Step 1: Find out the winding ratio of the trannie.
e.g. 20dB = 1:10; 26dB = 1:20; 30db = 1:31.6; etc.
(there are more on the web, no need to calculate...)
Step 2: Find out the 'natural impedance' of the trannie.
Ratio^2 * phono-pre input impedance
e.g. 31.6 * 31.6 * 47k ohm = 47 ohm for a 30dB trannie
(this will be what the cart sees if you do not use secondary or primary loading of the trannie)
Step 3: Check for phono-pre over load
e.g. of cart V = 0.25mV; cart DCR = 6 ohm
[cart output V * ratio]* [load impedance/(load imp + cart DCR)] =
e.g. [0.25 * 31.6] * [47 / (47+6)] = 7.9mV * 0.887 = 7.0mV
try for max. input value not to exceed ~ 7.5mV
With a 26dB, you will have a higher 'natural impedance' i.e. 117.5 ohm (and not 47 ohm as with a 30 dB item) which in turn might require secondary or primary loading to lower the impedance that the cart will see.
Secondary trannie loading is reflected 'back' to primary, by dividing the secondary R by the square of the winding ratio.
But first you have to get the correct value e.g. if 18k ohm is loaded on the secondary then:
47k*18k / 47k+18k = 13.01k ----
then reflected = 13.01k / 31.6*31.6 = 13ohm that the cart will now see.
But secondary loading will ALSO have the effect of damping the trannie, as more current is pulled on the secondary side, this may be just what is needed -- but it may be also not a good thing --- only testing & listening will tell.
If primary loading is indicated (if e.g. secondary gets too dull) then:
e.g. 18ohm load R with 47ohm 'natural impedance' =
47*18 / 47+18 = 13 ohm again what the cart sees.
I hope this will be enough to give you a starting point for checking the various option.
Cheers, |
Axel, thanks very much for the primer. I now understand these SUT impedance calculations. Using your formulae, I was able to recalculate the 3 Ohm input impedance for the Auditorium A23 Hommage T1 MC transformer that is referenced in Art Dudley's recent column based on the winding ratio (1:30) and the secondary coil impedance (2.8k ohms) specified in the column.
Very educational.
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Cincy Bob -- Art Dudley's writeup was perhaps worded a bit misleadingly, and I want to make sure it's clear that the load impedance seen by a cartridge working into that transformer is NOT 3 ohms. It is, to a close approximation, the input impedance of the phono stage or preamp which is connected to the secondary side of the transformer, divided by the square of the turns ratio. In other words, typically 47,000 ohms divided by 933 in this case, or 50 ohms.
The voltage gain would correspond to the turns ratio, which is just over 30 times (actually 30.54, the square root of 933). The voltage gain expressed in db is 20 times the log of 30.54, or 29.7db.
I'm not certain, but I suspect that the 3 ohm and 2800 ohm figures he mentions are dc resistances, which will result in a slight but insignificant attenuation of the signal.
Axel -- I second the thank you which Bob offered for your excellent post.
Regards,
-- Al |
In other words, typically 47,000 ohms divided by 933 in this case, or 50 ohms. I should have added, of course, that the 50 ohms would be less if load resistors are added. Regards, -- Al |
Here is the excerpt from Art Dudley's write-up: Aschenbrenner's Hommage T1 measures somewhat differently, with a primary coil impedance of about 3 ohms and a secondary coil impedance of 2.8k ohms I assumed that the secondary coil impedance was equivalent to a secondary trannie loading that is reflected back to primary as discussed in Axel's post above. The math seemed to confirm this: 47k*2.8k / 47k+2.8k = 2,643 / 933 = 2.83 ohms Is it coincidence that the mathematical result agrees with the "primary coil impedance of about 3 ohms" that is cited in the write-up? |
Art Dudley was using "impedance ratio," in this case 933 to 1, as a round-about way of indicating the turns ratio (which is the square root of 933, since impedance is reflected between the two sides of the transformer in proportion to the square of the turns ratio).
Hence the primary and secondary coil impedances he refers to (3 ohms and 2800 ohms) differ by a factor of 933. But I'm not sure how either of those numbers are defined -- as dc resistance, or as impedance at some frequency with the other side of the transformer connected to some load, or what?
The reason for the 2.83 ohm vs. 3 ohm "coincidence" is that the 47K load impedance is an insignificant load in comparison to 2.8K, which in turn reflects back to the primary as 3 ohms, while 47K in parallel with 2.8K reflects back to the primary as 2.83 ohms.
However, it is incorrect to consider the 2.8K as being in parallel with the 47K. Within the range of its intended operating conditions (frequency and voltage), a transformer has (to a very close approximation) no impedance of its own. In Axel's example, the 18K that he assumed was in parallel with the 47K preamp impedance represented an external 18K resistor, not the transformer's secondary coil impedance.
Regards,
-- Al |
Al, thank you for clarifying this:
>> In Axel's example, the 18K that he assumed was in parallel with the 47K preamp impedance represented an external 18K resistor, not the transformer's secondary coil impedance.<<<
The example aught to be clear, by looking at the basic calculation of parallel resistance: 47k input impedance parallel with an 18k load resistor on secondary.
It true, that the secondary coil DCR is ALSO reflected back into primary, again by dividing it by the square of the winding ratio.
