The white papers by Atma-sphere, who sometimes posts on the forum, can be found here. You want the third and the fourth.

"Watts" Versus "Current"

Can someone discuss, in layman's terms, the differences here? I've seen some high wattage amps that do not produce much peak current and some low wattage amps that produce a lot of current. Which stat is more important--watts or current--in terms of deciding on a match with speakers? If current is more significant why to we (and manufacturers) talk mostly in watts?

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Both are important depending on the speaker. Watts will determine how loud the speakers will go cleanly. Current will determine how balanced top to bottom the sound is on speakers whose impedance loads vary greatly at different frequencies, particularly at lower to moderate listening volumes. In my case, with large Ohm Walsh 5 speakers, I swapped a 360 w/ch but low current Carver amp for a 120w/ch Musical Fidelity that delivers a lot more current/watt for this reason with very positive results. The switch also produced similar positive results with my Dynaudio Contours. Both Dyns and OHMs like lots of power AND current. |

Watts = Current * Voltage Watts = Power = W Current = Amps = A (current is similar to water flow) Voltage = Volts = V (voltage is similar to water pressure) So you can see that both current and voltage are important in making up a power/watts rating. A good example you may be familiar with is tube vs solid state amps. Tubes are voltage driven devices. They operate at low current but high voltage and control/amplify it easily. Transistors operate at low voltages but control/amplify current easily. So you can see that to get 1W you can have 1000V * .001A or .1A * 10V As Mapman says, which is better depends on your speakers. Everything above is in general, and in general, woofers sound better with lots of current. So SS is generally considered a better choice over a tube amp with a similar rating. Tweeters don't use much current so they tend to sound better with tube amps. And yes, midranges are in the middle so it's your choice. Now marketers say just about anything and everything so you just have to take all that with a grain of salt. It's really not which is better (wattage based on high voltage or high current), it's which sounds better in your system to you. |

Forgive the intrusion, but I've got a pratical question about the use of the two different amps, i.e. low power high current and high power low current. When driving a speaker at say 10 watts max (including peaks) and your two amps are, say 200wt, one which has 'high current' and doubles down to 2 ohms, and the other which can only produces say a 50% watt increas at 4 ohms, will you hear a difference because of the difference when you play speakers with a mean impedence curve, or does that only happen when you start to approach the power limits available at any particular frequency/impedence demand? Just curious. |

Current is a component of watts (power) as Dan just indicated. Therefore, all watts are not created equal. 1 amp (of current) X 10 volts = 10 watts of power. But 10 amps X 1 volt also = 10 watts. Further, the basic W =A x V really works for direct current (DC) only. The formula is a bit more complicated for AC (as in: music signal). But leaving that aside for now, how the power is created will either be better or less good for certain kinds of speakers. Why? Because every kind of driver generates a reactive signal BACK TO THE AMP. When the signal from the amp moves a cone back and forth, for example, the action of the cone's voicecoil in the magnetic field actually GENERATES a reverse electric current BACK to the amp! This reverse current adds to the impedance (load) the amp "sees". What's CRITICAL (in amp selection) IMO, is HOW how that (let's call it 'phantom') impedance is created. If the amp sees a capacitive type of load (an electrostatic speaker) it needs for its available watts to consist of lots of amps (current). If it sees a resistive type load (like cones, domes, ribbons and planar magnetics) it needs for its available watts to consist of lots of volts (to overcome the reverse voltage created by those kinds of drivers.) I'm going to generalize here, but 1 tube amplifier watt (a typical push-pull circuit with an output transformer, nothing exotic!) will be rich in the current it can make available (for that 1 watt of output.) 1 solid state amplifier watt (a typical push-pull circuit with the output transistors directly coupled to the speaker) will be rich in the voltage it can make available (for that 1 watt of output.) You can guess where this is headed ;-) Tubes are better than silicon for capacitive loads, and silicon is beter than tubes for resistive loads. All right -- this is a sweeping generalization! HOWEVER, because the impedance of all speakers varies widely depending on frequency (the rated impedance of speakers is only a nominal figure) there could be times when you'd need a HUGE tube amp to drive those dynamic drivers (God forbid acoustic suspension woofers!) properly. Or if you are driving electrostatics (panels only, we're talking about) you might need a HUGE solid state amp to give those stats the current they need when a tube amp of far less output would sound as good (and likely better IMO ;-) People into fine audio are generally at the 3-digit Stanford-Binet level. But many (probably most) have had little occaision to become familiar with the nature of (yikes!) circuit design. But if you will take a little time (good bathroom reading for instance as it takes several passes to sink in ;-) and go through this excellent and (reasonably) clear, easy to understand online tutorial, you will find yourself more informed than many of the self-appointed/annointed 'know-it-alls'/'experts' that haunt the halls of audio-underworld ;-) http://www.allaboutcircuits.com/vol_1/index.html . |

