Plus, db are units expressing the ratios of voltages, having nothing to do with current. Anyone?
DeciBels refer to voltage, yes.
The higher gain settings must invoke downstream gain stages that add to signal voltage output by the indicated db’s. Is that correct?
Yes. Transimpedance phono sections have cartridge-dependent gain as you point out. There's a limit to this, for example you really don't want to push most opamps past about 20dB of gain and since the cartridge is an uncontrolled variable in this, you will need to change the gain somehow downstream.
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+1 Dave. Thanks! Your description also supports my prior assertions- so lookout- Raul will be after you next 😁
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Incidentally, I don't intend to further participate in this exchange.
@wynpalmer4
That's unfortunate.
I was/am familiar with Lenz's Law. I was simply hoping you could elucidate on how the input of the phono preamp (where the load is) provides a back EMF, or it I simply was misunderstanding what you wrote. I'm perfectly open to learning something new.
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The distinction between virtual and actual ground is a subtle one.
When an opamp inverting input is used, the virtual ground is extremely similar to a real ground as the action of negative feedback makes it so. The difference is in the error term- i.e. the output voltage divided by the open loop gain of the amplifier, together with any impairments added by the amplification system.
This difference can be extraordinarily small, so arguing that the virtual ground is not a real ground is largely facile.
As far as the source is concerned there can be essentially negligible difference between a real ground and a virtual ground.
Claiming, arbitrarily, that it is prima facie audibly different as far as the source is concerned is not reasonable.
@wynpalmer4 The virtual ground is one thing when it is driven by a resistance in series with the input (which might be a cartridge). Its a bit different when the cartridge itself is that resistance. WRT actual ground, I think we can both agree that a cartridge driving a dead short will not produce anything that can be amplified. A virtual ground is different in that manner 😉
One thing that has come up in this thread is the mention of Lenz's Law, which has to do with the radiation of a magnetic field from a conductor when conducting. Its been brought up in the context of 'back EMF'. I've not been able to discern how this is supposed to work; if driving a loudspeaker (which has significant inductance interacting with a magnetic field) you do get back EMF but in the case of the cartridge an impedance (for the most part) very little inductance in the load is present. So I would appreciate an explanation; as best I can make out the back EMF would be insignificant WRT the source.
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For the record (if you see what I did there) I've never stated that loading a cartridge would cause mistracking! That is simply Raul with his usual logical fallacies again (in this case, the classic Strawman).
What I have stated is that loading the cartridge will cause the cantilever to become stiffer. This fact is unavoidable as anyone grounded in generator and alternator theory knows. This does not mean that the cartridge will mistrack. It means what I said: the cantilever will be stiffer; whether that affects the performance of the cartridge is another matter and other than suggesting that it might affect high frequency tracing ability. I've also been careful to not state what frequency, which may well be ultrasonic; at any rate obviously is an unknown.
Raul, in his on-going quest to simply make me wrong, has been trying to put words in my mouth. Fortunately I'm way to far away for him to actually do that 😁
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That is not the case in this instance.
This gives me confidence in my belief that the load on the cartridge is responsible for the sonic changes that I reported above.
I don't doubt that it could be affected. Loading the cartridge causes the cantilever to become stiffer (more work is being asked of it and that has to come from somewhere: the cantilever is thus harder to move)- and thus can introduce the possibility that even though the bandwidth of coil is unaffected, the mechanical aspect of the cartridge will be affected by that added stiffness- possibly making it less able to respond to higher frequencies.
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Is there any chance the major contributors to this thread will be designing, producing, and selling a new type of phono stage in the near future?
We've been making phono sections since 1989. Our MP-1 had the first fully differential balanced phono section made.
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in an ideal current driven phono stage that uses an op amp to sense current, is the coil of the cartridge connected to that virtual ground that you describe? One end of course. If that is the case, where do they connect the other end of the coil?
@lewm
Ground.
The result of this experiment was interesting. Even thought the gains of both situations were the same, the 4.7kΩ load through the 1:8 sounded a good 2dB louder. When the gain of the 2.2Ω load was bumped 2dB suddenly it was preferred and then going back to the 4.7kΩ 1:8 at the same +2dB level started to hurt my ears.
@intactaudio
Transformers transform impedance. Further, to prevent ringing they must be properly loaded at their output, to something called 'critical damping' where a squarewave input to the transformer results in minimal overshoot.
