Cartridge Loading.....Part II

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

(1) what if the input device is a discrete transistor or a tube, not an op amp?

@lewm 

Then there won’t be a virtual ground. So right away its a voltage amplifier not a current amplifier.

(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?

Opamps have nearly infinite gain when open loop; the feedback resistor and the input resistor thus define the gain of the circuit and the virtual ground is formed at the intersection of the input resistor and the feedback resistor (see my prior posts for more information).

There is no connection between actual ground and virtual ground; the latter is created as a result of the feedback meeting the input signal. So there isn’t (as in the case of 5 Ohms) 5 Ohms between the ground and the virtual ground. In fact the actual impedance is much higher.

The ’0 Ohms’ value you see in so many phono sections that have transimpedance inputs probably isn’t helping people to understand what is going on. That value is probably the marketing department talking since they probably didn’t understand what a virtual ground is.

The tricky bit is that in a transimpedance input, the cartridge itself is the input resistor. This means that the actual impedance load on the cartridge varies with the impedance of the cartridge itself- and with it, the gain of the circuit. As I pointed out earlier, the lower the impedance of the cartridge the higher the gain of the circuit.