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pjpoes 6th July 2007 05:59 PM

How much power to produce 10KW or more
I dont want or need this, its a curiosity question. I was talking to someone on another forum about my amp project. He was argueing that I wasted my money and should have purchased a used QSC or Crest amplifier, becuase I could have had 10,000 watts for 500 dollars, or whatever. I said that, first, I did it because I like to DIY, second, I don't like proaudio amplifiers because of the fans, and third, I doupt highly he was getting 10,000 watts from his amplifiers. I'm curious, because he is now saying he tested his amp to 18,000 watts and is running more than one of them, is that even possible for peaks? I mean, it would require over 150 amps of current at 117 volts. More current at 100 volts, which is probably closer to the rails of such an amp, but I don't know. Last amplifier I used rated at even 5,000 watts RMS used 240 lines because of the current problems. His amp uses a switchmode supply, now those don't ever seem to have much reserve capacitance, so am I right to think that the outlet would need to supply most of the current? Wouldn't even a brief peek at 150-200 amps be enough to not only trip the breaker but potentially melt the insulation on 14 guage Romex?

Now this probably is a very flawed test, but I have used an AC Inductive current measuring meter from Fluke to measure the amount of current that is drawn at startup before of many of my amplifiers. This method may be too inaccurate or too slow to get an accurate reading, but doing this, I've never even seen 100 amps through the A/C lines when it does trip a breaker. It seems like physics alone dictates that 10kw from a normal household outlet is next to impossible. I would even question if some of the amps he is mentioning have enough output devices to support that.

tiltedhalo 6th July 2007 06:20 PM

I used to run two Carver 4.0t's bridged on the same 120V AC line, they could easily put out 1200W into my 8 ohm speakers all day and would pop the 25amp breakers long before the peak lights would ever flicker. Lets just take some really simplified math and basic amp designs, said 10KW amp is bridged using 120VDC+/- rail for a total of 240VDC * 2 for a bridged swing of 480VDC. 10,000/480=20.83amps. Now the amp is stereo so you have two channels 41.66 amps, now factor in the power loss of the amp we will even give it a very efficent class H at a thirty percent loss total 54.158amps!! and thats very conservative.
The kind of power exceeds most home mains in the US let alone standard wiring.

pjpoes 6th July 2007 06:37 PM

Using the formula on Wikipedia for peak power, no idea if this is a correct formula, it says, take the peak to peak voltage, then find the peak amplitude it can drive, I assumed it was half the peak to peak, square then, then divide by the impedance, and multiply by the efficiency rating.

120 volt rails X 2, 240 volts P-P, bridged as you put it is then 480, but then divided in half for peak amplitude. Thats 240 squared divided by 8 ohms is 7,200 watts, or 14,400 watts rms. 14,400 watts for 1 second equals 14,400 jouls. Now I understand that a peak impulse could last less time yet, and the only thing I could find was that a single milisecond would multiply that times 250. I couldn't find any goo d charts, just an article that said 2 jouls equals 500 j/ms. If an outlet can produce a maximum of say 3000 jouls, and we reduce this to 1 ms, then I do start to see how an amplifier could potentially produce some insane numbers, if it could utilize that much energy. I mean, thats 750,000 watts per millisecond then. However I have no idea if thats correct or the right way of looking at it. I can't find any good way of calculating an amplifiers instantaneous peak, and because its looked at as a useless number, it seems to be difficult to find.

MaxS 6th July 2007 06:46 PM

Hi dudes,

Have a look about Labgruppen amps. Maybe you'll fine some answers in it.

pjpoes 6th July 2007 07:12 PM

Thanks, they don't seem to have any technical white papers, which is what I thought you were telling me. However, in the specs they do mention the peak output voltage and the peak current, which implies over 11,000 watts per channel in the largest FP13000. However, nothing is given to suggest how this is accomplished. I see that it can run on 120 volts, but can it produce that much output from 120, and if so, how much current draw. I do see however that its power supply is much more robust than is typical of such a supply. It has quite a bit of capacitance it looks like, which would probably help for peaks, though I still don't see enough capacitors, not with such high rails, to store enough joules of energy that much power.

pjpoes 6th July 2007 07:20 PM

Ok reading the manual did enlighten me a bit, they rate it for use with a 115 volt twist lock nema plug rated at 30 amps. They also say that to reach just 1/4 of its rated power requires 58 amps from the outlet. However you can run it on a standard 220 line rated at 15-20 amps and it will only draw 28 amps to reach 1/4 its rated power. Still doesn't give me much info in regard to understanding or measuring an amps peak output, other than it does look like using even a 30 amp outlet, these amps could not possibly reach there peak output other than for just brief moments.

MaxS 8th July 2007 06:49 AM

myhrrhleine 8th July 2007 03:15 PM

It's normal to derate a circuit breaker to 80%, so:
15A=12Acontinuous=1440Watts (120*12)
20A=16Acontinuous=1920Watts (120*16)
30A=24Acontinuous=2880Watts (120*12)

30A@240v=24Acontinuous=5760Watts (240*24)

So, you'd need about a 60 amp breaker@240V for 10kw continuous (100% amp efficiency)

imix500 8th July 2007 04:26 PM

Sure, 10,000W is possible. It just requires 208 3 phase at 30 amps. I'm sure many know of this beast:

BlueWizard 8th July 2007 05:08 PM

I think you are making this more complicated than it needs to be. This is a matter of simple arithmetic.

Here are the power formulas -


P = Power in Watts
I = Current in Amps
E = Voltage in Volts
^2 = a number squared
sqrt() = square root of number in parenthesis

P = IE ...(Current times Voltage)
P = E^2 / R ...(Voltage squared divided by Resistance)
P = I^2 R ...(Current square times Resistance

(link to Ohms Law Pie Chart, scroll down to bottom of link)'s%20Law.htm

[sorry the link won't work because it has a 'space' in it (%20), but I think if you cut and paste the whole thing into your browser, it will work.]

To find the necessary Current applied to an 8 ohm load to produce 10,000 watts -

I = sqrt(P/R) = sqrt(10,000/8) = 35.355 AMPS

To find the necessary Voltage applied to 8 ohms to produce 10,000 watts -

E = sqrt(P x R) = sqrt(10,000 x 8) = sqrt(80,000) = 282.84 VOLTS

That is not from the power grid or electrical outlet, you have to apply that Voltage and that Current to an 8 ohm load to get 10,000 watts.

To find the current necessary from the electrical outlet, assuming 120 volts, we again use I = P/E

I = P/E = 10,000/120 = 83.33 AMPS

83.33 amps from the wall outlet, I don't think so.

Even if we increase it to 240V, that's still 41.667 AMPS, and again, I don't think so.

I'm not even aware of the existence of a 10,000 watt amp. Yes, certainly Rock Concerts use more that 10,000 watts to drive their PA systems, but that is a combination of many amps working together, not a single amp.

The Lab Gruppen amps referred to by someone else are professional concert power amps. One of their larger amps is 2 Channels at 3200 watts per channel for 6400 watts total. That is a rare and expensive amp, I just don't see it going for $500. Their biggest amp is 4 Channels at 2500 watts per channel for a total of 10,000 watts. Both amps into a 2 ohm load.

I like to know how your friend determine his amp was 10,000 watts. No offense, but personally I think he is full of crapola. Especially, when he claims you can buy it for $500. I don't think so.

How about giving us the Brand name of this Magic 10,000 watt amp that sells for a magic 500 dollar bill?


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