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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
almarg's avatar
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almarg
9,670 posts
 
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
bobsdevices
100 posts
 
The Analog Dept link describes the process I use. Here is another link:

http://www.bobsdevices.com/What-about-Impedance.html
dfel OP
62 posts
 
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.

dfel OP
62 posts
 
For that last post I am referring to the chart HERE:

http://s28.postimg.org/9kno6o95p/Denon_Chart.png
almarg's avatar
Ag insider logo xs@2x
almarg
9,670 posts
 
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
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