The correct internal-inductance of Windfeld cart.?

That sounds so impossibly high that I agree it MUST be an error. There's just not enough wire in there to generate that much inductance. I will bet that they meant to write "micro-" Henries, not "milli-". Even that much seems a tad high. Of course, I am only guessing.

Dear Axelwahl: I don't know to what brochure you are reffering but here you can read that >10 ohms:
http://www.ortofon.com/index.php?option=com_content&view=article&id=230&Itemid=225

I don't know too why is so important to you the internal-inductance but you can ask directly to Ortofon here:
http://www.ortofon.com/index.php

Regards and enjoy the music.
Raul.

According to the Ortofon specification sheet, the PW has an internal resistance of 4 ohms, and an output of .3mV. If you are trying to work out what kind of step up transformer will work for this cartridge, those numbers are what you need. These are pretty much normal numbers for a relatively low output cartridge.

As for the inductance, I have no idea. Why would you need to know the inductance? I would assume that the PW is like any other MC cartridge, which means an insensitivity to capacitance loading.

Thanks for the help, why I asked for the correct impedance?
To determine the bandwidth of my system after SUT loading!
f/3dB = R / 2pi*L ...that's why :-)

Got it right now by primary loading 13ohm! Fidelity Research XF-1 type M (1:31.6 ratio) ergo 47ohm natural impedance with 47k phono-pre input impedance. (47k/31.6^2 = 47ohm)

So, 13*47/(13+47) = 10.18 ohm primary input impedance to match PW >10ohm recommended impedance.

The brochure I was referring to is the very one THAT CAME WITH THE CART! Mr. Rauliruegas
(Ortofon do know about their miss-prints, why ask?)

Meanwhile NOBODY at Ortofon commits to a correct impedance figure, other then 10 - 100uH -- Not so professional I say --- the reason? "We don't measure it..." well, well, well

Axelwahl, You originally asked about inductance, not impedance. This naturally led some people to ask why you wanted that information, since the proper loading of your cartridge is easiest to calculate based on the square of its turns ratio of the SUT you are using, if you know the desired load. For example, if you want the cartridge to "see" a 100-ohm load (a reasonable guess for a cartidge with a 4-ohm internal impedance, altho I know the Ortofon likes to see a higher value), and if your SUT has a turns ratio of 10:1, then the proper value of the load resistor on the secondary side of the SUT is 10,000 ohms. (The square of the turns ratio = 10^2 = 100; 100*100 ohms (the desired load) = 10,000 ohms.) If you want the Ortofon to see 500 ohms, then you need a 50K ohm load resistor. Maybe you know all this, but the fact remains that inductance of the cartridge is not needed to make this calculation. Folks were trying to help you.

Hi Lewm,
thanks for sharing the calculations.
1) Fidelity Research XF-1 type M, is 1:31.6 xfactor i.e. 30dB.
2) That Windfeld is a bit 'funny' since with no SUT it seems to like at least 500 to 1k ohm load and with some phono-pre's it's even better with 47k!
(I and also some reviewers came to this same finding)
3) I use PRIMARY loading, because if you get it right, it sounds more open --- using secondary loading is something else since as you pointed out the || of your sec. load with the 47k pre-input imp / by the square of (in this case) 31.6 is what the cart sees. Therefore with NO sec. loading the cart sees 47k / 31.6^2 = 47 ohm, didn't I go into that alread?
So any sec. loading now goes || with that 47ohm 'natural-impedance'(I'm neglecting the primary DCR (in the XF-1 case only 1.2 ohms which is also part of the deal)
PRIMARY loading with 100 ohm makes (in this case) the PW cart go beserk, higher like 250, 500, etc is also unworkable.
So you check the minimum spec. loading (usually 2.5 x DCR = 4ohm DCR x 2.5 = 10ohm, it happens to be min. spec by Ortofon, right)
What's left is to get a 1:1 impedance match to this! i.e. 47x13/(47+13)=10.18ohm ---- and as it turns out this sounds the best of the 1001 different versions I've tested.
So it seems to me, that you are working with loading that is non SUT relevant --- the big difference is that with an impedance matched SUT, that cart is working in 'current mode' (not 'voltage mode' as in a non-SUT set-up) and THAT changes EVERYTHING as I had to learn. You can not, I repeat NOT, compare 'current mode' to the 'normal'-voltage mode- loading i.e. with out SUT. Even when using secondary loading -current mode rules?- apply, just that we are now looking at ~ 18k ohm on the seconary producing [47k*18k / (47k+18k)]/ 31.6^2 = 14.1 ohm what the cart sees. Going down to e.g. 13k to make the cart see ~ 10ohm makes it to bright and brittle --- so not all things are quite equal between primary and secondary SUT loading either.
Also using secondary 18k ALSO damps the secondary SUT side --- not necessarily what is best, depending on the circumstances.
I actually just wanted to figure out what bandwidth I can expect by asking an apparently simple question about INDUCTANCE of the PW cart (not impedance as you rightly pointed out, I can measure DCR, but NOT uH!) So... I am still guessing we are looking at ~ 10uH and that gives me ~ 23kHz bandwidth.
Thank you for careing,
Axel

