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3rd April 2008, 10:36 PM  #41  
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Location: East Coast of South Africa

Quote:
Sigurd introduced you to me and I have been on yopur website. Very nice clear and concise, it would seem you are also a headfi man. Kind regards and nice meeting you through our mutual friend, Nico 

4th April 2008, 06:44 PM  #42 
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The link at the beginning of the thread still states that the amp puts out 240 watts into 8 ohms and 380 watts into 4 oms?

5th April 2008, 02:09 AM  #43 
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Join Date: Nov 2005
Location: East Coast of South Africa

Hi Bruce,
it is not that easy to explain the output power of an amp since you never listen to sine waves and you seldom have resistances as speakers. But for you this might be a good place to start. Root Mean Square, is the value assigned to the equivalent heating effect (power) produced by a pure sine wave with voltage and current in phase as would be by a DC source. To make this simple, if 1VDC is applied across a 1 ohm resistor and 1 Amp DC flowing through the resistor the resistor 1 watt of heat will result. In order to produce the same amount of heat in this 1 ohm resistor 1.414 VAC potential across it and 1.414 AAC must flow through it in the same period. This means that you would need 1.414 VAC x 1.414 AAC = 2 watt rms in order to achieve this. I did not specify any of the figures that you mention earlier, what I did say on my web site was that you could get 180 watt and 240 w rms into an 8 and 4 ohm resistive load respectively. Now knowing that the typical volt drop across a mosfet is between 4  6 V, it becomes pretty straight forward to calculate the DC supply voltage needed to generate these sort of powers in a resistive load using a single pure sine wave. Now let us assume for a moment that the amplifier producing this sine wave is equally capable of producing a square wave, so one can say that the full DC producing 50% of the power during one half cycle and again during the negative half cycle. This formula is strictly VxI and the +V is equivalent to the positive rail and V, the negative rail. Similarly there exist +I and I and the power for the full cycle is still V x I for the period. In other words if your power supply was a battery and could sustain both the supply voltage and current under full load conditions then the output power of this amp is truly VI watts. Take they case of an 8 ohm load and 55V battery and connect it across the load, a current will flow of 6.8A the power that the amp thus can produce is 378 watt. If we change this stimulus to a sine wave, then of course it could be approximated by 0.707 x 378 watt = 267 watt. But integrating the area under this curve reveals only half that and an rms power equal to 133.6 w is achieved. Now in my web I dropped the 33.6 watt (because the MOSFETS don't swing rail to rail unless you operate the driver from a higher voltage) and I claimed that with 55V rails you will get 100 watt rms, for whatever this is worth. Doubling the output devices to increase the drive current and adjusting the supply voltage for 8 ohm and 4 ohm loads you can expect 180 watt and 240 watt respectively from this amplifier. If you care to read what I say in my web, not the hijacked version that was referenced in the first post the numbers are quite clear. Bruce, I am still keen to have your apology for the lies that you told about me earlier on the forum. Kind regards Nico 
5th April 2008, 06:18 AM  #44  
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Quote:
Pmax = U^2/(R*SQR(2)) 55 volt power supply, 8 ohms load 55*55/(8*1.414))= 267 Watts theorectical max! In order to reach near this the driver must have higher supply voltage than the output transistors. The gate voltage must be 1012 V higher than the drain.
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5th April 2008, 06:25 AM  #45 
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Bruce, if you don't have the theoretical background, I would advice you to use an humble attitude and then ask politely if you think facts seem to be wrong. As you can see here according to my BASIC calculations (ohm's law) Nico is in the right area of his claims. As others have pointed out, you can always have done a mistake or improve a design but this can be discussed in a mature way.
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5th April 2008, 10:41 AM  #46  
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I simply showed how to get to this standard formula, and what RMS means, few people know where it comes from. I also agree totally with the your second statement. To keep this amp simple I did not want to use a higher voltage on the drive stage. Nico 

5th April 2008, 10:49 AM  #48 
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Location: East Coast of South Africa

I may also add that if you do not stabilize the supply that under full load conditions the supply voltage will drop to close to the rms voltage of the transformer. Here the reservoir capacitors, line regulation and transformer ratings start playing a major role..
So when designing an amp that you want to be sure that it can deliver the power you intend then you must ensure that your power supply can maintain the rail voltage. This is very seldom the case and manufacturers would derate power supplies according to how they believe that the equipment will be used. Professional P.A. equipment are 100% rated (or should be). 
5th April 2008, 10:51 AM  #49  
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Quote:
there was a little sarcasm I admit. My apologies. Nico 

5th April 2008, 11:07 AM  #50  
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