MC Step Up Math


Hi all,

after posting a thread on here years ago and becoming exceedingly confused about cartridge step up maths, I gave up, embarrassing for a math major..perhaps I should have studied electrical engineering. Recently I have been reading up on this topic and would like to once and for all figure out how to run the math/electronic theory to find the correct step up to mate with a MC cartridge.

I have looked at 2 different links.

Link (1)

http://www.theanalogdept.com/sut.htm

and

Link (2)
http://www.rothwellaudioproducts.co.uk/html/mc_step-up_transformers_explai.html

Now, everything I read in link 2 falls apart after reading what is on link 1 and I am once again confused about what to look for in a MC step up.

In the second link the author explains that you simply apply a 2 step process: A. multiply the turns ratio by the cartridge output to find the voltage and make sure that it is not overloading the MM phono stage input (i.e/ between 2.5 and 10 MV) and then B. Perform the calculation to show you how much resistance the cartridge actually sees and apply a rule of thumb at least 3 to 10 times ratio between the source impedance and the input. The rule is for the most part out of thin air, though he does explain that matching to equate the 2 is a bad idea.

In the first link however, the author takes a different approach. He explains that a turns ratio cannot just be multiplied to give you the voltage on the other end. For example the cinemag 3440 cart used with the dynavector illustrates the point. The output is .30 MV and the turns ratio is 35.4 resulting in 10.6 MV out.

Now here is the bit I need help with. He explains that in reality the with this combination the output is really 5.1387mV NOT 10.6MV. He uses this equation to adjust the 10.6 MV to 5.1387MV:

(Vout / Vcart) = (R(Load_effective) / (R(Load_effective) + (Rcart)))

he finds Vout and then Multiplies by the turns ratio.

The parameters are as follows:

Rcart: is internal resistance of the MC cartridge
R(Load_effective): resistive load seen at the MC cartridge
Vout: Voltage output at secondary side of tranny
Vcart: Voltage output at MC cartridge

Hi all,

after posting a thread on here years ago and becoming exceedingly confused about cartridge step up maths, I gave up, embarrassing for a math major..perhaps I should have studied electrical engineering. Recently I have been reading up on this topic and would like to once and for all figure out how to run the math/electronic theory to find the correct step up to mate with a MC cartridge.

I have looked at 2 different links.

Link (1)

http://www.theanalogdept.com/sut.htm

and

Link (2)
http://www.rothwellaudioproducts.co.uk/html/mc_step-up_transformers_explai.html

Now, everything I read in link 2 falls apart after reading what is on link 1 and I am once again confused about what to look for in a MC step up.

In the second link the author explains that you simply apply a 2 step process: A. multiply the turns ratio by the cartridge output to find the voltage and make sure that it is not overloading the MM phono stage input (i.e/ between 2.5 and 10 MV) and then B. Perform the calculation to show you how much resistance the cartridge actually sees and apply a rule of thumb at least 3 to 10 times ratio between the source impedance and the input. The rule is for the most part out of thin air, though he does explain that matching to equate the 2 is a bad idea.

In the first link however, the author takes a different approach. He explains that a turns ratio cannot just be multiplied to give you the voltage on the other end. For example the cinemag 3440 cart used with the dynavector illustrates the point. The output is .30 MV and the turns ratio is 35.4 resulting in 10.6 MV out.

Now here is the bit I need help with. He explains that in reality the with this combination the output is really 5.1387mV NOT 10.6MV. He uses this equation to adjust the 10.6 MV to 5.1387MV:

Equation (*)
(Vout / Vcart) = (R(Load_effective) / (R(Load_effective) + (Rcart)))

he finds Vout and then Multiplies by the turns ratio.

The parameters are as follows:
Turns ratio: The turns ratio of the step up device
Rcart: is internal resistance of the MC cartridge
R(Load_effective): resistive load seen at the MC cartridge defined as 47,000/(Turns Ratio)^2
Vout: Voltage output at secondary side of tranny
Vcart: Voltage output at MC cartridge

for this example they using a denon 103 + cinemag 3440 are:
Turns Ratio: 35.4
Rcart: 40
R(Load_effective): 47,000/(35.4^2) = 37.5 ohms
Vout: to be solved for
Vcart: .30 MV

Putting it into equation (*) and solving yields
.1452mV for Vout.

He then takes Vout and multiplies by the turns ratio.