Axel, After I made my post, I re-read yours, and I realized I had slightly missed the point of your question. You might want to read the "white paper" on the Jensen transformer website, where they cover the complex issue of loading very well and describe the use of a Zobel network to attain flat fequency response. I've never used a SUT, so thankfully I have not needed to think it thru in as much detail as you have done. I would not have thought that the Ortofon or any other similar MC would be "happy" with only a 10-ohm load, but you have done the listening, whereas I have not. If you have a 47K resistance on the secondary side of you SUT, that will always reflect a 47-ohm impedance to the cartridge (if the tranny has a 10:1 ratio). So if you put 100 ohms on the primary side, the cartridge would "see" 100R in parallel with 47R (i.e., very low). For 10R, the net Z gets even lower. All of this interests me because I have a newly acquired Orto MC7500 direct-driving my phono section. By all accounts these Ortos like about 500 ohms. As regards bandwidth, does not the inductance of the SUT itself have much more effect than the inductance of the cartridge windings? Dunno.

Hi Lewen,
some "small" correction.
A SUT with 10 dB gain, 1:10 xfaxtor i.e. winding ratio, gives you 470 ohms on the cart side and NOT 47 ohm, since 47k / 10^2 = 470ohm -- agreed?

A 1:31.6 xfactor (30 dB gain) give you 47 ohm, since 47k / 31.6^2 = 47 000 / 998.56 = 47.067... ohm :-)
I hope we can agree on the maths, so I won't comment on your 100 ohms consideration etc.
As to the bandwidth:
The trannie(a good one) has about 200kHz bandwidth, so the cart is VERY well within that envelope AND thereby (to interested folks) truely to a degree of interest.
Since, if by some bad coincedence this bandwith goes well below 20kHz and you wonder about all the 'air' that's missing ---- you know where to look.
Greetings,
Axel

Sorry, you are correct about my math error. 470 not 47. Much better.

Your findings as to the sonic differences between loading the PW with NO trannie vs loading it either on the primary side or the secondary of a SUT are quite fascinating, and I don't know of anyone else who uses a SUT and loads the cartridge on the primary side. Tonight I have been listening to my MC7500, father of your PW, with no SUT and a 100R load. I don't hear any particular tonal imbalance, but the sound seems a tad closed in compared to my Koetsu Urushi (but tonearm and mat and tt are also different so not a good piece of data). Nevertheless, I am prepared to try 500R next, as this was advice I got from the previous owner of the MC7500. Can you once again describe the circuit you are happiest with? Reading two posts up from this one, I think you are using a 13R resistor on the primary side and a 47K resistor on the secondary side of your SUT, so that the PW sees 13 in parallel with 47 to give about 10R. Is that correct?

Hi Lewm,
That a trannie sounds better (ONLY MC!) is simply due to it using its "strength" - current delivery, rather then it's "weak point" - voltage (micro Volts).
It enables the cart to 'dig up' more detail / dynamics etc. More current through the cart's coil, not to forget, gives it superior damping, a mostly ignored fact these days.
However, there are A-PLENTY phono-pres that do it any way, since they use trannies as part of their design, EAR and Manley, come to mind immediately, it's just not too obvious.
EAR not only with their tube pres, but also with their SS-pre like the EAR 324. So it's not SO out of the ordinary after all.

Now to the 500ohm loading of your MC7500. Try it by all means, BUT if I estimate it right, the "mechanical damping" of the MCxxxx carts is NOT at all the same as Ortofon's newer cousins PW, Jubilee, Kontra-Punkt, Vienna, Venice, etc.

