# Impedance matching with transformer and resistor

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.

• Is that really the impedance ratio? The TME web page says "Transformation ratio" which usually means turns ratio. The impedance ratio is the square of the turns ratio. That transformer has a skimpy datasheet in Polish, which doesn't help much. But I see you're actually squaring 0,048 so perhaps you just didn't use the right terminology in your post. Also give that that transformer has numerous windings/tap on both primary and secondary sides (if the picture is correct), a single "Transformation ratio" doesn't make any sense to me. – Fizz Oct 30 '15 at 13:47
• The mfg has a page about that series: indel.pl/e_tgl.htm It seems at least this is a xmfr for valve/lamp outputs and the ratio given is the voltage/turn ratio indeed. – Fizz Oct 30 '15 at 14:00