The correct internal-inductance of Windfeld cart.?

I think Axel is spot on with his comments and approach. I'm not quite sure I follow his distinction between "current" and "voltage mode" but maybe that is best for a different thread.

Lewm wrote:

I don't know of anyone else who uses a SUT and loads the cartridge on the primary side.

I covered this in a short thread back in november which can be found here

dave

Hey Lew,

In a perfect world, The SUT would be ideal and hence invisible to the cartridge and all loads of equal value would sound the same. Experiences Like Axel's just show how far from ideal transformers really are. I do not think any sweeping generalizations can be safely made about loading with SUT's since in the real world so many parameters come into play. I do think they sound far better but as usual in this hobby, "better" is subjective and often comes with its own set of baggage.

dave

Hey Axel,

Last part: Therefore "Current mode" means that the cart delivers the most current it can, into the SUT's primary and the best / most when properly impedance matched.

I still don't get your "voltage mode, vs current mode" statements with respect to transformers. The amount of current a cartridge delivers into the primary of the SUT is fixed by the inductance of the SUT in parallel with the reflected load the transformer provides.

In in your case of the 1:30 and the 47K, the reflected load (~52r) dictates the current through the transformer primary. By adding the 13 ohm resistor in parallel with the primary you do not appreciably adjust the current through the transformer primary. The current output from the cartridge increases but all of the added current simply traverses the resistor to ground.

This doesn't mean that the primary load doesn't have an effect on the transformer behavior since it does. I would expect the output impedance plot of a cartridge to show a rising value with frequency due to the inductance of the coil with a peak at the point where the inductance and capacitance resonate and then a decrease beyond that. This typically happens close to or within the audio band. If you consider how the source impedance has an effect on the transformer behavior and understand that the primary loading resistor damps (and lowers) the output impedance of the cartridge it quickly becomes clear that this is a very complex relationship that doesn't lend itself to generalizations and ROT's

dave

Now to the current. A trannie like a 1:31.6 has a DCR of about 1.5 ohms on primary, and as we calculated 47ohm inductance and with a 13ohm in || a 10.18 ohm impedance.

what is a 47 ohm inductance? Also the 10.18 ohm is only the resistive nature of the reflected load. In order to call it the impedance you need to add in the inductance and capacitance of the SUT. To give an impedance you also need to give a frequency (or better yet a plot of impedance vs. frequency) and if just a fixed value such as 10.18 ohms is given, I would expect the word "Nominal" to be added.

Now try do the maths and see how much more current will go through the primary than in a non SUT set-up.

this confuses me, the non-sut setup will not have a primary. The only way to increase current through the primary is to apply the desired load across the secondary, by loading the primary you DECREASE the current through the SUT primary.

On the secondary you have MORE than twenty times the Voltage! Now this can ONLY come from a higher current, right? (ohms law)

actually it comes with less current, the current follows the exact inverse of the voltage. The volts * amps (assuming the ideal transformer) must be equal on either side so if you have 20X the voltage you have 1/20th the current.

Veff= [10/(10.18+4)]*0.3*31.6= 6.7mV that the phone-pre will see.
As I said, more than twenty times 6.7/0.3= 22.3

that is the voltage increase and why we use SUT's I don't see where current ever comes into the picture since it is purely a function of the V and Z.

So it is the SUT's primary that enables the cart to deliver 20 time more current than if you are offering 0.3mV to your phono-pre.

I'm sorry this is incorrect. The transformer allows the delivery of 20X more voltage but that comes at the cost of 1/20th of the current.

The cartridge outputs current based on the load. Assuming a 10R load, the cartridge will output the same current whether or not a transformer is in place. If the 10 ohms is provided by terminating a 1:30 with a 9K resistance then all of the current output by the cartridge will travel through the transformer to the load. The other extreme of this is a situation where a 10 ohm resistor is placed on the primary side of the transformer and the secondary is left open. Then all of the current generated by the cartridge will traverse the 10 ohm resistor and virtually no current will traverse the transformer. This is of course makes lots of assumptions but will suffice for a simple first order model. As you add in the parasitics your model becomes more complex (and more accurate) but you still need to adhere to the concepts of the ideal.

dave

The main point you are missing here is that 20 times the voltage i.e. 6.8mV instead of 0.3mV into 47k pre-input-impedance, WILL and DOES 'pull' more current on the primary side which has ONLY 10 ohm impedance! A virtual short circuit!!

I'm sorry, No it doesn't. In order to not break the first law of thermodynamics, applying the load to the primary of the transformer while keeping a fixed load on the cartridge must DECREASE the current through the transformer.

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.

From the transformer perspective, the secondary load will draw more current through the transformer.

the total "source impedance" for whatever is downstream of the 47K resistor is (assuming a 2.5R cart):

Primary load has a source impedance of 2000 ohms
secondary load has a source Z of 2000 ohms
the No SUT has a souce Z of 2 ohms.

dave

If I have said anything technically incorrect i'll gladly accept corrections and apologize for errors I have made.

dave

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

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

3) Why didn’t Dave step up to the plate and tell me he is a SUT manufacturer, maybe he thinks we all know this? Well, as you can see – I didn’t.

I didn't know it was expected. Many forums frown upon manufacturers hawking their wares so I simply used my company name as my moniker. If anyone here thinks me not disclosing my industry nature was inappropriate, I apologize since the last thing I want to appear as is a "sock puppet". The idea that I wind transformers has nothing to do with my perceived technical inaccuracies in your posts which is the only thing I have responded to. If you want to go back point by point to clear up any possible misunderstandings I may have had, I'll gladly have that discussion. I specifically asked you to properly quote me since many of your comments attributed to me are improper interpretations of my intent.

enough of that, now onto the meat of your post.

- Can a cart be generating more current when it operates into a lower impedance.

Absolutely, it is a simple oms law thing from the first order perspective. This isn't even worth discussion and hasn't ever been something I have disagreed with.

- Dave (our SUT Man) vehemently defends, that by the “First principle of Thermodynamics” that is in NOT POSSIBLE, and that therefore the term “current mode” (not my own as I also stated before) is NONSENSE to put it simple.

I simply used the first law of thermodynamics to state that a SUT cannot increase both voltage and current at the same time since that would increase power. Of course an increase of voltage into a fixed load will increase current but I did not interpret your writings as saying such. If I misunderstood your intent the proper response would have been to discuss the topic on point with proper quoting so we could clear up the misunderstanding.

He maintains a cart has a ‘fixed’ Voltage AND a fixed Current.

I have not suggested this. With both fixed voltage and current, there could be no AC (music).

The only thing I have taken issue with is your insistence that a SUT is required to make a Cartridge operate in "current mode". If you carefully read back through every post I have made in this topic, I hope that will be clear. If it isn't clear, tell me and I'll try to be more lucid in the future.

dave