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Phono rig capacitance

I have read up on LPFs (low pass filters) and corner frequencies. and found the following... this equation gives the -3db corner frequency: Fc = 1/(2*Pi*R*C), inductance is ignored but can be impleneted using the R-adjusted instead of R as SQRT(R*L), geometric average. Though the value may not be significant, which is why I usually see it omitted.

I am interested in:

1. how one computes the -0.5, or -1db or any db cut in frequency NOT just the 3db corner frequency.

2. How to compute the corner frequency for the cartridge to SUT, given the amount of capacitance in the interconnect. For the example I suppose using the all familiar cinemag 3440 makes sense and for the cart the denon 103.

3.Same as above, but to compute for the interconnect from the SUT to the preamp..

4. Same as above but compute for the interconnect from the preamp to the power amp.

5. And perhaps the same for loudspeakers as well.

The goal is to find a value that ensures there is no roll off taking place and to select a suitable wire for each interconnection in a phono based playback system using an MC cartridge->SUT->Pre->Power.

I know, less capactiance blah blah blah, buy a 4 thousand dollar cable blah blah blah is the usual answer, but I am looking for a more scientific and technical approach to selecting wires that are in the ballpark of what makes sense based on well understood engineering principles.

I know that there are several members with advanced degrees in electrical engineering or are technically apt (Almrag, Atma, Raph etc...) and I am hoping that one of you can find the time to chime in please.

Thanks guys, looking forward to hearing your take!
dfel
dfel OP
62 posts
 
Wait a Minute I had a dull moment let us redo this:

Too bad there are no edits, now the mistakes look silly forever.

Assumptions/Signal Chain

Cart[x uh,30Ohms , wire(100pf,2uh,1 Ohm)->SUT->(Same wire)->Preamp(47K)

Cart sees everything on its side and the other side of the SUT DIVIDED by 10^2 or 100:

-100Pf (from the Cart->SUT connection) + 100pr/(10^2) (From the SUT->Preamp) back through the SUT = 101pf

-The 47K load gets reflected back at 47K/(10^2) = 470 Ohms

-and the inductance of the wire is meaningless and even less so after it is divided by 10^2.

-The cartdrige inductance is x and can be substantial depending as explained earlier

Other side ofSUT sees everything stepped up by 10^2:

-100pf *10^2 +100 pf = 10,100pf (WOWZER)
-inductance from the cart x uh*10^2 = 100X uh
-and 40 Ohms from the cart *10^2 = 4K

driven into a pre with a input imp of 47K.

This should be correct, sorry for the mixup, and please let me know if I made a mistake here (again).

Note: I have ignored reactance (Capacitive and Inductive) though I do not think it is substantial and I am also not certain how that follows through the ends of a SUT, If I were to guess Z is altered as sqrt((Xc/10^2)^2 + (XL/10^2)^2) using a linear transform of variable...ditto for the other direction.
dfel OP
62 posts
 
Jcarr, Maybe I am off here but on the primary side/ what the cartridge sees: Do you not divide by turns ratio square i.e/ 10^2 or 100 in my example.

So:
47K becomes 470 Ohms
100 pf SUT to Preamp becomes 1 Pf

??

I would have thought that this implies that the connection from the cartridge to the SUT may be more capacitance critical than the connection from the SUT to the pre.

going to the secondary side however requires multiplication by turns ^2 or by 100.

Please let me know what I am missing here.
almarg's avatar
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almarg
8,673 posts
 
Dfel, what Jonathan said is of course correct. The cartridge does not see the capacitance on the secondary side of the SUT divided by 100 (the square of the turns ratio we are assuming). Since the SUT transforms impedance in proportion to the square of the turns ratio, the cartridge sees the **capacitive reactance** that is on the secondary side divided by 100.

Since as you indicated earlier Xc = 1/(2*pi*f*C), capacitive reactance is inversely proportional to capacitance, and so the cartridge sees the capacitance on the secondary side **multiplied** by 100.

This is all based on an assumption of ideal behavior by the SUT, of course. No transformer will behave in a completely ideal manner, due to many factors. So all of this is of course just an approximation, but it is a good approximation for practical purposes.

Regards,
-- Al
jcarr
341 posts
 
Dear Dfel: Here is an alternative way to envision what a SUT does.

As a passive device, a SUT cannot create more energy. Its voltage amplification abilities are limited to converting the MC cartridge's output current into voltage. But since a SUT cannot create more energy, the cartridge signal current that gets converted into signal voltage is no longer available at the SUT output.

And since current is what charges and discharges capacitance, inserting a SUT between MC cartridge and phono stage reduces the cartridge's ability to charge and discharge whatever capacitance is present on the SUT secondaries by the square of the primary-secondary turns ratio.

hth, jonathan carr
dfel OP
62 posts
 
I can say I only partly understand, I will have to learn a little more to fully wrap my head around how it works. However, for now:

I tried modeling the transformer, not sure why I am running into problems here. I will try to post the image later maybe you can offer some guidance on the schematic/model for simulation. again ,much appreciated thanks guys!
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