" 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! |
"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.
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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."
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. |
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 |
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.
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For that last post I am referring to the chart HERE:
http://s28.postimg.org/9kno6o95p/Denon_Chart.png |
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)
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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. |
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.
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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. |
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?
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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. |
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. |
Swamp...LOL was waiting to see a post like that in this thread. |
