09-12-11: Almarg
I would just add to what has been said the thought that if there are in fact mechanisms by which warmup beyond the point of thermal stabilization can produce audibly significant changes, I see no reason to expect those changes to necessarily be for the better. And I would certainly expect both the magnitude and the character of the changes to be dependent on the specific design.
a lot (in fact, i suspect the great majority) of high end audio is sold to people who know little about electronics. so in some regards, the most important feature of high end audio products is marketing: because it is through marketing that you make the suggestions that influence consumer perceptions about audio products. you can convince people to believe just about anything with enough suggestion (e.g., a surprisingly large percentage of people still believe that saddam heussein was involved in 9/11). to the extent that people believe that there are sonic benefits from keeping equipment on 24/7, then that's just what they believe. however, other than for the possibility of bad system design, i see no scientifically valid evidence to support such beliefs. first, while it is true that the electrical characteristics of semiconductor devices does exhibit some thermal dependence, it is also true that semiconductor devices are characterized over a temperature range that goes from well below zero to well above boiling point. so i question whether there really is that significant a difference in the sound of audio electronics over the operating ranges of these devices. second, even accepting that there might be some audible difference in audio electronics during the "warm up" period (and, btw, the testing methodologies to evaluate this are *highly* unreliable), what really counts is the junction temperatures of the devices, not the ambient temperature that you feel when you touch the heat sink. the junction temperature would rise much more quickly than the ambient temperature, so i would expect the devices to reach their "warm up" within 15 or 20 minutes at most - not over a period of hours. i would find a more persuasive argument that the sound character of the speaker changed (for largely mechanical reasons) during the course of operation. as to the concept of "break in": the term, as audiophiles use it, is really a psychological process, not an electrical process. it reflects the time that it takes for you to get used to the sound of a new component. when you install a component in an audio system, it will likely have it's own sonic signature. when you first hear it, you will likely be very sensitive to the differences in sound relative to that to which you may have become accustomed. however, over time you will get used to the sound and will become less sensitive to it. as an experiment, change the loading and/or gain on your phono stage. depending on how much of a change you make you should observe a noticeable difference in sound. if, for example, it sounds too bright at first, over the course of time, and as you get used to the sound, you may perceive that the sound is less bright. you've just done audiophile "break in" on an existing piece of audio equipment... |
09-12-11: Rower30
Pre amp to power amp leads need low cap. too. Big ass copper does ZERO at the lengths we're talking about. The attenuation is NOT the problem in leads with no current (E=IR). There is no real voltage drop across the leads. What there is, is an increase in capacitance as the leads get longer (say bye bye to higher frquencies at 3 dB per octave at the filter fo frequency which drops as the leads get longer). So all the gold and silver on earth won't do anthing except take your money. What WILL help, is short LENGTH and low capacitance.
Balance leads have HALF the capacitance as single ended RCA leads. The two wires are about TWICE as far apart, leading to LOW capacitance (about 8 Pf ft verses 17 PF/ ft). So the roll off is half the single ended RCA's roll-off. And, you get NOISE rejection added in for a bonus.
i don't quite understand why capacitance is of significance. yes, it is true that under the transmission line model, the wire induces signal delays that vary with frequency (i.e. distorts the signal), &c. but the transmission line model generally applied to signals that operate at microwave frequencies, not audio frequencies. as an example, resistance of audio cords is measured in tens of ohms per kilometer. so let's say that you have a cord with a resistance of 50 ohms/km; for a 1 meter cord, that means that the RC time constant is well into the megahertz range. what i don't understand is why roll off would even be a consideration in the audio frequency range. |
09-12-11: Rower30
What there is, is an increase in capacitance as the leads get longer (say bye bye to higher frquencies at 3 dB per octave at the filter fo frequency which drops as the leads get longer). So all the gold and silver on earth won't do anthing except take your money. What WILL help, is short LENGTH and low capacitance.
