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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
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
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almarg
8,673 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
dfel OP
62 posts
 
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.
dfel OP
62 posts
 
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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almarg
8,673 posts
 
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
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