With all of these new HE subwoofers and their need for large amps (to take advantage of the excursion), what type of power can the wall outlet safely put out. ..and for that matter, how many watts can typical surge protectors handle. My equipment is drawing a bit of juice as is and I am thinking about either a crown 1K or 2K.

Get the crown CE 2000, you won't regret it. I have a CE 1000 just laying around right now (I might sell it), because I found a smoking deal for a ce 2000, I got mine for 400 dollars because the store screwed up on the pricing, and their computer system was down so they wrote me up manually. The CE 2000 pulls about 10 amps of power when running off of 120v when it is putting out it's full 1950 RMS watts. Thats enough to dim the lights. A regular house outlet is rated at 15amps usually. I never have run mine at full power because my 4 Linkwitz transformed audiobahn alum12's won't take it. I just bought the amp cause I couldn't pass up that good of a deal. I am thinking of buying a BPD 1803 and going ported to handle all my bass below 40hz, or just use it by itself. I just need to get a job first since I have lost it for the 4th time this year due to the crappy economy hitting the IT sector so hard. Once I get a new job I will build a monster 1803 setup. I might not keep the 4 12's though.

Well, I am running 2 1503s and have a bunch of other equipment running through 1 outlet( 27"tv, HT reciever, sub amp BFD, and DVD...so far). I just didn't want to trip the breakers. My room is only 18x12 (bedroom), so the k1 will be plenty of juice.....but the extra clearence is nice. How much do you plan to sell your ce1000 for?

Jeremy.. 10x120 = 1200W. There is no way the amp could be putting out 1950W, when its only drawing 1200W...

hey pete, your absolutely right, it can't make more power than it recives. I looked on crowns site and found that the draw is different depending on source (16A on pink noise) and there is a note that says it could be higher on actual material. Not to be a smart ass but..... your calculation is assuming that I am using DC into a purely resistive load, I am not. This is a AC circuit going into a reactive load. The calculations are different and beyond me as of right now because I don't fully udnerstand the workings of AC. But I believe this explains why. Could someone please explain this more clearly to me and the rest of us because I would like to know the answer.

The UL sticker will indicate the max draw from the AC outlet. Assuming it say 10amps, then the rest is made up by storing extra power in one or more capacitors inside the amp. The recharge when the demand is low and discharge when the output exceeds what can be drawn from the AC power supply. Watts are watts and can be freely converted from AC to DC as long as you are not dealing with instantaneous power. That is because 120V AC is calculated RMS also.

I was under the impression that a 15amp circut was equal to about 1900watts and a 20amp, 2400watts. I don't rememeber where I got this info so please correct me if I'm wrong.

That's what the storage capacitors in Jeremy's amps are for. While it is correct that an amp cannot produce more power than it can consume from its source, it can produce short term peaks that may exceed the 120VAC/15A outlet that feeds it. In traditional amp designs, the manufacturer uses large capacitors (up to a 1/2 farad) to store a charge, and when the amp has a call for a larger than normal pulse (an explosion in a soundtrack, for example), the amplifier would then draw from the large reserve, stored in those caps, thus using more energy than the outlet may have been able to provide. In other designs (Carver comes to mind) they use switching type designs, that would draw ALL the current directly from the input AC line. These amplifiers would then require a seperate circuit for each amp. They store no additional power onboard whatsoever. Thanks, my .02... TK

Andrew, you can not freely convert from DC to AC when you at talking about a reactive load such as a amplifier that is using a transformer to supply it power. If I did, this is what I would get for my amp asuming DC. R= .3 Ohms, I measured the resistance between the plug terminals. V= 120V AC RMS So now we calculate I= 120V / .3 Ohms = 400 Amps, Well above the rated 15 Amps on the power outlet. Now if we use ohms law for AC Impedance in ohms = Voltage in volts / Current in amps The maximum current draw by the UL sticker is rated at 9.5 Amps so I will use that. Z = 120V / 9.5A = 12.63 Ohms Impedence, not .3 ohms. .3 ohms is the DC resistance and 12.63 ohms is the AC impedence which then of course comes out to the 9.5A rating. I guess the only real way to tell is to measure everything, but I don’t have a 2kw 4 ohm resistive load to do testing with. This was just an example how you cannot always inter change DC and AC ohms law. I really wonder if my amp can put out 1950 W into a 4 ohm load resistive or reactive when being powered off of 120V AC. Maybe we will get lucky and there will be an electrical engineer on the board that could explain this to us, because I have absolutely no friggin clue even if it is possible. And I can’t figure it out either because I don’t fully understand all of the AC ohms law equations.

Ok, I think I have the explination, I found it on crowns website. It's the formula on how they got their measurements. Watts used = (total output with all channels driven X duty cycle) / amplifier effeciency. The CE2000 is rated at 65% effecient. The amp can put out 1950 Watts into a 4 Ohm Resistive Load. Pink noise has a duty cycle of 50%, rock and roll roughly 30% and if you have compressed rock and roll it goes to 40%. So lets use pink noise since it has the highest duty cylcle. Pink noise is all frequencies at the same time (multiple sine waves) So all we have to do is plug in our numbers to the formula. (1950W X 50% duty cycle) / 65% Effeciency = 1500 Watts power consumption. So the amp really isn't producing more power than it is recieving. So now the question is what is the equivelant duty cyle of a sine wave? 50%? 70.7%??? Looking at crowns site I can't tell if the amp is rated at RMS or not.

