Calculating transformer winding ratio for 600 ohm > 8 ohm

[RockyRaccoon]

I have designed an amplifier with an output impedance is around 600 ohms. I want it to drive an 8 ohm speaker, so I am trying to calculate the proper transformer winding. I’m finding very little information on this so I’ve come here to ask for help — how do I calculate the proper transformer winding ratio to turn a signal with 600 ohms of resistance to a signal with 8 ohms of resistance? If it’s a common enough ratio I’ll order it but if not I have a machine that I use to wind guitar pickups that I’ll use to make it myself. Thanks!

[Benta]

The rule is that impedance transformation is the square of the winding ratio.
In your case sqrt(600/8) = N.

[RockyRaccoon]

The rule is that impedance transformation is the square of the winding ratio.
In your case sqrt(600/8) = N.

So I’m understanding you properly, the ratio should be 8.66:1?

600/8 = 75 and the square root of 75 is 8.66

Thank you.

[DaJMasta]

Exactly.  You can find RF transformer calculators around the net that will confirm this, also can give you an idea of losses for a given core material.  Here's one that gives you a bit more than just turn ratio, but there are many options: https://www.changpuak.ch/electronics/broadband_transformer_matching.php


For good low frequency response, you're going to want a good number of turns.

[floobydust]

You won't get a lot of power, maybe a few W at most unless your 600 ohm source is capable of a lot of output voltage, like a Pultec EQ.
To experiment, a 120V-12.6V power transformer is around 9.5:1 turns ratio. So 600 ohm input and 6.6 ohm output.