Cartridge Loading.....Part II

The topic of this thread is cartridge loading and it only makes sense to look at it from the two terminal perspective of the cartridge without concern of the "technology" used to provide that load.   There are two basic extremes of operation of a cartridge.  It can operate as a current generator where the load value is << the cartridge internal impedance or it can operate as a voltage generator where the load value is >> than the cartridge internal impedance.  There is also a fairly grey area inbetween these two extremes where the load value is ≈ the cartridge impedance.  For this basic discussion I think the two extremes need to conceptually be looked at from the ideal with respect to how the cartridge converts a mechanical movement into an electrical signal.  Since it is the source to load relationship that dictates whether a cartridge generates current or voltage it becomes important to determine if changing the load causes any mechanical or electrical change to the behavior of a cartridge.  

When operating as a voltage generator the load can easily be modified over a fairly wide range and still maintain the basic principles of operation.  I think most will agree that loads  of 10X the cartridge impedance and up have a cartridge operating squarely in the voltage realm and people will start to cry foul as your load approaches  the cartridge internal Z.  This doesn't have anything to do with actual cartridge behavior and everything to do with the type of amplification that follows.  For voltage amplification the unique case where Rsource=Rload nets a 6dB voltage loss and in the case of microvolt level signals that is huge.  Going to the case where Rload is 1/4 that of Rsource the voltage loss will be 18dB which immediately disqualifies that as an option for many.  The problem with that categoric disqualification is that you are trying to make a cartridge operate as a source of current into a voltage amplifier. The problem has nothing to do with the actual load and everything to do with using the wrong tool for the job.  If the goal is to actually load the cartridge with 1/4 the internal impedance then that load should simply be provided by a current amplifier.  If we want to discuss the effects of loading on the behavior of the cartridge we have to assume that the appropriate type of amplification is used.

The first question that needs to be addressed is in a perfect world with ideal amplification, will a 40Ω cartridge sound the same into a 5Ω load as a 47kΩ load? 

The first question that needs to be addressed is in a perfect world with ideal amplification, will a 40Ω cartridge sound the same into a 5Ω load as a 47kΩ load? 

Right up to this point I was in agreement with the prior text of this post. With this question we can safely say the answer is 'No.', assuming that the perfect amplification is voltage amplification. The question cannot be answered at all if current amplification is used.

If that is a static load (IOW, a resistor) then the cartridge will be making 4 orders of magnitude more work! That work has to come from somewhere, otherwise a new branch of physics is created 😉

So it will certainly cause the cantilever to be harder to move (stiffer) and that will affect how the cartridge 'sounds'. It won't affect the bandwidth of the coil in the cartridge at all- but will have an enormous affect on the mechanical aspect as the coil will have become a significant load with 5 Ohms loading it in turn. Also, the output level will be considerably decreased!

The question cannot be answered at all if current amplification is used.

Why not?  Consider the case to be an ideal current amp with the appropriate series resistance added so the input impedance is 5Ω.   In this gedanken world the ideal voltage amp and the ideal current amp sound identical.

If we can agree on this then we can get to the question I am really curious about and that is....  How much of the sound of current amps vs. voltage amps is simply due to the radically different load the cartridge sees? 

A simple experiment I have been contemplating playing with is to compare the loading extremes is to use a 1:10 sut loaded with 300kΩ to load an 8Ω cartridge with  3K.   Then take the same cartridge and feed a 1:20 and apply an 8Ω resistor as a load directly to the cartridge.  The extra 6dB of gain from the increased turns ratio will be offset by the ~6dB loss of the cartridge being loaded with its internal impedance.  One could even take it a step further and try a 1:40 with a 4Ω parallel load to see the sonic effects of the extremes.  

dave

Why not?  Consider the case to be an ideal current amp with the appropriate series resistance added so the input impedance is 5Ω.   In this gedanken world the ideal voltage amp and the ideal current amp sound identical.

It seems to me that you are still conflating virtual ground and actual ground as the same thing! As I said before this leads to confusion.

So the 'Why not?' is the same as before: Because that 5 Ohms is a virtual 5 Ohms instead of a real 5 Ohms. The cartridge is not loaded at an actual 5 Ohms. 'Virtual Ground', again, is opamp parlance for a point in the circuit that exists at the same potential as ground but isn't actually ground.

I recommend that you read up on opamp operation since this seems to be the hanging point. Here's a short tutorial opamp virtual ground.

If you don't want to do that, just keep in mind that 'virtual ground' isn't the same as actual ground. So the cartridge would not be loaded at an actual 5 Ohms even though the virtual ground is 5 Ohms.

Ralph, two questions:

(1) what if the input device is a discrete transistor or a tube, not an op amp? 
(2) the only way I can imagine two points separated by 5 ohms but at the same potential is if and when there is no current flowing. How does that work in this case?

Thx