This explanation will cover much the same ground as "John Smith"
did in article , but
in a different way.

In article RyIOb.2778$ZJ1.1138@lakeread01, "Craig Buck"
wrote:

I am baffled too., If the antenna is too short and therefore exhibits
capacitive reactance, why isn't the adding a coil (inductive reactance)
enough to balance it out? I guess I don't understand matching networks at
all. If the antenna is too capacitive or to inductive add the opposite to
get to zero. Seems like you add one or the other but not both to get a
match.

Well, as a rule, this won't be enough.

If you have an antenna which is a bit short of an odd multiple of a
quarter wave length, and has a capacitive reactance, then you could put
an inductor in series with it and cancel the capacitive reactance but
you would still have the same resistance, which might not be 50 ohms.
The same would be true if you tried to compensate for the inductive
reactance of an antenna a bit longer than an odd multiple of a quarter
wave length, just by adding a series capacitor.

Let's take an example. Suppose the antenna impedance is 50 - j20, or 50
ohms of resistance in series with 20 ohms of capacitive reactance. Then
adding an inductor with 20 ohms of inductive reactance will solve the
problem, giving an impedance of (50 - j20) +j20 = 50 ohms. (I am
assuming you want a 50 ohm impedance since you are going to use 50 ohm
coaxial cable and have a transmitter which is happiest if connected to a
50 ohm load.)

If the antenna impedance is 50 + jx instead of 50 - j20, you just add an
inductor or capacitor with (- x) ohms of reactance (note the opposite
sign). If the antenna is capacitive, use an inductor, and vice versa.

But what if the antenna impedance is 10 - j20 instead of 50 - j20?
Adding an inductor with 20 ohms of inductive reactance in series with
the antenna will cancel out the capacitive reactance all right

(10 - j20) + j20 = 10

but you will be left with a load impedance of 10 ohms and your SWR will
be 5 to 1, which will in all likelihood make your transmitter rather
unhappy. We need somehow to get the real part of the impedance up to 50
ohms.

The trick is to put the compensating reactor in parallel rather than in
series with the antenna.

Let's talk about current instead of resistance, since when loads are
placed in parallel, the currents add. Suppose for the sake of argument
your transmitter puts out 50 volts. You want the load to draw 1 ampere
to make the transmitter think it is seeing a 50 ohm load. Now let's go
back to the example of an antenna with impedance 10 - j20. If we connect
that antenna directly to this 50 volt producing transmitter, it will
draw 50/(10 - j20) amperes from the transmitter and not the desired 1
ampere. 50/(10 - j20) is too complicated to think about, as well as too
complex ;-), but we can do a bit of simplifying by multiplying both the
numerator and the denominator by (10 + j20). This won't change anything
since we're just multiplying by 1, albeit written in a strange way. We
get (50*(10 + j20))/((10 - j20)*(10 + j20)) = (500 + j1000)/(100 + 400)
= (500 + j1000)/500 = 1 + j2.

The reason this works is that when you multiply out the two factors in
the denominator, you get four terms: 10*10, 10*(+ j20), (- j20)*10, and
(-j 20)*(+ j20). The second and third terms amount to + j200 and - j200
and they cancel each other. The fourth term is + 400, since j*j = - 1
and therefore (- j)*j = + 1.

Note: The 10 + j20 quantity is called the "complex conjugate" of the
10 - j20 quantity. This trick will always work to make the new
denominator a pure real number (no j term) -- just change the sign of
the coefficient of j (the imaginary part) in the denominator and
multiply. But to be fair, you then have to multiply the numerator by the
same thing too, and we did that.