.1452 * 35.4 = 5.1387mV

NOW: If you take the simple method (from link 2 by multiplying turns with output) you get 10.6 MV, using this adjusted method with equation (*) you get 5.1387 MV. So my question is this. What is equation (*), is there some theory here that I am missing, is this voodoo? I would like a reliable way to select components that match, though I have trouble trusting the equation (*) method without knowing where why he is using it and what it is. I certainly want to get this ironed out before I start buying different transformers to play with, and any help with this would be greatly appreciated. Thanks.
dfel
The Analog Dept link describes the process I use. Here is another link:

http://www.bobsdevices.com/What-about-Impedance.html
I haven't taken the time to read the references you provided, but I believe I can shed some light on the discrepancy you cited. First,
Vout: Voltage output at secondary side of tranny
In the given context "secondary" should be "primary." Vout is being used to refer to the voltage appearing across the primary side of the transformer when the cartridge is being loaded via the transformer. As you indicated, that voltage is then multiplied by the turns ratio to derive the voltage appearing at the input of the phono stage.

Second, equation(*) is apparently based on the assumption that the output voltage of the cartridge is specified under conditions of negligible load, such as 47K. The equation then adjusts that spec to reflect the voltage that the cartridge would provide under the much heavier load conditions it sees when connected to 47K via the transformer.

I don't know whether output voltage specs for LOMC cartridges are typically based on load conditions that are essentially negligible (e.g., 47K), or under load conditions that are recommended for the particular cartridge, or on some other load condition. In general, though, it shouldn't make much difference, because in general the optimal load will be considerably higher than the cartridge's specified impedance.

In the given example the cartridge is being loaded at a value that is lower than its own output impedance. My understanding is that that is way too heavy a load to be optimal in most and perhaps nearly all cases. Therefore the step-up ratio of 35.4 is much too high. Reducing it to say 20 (26 db) would result in the cartridge seeing a load impedance of 117.5 ohms. That in turn would result in the cartridge's 0.3 mv specified output being increased to 6 mv if the Vout/Vcart correction is not taken into account, and 4.5 mv if the correction is applied (and if the 0.3 mv spec is based on conditions of negligible load). Which is not much of a difference either way. And I wouldn't be surprised if optimal loading for the 103 would often be found to be considerably higher than 117.5 ohms, which would narrow the gap between the two numbers even further (albeit at the expense of making phono stage noise performance more critical, due to the lower signal amplitude received by the phono stage as a result of the reduced turns ratio).

So as you can see, the application of equation(*) can be expected to make a significant difference only if the cartridge is being loaded excessively.

Regards,
-- Al
"So as you can see, the application of equation(*) will tend to make a significant difference only if the cartridge is being loaded excessively."

Please tell me if I am understanding what you mean by this?

equation (*) increases as Load Effective increases, this can be verified by the look test or the first derivative of (*) with respect to the Effective Load variable. Equation (*) essentially is a penalty function of sorts. When the Effective Load is low it gives a number below one, as the Effective Load increases equation star gets closer to 1. This is significant since it is multiplied by the cart output voltage and the turns ratio to give the "real" voltage seen by the MM phono stage input. As claimed by the author.

Does this not mean that as the load becomes bigger, and quation (*) approaches 1 that the using it becomes less useful since when it is one you are just multiplying the turns ratio * cartridge output * 1 ?

It seems like it may matter more when the load effective is LOW, because then the calculation changes. Turns ratio*Cartridge output * ( a number less than one). THis will decrease the final value. Is that called heavily loaded? Am I understanding you correctly? Or did I make a mistake.



Essentially what I am saying in a nutshell is this : Equation star appraches 1 as effective load increases. When it is 1 you can simply multiply the turns ratio * the cartridge ouput.

However equation (*) becomes less than 1 but greater than 0 as effective load becomes small. if equation star is .5 then the voltage is cut in half! This makes a big difference.
I get the math, it is a simple equation, However I still do not understand why or where it comes from.

1. What is effective Load. He is taking 47,000/(turnsratio^2). What does that tell you, what is that calculation?

2. What is equation (*)? Where does this come from?
As the load impedance seen by the cartridge becomes higher in relation to the cartridge's internal impedance, the difference between the two calculations (i.e., the calculations with and without application of equation*) becomes progressively smaller and less significant. That can be seen by running some calculations for various load impedances and turns ratios, such as the 117.5 ohm/20x example I provided.

The difference between the 10.6 mv and 5.1387 mv numbers is only as large as it is (more than a factor of 2) because the 37.5 ohm loading is undoubtedly much too low to be optimal for the particular cartridge, given the cartridge's 40 ohm specified impedance.

