Rwd, if you're just needing the amp to drive the sub/bass woofers, than that could well be less critical a balancing act between a need for power and some other, possibly more refined sonic qualities. Although a nominal 4 ohm impedance, if truthful, won't be a difficult load for many good quality amps rated into 4 ohms, driver size will also come into play. For bass drivers 10" - 15" in diameter, not only will a successful amp choice need to work into the low impedance if present, it will also have to be high-powered overall, ideally with a relatively high damping factor, to control the more massive cone's greater inertia, as well as contend with the back-EMF generated by the larger motor structure.
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Boy...did I get an ear full!!! Only joking folks! I must confess that most of what I read is very interesting but my exposure to technical knowledge is really not there! ( Business Major :)....any way...Sean answered this question by answering my reply to another thred elsewhere!! So Thanks again Sean and I will be on the look out for a Perreaux 3150 power amp (which has high current) needed to drive my sub woofer's (4 ohm load) Infinity's. |
BTW the most successful setup with SA100 I heard with Genesis 5, Sonic Frontiers SFL2, Basis 2000/Plinius Jarrah/Dynavector 20XH/; Analysis+ 9 speaker wire, AudioQuest Diamondblack interconnects almost everywhere. Listening to "Stabat Mater" by unfamous and forgotten italian composer Pergolesi, despite being so skeptical auditioning classical music, the voices were so real and 50y.o. vinyl made me forget about clicks and pops and dive into the music everything else simply dissapeared. |
Sean, Indeed Plinius gear(listened in several setups) is worm and musical. The situation you're describing I heard with Merylin speakers but than again I've never heard Merylins soft or musical. SA-100 has a large range of operating in class A(upto 25W/ch) and has a closes to the tubes midrange. I assume that it will do the best with low-impedance speakers such as JM Lab 921, Thiel 3.6 and might be bright and hard driving high impedance speakers such as Coinsident. |
Fbi: Hhang onto your hat. From what i hear through the grapevine, such a "feature" might not be that far down the road. The folks at A-gon have got some really "groovy" tricks up their sleeves from what i've been told. Marakanetz: I have very limited experience with Plinius products. I have heard a Plinius preamp driving an SA-100 ( don't know if it was an original, Mk II, etc.. ) on one occasion. I was not familiar with the recordings but the front end was a three piece Metronome transport & matching tubed DAC with external power supply. The speakers were custom built sub / sat's with a ribbon tweeter and dual mid-woofers arranged in an MTM. Cables were Kimber if i remember correctly. While i don't know exactly which pieces contributed to what i remember hearing the most, all i know is that the sound was phenomenally hard and sterile and gave me listening fatigue in a matter of minutes. I've been told that the Plinius gear is very smooth and warm sounding, so it makes me wonder just how "bad" the rest of the gear would have had to have been to make things sound the way that it did. Sean > |
~~~~Seandtaylor99's first post sums it up best as Zaikesman concurs. A chart of component synergy, preamps / amps / speakers, actually used by members, good results/bad results. Perhaps a starting point for all who wish to try different combinations and avoid proven mismatches. Would need a volunteer to do the spread sheet/chart. Audiogon could have this as a free access page (members only)? Updated regularly as combinations are tried. Anyone care to comment on this idea? Just The Facts By |
In defense of the Bryston Stereophile measured a Bryston 9B-SST (THX at the time) and said it had no problem delivering it’s rated 120 wpc with all five channels driven. I’m mentioning this amp because it was the only Bryston review on Stereophile’s web site. Say what you will about the musicality of Bryston products but I’ve read anyone complain that they did not meet their rated power. |
Sean, the equation I've mentioned is not as simple and yes, this equation is being used to specify the speaker impedance curve in responce to any particular freequency. The reactance is a directional part of an overall impedance and resulting vector of reactance can be negative and that's when the impedance falls can occur. Indeed the music is a complex signal having lots of harmonics and subharmonics but it's more arithmetics still since the numbers we work here with is finite and can be calculated. The theory states that the signal of any arbitrary form or complexity can be represented by the sum of sine waves of different order(Fourier theorem) which means that the freequency(to be more precise a sinusoidal freequency components) of an arbitrary form signal can be defined using Fourier theorem. As to your statement about the differences in sound with relatively same capabilities amps and the speakers first of all they're assume will be not too much audiable, second there is a specific formula but truly it doesn't make a sence to excersise it untill you find the family of infinitessimal values specifying these nano-ohms and nano-volts once again using a Fourier theorem along with Ohm's law. As to the good amp formula so not to be so exhausted calculating limits using a complexed engineering formulas(certainly applied in more sophisticated amplifiers than audio) you're completely right and I believe that Plinius SA100 falls in this category. |
