DAQ
The practical solution is to measure.
Put on a record and check the claim under real operating conditions.
😉
The practical solution is to measure.
Put on a record and check the claim under real operating conditions.
😉
The secret sauce is greatly reducing shunt capacitance, something uniquely possible for DIYers. If unwilling or unable to do that, it's just theoretical.
Phono cartridge / phono preamp electrical models have been very complete for half a Century, and modern Spicey modelling is more predictive than can currently be measured with existing test records, which are less believable. Lots of great threads within this forum on this and related topics. Plenty that were active within the past several months.
All good fortune,
Chris
Phono cartridge / phono preamp electrical models have been very complete for half a Century, and modern Spicey modelling is more predictive than can currently be measured with existing test records, which are less believable. Lots of great threads within this forum on this and related topics. Plenty that were active within the past several months.
All good fortune,
Chris
The initial question is loud: who's making more noise? The postulate (that's how I understood Nick's video message) : B is noisier. And that is wrong.
Live and let live
Chris,
how right you are - and we kids already built our EQ directly into the drive in the 1970s, even with a balanced input ... But the parasitic capacities are not Nick's only concern, are they ?
Give Nick Sukhov a try to take a stand for himself, I think that will shorten the topic.
Regarding the noise issue, I'll be happy to supply the PSPICE number this evening. I'm already quite excited myself- even though I already know the answer.
Bye,
HBt.
Chris,
how right you are - and we kids already built our EQ directly into the drive in the 1970s, even with a balanced input ... But the parasitic capacities are not Nick's only concern, are they ?
Give Nick Sukhov a try to take a stand for himself, I think that will shorten the topic.
Regarding the noise issue, I'll be happy to supply the PSPICE number this evening. I'm already quite excited myself- even though I already know the answer.
Bye,
HBt.
Incidentally, measurement technology is a specialized discipline
Well,
it's not about absolute values (in this case that is not necessary at all), but about the comparison ... comparative measurement!
Post #4 ?!
And the result is absolutely reliable.
However, we can pass the time for a few more hours by beating about the bush.
HBt.
To check frequency response I would suggest a sound card loop test and inject a test signal sweep through 1meg series resistance directly into the MMC input.
Are you unfamiliar with resistor thermal current noise (or Nyquist/Norton_noise)? - go to primary school [ https://en.wikipedia.org/wiki/Johnson–Nyquist_noise ] or simply calculate Thermal Current Noise of Resistor I=sqrt(4kTB/R) [ http://tomahola.com/xtron/noise.html ]: 150 k vs 47 k = 46.4491 p vs 82.98 p = 1.785 = 5 dB . Dont you khow what MarcelvdG`s "cooling" is for ? 🤔
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Dont you know how to use a multilanguage subtitles @ youtube ? 🤔 I wrote about it in the second line of the first subject message ...
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It's quite simple when you use the Norton model of a noisy resistor and assume equal transfers from the cartridge to the output.
The RMS thermal noise current in a bandwidth B is √(4kTB/R), so the higher the resistance, the smaller the noise current. When the resistor is in parallel with a cartridge with impedance Z, a simple noise transformation shows that it has the same effect as a voltage Z √(4kTB/R) in series with the cartridge EMF.
Hence, assuming that two phono amplifiers, one with a thermal-noise-generating 47 kohm and one with a thermal-noise-generating 150 kohm input termination, are designed such that the transfer from the cartridge EMF to the output is the same, then the one with 150 kohm termination will also have lower noise at the output.
When the transfers are not the same, then it gets more complicated. Then again, they can't be very different over most of the audio band, or the sound of at least one of them will be bad anyway.
The RMS thermal noise current in a bandwidth B is √(4kTB/R), so the higher the resistance, the smaller the noise current. When the resistor is in parallel with a cartridge with impedance Z, a simple noise transformation shows that it has the same effect as a voltage Z √(4kTB/R) in series with the cartridge EMF.
Hence, assuming that two phono amplifiers, one with a thermal-noise-generating 47 kohm and one with a thermal-noise-generating 150 kohm input termination, are designed such that the transfer from the cartridge EMF to the output is the same, then the one with 150 kohm termination will also have lower noise at the output.
When the transfers are not the same, then it gets more complicated. Then again, they can't be very different over most of the audio band, or the sound of at least one of them will be bad anyway.
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Dear Nick,
about your question: read carefully post #2,#4,#6,#12,#14 ...
I read the subtitle, in German, several times. I don't think I need to go to elementary school again.
By the way, according to my simulation, your mere 150kOhm termination is even 5dB worse at f=1kHz than with 47kOhm in relation to a total noise figure (nV/sqrt(Hz)).
Only a measurement will show you and others the true noise - and this is exactly why I have decided not to post any simulations for the time being.