Regards,
-- Al

What is effective Load. He is taking 47,000/(turnsratio^2). What does that tell you, what is that calculation?
A transformer transforms impedance in proportion to the square of the turns ratio. So the load impedance seen by the cartridge corresponds to the input impedance of the phono stage (usually 47K) divided by the square of the turns ratio.
What is equation (*)? Where does this come from?
See this Wikipedia writeup on the voltage divider effect. In the first figure, consider Z1 to be the cartridge's specified internal impedance, and Z2 to be the load impedance seen by the cartridge. Consider Vin to be the voltage the cartridge would output under conditions of negligible load (e.g. 47K).

Regards,
-- Al
"As the load impedance seen by the cartridge becomes higher in relation to the cartridge's internal impedance, the difference between the two calculations (i.e., the calculations with and without application of equation*) becomes progressively smaller and less significant. That can be seen by running some calculations for various load impedances and turns ratios, such as the 117.5 ohm/20x example I provided.

The difference between the 10.6 mv and 5.1387 mv numbers is only as large as it is (more than a factor of 2) because the 37.5 ohm loading is undoubtedly much too low to be optimal for the particular cartridge, given the cartridge's 40 ohm specified impedance."

Yes, however the cartridge impedance is fixed and I was hoping to vary the step up in the analysis. In the end it makes no difference effecitve/(effective+cart)->40/(40+40) will always be 0.5, clearly the voltage drops by half when the impedance are equal. But you can see that when the effective load goes up the ratio also goes up.

look at what happens when the effective load goes to 100..100/(100+40) = 0.71....

or if it goes to 470...470/(470+40) = 0.92

or look at what happens when effective load falls to 5...5/(5+40)...=0.1111

Increasing the cartridge impedance is the flip side of the equation. Holding the effective load constant, as the cartridge impedance increases the equation falls from 1 towards zero reducing the output voltage.

All this to say, We are looking at the same thing, I just wanted to understand what exactly you meant by heavily loaded. Effective load close to or less than cart impedance is what I think you mean is a problem...or by heavily loaded.
Effective load close to or less than cart impedance is what I think you mean is a problem...or by heavily loaded.
Yes. And I'm also saying that having a load impedance that is heavy enough (numerically low enough) to approach the impedance of the cartridge will almost invariably not be optimal sonically. And if a much lighter (numerically much higher) load impedance is applied, that is likely to be more optimal sonically, whether or not equation* is applied becomes insignificant.
... however the cartridge impedance is fixed and I was hoping to vary the step up in the analysis
Everything I said is consistent with that. Keep in mind that as the turns ratio of the step up transformer is varied two things change: The load impedance seen by the cartridge (which is equal to the input impedance of the phono stage divided by the square of the turns ratio), and the input voltage seen by the phono stage (which is equal to the voltage applied to the primary of the transformer, referred to as "Vout" in equation*, multiplied by the turns ratio).

Regards,
-- Al
" What is effective Load. He is taking 47,000/(turnsratio^2). What does that tell you, what is that calculation?

A transformer transforms impedance in proportion to the square of the turns ratio. So the load impedance seen by the cartridge corresponds to the input impedance of the phono stage (usually 47K) divided by the square of the turns ratio.

What is equation (*)? Where does this come from?

See this Wikipedia writeup on the voltage divider effect. In the first figure, consider Z1 to be the cartridge's specified internal impedance, and Z2 to be the load impedance seen by the cartridge. Consider Vin to be the voltage the cartridge would output under conditions of negligible load (e.g. 47K). "

Wow! Thanks that is exactly what I was looking for. I will take detailed look at it so I understand where the way of calculating is coming from.

Assuming it holds, which at a glance of these electricity theories it does, the problem is still not solved.

HOW DOES ONE SELECT AN OPTIMAL STEP UP FOR THEIR CARTRIDGE.

That is really the point of this thread, and I hope it will be ironed out so it can be useful to others confused on this issue.

From what was mentioned so far:

1. It seems like equation (*) on the link can be used, and should be used, over the typical multiply the turns ratio to find voltage thing most guys are doing...especially when there is distance between the cartridge imped and the efective imped.

2.Matching cart to transformer impedance is just plain wrong is most cases. it steps down 50% basically and can crush the cartridge's character.

