The correct internal-inductance of Windfeld cart.?

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

Hi Dave,
let try this once more...
>>>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 who can NOT agree, that the load determines the current of ANY generator?
Of course it does do that! But you seem to still miss the main point.
Put that 10ohm load WITHOUT an SUT you get zilch! going to the phono-pre, as I stated. So I say it again, just to make sure.

Put a 10ohm load (a perfect impedance match on the SUT primary as in the given example) you get 20 times the Voltage for free on the secondary side! FOR FREE!
Why, because you tt motor does not even notice the difference in extra work it has to do, that's why.
It wiggles that cart generator just the same, 'cause it has 1000 times more torque/power than is ever needed in any way.

>>> 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? <<<

Call it 'zilch mode', 'cause you are pumping current through the cart coil and a RESISTOR to no aparent effect what so ever!
Why even ask, just put a short circuit...

The point you seem to have is with the 'mode' description I guess, right?

The cart will NEVER (with out a SUT) be able to pull an equivalent current as with an SUT set-up ---- unless you want now go and listen to a resistor, a piece of wire, 13ohm what ever!

Dave, you try to make some point here that is splitting hair over a naming convention that was NOT introduced by myself, please believe me.

You can take the tires of a car and go on arguing with me that you can still drive with it!
How silly, isn't it?
Greetings,
Axel

Call it 'zilch mode', 'cause you are pumping current through the cart coil and a RESISTOR to no aparent effect what so ever!
Why even ask, just put a short circuit...

a 10 ohm load will output the same voltage and current from the cartridge period. Of course the situation without the SUT would require 26dB more gain but that isn't what is being discussed.

the only difference between no load / loading the primary situation and loading the secondary is that the current delivered to the load with the resistor on the secondary must also traverse the transformer. The case of the Primary loading resistor and resistive loading of the cartridge without the SUT are identical so if one is "pumping current through the cart coil and a RESISTOR to no aparent effect what so ever!" then so must the other because they are equivalent.

The cart will NEVER (with out a SUT) be able to pull an equivalent current as with an SUT set-up

Maybe this is just a naming thing. First cartridges do not "pull" current, they deliver it. Secondly whether there is a SUT or not has no effect on the amount of current delivered by the cartridge. Yes a SUT will give you voltage gain, but that comes at the cost of the ability to deliver current.

dave

I admit I haven't read through all of these posts, so I am not clear as to the controversy here, but, I don't get the concept of "20 time the voltage for free." The cartridge's output is fixed -- it is a certain amount of mv per the particular modulation supplied by the record groove. As a low impedance source, it is delivering this output as, relatively speaking, a high current, low voltage signal. What phono stages do is amplify/convert the signal to a high voltage/low current signal. An input SUT converts the high current/low voltage signal to a low current/high voltage signal -- nothing is "free." A loading resistor of any value across the primary acts as a voltage divider which will dissipate some of the signal as heat (i.e., a loss). Of course, the higher values used means that little is lost. The voltage gain (at the expense of current) is determined by the turn ratio of the SUT, I don't see how it has anything to do with loading.

My phono stage has a loading resistor across the primary. The recommendation by the distributor of the phonostage is to experiment with the value of this resistor to optimally load the chosen cartridge. The distributor does not recommend changing the value of the resistor on the secondary side. This makes sense to me. That resistor provides the optimal loading of the SUT itself (these things will have their own electrical resonance properties).

Any one around this place that understands what is a generator, or how it works? Help!

Switch your car light on and the dashboard lights dim?
No need for a regulator either...

A cart going into a lower than usual impedance will output more current, NOT Voltage yes. But instead of frying the current through a resistor, you pass it through a transformer coil and as a result get that current (in the relevant ratio) transformed into Voltage on the secondary.

I find it absolutely staggering that this seems such a unbelievable concept to grasp. We are not in the 1800s, or?

I think I'm done here, and that’s just fine too. Nobody is using an SUT, and everybody KNOWS why it can't work, truly amasing that is.

Let's agree to close this thread, not value added this.
Axel