The confusion with amplifier ratings is that amps do not "produce" watts. An amplifier is a voltage source, just like the 120 volt receptacle outlet on your wall. An easy analogy is using a light bulb. What is the difference between a 100W and a 60W light bulb? And, why is a 100W bulb brighter than a 60W bulb in the same lamp plugged into the same 120 volt outlet? You cannot say that with a 100W bulb in the lamp that the wall outlet "produces" 100 watts. The outlet supplies a constant 120 volts - that's the key. So, how does one design a light bulb with 100W and one of 60W? Ohms law: Voltage = Current x Resistance. Watts = Voltage x Current (simplified). The 60 Watt bulb in a 120 volt outlet can tell us the current. Rearranging the Watts equation gives Current = Watts/Voltage or 60/120 = 0.5 amps We need to pass 0.5 amps through the bulb to get 60 watts. We control amperage by resistance. To get that 0.5 amps we now go to the Voltage equation and rearrange to solve for Resistance: R=V/I or R= 120/0.5 = 240 Ohms. So, to design a 60W bulb requires an element with a resistance of 240 Ohms. That gets us 60 watts in a 120V outlet. For a 100W bulb: I = 100/120 = 0.833 Amps, R=120/0.83=144 Ohms. So, to design a 100W bulb requires an element with a resistance of 144 Ohms. That gets us 100 watts in a 120V outlet. The difference is the resitances of the tungsten elements WHICH CAUSES A GREATER CURRENT TO FLOW from the receptacle. That extra current is the extra wattage (W=Volt x amps). Ok, great, you say. What about amplifiers and current? Same thing. The only difference is that speakers, unlike light bulbs, have different resistances at different frequencies. To the amplifier, that presents a varying load. That is like constantly installing different wattage light bulbs in the same socket. That presents a similar varying load to the wall outlet. But the wall outlet has an infinite store of current, that is, you can lower the resistance and the outlet will supply the greater current to a maximum of 20-amps (the circuit breaker) or 10,000 amps (the maximum short circuit current of the utility transformer). But your amplifier does not have 10,000 amps of current at its disposal. The amperage that it can supply is limited by its power supply transformer (VERY simplified statement, there). If you have an amp that is "100 watts", you have a meaningless rating. If you have a 100 watt amp at 8 ohms, now that means everything. To design a 100 watt amp at 8 ohms you will need a 100 watt transformer for the power supply, so we buy a 100 VA transformer. Since the amp is rated for its power at 8 ohms, that means the transformer will have the following properties: 120 volts on the primary, to match the wall voltage. For the secondary voltage: Watts = Voltage(V) x Current(I). We're stuck because we do not know the secondary current. BUt we do know that the load is 8 ohms and Current(I) = V/R. So the power equation becomes: W= V x I = V x (V/R) = V x V/R Knowing R=8, the secondary voltage is 100 = VxV/8 = 28 volts. We specify a winding of 28 volts. The current "available" from the transformer is: Current = Watts/Voltage or I = 100/28 = 3.5 amps So the 100 watt amplifier is like a wall outlet at 28 volts with a maximum amperage capacity of 3.5 amps (not 15, not 10,000). The speaker is like a light bulb at 8 ohms. Now we have a "100 watt" amp. When the speaker drops to 4 ohms, the current required is I = V/R or I = 28/4 = 7 amps. That's twice the current. But wait - the transformer only gives us 3.5 amps. That means we cannot maintain 28 volts across a 4 ohm load. We need more current. But since Watts = VxV/R, at 4 ohms that power is 200 watts. Makes sense: twice the current means twice the watts. What to do? Well, we put in a bigger transformer - a 200 watt transformer at 8 ohms (Or a 200 VA transformer is more accurate). That gives us the doubling of current when the speaker load drops to 4 ohms. But wouldn't that be a "200 watt" amp? No, because at 8 ohms the secondary needs to be at V=40 volts for a 200 watt rating (200=VxV/8). In this case we keep the same 28 volts. That still makes it a 100 watt amp at 8 ohms with a 200 VA (volt-ampere) transformer, meaning you get 3.5 amps at 8 ohms still and 7 amps at 4 ohms. This is key: with a 200 VA transformer you either have a 100 watts at 8 ohms and 200 watts at 4 ohms when wound with a secondary of 28 volts OR an amp with 200 watts only at 8 ohms when wound with a secondary of 40 volts. Same reasoning if you want the power to double at 2 ohms. Use a 400 VA transformer, secondary at 28 volts, and that gives a 14 amp capacity to take care of the 2 ohm load. Basically, the power supply determines the current available and the speaker load at a frequency determines the power. A high current amp is an oversized beast that has a bigger transformer wound down to a lower secondary voltage to get the extra current. Kinda long winded, but this is not an easy answer. |

