Dave asked what if he puts a 10ohm!! resistor there
I asked no such thing. I made a statement of fact that paralleling the 47K resistor at the input of the phono stage with a 13 ohm resistor and DISPENSING with the SUT would provide a 10 ohm load on the cartridge. In the future, when quoting me, please try to be accurate.
This is my exact statement:
Furthermore, If you want a 10 ohm load on the cartridge, you can do one of three things,
-10,000 ohms on the secondary of a 31.6:1
-13 ohms on the primary and 47K on the secondary
-dispense with the SUT altogether and place 13R in parallel with the 47K input resistor.
In all three of these situations will draw the same current from the cartridge.
Do you agree that the above is factually correct?
If you do then you also have to agree that it is the load that determines the current delivered by the cartridge and not whether there is a SUT involved.
Now we are back with that bone that Dave is still chewing, we are operating a cart in current mode by using an SUT. This means the loading parameters are NOT comparable at all to voltage mode i.e. going straight into a phono-pre (with out a step-up trannie).
I don't see what there is to chew. The SUT merely reflects back whatever is across the secondary by the turns ratio squared. If the SUT is internal It can be connected directly to a tube grid (make it a pentode) and the secondary will effectively be an open circuit (call it 10meg) Given the 1000:1 impedance match the reflected load to the cartridge will be 10K. Again this is a statement of fact and assuming the 10meg is an accurate number, what "mode" of operation would your cartridge be operating in?
Looking at the situation without a SUT, if we do parallel a 13 ohm resistor with a typical 47K what "mode" will the cartridge be operating in?
dave