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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

Showing 5 responses by lewm

lewm
13,476 posts
 
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).
lewm
13,476 posts
 
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.
lewm
13,476 posts
 
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.
lewm
13,476 posts
 
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.
lewm
13,476 posts
 
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.
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