2. analog
  3. 1392765748

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
8,673 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
8,673 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.
almarg's avatar
Ag insider logo xs@2x
almarg
8,673 posts
 
Effective load close to or less than cart impedance is what I think you mean is a problem...or by heavily loaded.
Yes. And I'm also saying that having a load impedance that is heavy enough (numerically low enough) to approach the impedance of the cartridge will almost invariably not be optimal sonically. And if a much lighter (numerically much higher) load impedance is applied, that is likely to be more optimal sonically, whether or not equation* is applied becomes insignificant.
... however the cartridge impedance is fixed and I was hoping to vary the step up in the analysis
Everything I said is consistent with that. Keep in mind that as the turns ratio of the step up transformer is varied two things change: The load impedance seen by the cartridge (which is equal to the input impedance of the phono stage divided by the square of the turns ratio), and the input voltage seen by the phono stage (which is equal to the voltage applied to the primary of the transformer, referred to as "Vout" in equation*, multiplied by the turns ratio).

Regards,
-- Al
dfel OP
62 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). "

Wow! Thanks that is exactly what I was looking for. I will take detailed look at it so I understand where the way of calculating is coming from.

Assuming it holds, which at a glance of these electricity theories it does, the problem is still not solved.

HOW DOES ONE SELECT AN OPTIMAL STEP UP FOR THEIR CARTRIDGE.

That is really the point of this thread, and I hope it will be ironed out so it can be useful to others confused on this issue.

From what was mentioned so far:

1. It seems like equation (*) on the link can be used, and should be used, over the typical multiply the turns ratio to find voltage thing most guys are doing...especially when there is distance between the cartridge imped and the efective imped.

2.Matching cart to transformer impedance is just plain wrong is most cases. it steps down 50% basically and can crush the cartridge's character.

So what else is important

1. Every phono section is different, so using equation (*) to find the adjusted voltage will give you a MV output for any given turns ratio, but what is good for your phono stage 2.5MV,10MV or anything between. Running the calculations for a sample cart, I found that if I get between 2.5 MV and 5 MV using equation (*) it shows me that just about anything between 20-40 as a turns ratio is can be used. That is a HUGE range of step ups. There has got to be a better method than trial and error to get the right step up. I mean not everyone has thousands of dollars to buy 10 of them in those ratios to try them. The other thing is that simply spending more on a step may not put you closer to the ratio you actually need...I side with matching correctly over just spending 5k on a step up and hoping it works.

2. The effect of choosing an effective load <= cart is a no no, but what happens when it is chosen too high ? The "experts" recommend 8 to 10 times as a rule of thumb, what happens if you use 3 or 50? I do not know if there is a predictable result for chosing them too wide, or if it has any effect at all. When do the transformers being to ring?

3. The MV output from your cart is not stable, nor is it always at the specified rating. .30 MV...yeah at 1000HZ, and even that number may come out exactly as the manufacturer measured it.

so in the end...using this information really only puts you in the ballpark for what you need. Hopefully someone out there knows of a way to narrow this ballpark further.

Thanks again AL for being patient and contributing to this thread!
More to discover