If not loaded the transformer can 'ring' with excess harmonics- its making distortion. Your ears will respond to that as sensing it as loudness and yes, it might even hurt if the volume is up a bit.
Because transformers transform impedance its not necessary to load the cartridge directly with a low impedance in order to achieve a low impedance- you can do that on the output side of the transformer as well. IOW putting a lower impedance load on the output will reduce the load impedance the cartridge sees. This is because the only isolation a transformer offers is galvanic and DC; it does not offer impedance isolation.
This loading issue is one reason I avoid SUTs- you do have to manage their care and feeding. If the SUT is designed for a specific cartridge, the correct load is probably 47K (but will be different if a different cartridge is used). Its really important to keep interconnect cable capacitance to a minimum. But if you simply have enough gain and your phono section has no worries with the RFI that will be generated by the cartridge/cable interaction, then you have no worries- its plug and play.
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My belief is that the nature of current amplification vs. voltage amplification (namely how they load a cartridge with low vs high impedance) has a dramatic effect on the mechanical behavior of the cartridge and may be partially responsible for the differences heard between the two types.
It also has a lot to do with distortion and RIAA EQ differences. Separating those out might be a bit difficult.
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I haven't kept track of which current driven phono stages on the market use op amps or discrete transistors or tubes in the input I/V stage, although I don't know of any that use a tube or how you could use a tube in that fashion.
@lewm Transimpedance preamps have to use opamps since that is the only way to get a virtual ground. You can use discreet opamps as those do exist (a friend of mine designed his own) but you do need an opamp to pull it off.
Take a 30Ω cartridge and hook it up to an ideal voltage amp and then another one and hook it up to an ideal current amp. Now take two AC microamp clamps and monitor the current output of each cartridge. Will the currents be the same for each cartridge?
@intactaudio I doubt it but that depends on the actual impedance that the cartridge is driving in either case.
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I have been complaining about that on this forum over and over again. Only be direct questioning can one find out what the input impedance of these "current driven" phono stages actually is, and it's typically from a few ohms in the best case to as high as 20 ohms.
@lewm I don't know how you could specify the input impedance; it varies with the impedance of the cartridge! Again, the gain of the circuit is defined by the ratio of the feedback resistor vs the input resistor (which is literally the cartridge itself). So if the cartridge is 30 Ohms and the feedback resistor is 300 Ohms, the gain of the circuit would be 10 or 10dB. If the cartridge were only 15 Ohms that would mean the gain of the circuit is 20 (16dB). So the input impedance can only be defined by the fact that a virtual ground is present. But you have another problem, which is that with almost any opamp made you run out of Gain Bandwidth Product over about 20dB of gain or thereabouts.
This means that with cartridges that have a very low impedance the circuit may lose neutrality. Personally I would prefer to have the gain be a set thing so that the cartridge would not be able to affect the phono section in that manner.
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(1) what if the input device is a discrete transistor or a tube, not an op amp?
@lewm
Then there won’t be a virtual ground. So right away its a voltage amplifier not a current amplifier.
(2) the only way I can imagine two points separated by 5 ohms but at the same potential is if and when there is no current flowing. How does that work in this case?
Opamps have nearly infinite gain when open loop; the feedback resistor and the input resistor thus define the gain of the circuit and the virtual ground is formed at the intersection of the input resistor and the feedback resistor (see my prior posts for more information).
There is no connection between actual ground and virtual ground; the latter is created as a result of the feedback meeting the input signal. So there isn’t (as in the case of 5 Ohms) 5 Ohms between the ground and the virtual ground. In fact the actual impedance is much higher.
The ’0 Ohms’ value you see in so many phono sections that have transimpedance inputs probably isn’t helping people to understand what is going on. That value is probably the marketing department talking since they probably didn’t understand what a virtual ground is.
The tricky bit is that in a transimpedance input, the cartridge itself is the input resistor. This means that the actual impedance load on the cartridge varies with the impedance of the cartridge itself- and with it, the gain of the circuit. As I pointed out earlier, the lower the impedance of the cartridge the higher the gain of the circuit.
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Why not? Consider the case to be an ideal current amp with the appropriate series resistance added so the input impedance is 5Ω. In this gedanken world the ideal voltage amp and the ideal current amp sound identical.
It seems to me that you are still conflating virtual ground and actual ground as the same thing! As I said before this leads to confusion.