You're quite right, nothing really happens until those watts are called for, but when they are, the device appears to the amp as a certain kind of load. And it's the nature of the speaker load (or more accurately, the nature of the speaker's reactance) that will determine mix of amps x volts the amp must supply. So it's not just that speakers present different impedences at different frequencies, but that those impedances can be resistive or capacitive. Remember that current flows easily through a voice coil (so volts are called for), but almost not at all across the [air-space + step-up-transformer] assembly of an electrostat (so current is called for.) . |

Thanks for the posts guys (girls?). Once I get the camel dung out of my head, I will try understanding the current flow of the brightest shooting star I've ever seen (last night) while searching for planets with my Mead telescope. The best part is, my 6 year old nephew also saw the shooting star. We jumped around, slapped each other high fives. He plans on making a drawing. Seriously, I appreciate the explanations! derree aaars, and all. Carry on... |

One last analogy that helps understand voltage vs current is that of water pressure (voltage) and flow (current). Think of a great big water reservoir up on top of a mountain and you're at the base of the mountain (yes, this is your power supply :). Let's say you want to design the system to put out a forest fire should it get near your house. To do that you need a lot of water flow really fast. So you make sure to connect a really big pipe down to your house. Now you've got a system with high water flow (current) that will flow for a shorter time but put out a raging fire. Conversely, let's say you need to shoot a small stream of water over a large ravine next to your house. Now you can run a smaller pipe to build up the water pressure/voltage. The water at the end of the pipe will shoot out of the pipe, over the ravine and save the endangered species from drought. Both systems are doing work but with vastly different design parameters required for flow/current and pressure/voltage. Which is better? Well, it obviously depends on what you want to do with the water :) As many have pointed out, this ignores resistance, impedance, capacitance, or any other kind of system feedback/overload that can impact the system. But I hope it helps to get the general idea across very simply. |

Dodgealum - great question, thanks for posting. Now I finally "get" why all wpc are not created equal! BUT - someone please tell me how I can get to the next level of insight based on (most?) manufacturer specs...I see wpc ratings listed but I don't THINK I'm seeing info that would help me understand what an amp's current capacity is. Am I missing something? Bottom-line, how can we apply the info in this thread when trying to decide on possible candidates for audition or purchase. |

Great analogy from Danmyers. Most important part of his post, IMO is this "vastly different design parameters required for flow/current and pressure/voltage. Which is better? Well, it obviously depends on what you want to do with the water" IOW, depends on your speakers. Read Atmasphere's posts and white paper. Amp choice highly dependent on whether your speakers are current optimized or voltage optimized. Which strongly suggested that at least a part of "system synergy" can be predicted. |

Swamp, not to take away from Dan's similar explanation, but if you read my post at the beginning of this thread, you'll realize I said the exact same thing: Therefore, all watts are not created equal. 1 amp (of current) X 10 volts = 10 watts of power. But 10 amps X 1 volt also = 10 watts. Further, the basic W =A x V really works for direct current (DC) only. The formula is a bit more complicated for AC (as in: music signal). But leaving that aside for now, how the power is created will either be better or less good for certain kinds of speakers.. |

Ns- you bet. I've been watching this thread for a few net sessions and quite honestly had meant to mention it as well. I was mostly referring to the analogy which for non-tech types of math-phobes (I'm not one, but many are) its easier to grasp the concept that way and then on to the math and tech detail. I know I'm right about at the edge of my tech expertise w impedance. I know its the equivalence of resistance and I understand its important in matching amp and speaker. I know (or at least I think I do) that most amps operate better into higher impedance loads and that lower impedance loads will require/draw more "wattage" from an amplifier. I do get confused with input impedance and output impedance and matching source/SUT/phono stage/pre-amp/amp with respect to those parameters. And I have an advanced degree in a science (not a math oriented one, admittedly). A analogies help me make the conceptual connections that allow me to go from trying to remember "rules of thumb" to actual understanding. Then I can make judgments for myself. |

Not to throw water (pun intended) on danyers' analogy but the water pressure at the hose nozzle is independent of the size of the hose (WHEN THERE IS NO FLOW), but will actually be LOWER at the end of a smaller hose because of the extra resistance to the water flow offered by the smaller hose. The rservoir at a certain height is the Voltage and the amount of water flowing through the hose is the current and the size of the hose (pipe) can be considered the load. If the load is high, the current is actually lower. In danyers' analogy, the reason that the water is 'thrown' further from the shorter hose is not because the pressure is higher, but because less water is being supplied at the nozzle at a higher speed thus the water can travel a further horizontal distance before falling to the ground. The trouble with using water analogies (which is commonly done for electricity) is that one must also understand the mechanics of fluid flow for the analogies to make sense. In this case the analogy is not very usefull, except to misinform. Gs5556's reply is, by far the best analogy, because it is not an analogy but an explaination using an example. Salut, Bob P. |