Marakanetz: Since music is typically comprised of multiple instruments being played at the same time, the signal produced can be extremely wide in terms of frequency response at any given moment. All of these notes / instruments can be played simultaneously. As you know, most speakers are "multi-way" units i.e. have woofers, mids, tweeters, etc... and some may have multiple drive units within each range. Besides the amplifier trying to load into all of the various drive units simultaneously at any given point in time, it also has to deal with the signal dividing crossover network. As such, it is possible for an amp to see very different impedances and levels of reactance at different points within the audible bandpass at the same time. With that in mind, we are now using a complex signal ( multiple notes over a wide frequency range ) driving a complex load ( multiple drivers working within limited frequency ranges ). As such, each driver / frequency range as provided by the crossover network could present a different picture with various challenges to an amp at any given time. You simply can't "sum" all of those variables into one simple equation. On top of this, reactance of a speaker / individual drive unit can vary with different drive levels applied. Obviously, not all drivers saturate at the same point nor do they have the same amount of excursion capability. While better designs seek to eliminate such variables and / or reduce the potential for what would appear to be a "seam" between drivers in a multi-way system at any given drive level, such is not always the case. If the "lack of seamlessness" situation does arise, you can bet that part of what you hear is not only attributable to the speaker design itself, but also to that of the specific amplifier being used. The amplifier is responding to oddities in terms of loading within the "problem" region. As such, some amps may respond in various manners to loading conditions and the result is why we have audible differences. This is the reason that you may think that a speaker has a certain sound to it yet hear a very different presentation when you change amplifiers. The electrical constants of the speaker have not changed. The ability of the circuitry to deal with those electrical characteristics is what changed when you swapped amplifiers. Given that both amps were being driven within their range of linear operation and were not clipping, how can you mathematically calculate the discrepancies heard and explain why one amp sounds "better" or "different" than the other with the same load on them ??? You can't and there is no specific formula to do so. If i were to "pick a formula" for a "good" amp, the amp would have a helluva power supply. I am not talking about just a big reserve of staggered value capacitors placed at strategic points within the circuit, but a very powerful transformer with high rail voltages with extended duration current capacity. The circuitry would be very fast in terms of rise time. In order to to have a fast rise time, you have to have wide bandwidth as you can not have one without the other. This insures good linearity within the audible bandpass since the unit is fast enough to keep up with signals above and beyond that range. The amp would also have a very fast slew rate. This would mean that it could respond to changes in amplitude very rapidly. Between having a quick rise time to duplicate sharp directional and / or polarity changes in the waveform, the high slew rate would allow us to do that regardless of intensity or amplitude of those changes. On top of all of that, the amp would have a very low output impedance. This would minimize the ability of the speaker to actually "modulate" the output of the amp. Some call this the "damping factor" of an amp. The end result would be an amp that was quite stable and retained consistent loading and sonic characteristics into whatever you threw at it. Obviously, "passive" parts quality and circuit layout does count, so one would have to take such things into consideration. If you can get those basic things right, chances are, you'll have an amp that will work well into just about any load you can give it AND sound pretty good doing it. Sean > |
