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Effects of MC Loading

Jim Hagerman said:
Negative. A changing magnetic field induces a current in a wire. Same thing by moving the wire through a static magnetic field. If that wire is open-circuited there is no place for that current to go, hence (conservation of energy or something like that) a back-EMF creates an equal but opposite current, cancelling out the induced current. This is how an output voltage is generated by a cartridge. Leave it open circuit. Drive a short circuit (transimpedance stage) and you get maximum current. Everywhere in between is a mix, for example, loading with Rload = Rsource you get half of peak voltage and half of peak current, and maximum power.
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"Speaking" crudely. The emf generated by the cartridge coil is defined by Faraday's law of induction i.e. e=N*dPHI/dt- where e is the developed voltage, N is the number of turns, and PHI is the coupled flux.
The "back emf/potential" developed across the cartridge's inductance is governed by the equation v=-LdI/dt, where L is the inductance value and I is the current flowing, which is essentially a conservation of momentum (not energy, per se) constraint. i.e. no flow of electrons or energy transfer through the inductors field occurs when the inductor is open, but once it is shorted to ground the instantaneous back potential (ignoring any field propagation effects) must be precisely equal and opposite to the GENERATED emf as a consequence of these laws and of the application of Kirchoff's circuit laws.
As time progresses current will start to flow at a rate governed by these laws/equations, i.e. when e is constant, I will initially increase at a rate proportional to e and inversely proportional to L but will decrease exponentially with time to reach a steady state value that is proportion to the "loss" resistance in the inductor and whatever residual resistance actually exists in the "short".

The issue in this "problem" is not whether these trivial relationships hold but whether the current in the coil can create a restoring force in opposition to the motion of the cartridge that is sufficient to disturb the dynamics (signal output, tracking) of the system.
There is a form of reciprocity at work here that is extremely difficult to quantify.
However, whatever effect exists it is most certainly maximized by the use of a virtual ground (current input) for the loading/input node into the phono preamp, and there are those that claim the obvious superiority of such a configuration.
 
Jim Hagerman said:
Negative. A changing magnetic field induces a current in a wire. Same thing by moving the wire through a static magnetic field. If that wire is open-circuited there is no place for that current to go, hence (conservation of energy or something like that) a back-EMF creates an equal but opposite current, cancelling out the induced current. This is how an output voltage is generated by a cartridge. Leave it open circuit. Drive a short circuit (transimpedance stage) and you get maximum current. Everywhere in between is a mix, for example, loading with Rload = Rsource you get half of peak voltage and half of peak current, and maximum power.
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Jim Hagerman said:
Back EMF is the rated output voltage of a cartridge, typically specified at a certain velocity (cm/s). Use this rating to determine how much gain is required from phonostage.
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No, the back EMF is created by the current flowing through the coils. No current means no back EMF and no damping effect. You have it backwards.
 
wyn palmer said:
The issue in this "problem" is not whether these trivial relationships hold but whether the current in the coil can create a restoring force in opposition to the motion of the cartridge that is sufficient to disturb the dynamics (signal output, tracking) of the system.
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On another forum several years ago some chap did some calculations to show that the forces exerted by the rubber mount holding the stylus cantilever was orders of magnitude higher than the force created by any back EMF. He used the cartridge manufacturer's compliance figures in the calculations, but I can't remember the details.
Of course, a cartridge's compliance is specified at one particular frequency, and it's usually a low one. What it is at 1kHz or 10kHz might be very different. :dunno:
 
rothwellaudio said:
Thanks for the link. It was me who started that thread! I'd forgotten about that. The first post is attributed to "guest" because I was later booted off the site for allegedly promoting my own products.
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I'm curious: how does the end-user account for and correct the coil's back EMF or the effect of compliance with the type of rubber in the mount holding the cantilever onto the cartridge in the phono pre-amp setup?

Or, at least use this data when selecting a cartridge for the tonearm setup.

Thanks again! :)
 
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Mr Boochie said:
I'm curious: how does the end-user account for and correct the coil's back EMF or the effect of compliance with the type of rubber in the mount holding the cantilever onto the cartridge in the phono pre-amp setup?

Or, at least use this data when selecting a cartridge for the tonearm setup.

Thanks again! :)
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To be honest, I have no idea. The cartridge's compliance should be used in conjunction with the arm's effective mass to get the resonant frequency in the acceptable range, as is well-known. Any contribution from back-EMF is probably so small as to be insignificant. To me it's seems like more of an interesting theoretical proposition than a serious practical consideration.
 
rothwellaudio said:
To be honest, I have no idea. The cartridge's compliance should be used in conjunction with the arm's effective mass to get the resonant frequency in the acceptable range, as is well-known. Any contribution from back-EMF is probably so small as to be insignificant. To me it's seems like more of an interesting theoretical proposition than a serious practical consideration.
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Thank you for the candid answer.

I appreciate it.

:thumbsup:
 
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@JP :

One of the things I see in the chart below is there will be less difference in amplitude between two higher loads and I suspect that as the loads exceed 1kohm the amplitudes get even closer together and converge on the highest output possible. AT OC9XML R Load Comparison.png Do you think other MC cartridges also demonstrate this characteristic?
 
AvFan said:
Do you think other MC cartridges also demonstrate this characteristic?
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Yes, absolutely. The cartridge's impedance forms a potential divider with the load impedance. At a ratio of 1:10 the signal dropped across the source impedance (ie effectively lost) is just under 1dB. As the load impedance goes up you can only recover that 1dB loss, no more than that. If you were to do the calculations and draw a graph the curve would rise steeply from zero and then flatten out and approach 0dB loss asymptotically.
 
rothwellaudio said:
No, the back EMF is created by the current flowing through the coils. No current means no back EMF and no damping effect. You have it backwards.
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Oh my, you are correct, I did have it wrong. I had been thinking in terms of current in a wire in a magnetic field (Lenz) and F = i*LxB (Lorentz), which had me believing EMF drove an opposing current to maintain B. It does, of course, but my error was thinking EMF depended on whether or the termination of the wire (coil) was open or shorted. Wyn straightened me out (thank you), explaining EMF is solely a function of magnetic flux and time.

emf.png

So EMF is present during both open and closed circuit conditions. My results are correct (open circuit voltage = EMF, closed circuit current = EMF/R), but my interpretation of Faraday's Law was flawed.
 
mr_petit said:
So you don't have a set of decent headphones or an extended RCA cable that lets you plug in your computer in your stereo system?
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I have headphones. Not sure how good the DAC is on this Dell PC. Anyhow, I ran the test, but wasn't sure what I was supposed to choose. Am I supposed to select what I thought was 1k loading or something else? Does clicking on the ABCD change which file I am listening to? I randomly clicked on A for all of them and got 3 of 10 correct, which is what should happen statistically (1/4th).

results.png
 
Jim Hagerman said:
Ooooooh. I never clicked on X. I still believe the statistics are skewed, this is a game that cannot be won. Change it to two choices, 47k and 100R. Then it is fair.
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If the difference are as prevalent as some claim this should be child’s play even with the decreased sensitivity of more than two choices. You’re also still free to download the files and compare whichever two you want using foobar.
 
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