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Old 3rd April 2008, 10:36 PM   #41
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Quote:
Originally posted by peranders
I have had 100 pcs 2SJ50 and equally many 2SK135. The Vgsth goes from almost zero up to 2.2 volts so using a fixed bias value is not a good idea in the long run. In my case you were fource to have 0-6 Volts, adjustable. and also temperature stabilised. My first mosfet amp had to stable temperatures, one luke warm and one hot and the bias element was only a pot.

An another comment: If you have a large voltage swing on your BD139/140 (or any other model for that matter) you will get heavily increased distortion with the voltage swing is near Vce max. Max 25-30 V peak would be advisable in your case.
Hi Per Anders,

Sigurd introduced you to me and I have been on yopur web-site. Very nice clear and concise, it would seem you are also a head-fi man.

Kind regards and nice meeting you through our mutual friend,

Nico
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Old 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?
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Old 5th April 2008, 02:09 AM   #43
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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 hi-jacked 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
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Old 5th April 2008, 06:18 AM   #44
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Quote:
Originally posted by Bruce@bluff
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?
Power is calculated from this:
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 10-12 V higher than the drain.
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Old 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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Old 5th April 2008, 10:41 AM   #46
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Quote:
Originally posted by peranders

Power is calculated from this:
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 10-12 V higher than the drain.
Thanks Per Anders.
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
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Old 5th April 2008, 10:47 AM   #47
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For Bruce : RMS

For a minute there i thought Oompje Nico was the most coolheaded engineer in ZA, what a relief.
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Old 5th April 2008, 10:49 AM   #48
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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).
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Old 5th April 2008, 10:51 AM   #49
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Quote:
Originally posted by jacco vermeulen
For Bruce : RMS

For a minute there i thought Oompje Nico was the most coolheaded engineer in ZA, what a relief.
Sorry Jacco,

there was a little sarcasm I admit. My apologies.

Nico
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Old 5th April 2008, 11:07 AM   #50
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Quote:
Originally posted by peranders

Power is calculated from this:
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 10-12 V higher than the drain.
I must correct myself, it should be 2 not SQR(2) => 189 W (I thought my earlier result was a bit too much)
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