...now we're speaking in full engineer terms but let me clarify further: First, we need to understand further the concept of IMPEDANCE rather than RESISTANCE. In case with DC we call it RESISTANCE. In case with AC we call it IMPEDANCE since it has a reactive components such as capacitance and inductance. The resulting complex value of impedance consists from active resistance R and reactance which is a complex value: Z = R + i/2*pi*f*C + j*2*pi*f*L where R is active resistance C is capacitance L is inductance i, j are directional vectors since capacitive reactance with inductive reactance is out of phase by 90degrees, f is the freequency and pi is the famous constant. The sum of members with directional vectors in the formula is called reactance which is the value excluding the active resistance. Please note that reactance consists of vectors and doesn't follow the arithmetic adding and certainly the Impedance is a vector value and has a direction. So the Ohm's law for the alternated current will yeald: U = Z*I where U is voltage(volts) Z is IMPEDANCE(Ohms) and I is current (amperes) As to the Bryston discussion it realy has not enough current for continues power and clips! Their professional models don't clip but sound pathetic. If you have Bryston amp you'd better not use low-impedance speakers, it will clip. Bryston 4b-st for instance is not even capable to deliver 200W into 8Ohms as specified. Plinius SA100 rated 100W/8Ohms will rock much faster and easier and even louder with the speakers of the same sencitivity and impedance no $hit! To finalize this situation I believe that 80% of that issue is depending on power supply and the rest is to electronics capability to deliver the current. |
Rwd - Seandtaylor99 has it right, the first question you should be dealing with, and the one that is not answered here - because you have not supplied us with the info to do so - is: Why am I being told I "need" a "high-current" amplifier in the first place? What are your speakers (and room size, and listening tastes and volumes), and what is your present amp? Who told you that you will need a different amp, and why? Do *you* hear a problem? Can you listen the way *you* want to happily? If not, why not? You shouldn't be thinking about "numbers to look for". This will only lead to the kind of confusion you see above (not that the many worthwhile attempts to explain the whys and wherefores of the "current question" aren't valuable or interesting, or appreciated), and will not necessarily help you to choose a new amp you will be happy with, should that prove advisable. You should be thinking about sound results, and maybe also the agenda or qualifications of your advisor. The most important thing to realize, if you are indeed unsatisfied with the sound you're getting right now, is that nothing anyone tells you, either here or anyplace else, if going to be a proper substitute for listening to some options and deciding what you like for yourself. If you want some applicable perspective on your situation, please respond with specific answers to the background questions, so we can actually assess and address *your* audio needs, desires, and questions in a more pertinent fashion. The ability of any amp to output "high" current into a given load if called upon to do so is probably always a 'good thing' if looked at in a vacuum, but audio design always involves trade-offs (especially as cost, preference, and system-matching are concerned), and that attribute in and of itself will never alone dictate which amp you'd be most pleased with for your system, tastes, and budget. [P.S. - BTW, Sdcampbell, I have always been led to believe that the correct 'water' analogy - as far as it goes - is that voltage represents 'electrical (water) pressure' and amps of current the 'rate of flow'. Your description strikes me as the reverse of this. Care to comment?] |
There's one other factor with an amplifier that envolves its feedback circuit. This also affects how the amplifier "sees" its attached load. Some manufactures install small coils of wire (almost like a choke) on the output to prevent the amp from ringing into loads it doesn't like. As Sean said, this is much to complicated an issue to be summed up in a simplistic Ohms Law formula that was designed for DC circuits. You can't even apply it to the resistance in a piece of straight wire. Once you apply AC to it, the wire can react rather strange(depending on its composition) They use to use actual current limiting circuits to keep the amp from overheating. My best advice is to decide what type of speakers you are going to use an experiment around until you find a good amp to match. Way too many variables to say an amp is right based on specs. Amps that "Really" double down are a good place to start. Some of the best sounding amps---well you're not going to worry about current(eg. the Pass Labs Aleph 3)One heck of a good sounding amp at 30/ch into 8 ohms and about 45 into 4. It drove a lot of speakers well as long as you didn't want to rock a hall! |
