The correct internal-inductance of Windfeld cart.?

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

Hi Intactaudio,

I'm not making this up believe me. Firstly a 30dB trannie has a 1:31.6 ration that gives you a reflected 47ohm since your phono-pre input impedance is 47k
i.e. 47k/31.6^2 ~ 47ohm.

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.

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

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

Veff= [cart impedance / (trannie impedance + cart DCR)]* Vcart*winding ratio

So a cart like the PW has 0.3mV and 4ohm DCR, and 10ohm impedance.

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

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 hope this makes some sense.

Greetings,
Axel

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

Hi Dave,
uff, let me try this. You are right that a higher output voltage (at the secondary) would come at the cost of a lower current --- IF you would not be able to DRAW more current through the cart coil, but you can. That is the WHOLE point.

As to the 47 ohm 'natural impedance' (on primary), I though to have said impedance! Sorry that was a typo, need an editor, methinks. Thanks for pointing it out.

The 10.18 ohms is also impedance since it is in parallel with the reflected secondary impedance of 47 ohm on primary, calculation as we know.

The next item is explained also by drawing more current, as I said on top. You are going into a 10.18 ohm impedance rather than some 100 to 47k! Take your pick, what ever your 'normal' loading would have been.

A cart is actually a 'floating' device, XLR floating, NO ground. So NOTHING is shunted to ground only +/- at least in one of the better scenarios.

>>> 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 <<<
That is right on the money, next to nothing can then go through 47k pre-input-impedance...

Yes, and all the parasitic etc., of course but that is actually for the practical reason negligible.

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!!

Well, I could go on but let's see if that makes any sense now.
Greetings,
Axel

One more thing to help understand this concept: A cart is a GENERATOR! and not some fixed current source.
Axel