Effective load close to or less than cart impedance is what I think you mean is a problem...or by heavily loaded. Yes. And I'm also saying that having a load impedance that is heavy enough (numerically low enough) to approach the impedance of the cartridge will almost invariably not be optimal sonically. And if a much lighter (numerically much higher) load impedance is applied, that is likely to be more optimal sonically, whether or not equation* is applied becomes insignificant. ... however the cartridge impedance is fixed and I was hoping to vary the step up in the analysis Everything I said is consistent with that. Keep in mind that as the turns ratio of the step up transformer is varied two things change: The load impedance seen by the cartridge (which is equal to the input impedance of the phono stage divided by the square of the turns ratio), and the input voltage seen by the phono stage (which is equal to the voltage applied to the primary of the transformer, referred to as "Vout" in equation*, multiplied by the turns ratio). Regards, -- Al |
I haven't taken the time to read the references you provided, but I believe I can shed some light on the discrepancy you cited. First, Vout: Voltage output at secondary side of tranny In the given context "secondary" should be "primary." Vout is being used to refer to the voltage appearing across the primary side of the transformer when the cartridge is being loaded via the transformer. As you indicated, that voltage is then multiplied by the turns ratio to derive the voltage appearing at the input of the phono stage. Second, equation(*) is apparently based on the assumption that the output voltage of the cartridge is specified under conditions of negligible load, such as 47K. The equation then adjusts that spec to reflect the voltage that the cartridge would provide under the much heavier load conditions it sees when connected to 47K via the transformer. I don't know whether output voltage specs for LOMC cartridges are typically based on load conditions that are essentially negligible (e.g., 47K), or under load conditions that are recommended for the particular cartridge, or on some other load condition. In general, though, it shouldn't make much difference, because in general the optimal load will be considerably higher than the cartridge's specified impedance. In the given example the cartridge is being loaded at a value that is lower than its own output impedance. My understanding is that that is way too heavy a load to be optimal in most and perhaps nearly all cases. Therefore the step-up ratio of 35.4 is much too high. Reducing it to say 20 (26 db) would result in the cartridge seeing a load impedance of 117.5 ohms. That in turn would result in the cartridge's 0.3 mv specified output being increased to 6 mv if the Vout/Vcart correction is not taken into account, and 4.5 mv if the correction is applied (and if the 0.3 mv spec is based on conditions of negligible load). Which is not much of a difference either way. And I wouldn't be surprised if optimal loading for the 103 would often be found to be considerably higher than 117.5 ohms, which would narrow the gap between the two numbers even further (albeit at the expense of making phono stage noise performance more critical, due to the lower signal amplitude received by the phono stage as a result of the reduced turns ratio). So as you can see, the application of equation(*) can be expected to make a significant difference only if the cartridge is being loaded excessively. Regards, -- Al |
As the load impedance seen by the cartridge becomes higher in relation to the cartridge's internal impedance, the difference between the two calculations (i.e., the calculations with and without application of equation*) becomes progressively smaller and less significant. That can be seen by running some calculations for various load impedances and turns ratios, such as the 117.5 ohm/20x example I provided.
The difference between the 10.6 mv and 5.1387 mv numbers is only as large as it is (more than a factor of 2) because the 37.5 ohm loading is undoubtedly much too low to be optimal for the particular cartridge, given the cartridge's 40 ohm specified impedance.
Regards,
-- Al
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What is effective Load. He is taking 47,000/(turnsratio^2). What does that tell you, what is that calculation? A transformer transforms impedance in proportion to the square of the turns ratio. So the load impedance seen by the cartridge corresponds to the input impedance of the phono stage (usually 47K) divided by the square of the turns ratio. What is equation (*)? Where does this come from? See this Wikipedia writeup on the voltage divider effect. In the first figure, consider Z1 to be the cartridge's specified internal impedance, and Z2 to be the load impedance seen by the cartridge. Consider Vin to be the voltage the cartridge would output under conditions of negligible load (e.g. 47K). Regards, -- Al |
Re Lew's point about the possibility of adding a resistor if necessary, I'll add that the parallel combination of two resistances corresponds to an overall resistance equal to the product (multiplication) of the two values divided by their sum. So if 11K is placed at the output of the SUT or the input of the phono stage, and the input impedance of the phono stage is 47K, the combined impedance on the secondary side of the transformer would be (47 x 11)/(47 + 11) = 8.9K. The cartridge would see a load impedance equal to 8.9K divided by the square of the turns ratio.
Also, some people report good results applying a load resistor on the primary side of the transformer. As you'll realize, in that case the value of the resistor will typically be a good deal lower, and the load resistance seen by the cartridge will be the parallel combination of the input impedance of the phono stage divided by the square of the turns ratio and the value of that resistor.
