Higher Impedance MC Carts on Transimpedance Stages?

Nothing blows up. The mating would be inefficient as the internal impedance of the cartridge exceeds the input impedance of the BMC, which we think is around 2-3 ohms. Thus some of the signal voltage derived in the Kiseki would be lost to ground.  But also, the Kiseki probably produces very little signal current; you can estimate that by dividing its output voltage by it internal impedance.  I don't know the output of a Kiseki but for example if it is 0.4mV then its signal current is around 0.4mV/40 ohms = ~10 micro-amperes.  For comparison, even my Ortofon MC2000 which produces only .05mV of signal voltage makes 25 micro-amperes of signal current because its internal impedance is only 2 ohms. Nevertheless, one might take advantage of the BMC's adjustable gain (0, +7, +11, or +14db) and set the gain high using the internal jumpers. My prediction is it won't be satisfying but worth trying.

Mathematically, current (I) is defined as the Voltage (V) divided by the Resistance (R):

I = V/R

The current (I) is inversely proportional to the resistance (R) , i.e., the higher the current, the lower the resistance and vice versa, the lower the current, the higher the resistance.  Therefore if the resistance of your cartridge is high, you will get lower current and as a result also lower voltage.  High impedance cartridges will result in lower output and will sound like your phonostage doesn’t have adequate gain.

I worked this out. Bandwidth is where current is half of maximum. This occurs when impedance from cart inductance equals cart resistance. If the bandwidth is acceptable, cart will run fine.  Bandwidth is given as:

f = R/(6.3*L);

Or you can use the handy calculator I placed on this web page:

https://www.hagtech.com/loading.html

Hagtech, In your equation, I assume R is the internal resistance of the cartridge and L is the cartridge inductance. Since you speak of "bandwidth", "f" must be bandwidth. Further I assume R is in ohms and L is in henries. I must have something wrong, because if those assumptions are correct, then LOMCs with high internal R (>>10 ohms), like the Kiseki, would have much greater bandwidth than a typical LOMC with an internal R of ~10 or less. This happens because LOMCs have very low inductance, usually less than 50 microhenries, and therefore the denominator is always going to <<1.0. Further, I would think an equation to determine bandwidth resulting from a specific match between cartridge and phono would have to include parameters of both the cartridge and the phono stage.

For example, the Kiseki with an internal resistance of 40 ohms and an inductance of say 50 microhenries or .000050 Henries (cannot find the actual value on line) would have a bandwidth in excess of 100KHz. Whereas a more typical LOMC with an internal R of 4 ohms and similar inductance would have a ten-fold narrower bandwidth.  Even if such a cartridge has also a lower inductance, that would not help much; the predicted value for f would still be well below 100kHz.

Thanks to everyone who has contributed to the thread so far. The explanations and the equations provided are very helpful in understanding how to find a good match.

@hagtech, I'm curious to your response on lewm's reply as I can see where the way you're calculating bandwidth may not be accurate.

I did see on Steve Hoffman Forums that there is a user using a BMC with a different Kiseki cartridge also with 40 Ohm impedance and doesn't seem to be experiencing any issues with it. It may be that Kiseki has lower than typical inductance?

It's a shame that cartridge manufacturers do not provide all the specs an end user may need to make better-informed decisions.

The inductance of an LOMC is trivial, in the sense that it’s very low, on the order of 1000 times lower than MI which is already much lower than MM. Your acquaintance is probably happy because it works and he has yet to try an LOMC with truly low internal resistance. I feel that I must be misinterpreting Hagtech’d equation .

There is no real contribution from phonostage, since it appears as a dead short, we are talking transimpedance, right?

Nevertheless, first order bandwidth rolloff is from L/R ratio of cart (and tonearm wire). Think current, not voltage.

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

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.