So what else is important

1. Every phono section is different, so using equation (*) to find the adjusted voltage will give you a MV output for any given turns ratio, but what is good for your phono stage 2.5MV,10MV or anything between. Running the calculations for a sample cart, I found that if I get between 2.5 MV and 5 MV using equation (*) it shows me that just about anything between 20-40 as a turns ratio is can be used. That is a HUGE range of step ups. There has got to be a better method than trial and error to get the right step up. I mean not everyone has thousands of dollars to buy 10 of them in those ratios to try them. The other thing is that simply spending more on a step may not put you closer to the ratio you actually need...I side with matching correctly over just spending 5k on a step up and hoping it works.

2. The effect of choosing an effective load <= cart is a no no, but what happens when it is chosen too high ? The "experts" recommend 8 to 10 times as a rule of thumb, what happens if you use 3 or 50? I do not know if there is a predictable result for chosing them too wide, or if it has any effect at all. When do the transformers being to ring?

3. The MV output from your cart is not stable, nor is it always at the specified rating. .30 MV...yeah at 1000HZ, and even that number may come out exactly as the manufacturer measured it.

so in the end...using this information really only puts you in the ballpark for what you need. Hopefully someone out there knows of a way to narrow this ballpark further.

Thanks again AL for being patient and contributing to this thread!
Yes AL, I am know very little about electrical theory and the Higher (numerically lower) and Lower (numerically higher) is what confused me, you were consistent not mistake on your part it is the semantic I was confused with.
I mapped out a curve of the various step up turns ratios for a denon 103, for those that are into seeing the visual of what we are discussing here. I dont have one of those carts, but tons of people do, so I figure why not use it as an example. have a look, enjoy. I posted the image to the following link.

http://s28.postimg.org/9kno6o95p/Denon_Chart.png
What low output moving coil cartridge do you own or want to own to use in your system? Perhaps the easiest approach would be to make that information known and then just get advice on what step-up ratio will work best for that cartridge. It is helpful/necessary also to know the internal resistance (= output impedance) of the cartridge itself.

What Al is telling you is that the ratio of the signal voltage seen at the input to the phono stage to the signal voltage emitted by the cartridge will be equal to the turns ratio of the SUT. However, the ratio of the output impedance of the cartridge (= its internal resistance) to that of the input impedance "seen" by the cartridge will be equal to the SQUARE of the turns ratio. (I realize you've got most of this already; sorry to sound pedantic.) So, say your MC has an output of 0.5mv and an internal resistance of 10 ohms. Say your SUT has a 1:10 turns ratio and that your MM phono stage presents a 47K ohm impedance. The phono stage will receive ~5mV of signal voltage (10 times 0.5), if the impedance matching permits, and the phono cartridge will see an input impedance of 470 ohms (47K divided by 100 or 10-squared). That is a typical OK match-up. You want the input impedance to be a multiple of the cartridge impedance. Classically, a 10:1 of impedances is ideal, but it is not a problem to go down to 5:1 and even lower. However, in this case we have 10 ohms looking at 470 ohms. We are way into the safety zone with a ratio of 47:1. What Al and one of your references was saying is that if the input impedance gets down close to the same value as the output impedance of the cartridge, then the total 5mV of signal is no longer deliverable to the phono input stage; some of it goes to ground and is wasted.

To add to the mix, some might decide that the cartridge sounds a little tizzy with this 47:1 ratio of input to output impedance, so to dampen the high end response of the cartridge they find ways to reduce the impedance ratio, such as to introduce a resistor that is seen by the SUT to be in parallel with the 47K ohms afforded by the naked phono stage input. In the end, you might decide that this cartridge sounds best when it sees a 100 ohm load. In that case, you could add a resistor in the range of 11K or 12K ohms, in parallel with the 47K to produce a net resistance of 10K ohms on the secondary side of the SUT. Now the cartridge will "see" 100 ohms across the SUT/ (10K divided by the square of the turns ratio).
Sorry. Once or twice where I referred to the ratio of "input to output impedance" above, I should more clearly have written "output to input". That is the internal resistance of the cartridge itself (~output impedance) vs the input impedance of the phono stage, as seen by the cartridge across the SUT.
Re Lew's point about the possibility of adding a resistor if necessary, I'll add that the parallel combination of two resistances corresponds to an overall resistance equal to the product (multiplication) of the two values divided by their sum. So if 11K is placed at the output of the SUT or the input of the phono stage, and the input impedance of the phono stage is 47K, the combined impedance on the secondary side of the transformer would be (47 x 11)/(47 + 11) = 8.9K. The cartridge would see a load impedance equal to 8.9K divided by the square of the turns ratio.

Also, some people report good results applying a load resistor on the primary side of the transformer. As you'll realize, in that case the value of the resistor will typically be a good deal lower, and the load resistance seen by the cartridge will be the parallel combination of the input impedance of the phono stage divided by the square of the turns ratio and the value of that resistor.

