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TX-103 break in.

I recently purchased a pair of Mu MC Step Up transformers by Bent Audio. Does anyone here know the break in process for these. I was able to find this by Thorsten Loesch but I don't understand it, doesn't quite make sense to me. Maybe a more simple explanation would be helpful. 

From diyAudio; Break-In Instructions from Thorsten Loesch

This worked very well for me, so I recommend it highly. This is what Loesch wrote, with a little editing:

"(The TX-103)... will require a substantial period of "forced burn in" to give it's best, simply because the magnetic core is huge and will not see much magnetisation with normal MC signals. Please consider connecting a CD-Player to the secondary (Output) of the TX-103 and then terminate the input with a low resistance resistor (quality uncritical), I'd say 27 Ohm when connected for 14db gain, 6.8 Ohm when connected for 20db gain and 2.2 Ohm when connected for 26db gain. Leave with a highly dynamic, wide bandwidth signal CD to play for a week or two. I would use music, but I'd expect pink noise to work well too."
charliee

Showing 2 responses by lewm

lewm
13,508 posts
 
Chakster, With all due respect and meaning no offense, first you describe the input shunt resistors of the CPP-1, which is not a SUT but a pre-preamplifier designed to amplify the voltage of the signal generated from a cartridge, without adding RIAA correction.  That is not anywhere near to what the OP needs to know.  Then, you go on to recommend that he should buy a vintage SUT, but the point is that he is the new owner of a new SUT and only wants to know how or whether to break it in.  Whether you or I or anyone else "believes" in the value of burn-in is irrelevant.  The only question is whether such a break-in regimen enhances SQ, and that ultimately is a subjective judgement.  So, if it pleases the OP to break in his new SUT, he only needs guidance in how to go about doing it. After all, it cannot hurt to try.

By the way, totally noise-free and zero inductance (or zero capacitance) resistors only exist in advertising brochures.
lewm
13,508 posts
 
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.
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