Moving Coil Step Up Maths and Optimal Matching - diyAudio
 Moving Coil Step Up Maths and Optimal Matching
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 19th February 2014, 09:21 AM #2 diyAudio Member   Join Date: Nov 2003 Location: Melbourne (Oz, not Florida!) Mmmm, my belief is that using a SUT and attempting to get the correct LOMC cart loading ... is like p*ss*ng into the wind! If you want to get serious about LOMC cart loading and you have a MM phono stage, use an (active) head amp. If it's built right, this will have a parallel pair of input RCAs: * one pair is for the phono stage. * the second is for "R-loaded" plugs. Regards, Andy
 19th February 2014, 07:46 PM #3 diyAudio Member   Join Date: Dec 2010 Location: Ivanhoe Archiekaras your link 2 is the more useful for showing the effects that impedance loading has on the transformer.The only aspect missing concerns how the combined source impedance of the cartridge with the xformer effects the noise figures of the preamplifier.As a rule of thumb,as there are a myriad of variations in topologies, valve stages give lower noise with about four times the source impedance of transistor ccts which typically have their lowest noise with a source impedance of 5000 Ohm.
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archiekaras-

sreten has the most practical answer. If you are still wondering where the math is coming from, it has to do with how they rate the output of the cartridge. If .3mV is the open circuit voltage output of the coil, then the 2nd link is the most accurate. Transformers don't really have an impedance unto themselves, they transform the impedance from one side as seen by the other side by a factor of the turns ratio squared. So a 34.5 turns ratio xfmr will transform the 40K Ohm load impedance to 37.5 Ohms on the other (cartridge) side of the xfmr. The cartridge then sees a load of 37.5 Ohms which forms a simple resistive divider as shown in the attached image. The voltage developed across the cartridge side of the xfmr is then .3mV x 37.5/77.5 or .1452mV. The xfmr steps this voltage up by a factor of the turns ratio (x 35.4) so it delivers 5.186mV across 40K.

If the cartridge output is rated .3mV into a 40 Ohm load, then the 1st link is most accurate. In that case, the resistive divider is already factored in and the cartridge will deliver .3mV across 40 Ohms since it would have .6mV output open circuit (the 40 Ohm load and the 40 Ohm source imp form a voltage divider that will halve the open circuit voltage or 40/80). If the xfmr presents a 37.5 Ohm load instead of 40 Ohms, the voltage would then be .29mV since the voltage divider would be 37.5/77.5 x .6mV. The xfmr steps this up by 35.4x or 10.28mV.
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 21st February 2014, 11:34 PM #6 diyAudio Member   Join Date: Nov 2003 Location: Brighton UK Hi, The error in post #1 and #5 and point 1) is only considering one side of the transformer matching for alleged gain. Work out what happens with : 40-47, 33 gain and 40K-47K. Compare to : 40-120, 20 gain and 16K-47K And : 40-470, 10 gain and 400-47K. rgds, sreten. __________________ There is nothing more practical than a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow Last edited by sreten; 21st February 2014 at 11:40 PM.
 22nd February 2014, 01:47 AM #7 diyAudio Member   Join Date: Nov 2013 I do not understand what was said. The open circuit/drive into terminology has confused me. When I look at a cartridge, specification it is called output voltage measured at 5cm/sec. The impedance is called internal impedance. So I am not sure which of the 2 scenarios you are talking about is in play with a MC cartridge. So it leaves me confused how to apply voltage divider theory, also does it apply the same way when a passive transformer is in the circuit as it does with the typical resistor in circuit examples they give online for voltage dividing. If I were to guess I would think that it is putting out 0.3 MV from the output pins after being run though a 40 ohm (internal impedance) load. Is the voltage drop/adjustment equation correct ? Under what scenario will it correctly give you the real voltage seen by the MM phono section.
 5th March 2014, 02:35 PM #8 diyAudio Member     Join Date: May 2006 Location: Behind the Cheddar Curtain dfoe- It all depends upon how the output is specified (I actually don't know which method they use). Think of the cartridge as an AC generator with an internal impedance. If it has and internal impedance of 40 Ohms and you measure the output with no load, it will appear ~twice as high as if it were terminated into a 40 Ohm load. So the question becomes: Is the 0.3mV output measured open circuit or into a 40 Ohm load? Regarding the xfmr, per my previous post, xfmrs do not exhibit an impedance of their own when operated within their specified range. The 40 Ohm load presented by the xfmr to the cartridge is the result of the 40K termination on the other side being transformed by the square of the turns ratio. The xfmr does not have a 40 Ohm input impedance unless it is terminated on its output by 40K. You do not have to apply the voltage divider reduction again on the output side of the xfmr. If you run the xfmr with no output load, the input side will be essentially open circuit as well and the voltage seen by the input will double (and so will the output voltage). Terminating the output with 40K will restore the voltage ratio as it now properly terminates the cartridge with 40 Ohms. Not sure what sreten is alluding to... The question still remains then: How is the cartridge output voltage specified; open circuit or terminated into the specified impedance?
 5th March 2014, 02:44 PM #9 diyAudio Member     Join Date: Mar 2009 not to be rude but why use a step up transformer when you could build a nice MC phono stage that can be configured to any MC cartridge loading with the change of a few resistors. then all you do is plug the phono stage into any line level input of your pre/integrated amp. or am i missing an important point of step up transformers.
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 Originally Posted by Bibio or am i missing an important point of step up transformers.
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