Given those possibilities of adding resistance in parallel, it seems to me that in choosing a stepup ratio it would make sense to err in the direction of having too low a ratio/too high a numerical load impedance/too “light” a load, PROVIDED that there is confidence that the noise performance of the phono stage is good enough to support the correspondingly reduced input voltage without introducing objectionable levels of hiss.

Re the chart you prepared, nicely done! To be sure it’s clear to everyone, the numbers on the vertical axis at the left represent millivolts at the secondary side of the transformer, i.e., the input to the phono stage. Also, a useful enhancement to the chart, if readily practicable, would be to superimpose on it a second curve indicating the load impedance seen by the cartridge, as a function of turns ratio, with the load value scale indicated vertically on the right.

Regards,
-- Al
Here is the thing. It leads back to the main dilemma.

Let us stick to the example for to see what I mean. The Denon 103 with .3MV, and 40 Internal impedance. I used the voltage equation to calculate what happens at all the turns ratios.

You can see it go up, until it hits its max and then it begins to decay. The max is reached at 5.142 MVs, with a turns ratio of about 35. NOW, the effective load with this turns ratio is about 40. This is not a shock, since if you maximize the voltage function by taking its first derivative, with respect to turns, and setting it equal to zero, you find it is maximized when THE INTERNAL IMPEDANCE = THE EFFECTIVE LOAD.

Now here is the question: Again, the rule of thumb has been taking over the thread, the internal impedance must be smaller by a factor of at least 10

...i.e/ the effective load should be 400 to go with the internal impedance of the cart which is 40.

Why is this necessary if this cartridge is putting out 5.142 MV MAX with any turns ratio, never over loading the phono section. the input to ouput impedance ratio here is one to one, not a multiple of 8 or 10 why would this be a problem?

If just applying the voltage equation, it would lead one to believe that any turns ratio between 10 and 80 would work, in this example. What am I missing here? With those turns ratios I get 2.5-5.0 MV into my phono section, and the effective load swings from 470 ohms to 7.3 ohms. That range of effective loads is above, below and equal to the internal impedance of the cartridge, which makes me think that either the equation(*) is junk, or the rule of the thumb (10 times the input impedance vs the output) is junk.

The question really is, which one is junk? Which one can you trust?

I am really curious to hear more about this, so this sillyness can finally make sense to me and others trying to get a good match without having to buy 10 of these things. Thanks again.

For that last post I am referring to the chart HERE:

http://s28.postimg.org/9kno6o95p/Denon_Chart.png
To answer your questions just above, there is substantial anecdotal/empirical evidence that if load impedance is numerically low enough to approach the value of a LOMC's specified impedance, dynamics, resolution, and transient impact will usually (and perhaps almost always) be adversely affected.

One reason for that may be that the 40 ohm or other specified internal impedance is not constant as a function of frequency, and will rise somewhat at high frequencies, due to its inductive component (coils have inductance, and the impedance of an inductor is proportional to frequency). That variation of cartridge impedance as a function of frequency will be inconsequential in relation to a relatively high load impedance. But if the load impedance is numerically low enough to be comparable to the cartridge impedance uneven frequency response will result, due to the resulting voltage divider effect being different at different frequencies. For the same reason, high frequency extension (i.e., bandwidth) may be excessively limited. Phase response issues may also come into play. Also, differences in loading will affect the behavior of the transformer itself.

Simply put, there are more factors that are involved than just assuring that proper voltage levels are sent into the phono stage.

Regards,
-- Al
In addition to what Al has said above. Once you understand that there are two different things at play here that should have no fixed relationship to each other, it becomes a little easier to understand.

One is the load the cartridge wants to see. This is not a fixed value but something that needs to be selected by ear. There are all kinds of ROT's but ultimately your ears are the only thing to tell you what this value should be.

The other thing at play here is the voltage gain. This is also not a fixed number. Again ROT's abound on this but in order to make a proper choice you need to not only take into consideration the input headroom of your phono but the overall gain structure of your system and your listening habits.

My approach is to get the gain set up properly first then address the load. In the case where Rsource=Rload you will lose 6dB but that is rare.

The mistake I see most making is attempting to link the numbers together to find some sort of convoluted solution. Sure one has the effect on the other but that is only based on other assumptions like the 47K input resistance of a phono. Lets face it, that 47K has no real meaning for a MC cartridge. For MC it should be much higher in value anyway to avoid the situation where the needed turns ratio loads down the cartridge too much.

dave
Thanks again AL, great post.

