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In Reply to: Oh My God!! DL-103R & CineMag posted by JRags on October 3, 2006 at 15:31:23:
I also have the 103R and the Cinemag. Set at 1:36 with 30k Ohms loading resistor and been "soaring" to vinyl heaven for almost two months now.FWIW, make sure the loading resistor is place near the preamp side and not the SUT side.
Follow Ups:
Amandarae, very nice looking arrangement of the step-ups. I have seen your picture before, and you might have mentioned this before in another thread, but what case are you using for the box. Also, do you have any inside pictures available? Finally are these insulating washers between the trannies and the box?Thanks very much!
With best wishes,
Rene
Die gefährlichste Weltanschauung ist die Weltanschauung derjenigen, die die Welt nicht angeschaut haben.
Alexander von Humboldt (1769-1859)
Hello Rene,The case was sourced from AES and is the cheap Hammond 4x3x6 with cover. I painted it and drilled the holes. The trannies are sitting on teflon washers so that the tranny enclosure does not touch the metal box. I am sorry that I do not have a picture of the inside. I followed the wiring diagram from Cinemag's website and I recall that Garth posted an instructions on how to interpret them correctly. I do have pics of the SUT with the switch(one for each but can be done by just a single one) incorporated already to change from 1:36 and 1:18 on the fly. Here they are.
Regards,
Abe
Hey Garth! Of course, the transformer does not care. At ideal conditions(lossless), the electrical equivalent of both locations are the same.But here is my reasoning behind it and please tell me if it has merit.
With the loading resistor close to the SUT, let's assume that a 5 mV output was coming out of the transformer secondary. If we load the the cart with its specified resistance, the voltage will be half for example. Then this signal will travel the, say 1 meter ,interconnect. It is possible that it will be reduce further and S/N will be less if the transmitted signal was less to begin with than the opposite scenario when the whole 5mv signal will travel across the 1m interconnect first before it was halve.
Does it make sense? I do not know. Besides, I should have posted "try loading the SUT near the preamp side and see if there's a difference" instead.
regards
The cartridge "produces" half the voltage of the unloaded condition. It is not a case of the "full voltage" hitting a "resistor brick wall" and being reduced.The secondary "sees" the load impedance and reflects it to the cartridge through the transformer primary. The load on/at the coils of the cartridge is X or 1/2X or whatever the transformer presents it.
There may be some good reasons (that I don't know) to put the additional parallel resistor at the preamp but that is not one of them.
Sorry. :-)
At least I got straighten out. I was assuming maximum power transfer since a 14 ohm impedance source driving a 14 ohm load. So now I know that it is not really the case. How do you compute the resulting voltage depending on the load values?thanks
Hi Abe! (it's earlier than I thought in SoCal)here's the formula:
Va = (Vi / Is+Il) * Il
Va = voltage actual
Vi = voltage initial
Is = Impedance source
Il = Impedance loadSo you can see for instance 2 / 28 * 14 = 1
As you point out that is max power but not max voltage. But who cares about voltage? Voltage is only loudness.
I really appreciate it.
I still have not done it but it is only a two second change.Do you really think the transformer cares? In principle when placed at the SUT it is nothing but a resistor with long leads, IE the cable between SUT and preamp. I frankly can't imagine it makes a difference but I need to try it I guess.
...please read my response above. i posted it in the wrong place.
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