Watts divided by volts gives you AC amps the average house circuit is rated at 15 amps but you have to add up the load on the whole circuit, not just one plug, there can be up to 8 plugs on one 15 amp breaker, so every plug in the bedroom could and probably is on one ciruit, so you have to take into consideration whatever else in the room is plugged in and on while you are in there

12 gauge house wiring is used for circuits rated at 20 amps, 14 gauge wiring for circuits rated up to 15 amps. Somehow I don't trust drawing the full load through typical modern (1975 and newer) circuits because the outlets in the room are daisy chained and connections made by pushing the bare wire end into what amounts to a clip inside the outlet as opposed to screwing it on and the metal to metal contact might not be all that good. Every outlet box the current must go through to get to your equipment means four more joints, two for the hot lead and two for the ground lead. Faulty daisy chaining connections at electric outlets are the reason why an outlet with nothing connected to it might start sparking, it is something such as a heater drawing lots of current further down the line. Video hints: http://members.aol.com/ajaynejr/video.htm

Well I'm a 4th year EE student..so i'll have a crack. Amp designs can vary, but i'll just use general EE knowledge, I am sure there are exceptions in rare cases. There is actually no such thing as RMS power. There is average power (derived from RMS voltage or current) and instantaneous power, which is derived from the peak voltage or current. Think of it as a sine wave, shifted vertically upwards. The max value is peak power, the least value is zero, and the average value..is average power The peak voltage is set by the amps designer, when he sets the rails of the amp. RMS voltage is simply the rail voltage divided by root 2. The instantaneous max power is ((rail voltage)^2)/(impedance of the load). When driving a hard load, the rails tend to sag a little. The capacitors provide that little extra storage needed to maintain the rail voltage, and prevent clipping. Dont forget losses in the amp, and the fact that a transformer is rated in VA, not Watts, which are not the same. VA is apparent power, which consists of imaginary power (VAR) and Real Power (W). Few amps run over 50% efficient..but some of these switching amps can get it far better. Power is much harder to understand than what most people think...and very few people actually understand exactly how it works. I know I dont..and I have done entire units on it.

ok folks the answer to the original thread question is finding the rms value of 340Vp-p typical wall outlet voltage. yes, 340V is houshold voltage. Vrms=( Vp-p * .707 ) / 2 ===> (340*.707)/2=120Vrms then use P=I*E P=120V*15A=1800W P=120V*20A=2400W but to add a bit: 340Vp-p AC has the same effective power of DC at 120V. this is the "effective power". follow? thats why you simply use ohms law. just remember that AC is always referred to using its rms value unless it specifically says otherwise. so you wanna measure power? if its a resistive load, measure the voltage drop across the load and use Ohm's Law to calculate the current through the resistance, then its just P=I*E. inductive load (a voice coil for instance), you need its XL, and XL=2(pi)fL. so you need to know the frequency and the value of the coil in Henrys (and later its TS parameter of Re) to get its ohm value. once you know the XL you need its impedance and that is Z=sqrt(Re^2+XL^2). measure the voltage drop on the coil then use Ohm's Law for the current. then plug in for power P=I*E. anyone see any errors? hope i remembered this stuff right.

Peter, you don't know how good it is to hear you say that there is no such thing as RMS power. I have had this argument several times with people. A watt is a watt is a watt. But what figures you use to get those watts can vary. But RMS Watts are a convenient way of saying that the measurements were taken from RMS Volts and RMS Amps. When ever I start doing AC ohms law calcs that involve a reactive load I get a little sketchy cause of the phase shifts and all. BTW, how do you calculate the theta when dealing with reactive loads. I can't find that any where. AC is still some what difficult for me to deal with. Damn you Tesla for inventing AC, damn you!

hello jeremy, may i introduce you to ELI the ICE man? in an inductive RL circuit the (E)voltage leads the (I)current(ELI) and for a capacitive RC circuit the (I)current leads the (E)voltage (ICE). get it? now the phase angles are determined by very simple formulas: for capacitance tan deg = XC/R for inductance tan deg = XL/R i gave XL=2(pi)fL previously XC=1/[2(pi)fC] things get complicated when more than one frequency is present in a system (ie. music) which is why amps are rated at a certain frequency (like: 100W/channel @ 1KHz). although thats probably not the only reason.

Yep, as Jeff said..thats how you calculate theta. The power factor is cos (theta). If its a purely resistive load, VA = W. Think it as Watts on a horizontal axis, Volt Amps Reactive (VAR) on the vertical axis (so its a complex plane). Volt Amps (VA) is a line from the origin out into the plane. W = VA*cos(theta). VAR = VA*sin(theta). More riveting stuff

Thanks guys, this AC stuff is getting easier the more I am looking at it. It's not that tuff and it proves that there really is a real world use for triginomitry.