Cartridge Loading for a phono pre amp

Perhaps I'd have a different result with a different phono cartridge.

We have a loading strip on our preamps too, and I advise customers with LOMC cartridges to not use it. But if you have a higher output cartridge, the loading strip can be quite handy! Because of their higher inductance, high output cartridges have that electrical peak at a lower frequency, which can be at the high end of the audio band or just outside of it. The winding of a higher output cartridge tends to have a lower Q value so loading can be helpful to reduce ringing. 

@cleeds 

That's what I thought, and thanks to atmasphere confirmed it.

ARC and ML (No. 32) not only provide a range of loading resistance settings, they even have "custom Loading" option for user to fine tune their setting!

So no papers needed- just a grounding in basic physics (high school level), in particular how alternators and generators work. A 'Sheesh!' is in order here.

Ok I had HS physics and few more courses after HS.

Ohm's Law and the power formulas are not a figment of someone's imagination 😁

The cartridge is being asked to do just under 2 1/2 orders of magnitude more work with the example you provided:

47,000/100 = 470

-meaning 470X more current is drawn from the cartridge; because 47K is so high a resistance value relative to the impedance of the cartridge winding, the output voltage is essentially unaffected so the current increase also represents the wattage increase).

OK if the cart was a 0 ohms output impedance, then I think that the voltage would be linear, and inversely proportional to the impedance. 

As the cartirdges have an output impedance that is not zero, I am assuming that the produced current will not result in a linear relationship between voltage and loading, so that even with a 47k load the voltage will top out making the current either stall from flowing out of the cartridge, or ??

Anyone with an elementary school education can work out the math here. Perhaps, knowing that, you can tell me where that current is coming from??  If you can answer that, a bonus question: what is the consequence of that current flow?

The cartridge.

This is such simple math I don't see why a paper needs to be written about it, but maybe even though its basic, some people simply haven't thought it thru. Or didn't do so well in math.

I am hanging my head in shame, as I am assuming that the achieved current flow could be something like V/(R-cart +R-Load).
But I am not getting the simple ness of the math.

@rauliruegas 

Thanks for your test LP recommendations, I will look it up.

BTW, I already have Hi-Fi News Test LP- The Producer's Cut  and The Ultimate Analog Test LP by Analog Productions.

The cartridge.

@holmz Oh man, you're doing Raul's homework for him😄! I suspect you know how any generator will respond when asked to do say, 470 times more work? BTW you are correct that the output voltage isn't the same with the two different load resistances since the source impedance isn't zero. But since both loads (47K and 100 Ohms) are over 10x the source impedance, they will have little effect on the output voltage (which I simply ignored to make a point). No need to hang your head in shame, although I suspect that comment was facetious.

At any rate, the answer to Raul's latest query to me is inside the answer to the last question I posed to him.

atmasphere @holmz : " " Your target for mechanical resonance is between 7-12Hz. The mechanical resonance is a product of the mass of the cartridge in the arm vs the compliance of the cantilever of the cartridge. Changing the load from 47K to 100 Ohms can easily get you outside of this target window- and that can cause tracking problems.. " "

Where out side in specific: 5hz? 15hz? 6hz? . All you have to do is show it.

R.