I would love to know why? What is the electrical theory behind this? There got to be someone out there who came up with the 10X rule, I am certain that it has some kind of physical/mathematical explanation.

And if it does, it might shed some light on how match cartridges with step ups properly.

As I understand it, the vout equation and this problem in general is different than matching i/o imped between preamps and amps due to the fact that it is a passive transformer, however similar rules apply there as well. For that equipment it is the 25X rule of thumb. There has got to be a reason for these assumptions.
To extend what Al has said in the last post.

With the dynamic nature of the impedance characteristic, would it not make sense to pick a "flattest" segment of the curve to operate on. The X Axis turns ratio variable can be converted to effective load, then picking a impedance load such that its relative operational neighborhood is sufficiently "flat" would make sense. However, why that would be @ 10X is beyond me.

I still dont get why the author of that page advocates equation (*), clearly it is in direct conflict with 10X rule. In the example, you can clearly see that the voltage is fine with any turns ratio 10-80, however the i/o impedance ratio is negative, even, and greater than 10 times in that same range.

I will formulate the chart later so that this will be more clear to others reading this thread later...for now I will just explain:

Turn Ratios between 10-80 give 2.5-5 MV which seems perfect.
BUT the i/o impedance ratios in the same range go from 0.20 to 22.0 ! A range that has extremes FAR AWAY from 10X, and the equation says they are all fine. (booooooo)

With the dynamic nature of the impedance characteristic, would it not make sense to pick a "flattest" segment of the curve to operate on. The X Axis turns ratio variable can be converted to effective load, then picking a impedance load such that its relative operational neighborhood is sufficiently "flat" would make sense.
I see no reason that choosing a point on the flattest part of the curve would have any relevance, since the turns ratio, once chosen, remains constant. Choosing a point on the flattest part of the curve would improve the accuracy with which the input voltage to the phono stage can be predicted, but I don't see that improvement as being necessary or beneficial.
Turn Ratios between 10-80 give 2.5-5 MV which seems perfect. BUT the i/o impedance ratios in the same range go from 0.20 to 22.0 ! A range that has extremes FAR AWAY from 10X, and the equation says they are all fine.
The equation says that they are all fine in terms of the voltage level that is sent into the phono stage. But it says nothing about how sonics may be affected as a function of the turns ratio, and hence loading, within that range. As indicated in my previous post, the turns ratio will affect sonics in ways other than by its effect on voltage.

Regards,
-- Al
Al, what is the reason for this 10X rule ? I mean at the core of it, do you just take it on blind faith. The ideal match really hinges on this figure, is there any science behind it or is it just an intuitive thing ?

I am not trying to be rude, please do not misinterpret, I just don't get how this is used so commonly with no science or reason blindly...and MANY do this, just about every source I have read from so far.

If Equation (*) is correct, and needs to be ignored and tossed out in favor of something else like this 10X rule, should there not be an explanation for the 10X rule using some scientific/physics/math/electronic property and theory?
As for the turns ratio flatness...

Is the issue you are talking about the "corner frequency"?
You explained that there may be issues from the change in impedance all three variables are related. I can put the voltage in terms of Impedance, effective impedance, or cart impedance. Either way it will trace out a transformation of that same curve. You are chosing a point yes, however as the cart impmedance changes you now have a neighbourhood. suppose it is +-10%, from 40 ohms..you are operating on 36-44, and it has now become a variable. looking at a the tranformed curve would it not make sense to seek the nbhd that has the flattest voltage change given the change in impedance? Because I was thinking that this can be done by changing the location you build your neighborhood around...this can be done by changing the turns ratio since the nbhood is fixed for the cartridge.

Like I have said many times, I only know enough to be dangerous, so please correct me if I am wrong.

Is the amount of inductance relevant in this setting ?

In terms of phase and frequency, well those variables are not part of equation (*), and I suppose it is the bit I am missing so that the magic number 10X makes sense in my mind.
Al, if you don't mind my asking..where the heck did you learn all this stuff? Do you have a background in electrical engineering? I have struggled to try and really understand the science behind this hobby and the learning curve can be steep at times...depending on what you are reading.
Dfel, I'm not sure that I can add much in answer to your recent questions that hasn't already been said. If you haven't already seen it, I would highly recommend that you take a look at the writeup by Bobsdevices for which he provided a link in his post earlier in this thread. He is a well regarded manufacturer of SUTs. And I note, btw, that the models he offers provide ratios that are switchable between two values.
Al, if you don't mind my asking..where the heck did you learn all this stuff? Do you have a background in electrical engineering?
Yes, I am a retired EE, with extensive experience in analog and digital circuit design (not for audio). I've also been an audiophile for about 35 years, and I've read widely on the subject.

