# Transformer Impedance vs. Winding question

I have gotten a little confused on some transformer specs.
I notice that many transformers are rated at some impedance given some load.
For example: Input Impedance = 10K Output Impedance = 2K
Or, Input Impedance = 600 Ohm Output Impedance = 150 Ohm
It is obvious that there is some winding ratio involved, and I'm used to seeing the winding ratio, and calculating the winding to match a particular impedance: X = a^2 * Y (Where a is the winding ratio). If there is a ratio of 5:1, then the idea is to match the impedances.
Say you have a 600 ohm source, and an 8 ohm load: 600 = (a)^2 * 8 Then a = sqrt(600 / 8) a = 8.66
OK, fine.
Now, I'm looking for a transformer at "about" that winding ratio. BUT! I am finding transformers rated at certain impedance to impedance.
QUESTION: Can I use the same algebra to essentially calculate the winding ratio for that transformer? (Meaning - they specify some ratio only because certain transformers are designed specifically for applications - and other than power and current ratings, if I can figure out the winding ratio, I might hone in on what I want, right?)
Thanks!
Gary
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On 11/4/05 12:30 PM, in article Xns97049DB161689GaryG@216.196.97.131, "Gary

Your post is somewhat confusing. Nevertheless, I think that you understand the relationship between turns ratio and impedance transformation. I really do not know why either turns ratio or impedance is not used exclusively. Don't try to be more complicated than necessary. I think that there are certain combinations, like speaker or microphone combinations that show up often.
There are wideband transformers for which the actual impedance values rather than just turns ratio are important. The windings themselves form a transmission line transformer. The transmission lines, usually bifilar wire, are wound on a core in such a way that they act as a conventional transformer. The transmission line supports the impedance transformation at high frequency while the conventional transformer supports low frequencies. In such cases, the absolute values of the source and load impedances, not just their ratio, becomes important.
Bill
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Don Kelly @shawcross.ca
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So - the 10K:2K is just as equivalent as a different impedance that would work out to the same winding ratio, correct?
They probably spec 10K:2K just because that's some commonly used transformer.
Thus, If my source impedance were uaing the same winding ratio, but was 1K instead of 10K (no calculator in hand to figure out what that is at the moment), it would probably be ok?
Thanks!
Gary
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---------------------------- <GEG> wrote in message

--------- Theoretically- ignoring the transformer internal impedance-- but a transformer designed for a 10K primary may well have a higher internal impedance than one designed for 1K and this "may" have some effect (i.e. the internal impedance may be seen as an added ,say, 500 ohms which affects the matching). However it is worth a try. Depends on what you are doing. Also it is important to consider the frequency range for which the transformer is designed.
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Don Kelly @shawcross.ca
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On 11/6/05 9:36 PM, in article eJBbf.428627\$1i.31015@pd7tw2no, "Don Kelly"

The "internal impedance" is going to arise from the magnetizing inductance. As frequency increases so will the magnetizing reactance. The major effect will be on the low frequency coupling cutoff. As frequencies increase, that magnetizing reactance will not greatly affect such coupling. At sufficiently high frequencies, the winding capacitance reactance becomes small enough to load down the transformer. That will then affect the high frequency cutoff.
Bill
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The internal impedance typically arises from leakage (series) reactance rather than "magnetising reactance" (shunt). My point is that a high leakage (or short circuit) impedance, will add to the load impedance as "seen" by the source. I did not intend to imply an appreciable change in the coupling. For a half decent transformer the "magnetising reactance" is not a major factor(as you indicate) as it is high and in parallel with the load. The leakage impedance may be a dominant factor- particularly at higher frequencies. As to the effect of self and mutual capacitances- if you push beyond the design (frequency) range of the transformer these will have an effect- loading may be part of this effect at frequencies above resonance but certainly the terminal voltage ratio will change with frequency. In such a situation, the concept of an impedance ratio is probably of no discernable use.
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Don Kelly @shawcross.ca
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