Transformer Impedance Matching Equation

Discussion in 'Homework Help' started by SamR, Aug 1, 2019.

  1. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    I have a Triad F-25X xfmr. The PDF gives no turns ratio which I can assume is same ratio as the PDF ratio of 115VAC Vp and 12.6V Vs.

    115/12.6 = 9.127:1 for N1 and 1 for N2

    For ri there is no data on the PDF so measuring the Rp including the cordset I get 25.4Ω.

    I don't know the physical dimensions of the primary coil or what it's core permeability is.

    The %Z is not given on the PDF and I am not able to test for it.

    So... Do I use the measured 25.4Ω for Zp in the N1/N2 = √Zp/Zs equation to solve for Zs?
     
  2. nsaspook

    Expert

    Aug 27, 2009
    6,013
    6,797
  3. BobTPH

    Senior Member

    Jun 5, 2013
    1,847
    479
    Just curious, why do you want to know the impedance? I have never even thought about the impedance of a transformer.

    Bob
     
  4. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    I'm using the F25X as an example and I have one that I use to play with for rectifier circuits. I am working my way through Grobb's Basic Electronics and it is part of an exercise in Chap. 19 on impedance matching xfmrs. It threw out a number for Zp without giving any basis for it so I started looking at what I had to work with and came up blank and scratching my head. So I asked.
     
    mlsirkis likes this.
  5. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    The reason I have that particular xfmr is it is used for the lab exercises for several of Dr. Malvino's books. I also use it as a backup for my bench PSU when I need a 2nd voltage source. In particular, for OP Amps with the LM317 module I built to use with it.
     
  6. crutschow

    Expert

    Mar 14, 2008
    22,714
    6,689
    Transformer impedance value and impedance matching are of interest for signal transformers, not power transformers.
    Power transformers are not impedance matched.
    The limiting factor in power transformers is normally heating from the I squared R loss due to load current in the primary and secondary winding resistances.
     
    SamR likes this.
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,287
    1,368
    Hi,

    I am not sure what you are doing either but if you want to match something to a transformer primary you have to know what is connected to the secondary. The impedance of anything connected to that alters the input impedance of the transformer.

    The impedance reflected to the secondary is the impedance of the primary divided by the square of the turns ratio.
    So if the primary input impedance measured 16 ohms and the turns ratio was 2:1 (step down) then the secondary impedance would be 4 ohms.
    Likewise, if the impedance of the secondary was 4 ohms then the impedance measured into the primary would be 16 ohms.
    If the transformer was ideal that would mean that there was a 4 ohm resistance connected to the secondary and so it looks like a 16 ohm impedance looking into the primary. Now if the 4 ohms went down to 2 ohms the impedance looking into the primary would be 8 ohms.
    So the impedance are related by the square of the turns ratio (or the inverse square) but the change is related just by the turns ratio.

    Hope that helps.
     
  8. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    Precisely. I know what's on the secondary. So I do use the measured resistance of the primary as its impedance? That is my question since there is no impedance data on the PDF. It is certainly easy to adjust the output load to match the primary once I know what primary impedance I am matching it to. Thanks for the response.
     
  9. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,287
    1,368
    Hi,

    I am still not entirely sure what you are doing here :)

    But, how are you measuring that resistance ?
     
  10. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    OK I finally woke up and measured the Inductance of my transformer primary through the cordset at 2.44 H using my tester. Which at 60Hz give me the Impedance of 844Ω if this is the correct method.
     
  11. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    Intellectual exercise, learning about impedance matching. The leads are shrinkwrapped to the cordset so I am having to measure from the prongs on the cordset plug. The cordset is about 12" long.
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,710
    494
    The turns ratio is probably not 115/12.6, because the 12.6 volt rating of the secondary is given for rated current load.

    Here's a video showing measurement of turns ratio of a power transformer:


    Also, you might find these links useful:

    https://www.electro-tech-online.com...-ohms-on-an-unknown-audio-transformer.124189/

    https://www.radioremembered.org/outimp.htm

    http://www.rfcafe.com/references/po...ormers-popular-electronics-september-1956.htm
     
    SamR likes this.
  13. crutschow

    Expert

    Mar 14, 2008
    22,714
    6,689
    As a learning exercise that's okay, but as I stated, impedance matching has nothing to do with power transformers, and is only used for signal, RF, or audio output transformers.
    It also has nothing directly to do with the resistance or inductance of the transformer.
    The impedance of an impedance matching transformer refers to the input/output impedances you want to match, not the transformer impedance which is typically much less than the matching impedance.
    Read this for starters.
     
    nsaspook likes this.
  14. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,287
    1,368
    Hello again,

    If you want to measure the impedance looking into the primary you can try this:
    R=E/(sqrt(tan(ph)^2+1)*Im)
    L=-(tan(ph)*E)/(sqrt(tan(ph)^2+1)*Im)

    where
    R is the equivalent series resistance of the impedance (real part),
    L is the equivalent series inductance of the impedance,
    ph is the measured phase angle between current and voltage (will be negative),
    E is the applied AC test voltage in volts,
    Im is the measured AC current in amperes.