So the 'Why not?' is the same as before: Because that 5 Ohms is a virtual 5 Ohms instead of a real 5 Ohms. The cartridge is not loaded at an actual 5 Ohms. 'Virtual Ground', again, is opamp parlance for a point in the circuit that exists at the same potential as ground but isn't actually ground.
I recommend that you read up on opamp operation since this seems to be the hanging point. Here's a short tutorial opamp virtual ground.
If you don't want to do that, just keep in mind that 'virtual ground' isn't the same as actual ground. So the cartridge would not be loaded at an actual 5 Ohms even though the virtual ground is 5 Ohms.
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The first question that needs to be addressed is in a perfect world with ideal amplification, will a 40Ω cartridge sound the same into a 5Ω load as a 47kΩ load?
Right up to this point I was in agreement with the prior text of this post. With this question we can safely say the answer is 'No.', assuming that the perfect amplification is voltage amplification. The question cannot be answered at all if current amplification is used.
If that is a static load (IOW, a resistor) then the cartridge will be making 4 orders of magnitude more work! That work has to come from somewhere, otherwise a new branch of physics is created 😉
So it will certainly cause the cantilever to be harder to move (stiffer) and that will affect how the cartridge 'sounds'. It won't affect the bandwidth of the coil in the cartridge at all- but will have an enormous affect on the mechanical aspect as the coil will have become a significant load with 5 Ohms loading it in turn. Also, the output level will be considerably decreased!
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The two statements you quoted from me seem to align well with each other and I was unaware that there was a different versions of ohms law for static vs dynamic loads
At any rate, Moncreif threw off his numbers by using a 5 Ohm load
I agree it is an interesting choice but his measured results clearly show signal and if if anything a lower noise floor so I fail to see where the issue is or how it invalidates his results.
In the case of a transimpedance input, feedback is applied to the output of the cartridge. Its a fair amount of feedback too- the more output the cartridge has the more feedback. That is quite a bit different than a simple resistor- you can’t equate a virtual ground with a static resistance- for one thing, you need an opamp to even create a virtual ground.
Ohm’s Law certainly is in play (how could it not be), but the issue here is that while a cartridge has its output at a virtual ground, that is significantly different from having the cartridge drive 0 Ohms (IOW, if it were actually tied to ground). In the case of the former, the actual input impedance is dynamic and isn’t actually 0 Ohms, so amplification can occur because a signal is present. In the case of the latter, the load is actually a short and because the signal is shorted out, no amplification can occur.
’Virtual’ means ’almost or nearly as described, but not completely or according to strict definition.’ If the 0 Ohms of a virtual ground is conflated with the 0 Ohms of actual ground, confusion is the direct result.
Real ground is when a terminal is connected physically to the ground or earth, whereas virtual ground is a concept used in opamps in which a node is assumed to have the potential that of the ground terminal.
I say Peter threw off his numbers because he chose a value that no-one would ever use (and a static value at that, no opamp involved, so the load was causing the output of the cartridge to change in a significant way, whereas real-world loads have negligible effect) since its a reasonable expectation that if you are going to use a load on a cartridge, you’d likely start with one that is 10X the source impedance of the cartridge.
I can see using static values lower than that, but not one that’s actually lower than the the source impedance! Barring a good explanation for that, when I read that in his article I found I simply had to take his results in abeyance. I’d have to read the article again (Google defeated my attempts to locate it just now), but IIRC he had some variables left uncontrolled that I felt at the time might affect his results.
One way around this is to ignore that transimpedance inputs exist and simply focus on what happens to the output of any source when the load is a fraction of that of the source. That’s really what I’m getting at here, if Peter was actually suggesting that we use such loads both back then and now the extra gain needed would be impractical.
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I didn’t mention a dead short and was only referring to the two terminal impedance the coil of a cartridge sees. Any reference to ground be it real or virtual does not factor into the load seen by the coils. Surely there has to be an actual input impedance for a current amp and it has to be low otherwise the coils will not generate any current to amplify.
Do you see how this above does not jive with this:
that depends on what load the input of the following stage gives.
If it is a voltage amplifier with 47kΩ, a 30Ω cartridge loaded by 5Ω it will be 17dB down. When you replace that 30Ω cart with a 2Ω cart the output will only be down 3dB. If it is a current amplifier with a 1Ω input impedance a 5Ω parallel load will lower the current into the 1Ω input node by1.6dB with both the 2Ω and the 30Ω cartridge. The absolute currents will be different for the 2Ω and 30Ω carts but the relationship of how the 5Ω load affects a 1Ω input impedance stays the same.