Regards,
-- Al
The only thing to say really is where does 10X come from ?

The answer however from reading many sources, including bobs is: everyone else is doing it. No real science there, just audiophile blind faith I guess. I was hoping to get some insight into why.
I personally use 10x as a starting point. It is at the point where the cartridge sounds unconstrained by loading to my ears. I then adjust it lower until it sounds constrained than back it off a bit. With a SUT, use a formula that allows an output of between 2.5 and 10 mV that gives a reflected impedance of between 5 at 10 times the internal impedance to start. It is not an exact science and I adjust it to whatever sounds best. Some cartridges sound better loaded lower and some loaded higher. I wish there was a way to scientfically match it perfectly. From experience with several common cartridges, I know which ones sound best with different step up ratios. But there are so many cartridges out there, that without trying different ratios, you can't tell what is best.
Golly, Dfel! You are making this much much harder than it really is, and it's somewhat difficult to begin with. "10X" is not a "rule", as others have also written, it's merely a guide. For reasons I, Al, and several others stated, as the ratio approaches 1X and below, you will encounter a perceived loss of HF, a subjectively objectionable (to me) loss of sound quality, and eventually dramatic loss of phono gain. A 10X ratio keeps you safely above any of those phenomena. 5X would probably be fine, too. Depending upon the inductance of a given cartridge (but with an LOMC, inductance is quite low) you will get into trouble at different lower impedance ratios. Just what ratio will start to sound bad is dependent not only upon the cartridge and the phono stage but also on your other equipment, your system sound as a whole, and your ear/brain. For example, there are some who actually seem to prefer the sound of the Denon DL103 at impedance ratios that approach 1X. One guy I know of admits that he loses a lot of gain at that point, but he just likes the sound on other grounds.
Well said, Lew and Bob.

Best regards,
-- Al
Yeah I guess I just have to roll with it, the rule of thumb as a guide along with the formula + testing by ear.

However, applying the rule becomes tricky since the data spec quotations on several transformers are not quoted at 47,000. For example:

1.Cinemag 3440AHPC has.....37.5 or 150 : 50K

2. Another RCA has..... 600 or 150: 15K

For the above trafos do I have to adjust the calculations since they are not into 47,000 ohms, one is into 50K and the other is into 15K, to find the effective load?

To find the turns ratios please tell me if this is correct:
1. 50,000/x^2 = 37.5 solve x-> implies x=36.5 turns ratio

then to find impedance do I standardize to 47,000 in the following way?

47,000/36.5^2 -> 35 Ohms ?

I will repaeat the logic for the second cartridge as well which using the 150 tap with the quotation value 15K.

2. 15,000/x^2 = 150 solve x-> implies x = 10 turns ratio

then to find impedance 47,000/10^2 -> 470 ohms

My point is most transformers out there are not quoted into 47,000 ohms like the phono section is, they are often 50,000 or even 15,000. So if using those quotations, do you not have to solve for the turns ratio with their quotation values first, then use 47,000/turns^2 to find the effective load ?

If you look at the above example 2. it is quoted at

"600, 150: 15K"...so using the 150 tap I get a turns ratio of 10, and this provides and effective load @ 47,000 of 470 and not 150 the way it was quoted.

Then once this is done which set of turns ratios and effective impedances to I put into equation (*) to find the voltage?

You can see in example 1. it did not make much difference 35Ohms vs 37.5 quoted.

But in example 2 that 470 vs 150 quoted is a huge difference. If you were trying to match a 40 ohm cart 150 would be rejected since it is much lower than a 10 multiple, but after converting (if I am doing this right) it shows that it is in fact 470 which is ideal for a 40 ohm cartridge.

It is very hard to explain and ask, via the message board and it takes forever to type these things out. But I would like a workable way to find transformers that are suitable regardless of how they are quoted. I would also like for anyone reading this thread who does not know how to match correctly to understand as well.
Your calculations look correct to me, Dfel.

As you've found, and as might be expected, 47K vs. 50K makes little difference, and is probably not worth correcting for. Imprecision in cartridge output voltage specs, resulting in part because at least two different measurement standards exist, will often make more of a difference. Significantly larger discrepancies, though, certainly need to be taken into account.
Then once this is done which set of turns ratios and effective impedances to I put into equation (*) to find the voltage?
The effective load impedance will be the actual input impedance of the particular phono stage that is being used (often but not always 47K) divided by the square of the turns ratio.