My point was that Ohm's law ( used in a generic manner ) is applicable at very specific points. Obviously, one can measure and calculate what is taking place electrically at any given point in the operating curve and break it down mathematically. However, the measurements at any given point may have nothing to do with what is taking place elsewhere within the parameters of operation. This is what makes a loudspeaker a "complex" load. Even if one were to try to break the speaker / amplifier interface down using Thevenin's theory, which is far more complex, those parameters would still vary somewhat with frequency and amplitude. With the above outlook taken into consideration, the amplifier / speaker interface is one of a dynamic nature that changes with amplitude, frequency and impedance. One can't model a "simplistic" speaker / amplifier interface based on just a few measurements that would otherwise work fine in a circuit with set parameters and limited variables. As such, trying to break the entire amp / speaker interface down to something as simplistic as Ohm's Law would be next to impossible unless the speaker maintained a constant impedance across the entire frequency spectrum at any given amplitude and the amplifier acted as a true "voltage source". Does anyone know of such a speaker ? The parallels that one can draw using a resistor as a load and that of an actual speaker are far and few between. For each "benign" speaker load that you show me, i can show ten that are not quite so simplistic. As such, how an individual amplifier and the corresponding circuitry will respond to each load that it sees with varied frequency and amplitude becomes completely unpredictable. Hell, some amps even change frequency response aka "tonal balance" as the drive level increases on a dummy load ( non-reactive "perfect" speaker ) on the bench. Who could predict how such an amp would respond with the variables encountered with different speakers and their associated reactances during real world operation??? If you think i'm making this up, John Atkinson noted this in a recent review of a Rotel amplifer in Stereophile and Moncrieff had covered this 20+ years ago in IAR. While some of the variables can be minimized by having a true "voltage source" as an amplifier and a speaker that was linear in impedance, this still does not take into account how this combo would actually "sound". As i've tried to stress, there are just TOO many variables to try and sum things up "simply". While many folks don't have the technical background to know why such things take place, they have enough experience to know that you will never know exactly how various components will mesh until you try them out within the confines of your system. We can call it system synergy, complimentary colourations, etc... but it still boils down to the fact that there is NO set formula other than the old "trial & error" method that most of us have had to do to get where we are today. If things could be summed up easily using a simple formula, i think that there would be no need for forums such as this. Sean > PS... I'm not trying to be argumentative or step on toes, so please don't take it that way. I'm simply trying to say that you don't know what to expect until you try it. I've been "confounded" way too many times before to know that things that should work sometimes don't and vice-versa. |
This has probably progressed well beyond the original question, but it makes for an interesting topic of discussion. I disagree with the statements that ohm's law is not at all applicable, because with the substitution of the term "impedance" for "resistance", ohm's law is very much true for a single-phase AC circuit. As stated by both Sean and Bigtee, the AC impedance is made up of resistance, capacitive reactance, and inductive reactance. The resistive component is constant at all frequencies, but the capacitive reactance is inversely proportional to frequency, and inductive reactance is proportional to frequency. These properties, along with the phenomenon of mechanical resonance, described above by Sean, explain why any given speakers impedance plot can vary so wildly, both above and below the nominal impedance. However, in order to look at a "simple" model of a loudspeaker, one must have some way to express the relationship between voltage, current, and impedance, and ohm's law provides that relationship. The other relationship required to model the circuit is a basic power equation, which is also different for an AC circuit. For a single-phase AC circuit, Power=Voltage*Current*Cosine of the phase angle between the voltage and current waveforms. This difference in phase is, of course, a result of the net reactance at any given frequency. To apply the power equation without accounting for phase relationships, you must assume that the reactive component of the equivalent impedance is zero, yielding a purely resistive load. I assume (but don't know for sure) that this is how published amplifier ratings are derived, with a discrete frequency sinusoidal waveform applied to an 8 ohm resistive load. If you accept the above, then in a very roundabout way, ohm's law does in fact have an effect on whether