Why 24/7 warm-up period on amps?


The 24/7 warm-up period on amps seems excessively unsupported. Yes, an amplifier (pre-amp or power amp) will change it's circuit factors as the init heats up since the resistive and capacitive values stabilize...but for months on end? Do we still have a "warm" heart for tubes, that do indeed need to get "hot" to work right?

A capacitor charges up based on it's RC time constant, which is in the SECONDS range, not days. OK, if you add the heat sink area so the heat going out is stabilized I can see maybe an hour or so. My DNA-225 gets HOT in thirty minutes, at which point it's steady state. That even assumes it doesn't have temperature correction circuits to make it more stable, and less subject to change over time.

Break-in periods are hard to judge what people think is happening. Circuit P/N junction temps get hot pretty fast. A mechanical device like a speaker or phono cartridge, sure, they will work-in just like a well used rubber band. But silicone? Factory burn-in is designed to find weak components that degrade outside of SOP ranges, not to "center" their attributes in a normal stable circuit. Did someone forget to add enough heat sink compound to a PNP or NPN transistor, for instance?

Assumming break-in is real, not to be confused with the warm-up period, once it's done it's done. After that it would be warm-up only time. And, warm-up is a simple thermodynamic process. It only takes so long to warm-up and it isn't "days" on end. Maybe hours...if even. Once things are to temp the circuit constants are set. What else is changing? A heat sink is designed to warm-up and hold a delta temperature where the measured performnace is flat. A small amp (pre-amp gain stage) has smaller heat sinks for this reason. Heat and resistance are related, so you have to pick a temp and hold it. You design to THAT attrubute on the component.

Wire conditioning in the amp? ( go here - http://www.angelfire.com/ab3/mjramp/golopid/grain.html) As well as several other sites and textbooks.

The DC path is just that, DC. The magic is the purity of the DC, not the wire moving it around. You either have the right voltage and current capability (wire size)or you don't. Once the amp is on, the wires capacitance hardly matters. PP, PE or Teflon dielectrics only ionizes-tree and fail at break down voltages around impurities, not below that. You do not want to ever ionize the insulation in normal practice.

AC is an interesting issue. The AC complex signal is ALTERNATING differently at each and every frequency point, so the magnetic and electric fileds keep switching with respect to frequency. So the dielectric can not have polarity, or current "direction". The dielectric will not "align" to anything.

Grain structure in copper does not change unless you melt it. It's set when the rod is made. Annealing just resets elongation by improving homogeneous grain alignment, not the grain boundary characteristics since wire is resitive annealed at well below the temp that would fully reform the grain boundary around impurities in the copper. Oh, all modern 9/16" rod copper is made in induction ovens and is essentially OFC grade. All wire is drawn from that rod. Modern copper is also "high conductivity". Again, these terms are throw backs to days gone by with coke furnaces and open air annealing to critical temps where impurities could be picked up, changing the grain boundaries around impurities.

I also notice the people seem to tout TEFLON over Polypropylene or polyethylene dielectrics. Teflon costs more, it is higher temperature capable to 150C-200C (like 80C on polyethylene isn't enough in electronics) but Teflon has a worse dissipation factor and loss tangent. Using Teflon has a more NEGATIVE influence on electricals than olefins. Teflon's velocity of propogation at RF frequencies way above 1MHz is 70% verses 66% for solid olefin dielectrics. But that is at RF. And, you can nitrogen foam either to negate that advantage of Teflon at RF, but NOT Teflon's high price, loss tangent or dissipation factor. Capacitance adjusted Teflon is a poor choice. So the important factors are capacitance, dissipation factor and loss tangent. We can easily fix the velocity of propagation. PE and PP is superior across the board and cheaper (that's probably the problem!).

Good circuits are good circuits. Could you even make a circuit that had electricals parameters that were undefined till it ran, "forever"? Nope, can't be done. Design would then be a game of chance. I don't think that it is. Stabilized junction temps are used to set electrical componenet attributes with respect to temperature. You can design heat sink characteristics to place "hot" components where thet need to be temp wise to meet a circuit requirement. A poorly designed amp that allows thermal run-away under load isn't appropriate and isn't made...for long. There is indeed a circuit junction temp that rather quickly defines the measurable performance of the circuit, and a STABLE delta attribute approximation(s) when a circuit is designed. You know going in what they will be in operation steady state.

So, I hear my speakers and phone stage "break-in. And they don't go backwards once thet are broken-in. They can, in fact, get worse and simply break-down! But my amp sounds fine in short order. The circuit reaches a thermodynamic steady state and we're off to the races. I just can't see a circuit that needs 24/7 "on" period to stabilize...unless it just isn't stable. To me that's a poor design, and one subject to possibly serious load induced instability when the circuit falls outside of the stable design region(s).

I'd sure like to see MEASURED attributes that support 24 /7 warm-ups on sound. I have yet to see any measured data to support this. Show me components used in amps that take MONTHS to reach stady values. I have read PLENTY to support first to third approximation(s) on amplifier circuits ambient thermal temperature stability points. Many circuits are designed to run "cold" and have inverse circuit systems to keep changes due to temp deltas away. This way, you have a more stable circuit at all times. The opposite designis technically UNSTABLE till it gets to temp. This also limits what you can do as it can't blow-up when it is cold BEFORE it gets hot and stable. So the circuit is a compromise.

So just what are the resistive, inductive and capacitive break-in periods on quality components used in a circuit? In God we trust, all else bring data.- unknown
rower30
Warm up in SS amps after 24 hrs is mainly due to the temperature of power transistors stabilizing, (with music playing).

The electrical characteristics of transistors are very temperature sensitive, and the heatsink/transistor thermal lag can very between different types of heatsinks.

This warm up (an hour usually) is audible, even if the amp has been on (idling) for 24 hours.

Well designed amps can be left on 24/7 for years without problems.

Temperature cycling of power transistors reduces the power transistor life, and that should be taken into account.

Power supply capacitor life is effected with excessive heat, over many years, but they are easily replaced.

I have always left my amps on 24/7 for years, with NO failures.

Life is too short to "wait" for good sound. I hate to listen to a cold amp come up to speed. Why waste time?

Leave the amp on 24/7!

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...
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.
Yes, of course. I stated that in my post here.
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).
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.

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.

The db change in that situation if voltage is being considered is 6db; the db change if power is being considered is 6db; the db change is 6db, period. That is why the formula for db as computed from voltage levels includes the constant "20," while the formula for db as computed from power levels includes the constant "10." Otherwise the two numerical values wouldn't work out to be the same, as you'll agree.

Kijanki's post provides an elegant mathematical proof of the equivalency of the two formulas, 10log(P1/P2) and 20log(V1/V2), as does the Ott paper, in a different way. But until you recognize that a db is a db, regardless of whether power or voltage is being considered, the other differences in our positions, including the issue of filter rolloff which started the discussion, will remain unreconcilable.

Regards,
-- Al

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...