09-12-11: Almarg
Some minor factual corrections to your post:
"3db per octave" should be "6db per octave" (at frequencies above the 3db point).
rower 30 is correct, 3dB/octave is the figure of merit. frequency response charts measure a quantity (typically voltage) as a function of frequency. 3dB represents the point at which the quantity is reduced by a factor of 1/2. 6dB would represent a 3/4 reduction in the quantity. people look at 6dB when referring to power but that applies only to the situation where a voltage (typically an output voltage) is being delivered to a load. in that case, when the voltage is reduced by 1/2, the current is also reduced by 1/2 which means that the power is reduced by 3/4. but that does not apply in the case of a balanced input. in that case, you have a voltage divider, so when there is a 3dB reduction in voltage at a node, there is also a 3dB reduction in power at that node. take a look at a bryston schematic and you will note that they label the balanced input as corresponding to a 3dB signal reduction. the explanation that i just gave is the reason why. |
09-12-11: Almarg
Paperw8, what Rower30 is referring to is the fact that the capacitance of a line level interconnect will interact with the output impedance of the component which drives the cable, forming an RC low pass filter. The "R" corresponds to that output impedance, not to the resistance of the cable.
oh, i see. my initial reaction is that you would have to have a reasonably large output impedance, but i'll think about this some more. |
09-12-11: Almarg
Paperw8, on the 3db vs. 6db per octave question, see this.
the equation for dB measurement is: q_dB=10*log(q1/q2) where q1 and q2 are measured quantities and q_dB is the dB measurement for those quantities. when q1=q2/2, q_dB is -3dB. if you read closely the reference that you cited, you can see that what the writer is saying is clearly wrong. 6dB/octave is the falloff for a *second* order filter; the filter shown in the cited reference is a first order filter. |
09-13-11: Almarg
Paperw8, we had discussed the definition of db for electrical signals in another thread a while back, and as I indicated then, with all due respect you are simply wrong. Please do some further research, and I think you will see that:
db = 10log(P1/P2) = 20log(V1/V2)
where P1/P2 is the ratio of two powers, and V1/V2 is the ratio of two voltages. 6db is 6db, regardless of whether it is derived on the basis of power (where it represents a factor of 4) or voltage (where it represents a factor of 2).
you were wrong when we discussed this matter earlier and you are still wrong. one of the problems is that people cite equations that they saw somewhere by rote without understanding what the equations really mean. look at the equation that you cited for power_dB: 10log(P1/P2) = 20log(V1/V2) the left hand side is a ratio of *power* levels while the right hand side is a ratio of *voltage* levels. power and voltage are 2 different things. the right hand side expresses how power levels change in response to changes in voltage levels. however, the equation that you cite is only correct when you are referring to a configuration where a voltage is being delivered to a load; there are other circumstances in which the equation that you cite is *not* correct (i explained that earlier so i won't go through that explanation again). so to give some context as to how the equation that you cite above really works, take a given voltage, v1, delivered to a load, z. the current, i1, through z in that case is: i1=v1/z the power delivered to z, p1, in that case is: p1=v1*i1=v1*v1/z=v1^2/z now deliver a voltage, v2, to z be described as: v2=(1/2)*v1 now the current, i2, through z is: i2=v2/z=(1/2)*v1/z the power delivered to z in this case, p2, is: p2=v2*i2=[(1/2)*v1]*[(1/2)*v1/z] p2=(1/4)*v1^2/z since: p1=v1^2/z p2 can be expressed as: p2=(1/4)*p1 so what this all means is when the voltage delivered to a load is reduced to one-half it's original value, the power delivered to that load is reduced to one-quarter it's original value. so to use the equation for dB we have: v_dB=10*log(v2/v1)=10*log(1/2)=10*(-0.3)=-3dB p_dB=10*log(p2/p1)=10*log(1/4)=10*(-0.6)=-6dB that's how this stuff really works: there is no magic where somehow -3dB corresponds to a reduction to half its original value when talking about voltage but magically -6dB corresponds to a reduction to half its original value when talking about power. |
09-13-11: Almarg
I could perhaps further support the credibility of my statements by describing my academic and professional background, but we've already diverted the topic of this thread far enough.
here's the way it works almarg: i don't know your background and you don't know mine. that's fine with me because i don't care what you did in school years ago. that means that the basis for the credibility of any statements made here is the *substance* of our respective comments. if your argument has degenerated to an assertion of "just trust me" then i respectfully decline to do so... |
09-13-11: Almarg
Paperw8, let me supplement the reference I previously provided with this one, supporting my contention that db = 20log(V1/V2), rather than 10log(V1/V2) as you have claimed. It was written by someone possessing technical credentials that are utterly impeccable.