That simply means you will have to load higher (using a SMALLER R value!) simply to damp out the typical > 10-12kHz resonance frequency, please bear that in mind! 500ohm might get already too "loosy-goosy", 1k I can't imagine.
Now recall that a PW still works jolly fine with 47k (no loading at all, other than the phono-pre input impedance). The top Dynavectors do the same too, as their damping scheme(s) take care of this typical > 10-12kHz over-shoot / resonance.

So, the SUT secondary "sees" the 47k input-impedance of the phono-pre! If you put -another- 47k R (as you implied) and it will be || with the 47k pre input-impedance and so you'd have the SUT see: 47k*47k / (47k+47k) = 23.5k! on secondary.
This 23.5k, reflected back to primary, as we saw (taken a 1:31.6 xfactor (30 dB) SUT => 23.5k/31.6^2 = 23.5 ohm) is now in || with the primary load! AND is now WAY too low, (about 5ohm) if you'd have a 13ohm R for primary load!! So, NO secondary loading, the 47k input impedance of the phono-pre is your load.

Now, it MAY not end there... since (depending on the trannie's resonance behaviour) you might have to consider an RC (on the secondary) to damp any trannie resonance. It will be THEN when you want to know your cart's L! My original enquiry, to figure out the values for this RC.

My XF-1 does not have an issue (as it seems) but if you get some of the new fangled 'high-speed' core material trannies it could well be in need to damp out somewhere around 20kHz. It's a tricky lot of calculations and completely dependant on the trannie / cart combination, there are ball-park figures but I guess it would require some trial and error. This RC would act as a Notch-filter in case that makes it more clear.

The other SUT loading method: secondary loading, might take care of such resonance, I mentioned it earlier since e.g. 18k secondary load (for a 1:31.6 xfactor!) will dive you ~14 ohm the cart will see, AND it damps the trannie also. That is why primary loading (not damping the trannie) gives you more detail / clarity -- if you can make it work for your trannie / cart / phono-pre combination.
Cheers,
Axel

Axel, Now it is YOU who have misunderstood ME. You wrote, "So, the SUT secondary "sees" the 47k input-impedance of the phono-pre! If you put -another- 47k R (as you implied)". No, I did not mean to imply nor did I think that you have two 47K resistors in parallel on the secondary, I did mean to ask whether the sole parallel resistance on the secondary side of your SUT is the 47K-ohm input Z of the phono section. That's what I wrote. Your whole paragraph on the undesirable result of paralleling two 47K resistors is beside the point. I just wanted you to summarize what resistances you have and where they are in your circuit to give you the sound you like from your PW.

And thank you for the advice on SUT installation, but there is no way I am going to be buying and installing a SUT, when my phono pre has oodles of gain and sounds great. I am just interested in your findings.

Hi Lewm,
+++ I did mean to ask whether the sole parallel resistance on the secondary side of your SUT is the 47K-ohm input Z of the phono section.
+++
YES SIR, right as you say!

+++
...when my phono pre has oodles of gain and sounds great.
+++
Well, as with all things --- "sound great" that's relative, no? :-)
And "oodles of gain" got nothing to do with it. My Levinson phono-boards (through my 326S) has 78dB gain if I open it all up (the ML has adjustable input gain +6, +12, +18dB, and the phono-board does 40dB or 60 dB. Believe me, SUTs are only to some degree, and in some more 'needy of gain' situations about gain.
The point I'm making is, that a trannie will make it sound 'greater' still --- but that is a personal thing, like buying another speaker, component, or such. Some day you may recall this little conversation about SUTs, is all :-)
Take care,
Axel

I really think it's quite interesting and unexpected (to me) that with pre-loading of a SUT, the optimal impedance for the PW seems to be about 10 ohms net, whereas, as you say, without the SUT the PW seems to like 500 to 1000 ohms net. I am not close-minded about using a SUT; this is the first report that suggests there could be an advantage notwithstanding any need for extra gain.

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

Thanks very much, Dave. I bookmarked that thread for future reference. But in your experience with primary vs secondary loading does it often occur that the optimal net load for a given cartridge, when you load the primary of the SUT, is radically different from its optimal loading when there is NO SUT or when the load is in the SUT secondary? I can see how that could happen due to introduction of the SUT, thanks to the post you cited.

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

I learned something. To paraphrase Gertrude Stein, a SUT is not a SUT is not a SUT.