an amplifier can drive a given speaker. If we could build a "complex" mathematical model for a given type of speaker paired with a certain amplifier (and don't forget the role of our choice of speaker cables in this model) ohm's law could describe, at any particular frequency, how much current our voltage source (amplifier) could supply. We could then look at the phase relationship between the voltage and current, apply our power equation, and we would have our value of power at clipping for any frequency that we cared to look at. Since I didn't do very well in differential equations, I will leave this modeling process to the wonderful people who design audio electronics for a living. BTW, I noticed that while I was formulating this response, a couple additional posts were added. Thanks to clueless for thinking on the same line as me (kind of scary, huh?), and Seandtaylor99 hit the nail right on the head, although I think we are still just short of a full fledged pissing contest :-) |
I have to disagree with Sean and say Ohms law, and its derivatives, tells you a lot about An amp and speakers (of course mnf. specs can be doctored). Ohm's Law for DC circuits is the fundamental relationship between voltage, current, and resistance. It is very relevant. It is usually stated as: E = I*R, or V=I*R, where E or V = voltage (in volts. E stands for "electromotive force" which is the same thing as voltage), and I = current (in amps), and R = resistance (in ohms). The equation can be manipulated to find any one of the three if the other two are known. For instance, if you know the voltage across a resistor, and the current through it, you can calculate the resistance by rearranging the equation to solve for R as follows: R = E/I. Likewise, if you know the resistance and the voltage drop across it, you can calculate the current through the resistor as I = E/R. A related equation is used to calculate power in a circuit: P = E*I, where P = power (in watts), E = voltage (in volts), and I = current (in amps). For example, if you measure 20V RMS and 2.5A into a load, the power delivered to the load is: P = 20*2.5 = 50W. This equation can also be rearranged to solve for the other two quantities as follows: P = E*I, E = P/I, and I = P/E. You can also combine the power equation with the first Ohm's law equation to derive a set of new equations. Since E = I*R, you can substitute I*R for E in the power equation to obtain: P = (I*R)*I, or P = I2R. You can also find P if you know only E and R by substituting I=E/R into the power equation to obtain: P = E*(E/R), or P = E2/R. These two equations can also be rearranged to solve for any one of the three variables if the other two are known. For example, if you have an amplifier putting out 50W into an 8 ohm load, the voltage across the load will be: E = sqrt(P*R) = sqrt(50*8) = 20V RMS. There is a form of Ohm's law for AC circuits too. In AC you must deal with capacitance(C) and inductance(L) in addition to resistance(R). Together, any combination of the three is impedance (Z). If you simply substitute Z for R in the Ohm’s law formula they are accurate for AC too. E=IZ, I=E/Z, Z=E/I are the AC Ohm’s law equations. You can tell a great deal about how an amp will drive a speaker with these equations all based on Ohm’s law. Finding Z is a little more difficult because it involves current out of phase with voltage and it involves complex numbers, vectors, etc… I think I agree with Scot’s statements above but it is a little confusing on a point that is often confused. He says: “Imagine a river. The amount of water that is moving downstream is analogous to the voltage -- i.e., it's a measure of the size or quantity of the flow (say, 2500 cubic feet per minute). The current, or force, behind the water (usually due to gravity) is the other measure of actual or potential energy.” This seems to refer to Voltage as “water that is moving” and the “force, behind the water” at the same time. Well….. is voltage like the stuff that is moving(water) or the force moving it(gravity)? Not good to confuse the two. The amount of water moving downhill seems analogous to Power not Voltage and from the equations above you see that these two are not the same. No amount of gravity is going to make a river flow if there is no water. This is a very basic and important distinction in electronics too. - between matter (electrons) and force (photons). Voltage does not move. Voltage is like a pressure between two points. You might say current flows through a resistor but it is wrong to think of voltage as doing so. Voltage is “across” a resistor not through it. A battery (voltage supply) does not supply current. The electric charge preexists in the wire but lacks organization if you will. A wire is a conductor precisely because it has free electrons. If a voltage is applied across a wire, with a battery, it will push the preexisting electrons in one direction (toward the positive). But the electrons are not the “force” either. To think this would be the same as confusing sound waves with the air the sound waves travel in. The difference in voltage potential creates an electric field and the electric field is the force. The Electric force moves very quickly. The drift of electrons is not much faster than ketchup. Alternating current vibrates more than flows. At least I think this is the basic model. I remain, |