    So basically you apply a test voltage and measure the current level and phase angle, then calculate R and L.
    The impedance is then:
    Z=R+s*L

    and keep in mind this could change if you change the load on the secondary.

    The above formula were worked out carefully but always double check your results using a circuit simulator. The phase angle is sometimes hard to measure so you have to try to get as close as possible for best results.
    Also, the load current has to be high enough to dominate the current response.
     
    Last edited: Aug 3, 2019
  15. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    Thanks Al. I was doing good until the Im part which tells me that the Zp is not a fixed number and depends solely on the output load. That being the case then how would you know what impedance to match the load to? On the other hand I'm being told it is the square of N1 in the turns ratio x Rl which seems to indicate it is a fixed ratio. I am beginning to understand why this was skimmed over during the exercise where they presented the Zp as a given instead of explaining the basis for it. I think at this point in my studies I will accept xrmfr Z matching as a fact without delving too deeply into the basis for it and shelve it for later and move ahead.
     
  16. crutschow

    Expert

    Mar 14, 2008
    22,714
    6,689
    What about my explanation is not clear? :confused:
     
  17. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    Not a thing and thanks for your help as you have been exceedingly generous in your time and patience with my many questions.
     
  18. crutschow

    Expert

    Mar 14, 2008
    22,714
    6,689
    Okay good, because it seems like you still aren't completely clear about the purpose of transformer impedance matching and how it works. ;)
     
  19. SamR

    Thread Starter Active Member

    Mar 19, 2019
    813
    311
    No I do understand it in theory. The changing Z due to I pulled by the load are details that I was missing. The reason I asked to begin with is Grobb gave an example using a Z for the primary and gave no basis for it so I was off on a tangent thinking "where did that come from". All it was doing was to illustrate the concept of matching source and load impedance to achieve maximum power transfer of a known source impedance and adjusting the resistive load to match. Now I now that is not as easy a task as I anticipated due to the changing nature of current-induced impedance. Current does not directly come into the RC and RL calculations but R and I are inversely related. It still a bit new and somewhat confusing. I went through all this years ago, but Grobb is either exposing me to some new concepts or I just flat forgot them over the years since I studied AC/DC theory.
     
  20. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,287
    1,368
    Hi,

    Well see you are approaching this inquiry from many different angles and so every thing seems "variable".
    A more typical approach is when you are given an impedance of some load and given another impedance of some source, and you have to match the two. Let me give you a more typical example.

    The load is 50 ohms and the source is 200 ohms. To match these two with a transformer we need a step down. What will be the turns ratio of that transformer?
    Zin/Zout=200/50=4
    N=sqrt(4)=2
    so the turns ratio is 2:1 here.

    Now any transformer made for this job will have low enough wire resistance to provide a reasonable match without too much loss.
    For example a transformer with secondary resistance of 5 ohms and primary resistance of 20 ohms might not be too bad, but one with 1 ohm secondary wire resistance and 4 ohm primary resistance would be better, while one with 25 ohm secondary resistance and 100 ohm primary resistance would not be very good because it would eat up too much of the power. If that was the only one available it might be used anyway but only if the losses would be acceptable too.

    So see how this works. It's not about measuring the transformer itself so much as what the turns ratio is because that matches one impedance to another. The measurement of the transformer then would be the primary and secondary resistance and turns ratio. The primary and secondary resistance tells us what impedances would work without causing too much loss.
    As above, if we have 10 ohm secondary resistance we probably dont want to use that with a 50 ohm load. On the other hand, if we had 1 hm secondary resistance we could probably use that with a 50 ohm load because the winding resistance is low compared to the load impedance.
    On the other hand, with that same 10 ohm secondary resistance we could probably use it with a 500 ohm load because then the winding resistance is only 2 percent of the load impedance.

    The above would be a general guide in how impedance matching with a transformer works. Now to the power transformer.
    The power transformer chosen must also have a secondary resistance that is low enough to properly power the load without too much voltage drop. If we have a 10v output secondary with wire resistance of 10 ohms, then if we used it to power a load that draws 1 amp, we would lose too much voltage in the winding resistance. On the other hand, if we had a load that only takes 100ma, it might be ok because then we'd only lose about 1 volt which may be acceptable. The primary resistance also can not drop too much voltage.
    So there is some matching going on there too but it's not quite the same as matching two system impedances.
    Can we look at it the same way? Well, the line power source is usually considered to be very low impedance so there is not really a matching procedure for that because the resistance of the primary will often dominate the entire resistance of the primary circuit. So we instead go with a transformer that can provide the required secondary voltage at the required secondary current and that's about it.
    Also note that a power transformer often has to power a variable load so the current might be always changing.
     
Loading...