Hi all,
there is something that seems to 'fox' Lewm with the different loading values, and Intactaudio got some of it --but not all.
A cart working in 'voltage mode' MUST work into some 'highish' level of impedance, in order to work at all.
Most MCs need somewhere around 100 ohm loading but hardly ever less (with no SUT!).
Actually more ideal would be the 47k of the pre's input-impedance --- but there is this resonance issue with (still) most MC carts, starting at ~ 10kHz and producing just too much HF energy up to 16 - 20 kHz before roll-off.
To handle this the cart must be damped as we know.
This can be done either electrically or 'electro-mechanically' i.e. you are either loading it down electrically (pulling more current through that load resistor and thereby the cart-coil, which creates the damping effect) or, like more of late, built into the cart is some proprietary 'electro-mechanical' means of damping. Ortofon by example uses a mixed bag of 'tricks' to achieve that, I won't go into it here. Just check their Website and see what they do e.g. with the Windfeld.
Dynavectors has their own bag of 'tricks' for their top carts, etc.
Now, more current (I) also means even less voltage (V)(Ohms law) for the signal => pre-amp.
Try to load the cart with 10 ohm (no SUT) and there is a big drop in output added to the treble-roll off.
Newer MC carts are now often run into 47k (the pre's impedance only) and can sound great, it is the damping system in the cart that makes this possible.

The better the cart the less loading is required! (Because it is more 'well behaved', less HF resonance).

Now take the SUT situation (this where I say Intactaudio got it 'almost' right), there is such a thing as a PERFECT impedance-match. When this occurs NO 'reflections' are produced by the signal (check out why there are BNC connectors and the like). The result is maximum power transfer from cart to primary with no reflections. In order to get there, the input impedance of the trannie first needs to be matched to the cart coil's output impedance.
Now, it also is known by ROT that this cart-output-impedance is DCR*2.5 i.e. WP RDC 4ohm*2.5 = 10ohm and quoted as such (by Ortofon) as the minimum loading impedance, yes?

So now it might make more sense WHY 13ohm loading R || 47ohm the 'natural impedance' of the SUT = 10.18ohm is a good fit.
Forget the .18ohm mismatch since there is, as always, more to it than just that, but it is a good approximation and the rest is trial and error.

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.
This 'maxed out' cart-current produces the best possible damping of the cart --- BUT most of all, the current of the primary is converted into VOLTAGE at the secondary according to its ratio. So you find the best of both worlds, given a decent SUT of good bandwidth i.e. VERY low losses.

Maybe this helps to get some more better idea. Google Impedance matching if you want the 'low-down' and the details of the concept.
Best,
Axel

Axel, No one is arguing with your logic, and I am not foxed. I will leave it to Dave (Intactaudio) to comment on the second half of your post. Dave knows more about transformers than both of us. I did learn something from your story, so thanks.

Hi Lewm,
excellent all’s as clear as can be, jolly good show.
I, incidentally, didn't imply you where arguing my logic, hm. Sorry if it sounded that way.

One fact remains that I was 'foxed' at the start with SUTs, and there are a-plenty reviewers that seem to be utterly oblivious of the possibility of loading SUTs, never mind the issues surrounding primary or secondary loading... However, they merrily do review the product (SUT) and tell you how it sounds.

If I would have followed that route, the XF-1 I'm now using would be completely and utterly unacceptable to be used with a 0.3mV cart i.e. unloaded using only the natural impedance of 47ohm (what the cart sees) sounded far to phasey and rolled of, completely unacceptable.

Yet, if you get 'un-foxed' there is more than expected, as so often, and that SUT sounds just great – if you know what to do.

I'm glad to share some of that little I learned, and in that spirit - thank you for sharing.
Axel

Axel, I am often vexed by the same phenomenon-reviewers who comment on the "sound" of a SUT without regard to the myriad of other parameters that affect the result. Particularly egregious is the "SUT shoot-out" type of review, where many different SUTs are compared using the same cartridge and the exact same loads, often without regard to the differences in turns ratio, let alone other less important factors. The result is worthless information, but it fills pages in a magazine.

Dear Lew: That specific reviewer IMHO is a " shame " for the audio environment.

Unfortunately he ( like other " pro " reviewers. ) has the " power " of a free-pen with out anyone to " stop ", sending/making wrong/corrupted audio information given to the audio customers: mis-information, confusion, false myths, non know-how advise and non sense reviews. Far from help this kind of people IMHO do a lot of harm to the whole audio industry.

IMHO this is one part/reason why exist so much mediocrity in the high-end audio industry and these kind of reviewers are important part of that mediocrity.

Regards and enjoy the music.
Raul.