I think you asked the wrong question. The correct question is "What amp will drive XXXXXX speakers for XXXX dollars ? My musical tastes are XXXXX and I like to listen loud/medium/soft. My room size is XXXXX." After that I'd demo the suggestions. An amp with sufficient current will sound powerful, not loud, and will sound clean, not sibilant, not over-bright. It will drive comfortably at your desired listening levels without producing smoke. I'm an engineer, and that's the way I'd go about it. Asking how much current is like asking how long a piece of string. You'll end up with a technical pissing contest of a thread, rather than a straight answer. It reminds me of that old sketch "I'd like to buy a grammaphone ...." Does anyone remember where that was from ? Was it Python ? |
Marakantz Please reread my earlier post. In that post I was talking about the power supplies, not the amplification circuits. I tossed in some statements about assuming quality amp circuits and the like but the big point was power supplies. When I refer to amps that can double their power rating I am referring to continuous power. For example a Bryston 2B-LP is ~60wpc into 8 ohm but 100 into 4 ohm. This is continuous power not a peak or shot term power rating. I assume Bryston added enough heatsinking for long term use. I my previous post I looked amplifier output voltage and current. Output current = current from power supply – small amount to drive associated circuits. Voltage at the speaker terminal = Voltage at power supply – voltage drop in amp circuit (I don’t know exactly how much this is). Rather than talk about voltage and current into the amp circuits I was talking about power and voltage out of the amp circuit. Account for a few losses and you have the power and voltage demands of the power supply. We are both making the same point. |
All of the explanations above are technically accurate, but here's a very simple way for those not versed in science or engineering to think of current vs. voltage. Imagine a river. The amount of water that is moving downstream is analogous to the voltage -- i.e., it's a measure of the size or quantity of the flow (say, 2500 cubic feet per minute). The current, or force, behind the water (usually due to gravity) is the other measure of actual or potential energy. A river or stream with little current does not yield much energy (or force). Similarly, you can also have a relatively wide river that is very slow moving, and it does not exert much force. (If you have ever waded a river, you know from experience that it's a lot easier to cross a very slow moving flow than one moving rapidly.) Conversely, you can have a narrow stream of water moving at very high current, and it will produce quite a bit of energy. To carry this analogy to an extreme, think of the cutting tool known as the "Water Knife" -- it forces a very small stream of water (with an abrasive added) at high pressure (up to 50,000 psi) through a small nozzle, and the stream is capable of cutting through a variety of very hard materials (such as steel). So, to stretch this analogy, there are two ways for an amp to drive a speaker load: high current (the force), or high voltage (the amount of electricity). Yes, yes, I know this is a rather unsophisticated explanation from an engineering standpoint, but it has the virtue of being conceptually simple for those who didn't take physics or EE courses in school. |
Ohm's law has nothing to do with whether or not an amplifier can drive a speaker. While an amp may be able to sustain high power levels into low impedances, that tells you nothing about how well it deals with various amounts and types of reactance across a wide frequency spectrum. It is possible for a speaker to have sharp phase angles at several different frequencies all at the same time. If you can picture a combination of a "slalom course" AND a "torture test" for amplifiers, that is what some speakers present. As such, you can have a speaker that is highly capacitive at treble frequencies, highly inductive at low frequencies and generates a high amount of reflected EMF ( electromotive force or "voltage" ) all at the same time. All of this can be independent of the amount of "pure" resistance that the amp sees at any given time or frequency. This is why speakers are a VERY complex load and why we don't have specific "tests" that show whether or not an amp can drive every load known to man OR maintain consistent sonic characteristics doing so. Something else that is not commonly considered is that amplifiers produce LESS power as impedance is raised i.e. kind of the "reverse" of looking for high current into low impedances. After