The correct internal-inductance of Windfeld cart.?

Hi Raul,
I have to admit getting a bit exasperated by Dave's not understanding what a generator is. Yet he otherwise seems to ask relevant questions, but always sounding more like the ‘Doubting Thomas’ -- a saint never the less?

Don't know if it is so hard to understand, or just to know, that if you put more load (ask more current) from a generator you have to drive it harder! Whether on your bicycle or shovelling more coal, and raise more steam in the turbine house. If you don't, the lights go out, right?

I thought any technically informed person would know, - a wrong assumption? Hey, pork chops come from a pig, not a plastic bag from the super market. Even if that’s the way we always see it. Must be getting old, sorry.

If the MC which I explained is a generator, it has no issue to deliver more current, the energy comes from your tt motor, turning the platter, moving the stylus in the groove, moving the coil in the magnet (A GENERATOR!!) etc. etc. etc.
So an MC is NOT a constant current source, as I also explained, uff.

If we may finally accept that very fact, the rest aught to come easy. You practically get 20 times (in the example based on a 1:31.6 ratio 30dB SUT) the voltage free! So who cares about some smallish loss (inter-winding capacitance, leakages, thermal losses, etc.) it is a FRACTION of what you get for FREE.

So that is not the issue at all, the issue is more what regards sound quality!
Any how, Raul, it will need some work, AND understanding about the cart matching the SUT.
And NOT if the dang thing works in the first place. I don’t know why I get so rattled by this. Like go argue with me if a wheel can turn, please…

Back to the SUT match. EVERY different SUT winding ratio results in what acceptable terminology calls a ‘natural impedance’ on the primary side, the one connected to the cart.

This is of course also determined by the input impedance of the pre-amp, usually 47k. It can not easily be upped as you Raul, have done to e.g. 70k or even 100k because you have to change a phono-pre's build-in load resistor(s). It is lots easier to lower this impedance by simply putting some R value in parallel with the secondary SUT output, the one connected to the phono-pre.

If you do that, you will however start electrically damping the SUT, which can be a good thing if it give you a resonance problem --- BUT it actually is a crude / shortcut. Dave asked what if he puts a 10ohm!! resistor there, a practical short circuit, you just have damped the hell out of it, so that NOTHING (no current) will be left to go to the phono-pre. It was a rhetorical question, OK.

If resonance damping of a SUT is required (and most don’t it seems) than a RC (creating a notch filter) on the secondary is the way to do it. You'd better know the INDUCTANCE (the L yes) of the cart and the SUT and start figuring out the resonance frequency in the SUT so you can get values for R and C. Usually in the ~ 200-300ohm and ~ 60-160pF region. Don't take my word for it, it needs to be calculated and then tested for sure.

Raul, are you still with me? I go into that, since if you want to do some testing as you suggested this is where it’s at.

I mentioned ‘natural impedance’ at the primary, i.e. the cart end. It is 47ohm with 47k phono-pre input impedance(like most all pres) and a 1:31.6 winding ratio (30dB gain) that give you just that (47k/31.6^2)= 47ohm, that the maths.

But now watch! To enable the cart to ‘pump’ current (that ‘for free’ stuff!) it has to be impedance matched to the SUT. Impedance matching is NOT a must, but it will give you the best, absolute very best, power transfer from cart to primary (no reflections as they say, ‘cause we are dealing with AC, right?)

Next, a rough check with you cart's output is also required i.e. am I going to create an over-load in the phono-pre?!

DON’T even think to offer more than 7.5mV to the pre, even though most will state 10mV over-load margin. The midpoint target spec. is 4.7mV it’s why most MM and HO-MCs target output around there.
So how do you check that quickly?
Vcart * winding ratio, my example was 0.3mV * 31.6 = 9.48mV !!!!!!!

According to what I just said this suggests we are too high, i.e. this cart would need a 26dB SUT rather, than a 30dB, right? WRONG!!

Of course if you have a lower one it will work also, but you have to open the phono-pre volume control some more (unless you got some fancy beast like me, with variable input gain).

So why are we now wrong in this case?
We are wrong, because I know that this cart will be no where near at its best going into a 47 ohm impedance.
(STOP! You’ll say but we always would put about 100 ohm, and the PW in the example likes even more 500,... 1k!)

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 explained some about electro-mechanical cart damping and 'plain' resistive loading (pulling more current) earlier, so let’s just continue.

The closest value of output IMPEDANCE of a cart is somewhere 2.5*DCR. In our PW example 4ohm*2.5= 10ohm (usually also the quoted minimum loading by the manufacturer --- if that cart could be used with an SUT!).
More than 0.4mV cart output in NON-SUT territory!

Let’s look back at “voltage mode” loading, and you will find it is rather more like 25 * RDC i.e. our example 25 * 4 = 100 ohm, and pretty usual at that.

On top I’d said WRONG!! when looking at phono-pre overload and step-up ratio. So let’s look at that some more. It is because:
“When the resistive load is equal to the internal impedance of the cartridge, the cartridge output voltage is reduced by 1/2.”
It's one way to say it nicely, or:
"When the input-impedance is matched to a source's output-impedance, the output voltage is halfed and the current transfere is at it’s maximum.

That impedance matching, you can go look it up in the science book.

This said, let’s look at our apparent 9.48mV overload example again and see what gives:
Vpre = Rload / (Rload + DCRcart) * Vcart * winding ratio
Vpre = 10 / (10 + 4) * 0.3 * 31.6 = 6.77mV !!!!!!

So, that done we know the SUT is still a match. A bit on the high side but still under our self-imposed limit of 7.5mV.

And that’s it Raul. If you familiar with it it’s less involved than what it looks like, but just hitching a trannie to a cart without considering the above is no good at all.
Lastly, there is this by many beloved cart the DL-103 with most of them having about 40ohm DCR and 0.3mV output.
In this here case you might get away without resistive loading i.e.
DL-103 cart min. load impedance 2.5 * 40 = ~ 100ohm
1:22 ratio (20DB gain)= 47k/22^ 2= 47k/484 = 97ohm natural impedance.
Hey, this looks very well within the assumed 100 ohm load and should work just fine ---- BUT as always YMMV.
I have not tried it but at least the maths would suggest it to be a good match.
Greetings,
Axel

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).