Here is my latest project.
It is a stereo 1000watts per channel into 2 ohms or bridged 2000watts into 4 ohms.
It is a stereo 1000watts per channel into 2 ohms or bridged 2000watts into 4 ohms.
An externally hosted image should be here but it was not working when we last tested it.
class D amplifier
mmm how do you propose to obtain 1000w into 2 ohms from a split 60v rail?
Assuming a perfectly regulated power supply and a 100% efficient amplifier you can obtain 60 x 0.7071 = 42.42 volts across the load and this translates to 900w at 2 ohms. So maybe you know something none of us know, how to design a 100% efficient class D amplifier. Looking forward to seeing your schematic.
Even with a perfectly regulated SMPS supply, the losses in the output MOSFETs are huge driving 40 volts plus into 2 ohms
mmm how do you propose to obtain 1000w into 2 ohms from a split 60v rail?
Assuming a perfectly regulated power supply and a 100% efficient amplifier you can obtain 60 x 0.7071 = 42.42 volts across the load and this translates to 900w at 2 ohms. So maybe you know something none of us know, how to design a 100% efficient class D amplifier. Looking forward to seeing your schematic.
Even with a perfectly regulated SMPS supply, the losses in the output MOSFETs are huge driving 40 volts plus into 2 ohms
I calculated the bridged mode then divided by 2 which is clearly wrong as the power in bridged mode is not twice that in single ended mode.
The schematic was taken from a 1000 watt amp but with higher rails.
You are working in RMS power I was working in peak power.
60v single ended into 2 ohms is 1800 watts peak 900WRMS
60v bridged into 4 ohms is 3600 watts peak 1800WRMS
Who is the target self-oscillation frequency ? Unfortunately I have found the answer to this question if I could see the component values, but the values can not be seen. Should consider well whether power IGBTs or MOSFETs can work well at that frequency. In those 1000W (theoretically) flow per output, power dissipation by IGBTs can reach 50W.
And some theory: peak value is 1.41 * Vrms or Vrms = 0.707 * Vpeak (for a sine wave). And power has the same addiction.
And some theory: peak value is 1.41 * Vrms or Vrms = 0.707 * Vpeak (for a sine wave). And power has the same addiction.
The switching frequency is around 250KHz.
0.707 isn't that far away from half. I was being very rough with calculations.
Its standard practice on ebay to say RMS is half peak.
0.707 isn't that far away from half. I was being very rough with calculations.
Its standard practice on ebay to say RMS is half peak.
I am using irfb4227.
At 20 amps and 20 milliohms that is around half a watt or quarter watt per mosfet.
To get the actual maximum peak signal voltage output amplitude, you still have to subtract the maximum p-p ripple amplitude AND the amplifier's clipping voltage (i.e. the minimum-possible voltage between the power rail and the output, across the power output stage) at maximum output power.
So if you end up with 54 Volts max peak output (with about 75000 uF per rail per channel), that would be about 38 Volts RMS for the max output voltage.
A max output voltage of 38 V RMS into 2 Ohms would give a rated max output power of 729 Watts RMS
A max peak output voltage of 38 x 1.414 = 54 V pk into 2 Ohms would give a max peak output power of 54 x 54 / 2 = 1458 Watts peak.
So you were correct that peak power = 2X RMS power.
So if you end up with 54 Volts max peak output (with about 75000 uF per rail per channel), that would be about 38 Volts RMS for the max output voltage.
A max output voltage of 38 V RMS into 2 Ohms would give a rated max output power of 729 Watts RMS
A max peak output voltage of 38 x 1.414 = 54 V pk into 2 Ohms would give a max peak output power of 54 x 54 / 2 = 1458 Watts peak.
So you were correct that peak power = 2X RMS power.
(1) P_rms = V_rms x V_rms / R
(2) P_pk = V_pk x V_pk / R
(3) V_pk = √2 x V_rms
Substituting (3) in (2):
P_pk = √2 x V_rms x √2 x V_rms / R
P_pk = √2 x √2 x (V_rms x V_rms / R)
Squaring square root and recognizing (1):
P_pk = 2 x P_rms
(2) P_pk = V_pk x V_pk / R
(3) V_pk = √2 x V_rms
Substituting (3) in (2):
P_pk = √2 x V_rms x √2 x V_rms / R
P_pk = √2 x √2 x (V_rms x V_rms / R)
Squaring square root and recognizing (1):
P_pk = 2 x P_rms
Sorry... Vpeak = 1.41* Vrms and Ipeak = 1.41* Irms.