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

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

OK Dave,
if you would have ever worked with an SUT you'd give me less of a hard time.

I might misread what you try to tell me -- but it sounds like: bumble bees can't fly i.e. must be due also to the: "first law of thermodynamics"...

Please forgive my lack of stamina with all this, at least nobody can say we haven't tried. I will also not ask for a job as physics teacher either, now that I have so miserably failed with such a basic matter as a step-up transformer.

I feel like Galileo must have during inquisition trial: and say it still works...

I'm out of thermodynamics right now.
Greetings,
Axel
PS: your missing thermodynamics come from the rotation of the platter, that imparts energy into the cart, that imparts energy into the coil, that imparts.... don't go ask me now about why the platter is turning, please.

Thanks for your input, Dave.

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

dave

Dear Axel and Dave: This dialogue was/is a learning one for me and I'm sure that for other people too, thank you to share your knowledge about.

What I confirm through it is that the SUT option in a non perfect world ( like the audio one. ) has several trade-offs and by my experience with commercial SUT's ( many too many to name it: from Expressive Technologies to the lowest AT model. ) no one till today can beat a good active gain design at least not the one I'm using ( but I heard other today active gain stages that are really good too. ).

Axel, I don't have any doubt that for what you already heard through SUTs your bias are on them against an active Phonlinepreamp but I wonder if you already experienced the " right " active gain devices.

Nothing is perfect ( SUT or active gain devices. ) but my " bias " ( till today ) is on active gain Phonolinepreamps, I like/prefer its trade-offs over the SUT ones. Maybe is time to make a SUT design journey, I can't say if it is worth to do it.

Anyway again thank's to both of you.

Regards and enjoy the music.
Raul.

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).

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

Dear Axel, Intactaudio is a professional in the use of transformers, autoformers, chokes, etc, in audio. He is highly respected on other internet forums for his knowledge and unbiased advice. Rather than leaving in a huff, you might be better served by considering that what he has written in response to your posts is most likely correct. Then you can learn something new, which is what we are all here for. No one, including Dave, is saying that your particular choice of a topology for your cartridge and your SUT does not sound excellent.

Hi Dave,
I just got that idea to check on your threads.
Now look here: "Thoughts on cartridge loading with a SUT" 11-22-08 !!

You appear to be quoting a couple of other originators quite liberally in your essay, hm.
OK, you are not selling a book here, but I try to figure out why you appear to give a hard time every step of the way, yet have all these superior insights as to be seen in your essay?

What points are your actually trying to score?!

If you are using an SUT already, why all this production of yours around here?!

And if you don't use one -- and you want to, why not ask some salient questions, instead of arguing over every weird detail?! Or is that just your idea of eduction?

You mention Van der Veen in your 'essay', - we use his trannie! While you talk about his 'Gorillas in the corner' using all his comic strip allusions... That's geeky, man!

Are you just trying to be extra clever and get a kick to make the rest look like fools?

I truely start having my doubts about your sincerity, and I wish I am wrong.
I like to help, and you seem to enjoy some geeky game instead.
Maybe - what we have here, is failure to communicate...

Axel

Lew, AFAIK Dave Slagle of Intact Audio is a transformer manufacturer. He isn't merely a user or repackager of transformers made by other companies, but designs and winds his own transformers, including AFAIR, MC step-up transformers.

regards

Dear Axel: I think there is a misunderstood on communication. A long thread with so much information can help to understanding " troubles " like this one you " feel ".

IMHO I can't think that Lew has its own " agenda " looking to achieve some target different of what the thread is.
I know Lew and I can say " free " that is a trusty person in a whole mean.

I had and still have ( now less ) several troubles with the people brcause of the way I'm ( I mean so direct and not " politic ". ) and because problems to understand and to communicate due my not so good English.

This kind of things happen everyday in a forum like this but I'm sure that almost all ( including Lew ) the people that post here are " good meaning " and honest with them.

Regards and enjoy the music.
Raul.

Jonathan, You are correct. I just did not want to go into great detail on Dave's qualifications.

Blow my mind!
1) good to know, we have only sincere value seekers going on in thread. I'm genuinely relieved.

2) I must be getting confused between the different participants? If that is so, my sincere apology.

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.

4) To boil all this down, there is actually ONE, and only ONE simple disagreement here, that I can see.

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

- 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.

- Everything else that’s been said and argued here, always came back to this difference in our understanding. He maintains a cart has a ‘fixed’ Voltage AND a fixed Current.

