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:
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?
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.