i didn't read henry ort's biography, but i did read his explanation of the decibel. you may be surprised to read this, but i actually agree with with mr. ort has stated. but i don't agree with what you agree with what you have stated. the reason being that i think that you have misinterpreted what mr. ort wrote to support your conclusions. in fact, if you read mr. ort's explanation more closely you will see that my description follows what he wrote. |
09-14-11: Almarg
Kijanki, no problem with the beer. Thanks for your comment, which as I'd expect is totally correct, with P representing power and U representing voltage.
hold on, this goes to my previous comment: you can't just cite equations without understanding what the equations mean. first, the equation 20log(v2/v1) is a dB relationship in *power* not in voltage; a relationship which is true under certain conditions (which, btw, are articulated in the previously cited henry ort reference as well as in my previous comments). second, what the equation means is that when voltage is reduced to half it's original value, power is reduced to one quarter it's original value. |
09-14-11: Atmasphere
The reason power cords make a difference despite the limitations described in this statement has to do with voltage drop in the power cord. It also has to do with how DC power supplies work.
These effects can be quite measurable!! For example, I have seen a 3 volt drop across a 6 foot power cord cost a tube amp of about 35% of its total output power. If you want a reason to look for, that one is pretty basic!
i'm still not convinced of the significance of upmarket power cords, but at least your comments give me something to work with more than just "i believe". at least you articulate mechanisms which can be discussed. here my comments to which i would be interested in reading your reaction. first, i will address the issue of voltage drop across the power cord. while you didn't state the current draw the produced this 3 volt drop, i will assume a current draw of around 30 amps, a reasonable figure for a practical system that would be in a residential setting. in that case, the cord presents about 0.1 ohms of resistance; for a 6 ft power cord that would translate to a resistance of about 50-60 ohms/km. that to me seems like a realistic resistance for a wire. the thing is, even if you went to an upmarket power cord, that resistance is not going to go to zero. so even if an upmarket power cord improves the wire resistance by 10%, that amounts to only about a 0.3 volt difference. if your amplifier is sensitive to that small of a change, there are probably a number of problems with that amplifier. first, it would be sensitive to your turning on lights, or appliances, or a range of devices that would draw a current because the voltage coming out of your wall can be influenced by this kind of stuff. second, i would suspect that a bigger contributor (but more difficult) contributor to voltage loss would be due to resistances in transformers and diodes/active components within the power supply circuit. but real circuits don't operate under ideal conditions. that is why power supply regulation is so important. while your comments suggest that you are aware of the mechanisms for power supply regulation, the question in my mind is why wouldn't a designer of audio equipment have the same awareness? if you really are observing the dramatic changes in output power that you are reporting, then that would lead me to suspect that you have a real power supply reguation problem since it would appear that you have a extremely sensitive amplifier. if true, it just seems to me that your amplifier wasn't designed for the real world, in which case you would probably still have problems after you bought an upmarket power cord.
09-14-11: Atmasphere
But there is more. Most DC power supplies have a power transformer, a set of rectifiers and a bank of filter capacitors. The circuit draws its power from the filter caps, which are replenished by the transformer and rectifiers. Now its a simple fact that the filter caps are not seriously drained in between cycles, else the amplifier will not work very well. But the rectifiers will only turn on at a certain time- whenever the voltage from the transformers is higher than that of the filter caps.
This only happens at the peaks of the incoming AC power. IOW, the power supply is only doing its work in very short bursts of energy. Now in normal operation what this means is that the diodes are doing some fairly high frequency service; they may only be on for a few milliseconds per cycle. This is called commutation- the turning on and off of the rectifiers, and the current that might occur at these times can be quite prodigious depending on the circuitry of the audio device.
Meanwhile the power cord may be doing double duty, especially if the amplifier has a filament circuit.
