MM cartridges and capacitance


Can someone explain to me why an MM cartridge would "want" to see more than the minimum possible level of capacitance loading?

This question is provoked by a lot of commentary on "that" "MMs are great" thread, and a question someone just asked, and the fact that I have two phono stages (granted, older Japanese stages) which have multiple capacitance settings and I have never gotten a "better" result from being at the high end rather than the low end. In many cases, it doesn't seem to make a difference, but so far I have not gotten an improvement from raising the capacitance setting.
t_bone
T-bone, if we use units of henries for L, and units of farads for C (which results in F being in units of hertz), then both L and C will be far less than 1. In fact C will be on the order of 0.0000000001 farads (or 100 pf). Therefore we will end up with a number in the denominator that is much smaller than 1, with the numerator being equal to 1. Therefore F, in units of Hertz, will be a number that is much larger than 1.

As for Farads x Henries resulting in frequency, after application of the square root and reciprocal functions and some constants, yes that is not intuitively obvious. But the derivation is as follows:

Resonance occurs at the frequency at which the magnitude of the impedance of L becomes equal to the magnitude of the impedance of C. Since the impedances of L and C have opposite polarities in the complex plane (don't ask!), the equal and opposite magnitudes will cancel each other out at that frequency, resulting in a net impedance of zero (apart from the resistance that is present).

The magnitude of the impedance of an inductor, aka its inductive reactance, is measured in ohms and is equal to 2 x pi x F x L.

The magnitude of the impedance of a capacitor, aka its capacitive reactance, is also measured in ohms and is equal to 1/(2 x pi x F x C).

So to find the resonant frequency we set those two formulas equal to each other, and solve for F.

2 x pi x F x L = 1/(2 x pi x F x C).

Re-arranging that equation:

1 = 2 x pi x F x L x 2 x pi x F x C

Therefore F x F = 1/(2 x pi x L x 2 x pi x C)

Therefore F = 1/(2 x pi x (sqrtLC))

Voila!

Best regards,
-- Al
Al,
You are a scholar and a gentleman.
Thank you very much for that explanation.
Al, Thanks so much. This is the first clear explanation for the apparently paradoxical effect of added capacitance on the treble response that I have come across. Makes perfect sense, now. Your post above should be archived somewhere.
Perhaps a naive question:

If "higher than standard 47k" capacitance was known to optimize frequency response by an MM cart's designers, wouldn't they have included that higher capacitance value in the spec sheet for that cart?

I use MMs almost exclusively and find these discussions most informative, though I've no simple and reliable means to increase the capacitance loading via my MM phono stage.
T_bone & Lew, thanks very much for the nice words.

JB, in my experience MM cartridge specs do usually indicate recommended load capacitance.

The 47K number, btw, is not capacitance, it is the recommended load RESISTANCE (47,000 ohms) for many MM's, and approximately corresponds to the input resistance of many or most phono stages or preamp phono inputs, that are intended for use with MM's.

Recommended load capacitance for MM's is typically in the area of 100 to 400 pf (picofarads). The cabling will usually be a more significant contributor to that than the input capacitance of the phono stage or preamp phono input, if the phono stage or preamp does not have a provision to switch in additional capacitance.

As you may realize, capacitance can be added by soldering an appropriate capacitor to an rca plug, and connecting it at the phono stage or preamp phono input with a y-adapter. I have a commercially made kit that I purchased in the 1980's which provides a selection of rca plugs with capacitors of various values soldered onto them, but I haven't seen such a kit offered in recent times.

Best regards,
-- Al