P = UI Ppeak = Vpeak * Ipeak = 1,41* Vrms * 1.41 *Irms
Ppeak = 1.41 * 1.41 * Prms ... 2 * Prms.
Power
PS. There is NO such thing as "RMS Power"😱
Power is calculated as volts x volts divided by impedance OR amps x amps x impedance SO HOW can the root mean square end up in the answer?
What you are saying is root 4 x root 4 = root 4. I guess this is also Ebay mathematics????
PS. There is NO such thing as "RMS Power"😱
Power is calculated as volts x volts divided by impedance OR amps x amps x impedance SO HOW can the root mean square end up in the answer?
What you are saying is root 4 x root 4 = root 4. I guess this is also Ebay mathematics????
RMS
Not to be rude Nigel but you are out of your depth here.
Volts RMS x Volts RMS divided by impedance CANNOT give you an answer with Root Mean Square in the answer.
Simple mathematics.
Kind of like the Ebay watts you mention.
Not to be rude Nigel but you are out of your depth here.
Volts RMS x Volts RMS divided by impedance CANNOT give you an answer with Root Mean Square in the answer.
Simple mathematics.
Kind of like the Ebay watts you mention.
I learned in 1980 that RMS is peak * 0.707
So 1000watts peak is 0.707 * 1000watts.
Or more commonly known as AES.
Just look through hundreds of websites including ebay and AES or RMS is used if not peak.
Just because it doesn't make sense mathematically doesnt mean it doesn't have a real life use.
I understand the maths.
So 1000watts peak is 0.707 * 1000watts.
Or more commonly known as AES.
Just look through hundreds of websites including ebay and AES or RMS is used if not peak.
Just because it doesn't make sense mathematically doesnt mean it doesn't have a real life use.
I understand the maths.
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In a sine wave,
RMS is 0.707 X Peak.
PEAK is 1.414 X RMS.
It gets more complicated with Bridge output power but there we have it.
That's it. Simple mathematics to be found in every publication worth its salt and learned in the first year if not week of any technical college course.
RMS is 0.707 X Peak.
PEAK is 1.414 X RMS.
It gets more complicated with Bridge output power but there we have it.
That's it. Simple mathematics to be found in every publication worth its salt and learned in the first year if not week of any technical college course.
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RMS Power does exist, and can be calculated. It just has no physical significance.
However, in the audio world, the term has come into accepted use as a synonym for average power or mean power (maybe especially when the load is purely resistive).
Power
Simple RMS stands for Root Mean Square and it is absolutely a stupid term and has no mathematical or electrical meaning.
A root multiplied by a root cannot give you an answer with a root in it.
Simple RMS stands for Root Mean Square and it is absolutely a stupid term and has no mathematical or electrical meaning.
A root multiplied by a root cannot give you an answer with a root in it.
"RMS Power" certainly has a mathematical meaning. It's simply the square root of the mean of the square of the power over some specified time period. Why do you keep saying "a root multiplied by a root"? RMS doesn't only apply to sine waves. See here: Root mean square - Wikipedia, the free encyclopedia
In the attached equation for the RMS value of ANYTHING, just set f(t) = power(t). As I already said, the true RMS Power, as calculated with the attached equation, has no physical significance. But it is obviously mathematically definable and calculatable.
What most audio people call P_rms is actually P_avg, the average power. But the formulas like P_avg = V_rms²/R, and P_avg = I_rms²xR, are only valid if R is a pure resistance. They are not valid when R is a reactive load (i.e. with any capacitance and/or inductance).
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This is the argument that you were trying to make and it is (technically) valid:
Hi Fi Writer - Meaningless RMS power
So yes, we should say "Average" power instead of "RMS" power, when reporting power calculated with the RMS values of sine wave currents or voltages and pure resistances.
However, "watts RMS" is widely accepted, in the audio world, and everyone knows what it actually means, EXCEPT, many of them don't seem to realize that the factors for its conversion to and from peak power are not 1.414 and 0.7071 (sqrt(2) and its inverse), like they are for peak and RMS sinusoidal voltage and current. They are 2 and 1/2, instead.
Hi Fi Writer - Meaningless RMS power
So yes, we should say "Average" power instead of "RMS" power, when reporting power calculated with the RMS values of sine wave currents or voltages and pure resistances.
However, "watts RMS" is widely accepted, in the audio world, and everyone knows what it actually means, EXCEPT, many of them don't seem to realize that the factors for its conversion to and from peak power are not 1.414 and 0.7071 (sqrt(2) and its inverse), like they are for peak and RMS sinusoidal voltage and current. They are 2 and 1/2, instead.
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