It's not about anything here, nobody is interested in whether a Sukhov EQ is the only truth.
Thanks God' we have the blessing of measurement technology.
Merry Christmas,
HBt.
Psst.
6,966µV (150kOhm)
3,899µV (47kOhm)
bw = 20kHz
T = 293°k
about your question: read carefully post #2,#4,#6,#12,#14 ...
I read the subtitle, in German, several times. I don't think I need to go to elementary school again.
By the way, according to my simulation, your mere 150kOhm termination is even 5dB worse at f=1kHz than with 47kOhm in relation to a total noise figure (nV/sqrt(Hz)).
Only a measurement will show you and others the true noise - and this is exactly why I have decided not to post any simulations for the time being.
It's not about anything here, nobody is interested in whether a Sukhov EQ is the only truth.
Thanks God' we have the blessing of measurement technology.
Merry Christmas,
HBt.
Psst.
6,966µV (150kOhm)
3,899µV (47kOhm)
bw = 20kHz
T = 293°k
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Depends on where you look. The noise voltage at the output of the voltage divider will go down when R2 is reduced far enough (#), but the equivalent noise voltage at the input goes up, because the transfer from input to output will go down faster. When you design the amplifier to make up for that, the noise at the amplifier output goes up.
(#): As long as R2 is much greater than the impedance of the cartridge, the magnitude of the parallel impedance will stay nearly the same, while its real part increases due to the extra losses. The noise voltage at the output of the voltage divider, √(4kTB Re(Zparallel)), may then go up rather than down.
Dear Marcel, dear Nick!
I have the suspicion that you don't read other users' posts properly, but just keep repeating the same thing over and over again.
That's totally boring.
You seem to be ignoring two important points: 1. nothing correlates here, 2. the transducer is not modulated with a series circuit consisting of a fixed inductance and a fixed resistor in series to form an ideal voltage source.
Please measure.
Merry Christmas,
HBt.
I have the suspicion that you don't read other users' posts properly, but just keep repeating the same thing over and over again.
That's totally boring.
You seem to be ignoring two important points: 1. nothing correlates here, 2. the transducer is not modulated with a series circuit consisting of a fixed inductance and a fixed resistor in series to form an ideal voltage source.
Please measure.
Merry Christmas,
HBt.
Marcel,
is there any point in repeating in your own words what someone else has written and knows perfectly well?
🙄
is there any point in repeating in your own words what someone else has written and knows perfectly well?
🙄
You seem to be taking the magnitude instead of the real part here. Reactive impedances and admittances don't generate noise, only the real parts do.
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Simply measure, or have it calculated from the raw data set. Just as I suggested in my post #4.
And we are at peace.🙂
And we are at peace.🙂
That is correct, is it?
1k||47k < 1k||150k
And how big are the losses at 20kHz, for example? The real part is (in your sense) the DC resistance and that is even smaller than 1kOhm, right?
Why does the noise power increase with frequency for your two-pole (fixed inductance in series with an ohmic resistor) in the simulation? [V^2/Hz] increases with the frequency and so does the impedance. If you make the scenario real / "reell", it does not increase.
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Suppose you have a 1 kohm and 500 mH cartridge with frequency-independent ESR (which is unrealistic) at 1 kHz.
Cartridge impedance: 1000 ohm + 1000 π j ohm
Cartridge admittance approximately 91.99966835 uS - 289.0254822 j uS
Admittance with 47 kohm in parallel: 113.2762641 uS - 289.0254822 j uS
Impedance with 47 kohm: 1175.464426 ohm + 2999.20884 j ohm
Admittance with 150 kohm in parallel with the cartridge: 98.66633502 uS - 289.0254822 j uS
Impedance with 150 kohm: 1057.848345 ohm + 3098.778606 j ohm
Hence, the noise density at 1 kHz at the loaded cartridge output is somewhat higher with 47 kohm than with 150 kohm. For a fair comparison, you would have to look at the input side, where the difference is slightly larger.
Cartridge impedance: 1000 ohm + 1000 π j ohm
Cartridge admittance approximately 91.99966835 uS - 289.0254822 j uS
Admittance with 47 kohm in parallel: 113.2762641 uS - 289.0254822 j uS
Impedance with 47 kohm: 1175.464426 ohm + 2999.20884 j ohm
Admittance with 150 kohm in parallel with the cartridge: 98.66633502 uS - 289.0254822 j uS
Impedance with 150 kohm: 1057.848345 ohm + 3098.778606 j ohm
Hence, the noise density at 1 kHz at the loaded cartridge output is somewhat higher with 47 kohm than with 150 kohm. For a fair comparison, you would have to look at the input side, where the difference is slightly larger.