The turns ratio that should be used, if it is not explicitly specified, is calculated per the examples you cited.

Regards,
-- Al
Those data you start with are unnecessarily complex, and are paradoxically offered to make life simple. Just know the turns ratio of the transformer. (Call up the maker and ask, if it's not published.) Then know the input impedance of your MM phono stage and the internal resistance of your cartridge. All else flows from those numbers.
Good call lewm, sure simplicity is good and when possible that is the best path. How about if it was made in 1930-1950 by a company that is long gone, or does not have the specs available any more? By applying the simple calculation, you know what you are dealing with even if that information is not available...figuring out the which tap does what (ground, phase, impedance etc..) can still be a pain.

Al you are right, it makes little difference for the 50K vs 47K it is not substantial, however there are still a lot of them quoted at 15K so I wanted to make sure I am doing the right thing, as it does make a difference in that scenario. Thanks for verifying that that method for adjustment is correct and what needs to be used in equation (*)..I was assuming it was as you said but wanted to make sure. You have been very helpful! Appreciate it, really, thanks for taking the time.

Heads up to those looking on the website for the first link as there are some calculation errors on that site with turns ratio and impedance on the cinemag example, and he also ignored the 50K bit (not as critical, to be fair)

At some point, when I am done reading on this, I will post a summary/dummy guide to these calculations that lead you to the ballpark, so anyone interested does not have to read the whole thread.
And this is one of the reasons I prefer an active gain stage over a xformer.
I agree with Swampwalker, an active gain stage, properly done is the best way to go. Unfortunately, in my case, my LFD phono stage did not offer a high enough gain for my Dynavector xx2 cartridge.

At 0.28v output and 43db gain, I was right on the noise edge. Some classical guitar albums that I have had in my collection for years were too noisy at the proper volume level.

I have to hand it to the op, I tried to understand SUTs but gave up and researched some contenders. I went with CineMag (Bob's Devices) 1131 because, he responded quickly, has matched my cartridge / phono stage before and was willing to take it back if I did not like the results.

The package is somewhere in transit now and delayed because of the horrible weather patterns we've been having in the midwest.

I really avoided as long as I could going the SUT route, but in the end I liked my XX2 / LFD combo too much to consider a higher output cartridge.
Swamp...LOL was waiting to see a post like that in this thread.
Dfel- It's not only the math and ee theory, it's also three other things:
1. No matter how well you understand the theory and do the math, the results are unpredictable.
2. Which means that unless you get really lucky, an internal step-up may not be optimal sonically, even if it is a good match electrically; which then leads to use of an outboard SUT, which requires additional interconnects. Which leads to issue #3:
3. It's my understanding that the step-up property of the transformer not only multiplies the voltage but also "multiplies" (in quotes because I don't think it's an exact relationship) the effect that the cabling has on the sound.

I'm pretty sure 1 and 2 are true, not sure about 3, but since my pre has in active gain stage that works just fine w a LO MC, I'm good to go, w/o math headaches, or trying to remember whether a higher impedance number means you are loading it down more or less, and whether you are matching the output impedance on the output device to the input impedance on the input device or vice versa.
The math required is high school algebra. The concepts are a bit more difficult to grasp, but one really need not grasp them exactly, if one grasps the math relationships.

The results are "unpredictable" only in the sense that mating any two disparate audio components with a human is unpredictable.

And believe me, I am no EE. I think the manufacturers are to blame for the confusion surrounding SUT use, because there apparently is no agreed upon standard for how to describe what the SUT does electrically. Some talk in terms of db, some in terms of ohms, few in terms of turns ratio. The latter however is most important.
Lew's post reminds me that one thing that should be described in this thread is how to convert SUT gain that is specified in db into turns ratio.

The relevant relations are:

Voltage out of SUT (Vout) divided by Voltage in (Vin) = turns ratio.

db = 20 x log(Vout/Vin), where "log" is the base 10 logarithm.

So the procedure would be to divide the specified number of db by 20, and raise the number 10 to a power corresponding to that result.

For example a 30 db SUT would have a turns ratio of:

10^(30/20) = 31.6 (^ denotes "raised to the power of")

Similarly, a 20 db SUT would have a turns ratio of 10.

The calculation is easily performed with any scientific calculator, including the one that is built into Windows. If using the one in Windows, which can be found under Start/All Programs/Accessories, set it to "scientific" mode under its "view" menu.

Regards,
-- Al