Consequently you have two effects: voltage drop at 60Hz, and the current ability at a fairly high frequency. The greater the demand on the cord the greater the likelihood that its effects will be audible on this basis; OTOH the lower the current and the more regulation employed by the audio device the less audible it might be.
i get the part about the diode switching on and off, and i get the part about the on period being very short. but for a 60Hz ac line voltage, that on/off cycle should only happen once/second. so i don't see where the "fairly high frequency" stuff is coming from that you described. as i see it, for the diode on/off cycles to occur with the frequently that you suggest would imply that while the line voltage is in the declining phase, the capacitor is discharging faster than the line voltage is decreasing; that sounds like extremely bad circuit design. as far as the amount of current that is pumped through the diode to charge the capacitor, it depends on how tightly you need to limit ripple in the dc voltage. but even still, the current pumped through the capacitor is not the current drawn from the wall. i mean, it's not like the power cord is jammed into the circuitry straight-on; it goes through a transformer. i would expect that the transformer is going to do something for you such that the amount of current drawn from the wall is somewhat less prodigious than the current through the diodes. which would mean that the current through the power cord would be less than the current through the diodes. so if you used diode current as the basis for an estimate of voltage drop across the power cord, you would have an exaggerated figure.
09-14-11: Atmasphere
Meanwhile the power cord may be doing double duty, especially if the amplifier has a filament circuit.
i don't understand the "double duty" comment. this might be a concept related to tube amplfier designs, but i don't know much about tube circuits. i'm old enough to remember how great is was when they came out with transistor radios, so for me it's ridiculous to go back to tube devices. maybe younger people have a different perspective... |
09-14-11: Almarg
Paperw8, the fundamental misconception you have, which ultimately leads you to incorrectly assert that first order filters roll off at 3db/octave, is the notion that the numerical db value describing the ratio of two signals applied to a given resistive load will be different depending on whether voltage or power is being considered.
while you have posted many informative comments on this forum, one of my criticisms of you is that you post references that you seem to have only half-read, and then draw conclusions that don't stand up to a closer reading of the reference. as to the first reference you cited, the writer presented a first order RC filter. he also showed the transfer function for the filter. so if you really do want to know the behavior of the circuit, don't take my word for it, just look at the equation and figure it out for yourself. if you look at the *asymtotic* behavior of that circuit shown (and by "asymtotic" i mean frequency>>RC), the transfer function is: v_out=(C/f)*v_in where C is a constant and f is frequency. let me first establish that an "octave" means a doubling of frequency. given that, for each doubling in frequncy, the output voltage v_out is reduced by half relative to v_in. so if you want to compute the decibel change in voltage for each octave: v_dB=10*log(v_out(f)/v_in(f)) v_dB=10*log(1/2)/octave v_dB=10*(-0.3)/octave v_dB=-3dB/octave that's the result; it's not subject to your opinion or my opinion, that's just what it is.
09-14-11: Almarg
I'll mention, btw, that early in my career as an electrical design engineer, a great many years ago, I had the exact same misconception, until my boss enlightened me.
The value of that number is one and same, whether voltage or power is being considered. For a given resistive load, reducing the applied voltage by a factor of 2 reduces power by a factor of 4 (as you agree), and the change in signal level is 6db. Period.
what "signal level" are you referring to? i've got a deal for you almarg, if you're so convinced that i am wrong, then rip an *unedited* copy of my postings and send them to henry ort and ask for his comments on them. that will be interesting... |
09-14-11: Almarg
The number "10" in all of your equations that I've quoted above should be "20," since you are computing the number of db based on a voltage ratio. I've explained it, Kijanki provided a proof of it, and Ott provided a proof of it.
kijanki, did you say what almarg says you said? if power is reduced by half is that a 3dB reduction or a 6dB reduction. the problem is that you're not clear on what ort is talking about. as to ort, when ort talks about dB he is talking about *power*. in the webite that you cited ort says "The dB is a logarithmic unit expressing the RATIO of two powers". what ort also says as an aside "Although the dB is defined with respect to power, it has become common practice to also use it to express voltage or current ratios". so while he is saying that some people compute dB for voltage, or current, ort feels that, properly stated, dB is a ratio of power levels. so when ort talks about dB, he is talking about a ratio of power levels. so when he refers to the 20*log(v2/v1) equation, the dB is a dB in *power*; that's why he calls it a "Derivation of dB as a Voltage Ratio". ort isn't saying that 20*log(v2/v1) is a voltage dB, what he is saying is that it is a power dB expressed as a ratio of voltages. in ort's view, it is not proper to talk of dB as a ratio of voltages, but he recognizes that many people do so. so what i would expect ort would say about my postings is that it is not strictly proper to talk of a dB in voltage but that it is commonly done even though the only proper dB computation is for power. thus, when you refer to the 20*log(v2/v1) equation, you have to be clear that, in ort's view, "This is only correct, however, when V1 & V2 (I1 & I2) are measured across the same value of impedance". |
09-14-11: Kijanki
Voltage ratio in electronics is, and always has been 20log(v2/v1). Pretty much anything other than power is always 20dB(k2/k1) including sound pressure, sound level etc.
