Topband: Transformer Question

[Dave Bowker]

It's not that simple.  The $:1 transformer (or any transformer) can only
transform a pure resistance (Z = R + j0) in a direct division ratio, i.e.,
if you have a pure 200 + j0  resistance across the high impedance side, it
will be transformed to 50 + j0 resistnace on the low impedance side.  Any
reactive component on the high impedance side introduces a complex impedance
with a complex solution which must be addressed in the transformation, and
it is not a simple 4:1 ratio of EITHER resistance OR reactance.  This has
long been a misconception about baluns, choke baluns, etc., in the ham
community.  They simply cannot by themselves, make a direct impedance
transformation to Z = R +j0 when reactive components of impedance are
involved.

There are three solutions to your posed problem: (1) convert the complex
impedance to a pure resistive impedance with an L-network, (2) create a
design using a single-tuned transformer with 4:1 ratio, (3) use a simple
impedance matching network (L/Pi/T) and eliminate the transformer.

The first solution would transform your Z = 260 +j330 to Z = 200 +j0 on the
high side (using an L-network (or other impedance matching network such as a
Pi, or T).  The transformer then transforms Z = 200 + j0 this to Z = 50 +
j0.  Thus your overall transformation from Z = 260 + j330 to Z = 50 +j0
would involve a minimum of three components (4:1 transformer, a capacitor
and an inductor in the case of the L-network).  There are two solutions to
the L-network which would achieve a solution to the problem you posed; (1)
series 280 pF capacitor and shunt 212 uH inductor, or (2) series 27 uH
inductor and 360 pF capacitor.    The Z = 260 + j330 (higher Z) termination
is placed across the shunt component in each case.

The second solution would be to design a single tuned transformer with two
components, the transformer and a capacitor.  The design of such a circuit
would require you to know the primary (Lp) and secondary (Ls) inductances of
the transformer, the mutual inductance (M), the coupling coefficient (k) of
the windings, and the desired circuit Q.  Q and k are an arbitrary selection
and one affects the other.  Generally Q is specified and k then determined
by calculation.  The procedure for designing the actual circuit is well
covered by Terman (Electronic and Radio Eng'g, McGraw-Hill 1955) and Kraus,
Bostian, Raab (Solid State Radio Eng'g, Wiley 1980) and others and is a
rather lengthly process.  The second reference gives some worked out
examples for Q's > 10 and Q < 10.

A double tuned transformer would also be applicable, but adds another
component.  Again the design of such is covered in the above references.

The third solution can be implemented with two components to make the
transformation directly to 50 + j0.  There are two L-network solutions in
this case:  (1) series 490 pF capacitor and shunt 26 uH inductor, or (2)
series 15.5 uH inductor and shunt 617 pF capacitor.  Again, the Z = 260 +
j330 (higher Z) termination is placed across the shunt component in each
case.  If you are looking for primary/secondary isolation, this third
solution is not applicable.

In all cases, the resultant transformation is narrow-band, and dependent
upon either the Q of the L-network or the tuned transformer.

Hope this helps...my recommendation is to go with the first solution...it is
far easier to design!

73, Dave, K1FK
Fort Kent, ME