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
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
The Analog Dept link describes the process I use. Here is another link:

http://www.bobsdevices.com/What-about-Impedance.html