It's not immediately intuitive, so I did a quick sim:
@hbtaudio, one way one can think of it is that at lower frequency the cart's DC resistance loads down the "load" resistor's own noise and 150k are easier to load down than 47k. Up to a point, that is, where the reactance gets large enough to reduce the effect to ultimately zero at very high frequencies... if there weren't for the intrinsic+cable+load capacitances. With 150k and my 100pF cap load sim we'll have significant frequency response peaking which skews the 150k load's total noise for worse, accordingly.
Note: The lower two curves are for "electrically cold" load resistors (10x).
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This is how it becomes a shoe
bw=20000 1/sec & T=293°k
Excuse me, please ... there is no capacitive reactance in the model in question. Accordingly, there is also no second-order low-pass filter, and there is also no resonance point. And now let's be clear: we can imagine the voltage source in the video on the far left as a dynamic short circuit. Now both resistors are in parallel.
R1||R2 with
a) 1k||47k=0k9792 and b) 1k||150k=0k9934
UNRa = 562.8nV; 574.8pA
UNRb = 566.9nV; 570.7pA
Now the TP1 with L1 and R2 comes into play:
Unoise= sqrt[UNRa^2 * ( 1 + Omega^2 * (L1/47kOhm)^2]
=939.6nV
Unoise= sqrt[UNRb^2 * ( 1 + Omega^2 * (L1/150kOhm)^2]
=614.63nV
-SNR 3.687dB in favor of the 150k termination. Only this model does not correspond to the actual generator.
Let's assume a current source model:
Inoise = sqrt[INsum^2 *(1 + (R2 / Omega *L1)^2)]
a) 717.82pA*979.2Ohm = 703nV
b) 1.477nA*993.4Ohm = 1.467µV
Now the advantage is extremely clear with the correct termination with 47kOhm, namely 6.3dB less noise in relation to the SNR.
And all this is also confirmed by the simulations. Only the measurement shows now which model and which simplification was permissible!
I maintain that you are making a mistake in your thinking. Your model and your idea of a model do not correspond to the generator at hand in reality.
So who is right? No measurements, no result, even if I have typed total "Bullenkot" into my calculator because your model is not correct.
The simplified and well-known model can only be used for a TP2 to set a suitable Q. No more and no less.
Without measurement, we're going round in circles forever; my simulations could also confirm your assertions, and they do. If I model the transducer as a currentsource, they confirm my view.
Now I can only ask everyone to carry out a very simple measurement or recording and then compare the two data sets with each other. But be careful, parasitic capacitances should be completely excluded, i.e. the jFet belongs in the headshell and is also terminated there, with Rg1= 47k and Rg2=150k.
Bye and merry christmas,
HBt.
Psst.
Without measurement, i.e. without proof, I merely take note of the points of view.
bw=20000 1/sec & T=293°k
Excuse me, please ... there is no capacitive reactance in the model in question. Accordingly, there is also no second-order low-pass filter, and there is also no resonance point. And now let's be clear: we can imagine the voltage source in the video on the far left as a dynamic short circuit. Now both resistors are in parallel.
R1||R2 with
a) 1k||47k=0k9792 and b) 1k||150k=0k9934
UNRa = 562.8nV; 574.8pA
UNRb = 566.9nV; 570.7pA
Now the TP1 with L1 and R2 comes into play:
Unoise= sqrt[UNRa^2 * ( 1 + Omega^2 * (L1/47kOhm)^2]
=939.6nV
Unoise= sqrt[UNRb^2 * ( 1 + Omega^2 * (L1/150kOhm)^2]
=614.63nV
-SNR 3.687dB in favor of the 150k termination. Only this model does not correspond to the actual generator.
Let's assume a current source model:
Inoise = sqrt[INsum^2 *(1 + (R2 / Omega *L1)^2)]
a) 717.82pA*979.2Ohm = 703nV
b) 1.477nA*993.4Ohm = 1.467µV
Now the advantage is extremely clear with the correct termination with 47kOhm, namely 6.3dB less noise in relation to the SNR.
And all this is also confirmed by the simulations. Only the measurement shows now which model and which simplification was permissible!
I maintain that you are making a mistake in your thinking. Your model and your idea of a model do not correspond to the generator at hand in reality.
So who is right? No measurements, no result, even if I have typed total "Bullenkot" into my calculator because your model is not correct.
The simplified and well-known model can only be used for a TP2 to set a suitable Q. No more and no less.
Without measurement, we're going round in circles forever; my simulations could also confirm your assertions, and they do. If I model the transducer as a currentsource, they confirm my view.
Now I can only ask everyone to carry out a very simple measurement or recording and then compare the two data sets with each other. But be careful, parasitic capacitances should be completely excluded, i.e. the jFet belongs in the headshell and is also terminated there, with Rg1= 47k and Rg2=150k.
Bye and merry christmas,
HBt.
Psst.
Without measurement, i.e. without proof, I merely take note of the points of view.