I know it may feel counter-intuitive, but what is happening here is a sacrifice between output amplitude (maximum current) and bandwidth.

Higher Impedance MC Carts on Transimpedance Stages?

Can anyone explain what happens if one pairs a transimpedance / current injection phono stage with a moving coil cartridge whose impedance may be higher than optimal? What would the result be?

This question arose from someone who wanted my thoughts on the BMC MCCI Signature ULN phono stage that I use as my reference, but that individual is using a Kiseki Blue which is spec’d to have an internal impedance of 40 Ohm, which I’ve found is higher than typical MC cartridges.

@lewm and @rauliruegas, you guys likely can answer this easily, but of course open to anyone else that can explain.

Thanks!

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Similarly, the gain you will need from transimpedance stage (assuming it's like an SUT into MM phono) is given by: G = 5R / V; where R is cartridge resistance (plus tonearm wire) and V is cartridge output in mV. For example, an 0.3mV cartridge with 8 ohms gives G = 133 ohms. Yes, gain for a transimpedance stage is given in ohms... |

@hagtech +1 Transimpedance works nicely if you have a nice low output. Not so well as the impedance of the cartridge goes up. In a transimpedance circuit, the cartridge replaces the input resistor of an opamp. The gain of an opamp circuit is the ratio between the input resistor and the feedback resistor. Since the feedback resistor is fixed, this means that the lower the impedance of the cartridge, the more gain the circuit will have (and also greater distortion since there is a trade-off between the two). The loading is less than ideal. The cartridge has a certain compliance which interacts with the mass of the cartridge and tonearm, resulting in a mechanical resonance. This resonance should fall between 7 and 12 Hz for best tracking. When you load the cartridge with a low impedance, the result is the cantilever becomes harder to move (so has less compliance). This results in a very measurable loss of ability to trace higher frequencies. So at best the cartridge choice has to match the transimpedance circuit so this doesn't happen in the audio range. As Deep Thought once said, 'Tricky'. |

Hagtech, You wrote, "There is no real contribution from phonostage, since it appears as a dead short, we are talking transimpedance, right?" I learned my audio electronics from a few good books and from reading what others say on the internet, always a risky way to learn anything. So I do not mean to be snotty; I am only trying to learn, and I realize your level of knowledge is far superior to mine, assuming you are THE Hagerman. I am not ashamed of being ignorant, but in fact don’t all "transimpedance" phono stages have some finite, albeit very low, input impedance? Else the signal from the cartridge would be lost to ground, as in a mute switch. I do realize that with an op amp, you can wire it to present a "virtual ground", but even in that case the signal has to get past the input. There are also a few current driven phono stages that do not use an op amp input, the BMC MCCI being one of those. I suppose you can derive a virtual ground using discrete transistors, as the MCCI must do. I think of current driven phono stages as devices that have an I/V converter at the input which drives a conventional voltage amplifier that includes the RIAA correction circuit downstream. All of that said, how can it be the case that the phono stage characteristics are not important? By the way, I hate the term "transimpedance". It leaves the false impression that such a device is unaffected by the internal resistance of the cartridge, which is not true. In your equation, G = 5R/V, G would be in terms of ohms per volt, not ohms alone. You can’t just throw away units like that. An ohm/volt is the inverse of current, in amperes. Likewise, in your equation, f = R/6.3L, which I still don’t understand, f would be in ohms per Henry (or millihenries, or whatever henries). I cannot make sense of that. How do you get f in Hz from ohms per Henry? |

Input impedance of my transimpedance stage is a function of opamp gain. Let's say we have a 100 ohm feedback resistor and an output voltage of 1V. That means input current is 10mA. Forward open loop-gain of opamp in question happens to be 85dB at 1kHz (IIRC), or roughly 18,000. Hence, voltage across input terminals is 1 divided by 18,000. This defines an input 'resistance' of 1 divided by 10mA and 18,000, or about 6 milliohm. In reality, this worsens with trace width, solder joints, connectors, etc. As for the gain equation, it was a simplification. The '5' represents 5mV, as a target output for an SUT-like stage. So units are R*V/V = R. For bandwidth, the criteria I set was for 1/2 signal (-6dB voltage). Sorry if I had typed -3dB earlier. It's just a simple empirical rule-of-thumb I created to determine the likelihood of an LOMC operating into a transimpedance stage (or equal resistance loading). Half output current occurs when impedance from inductor equals winding resistance. Impedance of inductance is X = 2*pi*f*L. Set X = R and solve. |

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