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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
 
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
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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

almarg's avatar
Ag insider logo xs@2x
almarg
9,670 posts
 
What is effective Load. He is taking 47,000/(turnsratio^2). What does that tell you, what is that calculation?
A transformer transforms impedance in proportion to the square of the turns ratio. So the load impedance seen by the cartridge corresponds to the input impedance of the phono stage (usually 47K) divided by the square of the turns ratio.
What is equation (*)? Where does this come from?
See this Wikipedia writeup on the voltage divider effect. In the first figure, consider Z1 to be the cartridge's specified internal impedance, and Z2 to be the load impedance seen by the cartridge. Consider Vin to be the voltage the cartridge would output under conditions of negligible load (e.g. 47K).

Regards,
-- Al
dfel OP
62 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."

Yes, however the cartridge impedance is fixed and I was hoping to vary the step up in the analysis. In the end it makes no difference effecitve/(effective+cart)->40/(40+40) will always be 0.5, clearly the voltage drops by half when the impedance are equal. But you can see that when the effective load goes up the ratio also goes up.

look at what happens when the effective load goes to 100..100/(100+40) = 0.71....

or if it goes to 470...470/(470+40) = 0.92

or look at what happens when effective load falls to 5...5/(5+40)...=0.1111

Increasing the cartridge impedance is the flip side of the equation. Holding the effective load constant, as the cartridge impedance increases the equation falls from 1 towards zero reducing the output voltage.

All this to say, We are looking at the same thing, I just wanted to understand what exactly you meant by heavily loaded. Effective load close to or less than cart impedance is what I think you mean is a problem...or by heavily loaded.
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