all, impedance raises at the point of resonance on a woofer. As such, power transfer is reduced right at the point that you need it most. Since vented systems typically have MUCH higher impedance peaks at resonance, you have even less power available to control a driver that is already lacking "damping". Even though most manufacturers do not offer power output specs into 16 and 32 ohm loads anymore, one might be able to make a more informed opinion about the overall build quality of an amp if we did know such things. If you think that this sounds "crazy", take a look at the impedance curve of a vented speaker with a large woofer(s). It is not uncommon to see impedance peaks at resonance along the magnitude of 40 - 100 ohms. What kind of power transfer do you think an average SS amplifier is going to produce into a 50 ohm load ??? Let me tell you, not much. Most "good" sealed designs keep the impedance of the woofers below 20 ohms, which results in much more accurate and controlled bass. This is due to the increased ability of the amp to transfer power and literally "muscle" the cone when it does not want to see any type of signal at all. After all, resonance is nothing more than the speakers' point of self oscillation. If excited at that frequency, it is literally contributing sound on its' own. It is up to the amplifier to "force feed" it at that point and damp / control the ringing that is taking place. Obviously, a lower impedance at resonance allows the amp to generate more power. This in turn can effectively work to control the speaker and produce greater accuracy. Having said all of that, rms power ratings are WAY to easy to fudge given the way that the FTC has things set up. I think that measured power output at CLIPPING at various impedances is FAR more revealing of how "sturdy" an amplifier is. After all, this is the point of maximum long term stress for an amp. As such, it is the ultimate test in terms of how much TOTAL power the output devices can pass and how much current the power supply can sustain. If an amp can come close to "doubling down" at the point of clipping from 32 ohms down to 2 ohms, it can probably drive just about any load that you can throw at it. To do so would mean that the amp was as close to a pure "voltage source" as we are currently capable of making. Even if the amp is sturdy enough to do something like this, there is no guarantee that you will like the tonal balance or level of refinement & detail that the amplifier produces. Passing the aforementioned test simply shows that it is capable of "brute force" into just about any given load and does not necessarily mean that it will "sound good". It will however, drive the load that you connect it to. Sean > |
Ohms law is not really a good way to figure current output of an amplifier. Ohm's law works in a purely resistive DC circuit. The output of an amplifier is AC. You generally will be required to factor in inductance, capacitive reactance and pure resistance (this makes up impedence)to come up with the current output into any given load. Amps that "Double down" are always better for driving low impedence loads. However, things can happen to an amps output when the load has steep phase angles and high reactance. Look at the dbw that Stereophile uses in rating amps for some reference. The bottom line here is that each amp will behave differently into the reactive load of different speakers. A resistor on the test bench doesn't mean a lot. With this said, current is the ability of the amp to deliver actual power expressed in watts into a given load. Power(in watts)= volts x amperage This is a simplification of a more complex calculation. |
Nikki..., Let me be straight forward on how good watts are different from bad ones... If you for example take a look on professional power amps such as Carver highly regarded by DJs available at RadioShack stores, you can see and feel that it has a plenty of boost and the power(~300W/side) to any impedance load and pretty darn cheap(aka $250). The Ohm's law mentioned up above states that ALL watts are "created" equal. The double wattage of the good amps into the lower impedance loads cannot be rated as a CONTINUES power and needs lots of lines to explain. Shortly I can say that price of an amp is not an explaination why one has a reserve power and the other one hasn't. To increase the power you either have to increase the current or the voltage. The main problem in this issue is our 110V wall outlet power that realy limits engineers to work on high-power-quality amps where designing a proper power supply is the most essential issue. So to correct you in both threads I must state that the main design difference between "bad watts" and "good watts" is that the power supply is OK to handle the large current. The quality of sonics (I must say here that it's completely different issue from wattage) is the quality of an active amplification elements(tubes, transistors, diodes) and also passive elements. The working area of an amplification elements in high quality equipment is selected so that it covers the widest-possible freequency bandwidth rather than working in the peak values. Thus more transistors or tubes is required to deliver the signal to desirable level with good output characteristics. |
