impedance matching transformer

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  #1  
Old 07-17-04, 06:58 PM
wabwan
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impedance matching transformer

hi! I made a transformer for impedance 7 ohm to match 50 ohm.
use transformer wire around the toroid, and the turn ratio is 2.67.
I tested find the voltage ratio between primary and secondary windings
is equal to the turn ratio when the load resistance connected with
the secondary winding is greater than 50ohm, this is basical circuit rule.
but if the load resistance is smaller than 50 ohm, the ratio is very greater
than the turn ratio, the voltage on second windings is much smaller than
the theoretical value for satisfying the basic circuit rule. I used the power
aplifier, so the source absolutely can drive this load resistance. So
what problem happens with my transformer? what can affect this voltage according to my knowledge which only depends on the turn ratio.
 
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  #2  
Old 07-18-04, 01:23 AM
lagunavolts
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I bet you are asian. I have a web site for you. http://electrical-contractor.net/cgi...000&LastLogin=
 
  #3  
Old 07-18-04, 05:40 AM
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Posts: 510
You are correct that in a "no-load" situation (load resistance >> 50 ohms) the voltage ratio = the turns ratio. However, once you put a load on the transformer the combined (1) output impedance of your source (amplifier), (2) primary winding resistance, and (3) secondary winding resistance are in series with the load impedance. The voltage across you 50 ohm load will be:

Vsec = TR*Vsource*Rload/(TR^2*(Rsource+Rpri) + Rsec + Rload)

where Vsec is the voltage acorss the secondary, TR is the turns ratio, Vsource is the source voltage (amplifier output), Rload is the load resistance (50 ohms), Rsource is the output impedance of the soruce, Rpri is the primary resistance, and Rsec is the scondary resistance. Note that when Rload=infinity (no load) then Vsec = TR*Vsource, which is what you measure.

Another possibility is that you are operating outside the frequency range for your transformer. At too low a frequency the core will saturate, at too high a frequency parasitic capacitances will shunt the windings, reducing voltage.
 
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