I want to drive a 8Ohm load with a tube amp(Push-pull 2xEL84) that needs a ~8kOhm impedance on its outputs.
I have a transformer that has a impedance turns ratio of 0,048. link
So if I just plug in my 8Ohm speaker to the output side, I will get 8 / 0,048^2 ~= 3,5kOhm. This might damage my amp so I am looking for a solution.
My idea until a proper transformer arrives, is to series connect a 10 Ohms resistor to the speaker like this:
simulate this circuit – Schematic created using CircuitLab
So now the impedance on the tube side will be 18 / 0,048^2 = 7,8kOhm.
Is my understanding of impedance matching transformers correct?
Apart from burning more than half of my output power into the resistor, does this have any negative effects?
EDIT:
The transformer has many taps on its primary for things like ultralinear output, but I use it as a simple center-tapped transformer. Also the other side seems to have 4 windings, but if you look closely, they are all connected.
DC resistance of primary: 18,2Ohm. Secondary: <0,1Ohm - Couldn't measure.
I assume this low DC resistance will have negligible effect on impedance, so my previous assumptions are correct.