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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
Ag insider logo xs@2x
almarg
9,670 posts
 
I haven't taken the time to read the references you provided, but I believe I can shed some light on the discrepancy you cited. First,
Vout: Voltage output at secondary side of tranny
In the given context "secondary" should be "primary." Vout is being used to refer to the voltage appearing across the primary side of the transformer when the cartridge is being loaded via the transformer. As you indicated, that voltage is then multiplied by the turns ratio to derive the voltage appearing at the input of the phono stage.

Second, equation(*) is apparently based on the assumption that the output voltage of the cartridge is specified under conditions of negligible load, such as 47K. The equation then adjusts that spec to reflect the voltage that the cartridge would provide under the much heavier load conditions it sees when connected to 47K via the transformer.

I don't know whether output voltage specs for LOMC cartridges are typically based on load conditions that are essentially negligible (e.g., 47K), or under load conditions that are recommended for the particular cartridge, or on some other load condition. In general, though, it shouldn't make much difference, because in general the optimal load will be considerably higher than the cartridge's specified impedance.

In the given example the cartridge is being loaded at a value that is lower than its own output impedance. My understanding is that that is way too heavy a load to be optimal in most and perhaps nearly all cases. Therefore the step-up ratio of 35.4 is much too high. Reducing it to say 20 (26 db) would result in the cartridge seeing a load impedance of 117.5 ohms. That in turn would result in the cartridge's 0.3 mv specified output being increased to 6 mv if the Vout/Vcart correction is not taken into account, and 4.5 mv if the correction is applied (and if the 0.3 mv spec is based on conditions of negligible load). Which is not much of a difference either way. And I wouldn't be surprised if optimal loading for the 103 would often be found to be considerably higher than 117.5 ohms, which would narrow the gap between the two numbers even further (albeit at the expense of making phono stage noise performance more critical, due to the lower signal amplitude received by the phono stage as a result of the reduced turns ratio).

So as you can see, the application of equation(*) can be expected to make a significant difference only if the cartridge is being loaded excessively.

Regards,
-- Al
dfel OP
62 posts
 
"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.



dfel OP
62 posts
 
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.
dfel OP
62 posts
 
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?
almarg's avatar
Ag insider logo xs@2x
almarg
9,670 posts
 
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.

Regards,
-- Al

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