B U T, it would be negligible in the case of my example and would be about 0.06ohm added to the calculated 13ohm of the example.
In the case of an FR XF-1 type M, (30dB) primary DCR = ~1.6 ohm, and secondary = ~ 60ohm (if I recall it correctly). |
The secondary coil DCR is ALSO reflected back into primary, again by dividing it by the square of the winding ratio. B U T, it would be negligible in the case of my example and would be about 0.06ohm added to the calculated 13ohm of the example. Yes, with the slight further clarification that the dc resistance of the secondary coil is reflected back into the primary by dividing it by the square of the winding ratio, in terms of its effects on the range of signal frequencies and voltages over which the transformer acts like a transformer. Which is why the 1.6 ohm and 60 ohm dc resistance numbers (dc being zero frequency) do not differ by a factor of 900 or so, in case anyone wonders. Best regards, -- Al |
Wow, Axel thanks a million. Now I am just having a little trouble calculating. What is DCR?
Could we work through a case together so this crystallizes?
Data: Denon Au310 Step up, and 2 cartridges.
#1 Denon Au 310:
Winding wire ratio (pressure-up ratio) 1:10
Primary impedance 40ohms
Secondary impedance 4kohm
Frequency response 20Hz - 40kHz
Deflection (as opposed to 1kHz) Less than ±1dB
Magnetic shield Double
Chord length About 1.2m
Dimensions Width 51x height 53x depth of 181mm
Weight 0.65kg
#2 First Cartridge Supex Super 900
Output .2MV
Impedence 3.5 Ohm
#3 Second Cartridge Madrigal Carnegie one
Output Sensitivity: >0.26mV
Load Impedance: 30 to 50K ohms
So this is the data, now the calculations:
For the first cart:
winding ratio 1:10 of step up,
then Step #1 says-> since 1:10 = 20dB
then step #2 says-> 20*20*47kohm = ?
then step #3 says-> Well I need DCR, Don;t know what it is
now I am lost from here on.
Please help me fill in the blanks. Thanks again for helping me understand. It is greatly appreciated.
|
31.6 * 31.6 * 47k ohm = 47 ohm for a 30dB trannie
Please explain also, how you get this equality ?
=998.56*47kohms
=46932.32 kohms how is this = 47 ohm ? |
Sorry for the silly typo!!!!!
It must read of course:
47k ohm / 31.6 * 31.6 = 47 ohm, for a 30dB trannie |
Dfelkai,
"DCR" = "dc resistance," which is usually just a few ohms for low output moving coils, and can be neglected without introducing significant error unless the cartridge is to be loaded to very low values (say below 30 or 40 ohms, or even less if the dcr is particularly low).
Your xfmr has a ten to one turns ratio, so the 0.2mv and 0.26mv cartridge outputs will be stepped up to 2.0mv and 2.6mv respectively, neglecting the dc resistance effect.
That is a little bit less than ideal (ideally I would aim for around 5mv, which a 26db xfmr would provide), but it should be ok if the gain and/or noise performance of the phono stage is not marginal to begin with. You would have to turn the volume control up by 6db, relative to the setting you would use with a 26db xfmr, which is not a huge increase, and typically would not raise noise (background hiss) levels objectionably.
The square of the turns ratio is 100, so the 47K phono stage or preamp input impedance will be seen by the cartridge as 47K/100 = 470 ohms.
The Carnegie One's very wide load impedance specification (30 ohms to 50,000 ohms) is probably intended to indicate that it would be within reason to drive it straight into a 47K phono stage (that provides adequate gain and adequately low noise) with no external resistor, as well as to run it into a high gain step-up xfmr. The 470 ohm reflected load obviously falls within that range, but you may want to try adding resistors of various values to fine-tune the sound to your liking. Keep in mind that adding a resistor in parallel can only lower the 470 ohms; it cannot increase that value. (And adding a resistor in series, which would increase that value, should not be done because it would attenuate the signal seen by the preamp).
According to the data at cartridgedb.com, the SD-900 Super is specified for a load impedance of 2 to 20 ohms, so adding a 20 ohm load resistor on the primary side, or a 2000 ohm load resistor on the secondary side, would appear to be a reasonable way to go. The 47K load impedance of the phono stage or preamp (or 470 ohms as reflected at the primary side of the xfmr) would cause only a negligible lowering of the 2K (or the 20 ohms), so there is no point in bothering to calculate its effects. (OK, I did it anyway: The 20 ohms would be lowered to 19.18 ohms, and the 2000 ohms would be lowered to 1918 ohms).
The Supex's "impedance" of 3.5 ohms probably represents dc resistance, and is small in relation to the 20 ohms so it will not affect the signal level and overload calculations to a significant extent.
Hope that helps,
-- Al |
Hi Al,
I think you do make the better teacher then I feel myself, as you have it so well explained.
Let's see if this gets Dfelkai sufficient clarity to proceed.
Thanks again, and let's see what comes up next. |
I still can't complete the calculation. This is problematic as I am looking to get myself a decent matched step up in the very near future. Seems like there are some errors in the above calculations. |
Were you able to follow all of my previous post? If so, why can't you complete the calculation (which I think my post essentially does), and what are the "errors" you are referring to?
Regards,
-- Al |