- So who’s got it wrong?

- Can we maybe agree that a GENERATOR, when demanded to operate into a lower impedance CAN deliver more current? If NOT, the voltage would drop and the light go out? If that is not the case then I am truly misinformed.

- Next, is the question what happens if we ‘down-load’ a cartridge (no SUT involved)? We lower the phono-pre’s input impedance from 47k to what ever our parallel resistor we picked will produce i.e. Rpre * Rload / (Rpre+Rload); just to make sure we are still on the same page, I know it’s what (most) everyone knows or at should know.

- Now we get back to the current issue. What will happen in that cart coil?! Will the current drop? Stay constant? Or rise? Remember we are talking about DAMPING the cart. So how does that damping work? How do we damp an e.g. woofer coil? By making MORE current available due to a higher ‘damping factor’ of an amp i.e. nothing else than simply having a MUCH lower output impedance for that amp, so it is less ‘resistive’ to the flow of current. Can we agree on that! If not, --- well let’s see.

- So damping happens, if MORE, and not LESS current runs through a coil – it kind of ‘clamps’ it – as is often said, so as not to perform to its own ‘inclinations’.

- Back to the cart. We ALSO have seen a impedance lower than 47k is used for damping a cart (some MC carts are fine with 47k since they provide their own version of damping) So we ‘clamp’ it down and thereby not let it have it’s own behaviour and go wild from 10 – 12kHz on until it has to finally roll-off.

- So now back to my point with Dave. Given the cart is down-loaded, can it produce more current-output or not? If it can not --- how then will it be damped?

Let’s just leave there and see where this takes us.

Greetings,
Axel

A very interesting thread (and thus my bookmark)...

:-) T_bone,
heck, there is someone that seems to enjoy this, hm - kicking cans...
Makes it all worth it now, doesn’t it?
Thank you for sharing,
Axel

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

Hi Dave,
this is getting us somewhere, I think.
I have no issue with what you say, so let's put this down to a number of 'crossed wires' on my side -- let peace be with us, it that is fine by you.

I stated somewhere, that you get 20 times the output voltage for FREE! (in that given example 30 dB SUT), and I think it was you, and some other contributor that took issue with this.

Well, you do get it for FREE. Since there is NO additional power input required such as in a gain-stage, to get just that. Never mind ohms law right now... You do get the additional ~ 30 dB gain for free! And that without added tube or SS derived noise. Seems I am selling your own stuff to you, it’s my pleasure (-:
Now please tell me where this statement is factually incorrect, Dave.

Jolly good, now how about the cart damping by increased current through the cart coil?
Tell me it this is correct? If not, how is the cart being damped then?!

Next, let us hear what is the difference in current when the cart goes into 47k impedance, and when it goes into an SUT with per example 10 ohm impedance?
More current? Well lots more, unless we want to argue about THAT some more.
If we do agree that it is more --- does it make sense to use the term "current mode"?

Since we are still with ohms law. If a cart works into a 47k impedance and then into a 10ohm impedance, in which situation do we expect the higher Voltage?
Will it be the same? Not according to the "First Law of Thermodynamics" as you quoted, since more current = less voltage, so the let's agree we have the higher Voltage at 47K impedance (i.e. into a phono-pre straight, no SUT) does the cart operate at a higher Voltage?
You said so, thereby what we have, "Voltage mode".

Maybe, just maybe, we can get it together...

Thank you for still being with us.
Greetings,
Axel

Axel,

I originally misinterpreted what you were saying about "free" voltage multiplication.

What is interesting about your experience with primary loading of the PW cartridge is that your ideal loading (the reflected load seen by the cartridge) is radically different when using that SUT than the recommended loading when using only an active phonostage or secondary loading.

I was wondering if you have any speculation as to why this is so. I know that some experts maintain that there is a specific "critical" load that effectively dampens the high frequency peak without excessively rolling off the high frequencies. I was wondering why it is not the same load (as presented to the cartridge) regardless of the type of downstream amplifying device.

I am asking this because my phonostage (Viva Fono) utilizes a SUT as the first stage voltage gain device (feeding a tube active stage). I don't even know the gain of the device, but, it is loaded at the primary with a 460 ohm resistor. This resistor is soldered across the back of the RCA input jacks. Because of the difficulty in accessing and changing the resistor value (tight space, a lot of nearby wires), I've limited my experiments to connecting and disconnecting one leg of the resistor. I am now wondering if I should try a MUCH lower value resistor.