Transimpedance phono sections have a dynamic load whereas a resistor is a static load. This is because the so-called 'virtual ground' (which will be 0 Ohms) occurs where the feedback resistor of an opamp meets the input resistance, which in this case will the cartridge itself.
At any rate, Moncreif threw off his numbers by using a 5 Ohm load, since that is not only not a real-world value that no-one would ever use, but is also one that would significantly decrease the output of any LOMC cartridge to the point that its output would be unusable (which is why its not real world...). Keep in mind that transimpedance phono sections didn't exist back then. As I mentioned previously, if he really wanted to make his point valid he needed to show the results using real world loads that are actually in use.
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When you consider the proliferation of "current mode" phono inputs in vogue the 5Ω value isn't so strange.
Transimpedance phono sections have the cartridge driving the 'virtual ground' of an opamp circuit. That's quite a bit different from driving an actual short (0 Ohms)! For example, if the cartridge isn't present, the virtual ground does not exist. They really aren't an example in this context.
Try placing a 5 Ohm resistor across the output of any LOMC cartridge and see what happens 😁
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I would say no since the two loadings PM shows for his IMD are 5Ω and 100Ω both of which will surely damp any possible LC resonance which would then eliminate the whole RFI aspect.
@intactaudio FWIW 5 Ohms as a load is so low that most cartridges will produce measurably less output whereas at 100 Ohms the output would hardly be affected. At least this is true of any LOMC cartridge I've measured (a cartridge of 30 Ohms would be affected by both). Knowing that, when I saw the 5 Ohm value in Moncreif's measurements I was forced to take his conclusions with a grain of salt. I think he would have produced better science with 75 Ohms as his lower limit.
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Isn't that essentially suggesting that compliance has no effect on the sound of a I cartridge?
I need to clarify: It has no effect on the output of the coil. It certainly has an effect on the cantilever, and if you look at my prior posts you'll see that I suggest this may affect its ability to trace higher frequencies.
did you ever get a chance to look at the copy of IAR #5 I sent you? I find it entirely plausible that making the cantilever easier / harder to move would have a sonic impact on the sound of a cartridge.
I did- thanks- and agree, since this likely has an effect on how it tracks. IMO what Moncrief did not show is where the IMD was coming from; IMO it is caused by the phono section rather than the cartridge directly.
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The lower value 200 ohm vs. 47k ohms means more loading which tends to attenuate higher frequencies (because we hear things in terms of overall balance, attenuating highs can also be perceived as more bass).
If light loading is entirely added intermodulated distortion and the "real" and "accurate" sound is what you get with high loading, I will take distortion.
The loading has no effect on the cartridge other than making the cantilever harder to move. You can take any LOMC cartridge and put it on the bench. Using a squarewave generator (at a very low loutput, so as to not damage the cartridge), you can put the cartridge in series with the squarewave and then measure the output. With no resistor in parallel this is an unloaded inductor. What you will see on the oscilloscope is a nice squarewave, looking really quite a lot like the input, neither attenuated or rounded and with no overshoot! This is simply because the inductance of the coil is too tiny to ring at audio frequencies.
So the 'brightness' we often hear is coming from somewhere else!
Putting a resistor in parallel will not affect that. So the resistor is not affecting the bandwidth of the cartridge. But it is interacting directly with the electrical resonance, which is a product of the inductance of the cartridge and the capacitance of the tonearm cable.
As we can see from the page http://www.hagtech.com/loading.html
the peak can be substantial. The reason is something called 'Quality' or simply 'Q'. Inductive coils can be long and narrow, having a low Q value, or short and wide, having a much higher Q value. The higher the Q value the higher the peak and the narrower the band of frequencies it covers. LOMC cartridge have high Q coils in them for lowest mass.
Some preamps don't like having all that RF noise caused by the peak at their input. So they do weird things and one of them is distortion. Loading knocks out the electrical resonance and if the preamp has a problem with the RFI, the brightness along with it.
One would have to compensate for the loss of gain from using such a resistor, but, one will hear quite a difference in sound
This statement is only true if the preamp is sensitive to the RFI at its input. If not, no difference will be heard. Most designers simply don't take the implications of this electrical resonance into account in their phono designs, which is why this loading conversation persists.