Now what is it that we just calculated? We've calculated the *current*,
in amperes, that our antenna will draw from the 50 volt transmitter if
connected directly to our 10 + j20 ohm antenna load. The 1 portion is
fine -- that's the 1 ampere we want. But in addition to the 1 ampere
there's that + j2 ampere quantity. If we connect a reactor that will
draw - j2 amperes in *parallel*, not in series, with the antenna, the
total current drawn will be (1 + j2) + (- j2) = 1 ampere and the
transmitter will no longer be likely to go up in flames. What reactance
should that reactor have if it is to draw - j2 amperes when fed with 50
volt? We want the its impedance to be 50/(- j2), and again to see what
that means we use the same trick we used before, multiplying both
numerator and denominator by the complex conjugate of the denominator.
Think of (- j2) as actually being (0 - j2), and you'll see that the
complex conjugate is (0 + j2), or in other words just j2. So the desired
impedance for the reactor we are going to put in parallel with the
antenna is (50*j2)/((- j2)*(j2)) = j100/4 = j25. This means we should
put an inductor with 25 ohms of inductive reactance in parallel with the
antenna. It will draw 50/(j25) = - 2j amperes (this time do the complex
conjugate trick yourself to verify that I divided correctly), and that
added to the 1 + j2 amperes that the antenna itself draws will give a
total current drain on the transmitter of 1 ampere, so all will be well.

You could assume the transmitter put out a different voltage from 50
volts, and you'd get the same result as I did, but the arithmetic would
be uglier.

If you look at what has been calculated, you'll see that it depends
critically upon the antenna drawing a current of 1 + jx amperes from the
50 volt transmitter, so that we can compensate by putting something else
in parallel that will draw - jx amperes. The "1" part of 1 + jx is
essential, since putting a reactor in parallel with the antenna can only
change the imaginary part of the current drawn, not the real part. Our
10 + j20 impedance antenna drew a current whose real component was 1
ampere. Provided the real part R of the antenna impedance R + jX is less
than 50 ohms, it will always turn out to be the case that you can make
the real part of 50/(R + jX) be exactly 1 if you just arrange for X to
be a square root of R*(50 - R). (Guess how you can verify this?) I say
*a* square root and not *the* square root since there are two of them,
one negative and one positive. If the X that you need and the X that you
have in the actual antenna impedance R + jX differ, you can fix that by
putting another (second) reactor in series with the antenna and
pretending that the "antenna" consists of the series combination of the
actual antenna and this added (second) reactor. For example suppose this
calculation of the square root of 50*(50 - R) shows that, for the
particular value of R that your antenna provides, you need either X = 49
or X = - 49, but you actually have X = 20. Then you can either add a
(second) inductive reactor with impedance j29, or else a (second)
capacitive reactor with impedance - j69, in series with the antenna.
Once you have arranged for the antenna, augmented if necessary by such a
series reactor, to draw exactly 1 + jx amperes (the value of x being
irrelevant) then you put your first reactor in parallel with the
(antenna + second reactor in series) combination, and make sure that the
first (parallel) reactor draws exactly - jx amperes from the 50 volt
transmitter.

If the real part R of the antenna impedance R + jX is exactly R = 50
ohms, then a single reactor of impedance - jX placed in series with the
antenna will do the job. Note that in this case R*(50 - R) = 0 and
X = 0 = sqrt(R*(50 - R)) as above. In this case there is no first
(parallel) reactor, just a second (series) reactor.

It's a little different if the real part R of the antenna impedance
R + jX exceeds 50 ohms. In that case, the current drawn by the antenna
will again be 50/(R + jX) but the real part of that quantity will be
less than 1 ampere. How much less depends upon the value of X, but the
larger numerically X is, the smaller the current will be, and even if
X = 0, it will only be 50/R amperes which is less than the desired 1
ampere. Putting a reactor in series with the antenna, as we did above,
will only make it worse. The trick is instead to put the second reactor
in *parallel* with the antenna, and then to put the first reactor in
series between the transmitter and the parallel combination of the
antenna and second reactor. The calculations are somewhat more
complicated than for the case where R 50, but you now have all the
machinery with which to do them.

David, ex-W8EZE

--
David Ryeburn

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