected power,   but if there are minimal transmission line losses the reflected power is just added to the power delivered by the transmitter.

    The power delivered to the load,    again ignoring possible losses,   is simply the forward power minus the reflected power, and if the losses are minimal then that is the power delivered by the transmitter.

  "I have just replaced my coax and now the SWR is worse, I think the coax is faulty",    maybe!   but lossy coax will almost always give an apparent improvement to the indicated SWR,   remember the reflected power has traveled the length of the coax twice!   if there are losses then the reduction of the reflected power will show a lower SWR at the transmitter end of the coax than if it were measured at the antenna.

    A SWR bridge can be used to check the loss of a long length of coax by putting the SWR bridge between the transmitter and one end of the coax,   leaving the far end unterminated,   the higher the indicated SWR the lower the loss, e.g. if the SWR was 10:1 the loss would be 0.87dB if the SWR was 20:1 the loss would be 0.43dB and if the SWR was 5:1 then the loss  would be 1.76dB.

    The general formula to calculate the loss is:-

          loss dB  =  10 x log10((SWR+1)/(SWR-1))

    It should be obvious from this that SWR readings quickly become meaningless when the loss on the transmission line becomes significant,   for example;  if the loss on a length of coax is 3dB then the SWR would only be 3:1 even if the far end was either open or short circuit.

    To be certain that an antenna is providing a correct termination to a transmission line it is prudent to measure the SWR without the antenna connected first, before measuring it with it connected,   a big difference in the readings indicates all is well, a small difference means there is some checking to be done.

    And finally the one that catches everyone out,   "I can reduce my SWR by running less power",   it is true that the majority of SWR bridges whether commercial units or home brew, will indicate a lower SWR when the power used to do the measurement drops below a certain level.

    So why?   in this case it is a shortcoming of the SWR bridge itself or more accurately the detectors used to convert the RF signals from the bridge into a measurable DC voltage,     all diodes have a threshold voltage below which they become quite inefficient,  when operating at these low levels the reflected signal being less than the forward signal,  appears to be much less than it really is,   giving a more optimistic SWR than that obtained at higher power levels.

    The SWR bridge still perhaps one of the most useful pieces of test equipment in your shack,   particularly in this age of solid state transmitters with 50 ohm outputs,   hopefully this piece will help you guard against its eccentricities and extend its usefulness.