Hello Ford,
Unfortunately, it is not quite that simple. The circuit has to be analyzed such
that it includes the transformer's primary, secondary, and mutual impedances,
such that:
Zin = Z11 + ((w^2)*(M^2 ))/ Z22
where: Z11 is the total primary impedance (Rpri+/-jXpri), let's assume the
primary resistance is zero
Z22 is the secondary impedance (Rsec + Rload +/-jXsec +/-jXload)
w is the frequency in radians, 2*pi*F
M is the mutual impedance between the primary and secondary, M =
k*sqrt(XLpri * XLsec)
k is coupling coefficient between primary and secondary, usually
high for wideband torroids
No without going through all the math (here) and making some assumptions,
Given ZL = 260+j330 (Z = 420.1 Ohms)
F = 1.8MHz
XLsec = 5*ZL = j2100 Ohms
XLpri = 0.25*XLsec = j525 Ohms (1:4 ratio)
Lsec = 1.82E-4 H, Lpri = 4.56E-5H
if k = 0.98, then M = 8.66E-5H
I get: Z11 = j515.7, Z22 = 260+j2430
working all of this through the Zin equation above:
Zin = 41.78 + j 125.4
Now making different assumptions about the transformer will result in quite
different results. What I always find striking about such an analysis is that
the resistive load term does not reflect to the primary by the exact 4:1
impedance transformation ratio.
I am a bit rusty at this, but I think the analysis is correct (25+ years since
sophmore networks class in college).
Best Regards, Thomas - AC7A (Tucson)
P.S. To tune out the inductive reactance, a series capacitve reactance of
appropriate value can be placed in either the primary or secondary. A quality
capacitor placed in either circuit should not have appreciable loss, but do note
the primary current will be the higher of the two.
Ford Peterson wrote:
> I need to know if there is a correct solution to transforming complex
> impedances.
>
> Assume a perfect 4:1 transformer at 1.83MHz. A load with R+/-j of 260 +330j
> I am fairly certain that the R term just divides down to ~65ohms
> (260/4=65). What happens with the j term?
>
> To tune out the 330 ohms, a 264pf series capacitor is in order. If the j
> term divides down to 82.5j, then a capacitor on the input side of 1000pf is
> in order. If I had to choose, the 1000pf is less lossy and would be best
> placed on the input side of the transformer. Right? Or am I missing
> something...
>
> Thanks for the feedback in advance.
>
> Ford-N0FP
> ford@cmgate.com
>
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