Swamp, in the beginning, most of us (I know I did anyway) conceptualized 'impedance' and 'resistance' as being essentially the same thing; because they are both expressed as ohms. and we were taught (first) that ohms are a measure of resistance. Strictly speaking, that is NOT TRUE! It is IMPEDANCE that is measured in ohms, and RESISTANCE is simply ONE KIND of impedance ;-) Impedance is indeed the better term, because it refers to any condition that IMPEDES the flow of current (resistance being just one such condition.) So, for instance, when a manufacturer specifies their speaker is "4 ohms" that's NOT what you will read if you just put an ohm meter across the speaker terminals ;-) That 4 ohm "nominal impedance" is in fact a "weighted average" of the impedance (to current flow) that the amplifier "sees" at EACH frequency across the entire spectrum. But that speaker is by no means 4 ohms across the board! Further, when a speaker manufacturer specifies "power handling capacity" (like, 50W - 200W) they're not just saying the minimum watts you need to 'make it play', or the maximum watts before you 'fry' your speaker ;-) What they are really getting at is the RANGE of (amplifier) power ratings in which the speaker will produce the whole frequency spectrum (and without distortion.) For example, below 50 watts, some parts of the frequency spectrum (associated with hi impedance) may not reproduce at all, or only with distortion. On the other hand, driving the speaker above 200 watts, at certain (other) parts of the frequency spectrum associated with lo-impedance, you could indeed fry something (like a dome tweeter's voice coil, or burn a hole in your electrostat's diaphragm!) Did you know that a MartinLogan panel's IMPEDANCE can drop to less than 1 ohm at 15KHz?! So if you were using a ss amp with a rated power of say 200W/ch @ 8ohms, then @ 4 ohms (following Ohm's Law) it could deliver 400W, @ 2 ohms, 800W, and @ 1 ohm, 1600W!! But at each stage, the VOLTAGE part of those bigger watts would be increasing more and more, compared to the current part. So if you were playing a LOUD 15kHz with a solid state amp into your MartinLogan, huge voltage would build up across the air gap in the panel, and pretty soon ZAP - lightening! (better known as arc-ing) and you've burned a hole in your diaphragm! OTOH, due to the "impedance matching" capability of transformers, a tube amp won't produce this result with an electostat (unless it's a mutha of a tube amp, and your hand slipped on the volume control!) Tubes have a high internal impedance (think of a high revving 4-cyl engine) and speakers generally have a low internal impedance. Solution for getting the power to the road? A 6-speed gear box! Better known as an output transformer. If you select the correct 2, 4, 8, 16 ohm tap (gear) then you can match the amp's power (torque) to the speaker's load (incline of the hill) for maximum power transfer! (Don't ya just love physics?!) What this means (broadly generalizing ;-) is that a tube amp can supply the same watts, comprised of the same proportion of amps and volts, into any impedance. Only one solid state amplifier I know can do this, and that is McIntosh; BECAUSE they ALSO use output transformers on their SS amps, even though they don't really have to -- but it keeps THEIR clientèle from making bad amp/speaker choices ;-) I just happen to be a McIntosh "clientèle" but not because I don't know how to select an amp! ;-) . |

Wow, an old thread but a topic I wanted to understand. Gs5556 really did a fantastic bit of education with his description. So I'm guessing that the JVC A-X9 amp I have, with a 390VA toroidal transformer delivers plenty of current to dip down to 4 or even 2 ohms. I finally get the idea how watts isn't the easy answer to what an amp can drive. |

Thamnosma and other contributors, great job. I've started a new OP which takes this OP to the next level. Specifically, is there any way to meaningfully and intelligently match an amp, be it tube or SS, to Speaker X. And I'm not even touching electrostats, maggies or horns -- just plain old dynamic cone jobs. Various reviews of my Paradigm Sig 8 v2 speakers contain performance graphs re impedance and phase angle that look like a roller coaster even though the speakers are "nominally" rated at 8 ohms. I drive the S8s with an ARC VS-115 120wpc tube amp. Could be just blissful ignorance, early stage deafness, or whatever, but the rig sounds pretty good to me. Plenty of bass slam, with a little help from a sub woofer, and extended sweet highs. Would the S8s sound better with a high current SS amp, or a different model tube amp?? Haven't a clue. So, is there a Guide for the Perplexed (pun intended) in audiophile-land regarding amp and speaker matching? P.S. The Guide for the Perplexed was written by a very famous Rabbi about 1000 years ago. Nothing in the Guide about amp and speaker matching. I checked. |