Cartridge Loading.....Part II

(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.

Ralph, you wrote, "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."

I have been complaining about that on this forum over and over again.  Only be direct questioning can one find out what the input impedance of these "current driven" phono stages actually is, and it's typically from a few ohms in the best case to as high as 20 ohms.  Even the term "transimpedance" is a marketing ploy.  Almost as bad as "quantum" used to describe an interconnect or a fuse.

Your other responses are helpful but raise other questions in my mind that I will hold in reserve for now. Thanks for your patience.

I have been complaining about that on this forum over and over again.  Only be direct questioning can one find out what the input impedance of these "current driven" phono stages actually is, and it's typically from a few ohms in the best case to as high as 20 ohms. 

@lewm I don't know how you could specify the input impedance; it varies with the impedance of the cartridge! Again, the gain of the circuit is defined by the ratio of the feedback resistor vs the input resistor (which is literally the cartridge itself). So if the cartridge is 30 Ohms and the feedback resistor is 300 Ohms, the gain of the circuit would be 10 or 10dB. If the cartridge were only 15 Ohms that would mean the gain of the circuit is 20 (16dB). So the input impedance can only be defined by the fact that a virtual ground is present. But you have another problem, which is that with almost any opamp made you run out of Gain Bandwidth Product over about 20dB of gain or thereabouts.

This means that with cartridges that have a very low impedance the circuit may lose neutrality. Personally I would prefer to have the gain be a set thing so that the cartridge would not be able to affect the phono section in that manner.

I haven't kept track of which current driven phono stages on the market use op amps or discrete transistors or tubes in the input I/V stage, although I don't know of any that use a tube or how you could use a tube in that fashion.  However, I would guess that many do use a discrete transistor and so by your definition cannot be current amplifiers.  If they all use op amps, then I stand corrected, and right there I have learned something about them.  However, every one that I have investigated was said to have some finite input Z, greater than zero and notwithstanding the cartridge with which it might be matched. (Especially because most of the advertising suggests you can match them with any cartridge having a "low" internal resistance, low being better than higher.)  But I do get that to be a pure current amplifier, the input impedance ought to be zero. Sticking my neck out in saying this, but I would rather gain understanding than worry about negative feedback, no pun intended.

OK... lets try a different tact since i do not think ground (be it virtual or real) matters in this situation.  I am simply relying on Ohm and Kirchoff for this ideal case.

Take a 30Ω cartridge and hook it up to an ideal voltage amp and then another one and hook it up to an ideal current amp.  Now take two AC microamp clamps and monitor the current output of each cartridge.  Will the currents be the same for each cartridge?

dave