Given those possibilities of adding resistance in parallel, it seems to me that in choosing a stepup ratio it would make sense to err in the direction of having too low a ratio/too high a numerical load impedance/too light a load, PROVIDED that there is confidence that the noise performance of the phono stage is good enough to support the correspondingly reduced input voltage without introducing objectionable levels of hiss.
Re the chart you prepared, nicely done! To be sure its clear to everyone, the numbers on the vertical axis at the left represent millivolts at the secondary side of the transformer, i.e., the input to the phono stage. Also, a useful enhancement to the chart, if readily practicable, would be to superimpose on it a second curve indicating the load impedance seen by the cartridge, as a function of turns ratio, with the load value scale indicated vertically on the right.
Regards,
-- Al
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To answer your questions just above, there is substantial anecdotal/empirical evidence that if load impedance is numerically low enough to approach the value of a LOMC's specified impedance, dynamics, resolution, and transient impact will usually (and perhaps almost always) be adversely affected.
One reason for that may be that the 40 ohm or other specified internal impedance is not constant as a function of frequency, and will rise somewhat at high frequencies, due to its inductive component (coils have inductance, and the impedance of an inductor is proportional to frequency). That variation of cartridge impedance as a function of frequency will be inconsequential in relation to a relatively high load impedance. But if the load impedance is numerically low enough to be comparable to the cartridge impedance uneven frequency response will result, due to the resulting voltage divider effect being different at different frequencies. For the same reason, high frequency extension (i.e., bandwidth) may be excessively limited. Phase response issues may also come into play. Also, differences in loading will affect the behavior of the transformer itself.
Simply put, there are more factors that are involved than just assuring that proper voltage levels are sent into the phono stage.
Regards,
-- Al
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With the dynamic nature of the impedance characteristic, would it not make sense to pick a "flattest" segment of the curve to operate on. The X Axis turns ratio variable can be converted to effective load, then picking a impedance load such that its relative operational neighborhood is sufficiently "flat" would make sense. I see no reason that choosing a point on the flattest part of the curve would have any relevance, since the turns ratio, once chosen, remains constant. Choosing a point on the flattest part of the curve would improve the accuracy with which the input voltage to the phono stage can be predicted, but I don't see that improvement as being necessary or beneficial. Turn Ratios between 10-80 give 2.5-5 MV which seems perfect. BUT the i/o impedance ratios in the same range go from 0.20 to 22.0 ! A range that has extremes FAR AWAY from 10X, and the equation says they are all fine. The equation says that they are all fine in terms of the voltage level that is sent into the phono stage. But it says nothing about how sonics may be affected as a function of the turns ratio, and hence loading, within that range. As indicated in my previous post, the turns ratio will affect sonics in ways other than by its effect on voltage. Regards, -- Al |
Dfel, I'm not sure that I can add much in answer to your recent questions that hasn't already been said. If you haven't already seen it, I would highly recommend that you take a look at the writeup by Bobsdevices for which he provided a link in his post earlier in this thread. He is a well regarded manufacturer of SUTs. And I note, btw, that the models he offers provide ratios that are switchable between two values. Al, if you don't mind my asking..where the heck did you learn all this stuff? Do you have a background in electrical engineering? Yes, I am a retired EE, with extensive experience in analog and digital circuit design (not for audio). I've also been an audiophile for about 35 years, and I've read widely on the subject. Regards, -- Al |
Well said, Lew and Bob.
Best regards,
-- Al |
Your calculations look correct to me, Dfel. As you've found, and as might be expected, 47K vs. 50K makes little difference, and is probably not worth correcting for. Imprecision in cartridge output voltage specs, resulting in part because at least two different measurement standards exist, will often make more of a difference. Significantly larger discrepancies, though, certainly need to be taken into account. Then once this is done which set of turns ratios and effective impedances to I put into equation (*) to find the voltage? The effective load impedance will be the actual input impedance of the particular phono stage that is being used (often but not always 47K) divided by the square of the turns ratio. The turns ratio that should be used, if it is not explicitly specified, is calculated per the examples you cited. Regards, -- Al |
Lew's post reminds me that one thing that should be described in this thread is how to convert SUT gain that is specified in db into turns ratio.
The relevant relations are:
Voltage out of SUT (Vout) divided by Voltage in (Vin) = turns ratio.
db = 20 x log(Vout/Vin), where "log" is the base 10 logarithm.
So the procedure would be to divide the specified number of db by 20, and raise the number 10 to a power corresponding to that result.
For example a 30 db SUT would have a turns ratio of:
10^(30/20) = 31.6 (^ denotes "raised to the power of")
Similarly, a 20 db SUT would have a turns ratio of 10.
The calculation is easily performed with any scientific calculator, including the one that is built into Windows. If using the one in Windows, which can be found under Start/All Programs/Accessories, set it to "scientific" mode under its "view" menu.
Regards,
-- Al
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