-3dB of voltage means 0.708 of a value.
i see; so in your mind a 3dB reduction in power "has always" meant that power if reduced to 0.708 of it's original value and not 0.5 of it's original value. ok...i can see that when you drill down, there is much inconsistency among the dB equation citing crowd. you're not even consistent with almarg; at least he realizes (i think) that a 3dB reduction in power means that power has been reduced to 0.5 of it's original value.
09-14-11: Kijanki
As for power cord. What you describe is average value. Amplifier might take 10A on average but it will be taken in narrow spikes of 100A or more, causing 10V drop on your 0.1ohm power cord equivalent to 20% drop in max power.
you would blow your circuit breaker if you tried to send a 100A surge through a power cord. audio equipment is not designed for use in industrial settings, they're used in homes for the most part.
09-14-11: Kijanki
Capacitor inside is not discharging faster. The problem is that it discharges very little. If voltage drops from one peak of 120Hz full wave to the next only 50mV (ripple voltage) and amplitude is 50V then charging will be done only in arccos((50V-50mV)/50V)=2.6deg. Charging pulse will be 16.6ms*2.6deg/360deg=0.12ms. Pulses will be a little wider because of all inductance in the circuit but as Atmasphere said - in millisecond range.
i think you misunderstood my point. i was not commenting on the *length* of a pulse, i was commenting on the *frequency* of pulses. as i read atmasphere's comments, he was saying that these pulses occured at a high frequency. what i was saying is that the pulses only occur once per second - hardly what i would call high freqeuency. |
09-14-11: Atmasphere
Paperw8, the amp I mentioned in my example draws 500W from the wall at full power.
i've got light fixtures that draw more power than that, so 500w is not a ridiculous power draw.
09-14-11: Atmasphere
If you dig around on this forum, you will find that the idea that the stereo sounds better late at night is a fairly common experience. IOW you are right that other loads on the AC line do indeed affect the sound of many stereos.
if you're telling me that audio equipment makers are selling equipment where the power output swings wildly as a result of only a few tenths of a volt difference in power from the wall, that i would say that some pretty shaky equipment is being sold at top dollar prices. one of the things that i appreciate in your comments it that it helps me understand how very challenging it is to design a power amplifier - in fact, it seems to me that that might be the most challenging device to design - but all the same, i would have thought that the power supplies in these devices were a regulated a bit better than your comments would lead me to believe.
09-14-11: Atmasphere
Finally, with regard to the posts you are trading with Al, if you have ever heard of a first-order crossover in a speaker, that is a filter that is as simple as they get. A first-order crossover is of course 6db/octave. I don't think I have ever heard of a 3db/octave filter, but such a filter would actually need *more* parts to make it happen.
here's the problem: referring to the henry ort reference, ort is correct, dB is, strictly speaking, a ratio of power levels. the problem is that it is true that people have also (and admitted incorrectly if you want to be a strict constructionist about it) used dB to refer to ratios of other quantities. that has led to confusion. when you say that a first order filter falls off by 6dB/octave, that is a statement of how the power levels change. to give you an example of the confusion, if you look at a bryston schematic, where they have the balanced inputs, they use a resistor ladder to reduce the voltage by 1/2. they refer to the reduction as a 3dB reduction in the schematic. if you use the 20log(v2/v1) equation, you would get a 6dB reduction. i suppose the proper course of action is to maintain the strict construction of what a dB is; i.e. that it is only used to inform on what is going on with power levels and nothing else. |