Current is the rate at which charge passes a given point in a wire. If you were somehow able to count the charges as they went by in a wire carrying 1 ampere (1 A) of current, you would discover that every second an entire Coulomb of charge passes by. Additionally, the "formula" provided in the first post is for a DC current. Since loudspeakers are designed to use AC current that formula will be of little use. High current amps are typically big and heavy with large heat sinks. There are many but think Krell and Levinson. Talk to a dealer that sells your particular speakers. He can best guide you in finding a good, quality high current amp that will satisfy your aural needs |
A minor clarification to my earlier response: Ohm's law does not provide for a definition of power in an electrical circuit, rather it defines a relationship between the DC voltage, DC current, and DC resistance. When I stated "using ohm's law", I omitted that I was applying ohm's law, along with the basic power equation, to obtain the relationship of P=I^2*R, where P=power, I=current, and R=resistance. Sorry for any confusion that may have caused, Mike |
Rwd, What kind of speakers are you intending to drive? "High Current" is a relative term. For a load with a given, fixed impedance, a higher powered amp will always deliver higher current INTO THAT LOAD. In other words, for a fixed 8 ohm load, a 200W amp will supply more current than a 100W amp. Therefor, based solely on current requirements, it could appear that using an amp with a higher rated power output would be the answer. However, the rated power of an amp into an 8 ohm load is not really a good indication of the MAXIMUM current delivery capability of that amp. Most speakers do not have a constant impedance over the entire audio frequency range, and many have less than 8 ohms nominal impedance. As described in the post above, using ohm's law, into an 8 ohm load, a 100 watt amp would deliver 3.54 amps of current (assuming a 0 degree phase angle between the voltage and current waveforms, for you hardcore engineers :). If instead of an 8 ohm speaker, we were driving a 4 ohm speaker, then to deliver the same 100 watts, the amplifier would have to output 5.00 amps of current. If the impedance dropped to one ohm, to provide 100 watts of power to the load, the current delivered would have to be 10 amps, almost 3 times the current at the 8 ohm rated power output. The ability of an amp to deliver higher current into low impedance loads is influenced primarily by the design of the power supply, and the type and number of output devices. The ability to dissipate larger amounts of heat also becomes an issue as current delivery increases. If you have difficult to drive speakers, such as Martin Logans, where the impedance drops as low as one ohm at 20kHz, it is important to have an amplifier that is designed to drive low impedance loads. Therefor, it is probably more helpful for you to consider an amplifiers performance into low impedance loads, rather than zero in on the term "high current". As described above, it is possible to have an amplifier with a large power rating into 8 ohms that will have "high current" at 8 ohms, but may not be suitable for driving low impedance speakers. It is also possible to have an amp with a modest power rating into 8 ohms, and thus a modest current delivery at 8 ohms, that is capable of delivering much higher current, or even much higher power, into lower impedance loads. In general terms, if you have speakers with an unusual characteristic impedance plot, you probably want to look for an amp with the ability to drive a 2 ohm load continuously. If a manufacturer won't put into writing that their design is capable of driving a 2 ohm load, then it probably isn't. That said, if your speakers sound good to you with the amp that you are using today, and the amp does not appear to be running abnormally hot, don't mess with a good thing. |
Ray Dall, Electronics Theory.com site here is good for basics on ohms law, circuits about as briefly as is possible. About 50 pages (big print.) An overview of tubes begins on page 37. I remain, |
If your speakers impedance is rated (nominally) 4 ohms or less, or if certain frequencies dip much below 4 ohms, you will need a power amp that is rated into 4 ohm loads. The lower the impedance, the more current being drawn from the amplifier. Amps put out volts, the load draws current. I'm being very general here for the sake of brevity because there are many factors to consider, such as room size, how loud you listen, etc. There are other things to know about your speakers, such as minimum impedance dip, phase angle, electrical sensitivity (sometimes referred to as efficiency) and such. Ask the speaker manufaturer. |