Charliee, In one of your responses, you mentioned hooking the CDP to the INPUT of the SUT. I am guessing that was a momentary lapse; Loesch and all the others are talking about hooking the signal to the OUTPUTs of the SUT. Just to be certain.
While I do not necessarily share Robjerman's skepticism about the value of break-in, I do wonder at Loesch's comment suggesting that it is desirable to "magnetize" the core of the transformer in order to achieve proper break-in. I would have thought that magnetization is undesirable; you'd want to break in the wire coils without magnetizing the core. But I have never played with a SUT in my own system, so I don't have a validated opinion. And Loesch is no dummy.
Also, contrary to what Rob said in one post, a SUT has no "resistance" of its own. There is indeed resistance presented by the wire coil on the primary side, between hot and ground, and a second analogous resistance presented by the wire on the secondary side, but these have trivial effects. Transformers merely reflect the output and input impedances of the devices to which they are connected. So, if you have a SUT with a 1:10 voltage gain, that means the turns ratio of the SUT is also 1:10. (That's 20db, a little more gain than that of the SUT you own.) The impedance seen by the device on the primary side will be related to that on the secondary side (or vice-versa) by the square of the turns ratio, or in this case, 1:100. If you drive the secondaries of such a SUT, the device connected to the primary side will "see" an impedance equal to 100X the input impedance of the driven device. So, if it's just a 27 ohm resistor or whatever, the resistance/impedance seen by the CDP connected to the secondaries would be 2700 ohms. Most CDPs can drive that impedance, but I did wonder why he stipulated such relatively low values of R. Maybe so as to assure that everything heats up.
While I do not necessarily share Robjerman's skepticism about the value of break-in, I do wonder at Loesch's comment suggesting that it is desirable to "magnetize" the core of the transformer in order to achieve proper break-in. I would have thought that magnetization is undesirable; you'd want to break in the wire coils without magnetizing the core. But I have never played with a SUT in my own system, so I don't have a validated opinion. And Loesch is no dummy.
Also, contrary to what Rob said in one post, a SUT has no "resistance" of its own. There is indeed resistance presented by the wire coil on the primary side, between hot and ground, and a second analogous resistance presented by the wire on the secondary side, but these have trivial effects. Transformers merely reflect the output and input impedances of the devices to which they are connected. So, if you have a SUT with a 1:10 voltage gain, that means the turns ratio of the SUT is also 1:10. (That's 20db, a little more gain than that of the SUT you own.) The impedance seen by the device on the primary side will be related to that on the secondary side (or vice-versa) by the square of the turns ratio, or in this case, 1:100. If you drive the secondaries of such a SUT, the device connected to the primary side will "see" an impedance equal to 100X the input impedance of the driven device. So, if it's just a 27 ohm resistor or whatever, the resistance/impedance seen by the CDP connected to the secondaries would be 2700 ohms. Most CDPs can drive that impedance, but I did wonder why he stipulated such relatively low values of R. Maybe so as to assure that everything heats up.