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Well, played many favorite records at 47k for awhile. Didn't like it overall. A little to bright, if that's a good analogy.
@quincy It is. If you experience brightness, IME this is because the phono section really isn't happy with RFI at its input. So loading is really the best thing you can do.
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Recommended loading is between 91ohms and 47kOhms determined by listening.
@quincy I love that Lyra states things this way- Jonathon Carr is well aware of how differently phono sections behave when presented with the RFI generated by LOMC cartridges. Normally you'd expect a very specific value for something like that, not something determined by 'listening'. But until more phono preamp designers get the implications of the LOMC cartridge making RFI, about the only way to do will be by 'listening'.
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I don't know what implications there would be with a SUT.
I suppose this is all variable, depending on the capabilities of the phono stage, as Ralph @atmasphere seemingly explained in the previous thread.
Normally you have a rather prodigious electrical peak (resonance) with a LOMC cartridge. It might be as high as 5MHz or as low as 100KHz. Generally speaking if a lower frequency the peak will be less prodigious; this depends on the Q (Quality) of the coil in the cartridge. Lower output cartridges tend to have a higher Q since the windings are shorter and wider with respect to their length; this done to keep their mass down.
That peak can be a good 30dB (1000x) higher than the signal! So if you have a 0.5mV output, perhaps half a volt at 2-5MHz is presented to the input of the preamp. Some preamps don't like that (and might sound bright or weird as a result)! For example, if the phono section employs high quality opamps, its quite possible that the circuit will have some bandwidth that that frequency and so could be overloaded (if overloaded, you'll get ticks and pops) or otherwise affected by an RF signal that powerful at its input!
That's why the loading resistor can be so important (its used to detune that peak, thus eliminating the RFI it generates)! Its certainly not to change the bandwidth or frequency response of the cartridge; at least with all the LOMC cartridges I've tested, all showed no ringing at audio frequencies- quite simply their inductance is too low for that, which implies that the load isn't needed to roll them off to prevent brightness- therefore something else is causing the brightness.
When you use an SUT, it simply lacks the kind of bandwidth to pass RFI at 200KHz or 3MHz- whatever the cartridge might be generating. The only thing with SUTs is that since they have a fairly high inductance, they need to be loaded to prevent ringing ('ringing' is distortion, caused by overshoot of the signal and causes brightness and harshness). If you have an SUT designed for your cartridge, the 47K might be a proper load so no worries. 'After market' SUTs like those made by Jensen Transformers are a different matter. They require specific loading that varies according to the source impedance of the cartridge (since a transformer transforms impedance, the correct load at the output will vary depending on the source impedance). They publish the values needed for a particular cartridge in a pdf file available off of their website. IMO Jensen Transformers make some of the best SUTs available.
I suspect that since the RFI issue is conveniently eliminated when using an SUT that this is why some people say that cartridges sound better when played thru them. Personally I don't find this to be the case, but the preamp I use has no issues if RFI is present at its input and can't be overloaded by that RFI, even if its at 0.5 Volts.
So where this is going: If you have no SUT and 47K sounds bright, you need a loading resistor to eliminate the RFI at the input of the phono section. IOW the brightness isn't the cartridge being bright.
If you have an SUT you do need to investigate the issue of its loading, which might be satisfied by the stock 47K input load of your phono section. BTW, if the SUT is designed for that and you try to drive a 100 Ohm load, you'll find it quite rolled off! There is something called 'critical damping' that applies to things like SUTs (and other interstage or line level audio transformers). If you are loading above critical damping there will be ringing causing the transformer to be bright; if you load with a resistive value too low the transformer will be rolled off.
All phono cartridges are generators, having a coil that interacts with a magnetic field. If you have a regular generator making AC power, for example a portable Honda generator for backup use or camping, you'll notice that when you load it more the engine has to work harder to keep the generator (or alternator) spinning. This is because electrical energy has to come from somewhere ('free energy' is still the stuff of conspiracy theories) so the generator presents a harder load to the engine when its asked to make more current.
The same thing happens in a phono cartridge- the lower the resistive load it drives, the harder it becomes to move the stylus since that is where the mechanical energy is input to be converted to electrical energy. IOW the cantilever becomes stiffer. If you have a means of testing the mechanical resonance of your arm/cartridge combination, you can see that this affects the mechanical resonance since in essence you are reducing the compliance of the cartridge.
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