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
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
Power is calculated from this:Bruce@bluff said:
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.
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.
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).
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).
once come to
my shop for repair some british amplifier i will not recall the brand ....classic design some 2*60w but the owner had run it heavilly on a kids party ....
result: the all circuit performed the best at the abuse, transistors didnt blow eventhough boiling hot ...nothing went wrong but simly lost the all secondary of the trafo .....
from that you can tell that it wasnt any mains fault cause if it is you loose the primary winding ....
nothing else was broken ...replaced trafo all working perfect ....
strange ha ???
my shop for repair some british amplifier i will not recall the brand ....classic design some 2*60w but the owner had run it heavilly on a kids party ....
result: the all circuit performed the best at the abuse, transistors didnt blow eventhough boiling hot ...nothing went wrong but simly lost the all secondary of the trafo .....
from that you can tell that it wasnt any mains fault cause if it is you loose the primary winding ....
nothing else was broken ...replaced trafo all working perfect ....
strange ha ???
With 55 V supply voltage and the given design you should be able to get a peak voltage of 42-45 volts asuming a stable supply voltage. This will give you 126 W at 8 ohms. It's also worth noticing that the mosfets got better after some time. My first amp delivered 280 W with old mosfets and with just replacing the fets I got 380 W! The Rdson was much lower than 1.7 ohms according to the datasheet.Bruce@bluff said:
... so I think we can say around 100 W out at 8 ohm and 55 V.
It can sometimes be hard to just calculate the peak voltage because gm and Vth can vary. With BJT's is't easier.
Hello all i must appologise about my comments i made about Nico. He seems to be taking alot of flack about his design
and so the saying goes one mans meat is is another mans poisin.
Good reading would be Douglas Selfs article on Science and subjectvism. An example in point while working for the Playhouse
the forman, a classical music lover asked me to build a small
amp for his workshop at home he was in a rush for it so i built a circuit very similar to the 20 watt fairchild with tip 142/147 dalingtons and while the " hi-fi " afficinados would cringe with a bararge of criticism the forman was more than happy with
the sound and after seeing him the other day 10 years i asked
how his amp and he said still sounding good.
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