This is a very basic question and a topic I'm sure has been covered extensively, but somehow I can't find any information, discussions or guidelines.
The question is this: knowing the rated output power and input +/- supply voltage of a given power amplifier, how do we calculate the maximum current that the PSU for this amp must handle?
For example, I'm assembling the LJM L12-2 power amp. It's 120 W into 8 Ohms at +/- 48 V.
P = I^2 * R, so I = sqrt(P / R) = sqrt(15) = less than 4 A?
Now, the key point I don't understand: does each + and - rail have to supply 4 A, or both of them in total?
My logic and understanding tells me that at any time, the amp is producing either the positive or the negative half-wave of the output, hence, the current is only being sourced by one rail at any moment, so each rail must be 4 A even though on average such a PSU seems to be twice oversized for its load.
Right or wrong?
Thanks in advance for any advice and explanations.
The question is this: knowing the rated output power and input +/- supply voltage of a given power amplifier, how do we calculate the maximum current that the PSU for this amp must handle?
For example, I'm assembling the LJM L12-2 power amp. It's 120 W into 8 Ohms at +/- 48 V.
P = I^2 * R, so I = sqrt(P / R) = sqrt(15) = less than 4 A?
Now, the key point I don't understand: does each + and - rail have to supply 4 A, or both of them in total?
My logic and understanding tells me that at any time, the amp is producing either the positive or the negative half-wave of the output, hence, the current is only being sourced by one rail at any moment, so each rail must be 4 A even though on average such a PSU seems to be twice oversized for its load.
Right or wrong?
Thanks in advance for any advice and explanations.
Hi
Assuming a class AB amplifier without excessive idle current, the DC current from the PSU will be the average of the output current.
The average current, Idc (assuming a sineous signal) will equal Ipeak/pi.
Ipeak can be calculated as sqrt(Ppeak/R)
In your example Ipeak = sqrt(2*120W/8 Ohms) = 5.48A
Idc = 5.48A/pi = 1.74A and that will be the requirement for each power rail.
Sanity check: Total PSU power = 2 * 48V * 1.74A = 167.4W
Total power loss = 167.4W - 120W = 47.4W
Efficiency: 71.6% (120W out of 167,4W) - and the theoretical best is 78.5%.
However: I've already used the word "assuming" twice.
And when I then add that most simple power supplies will handle a temporary overload which in combination with the varying nature of music allows for an "under sized" power supply where in contrast several audio oriented people will opt for a power supply with plenty of reserve power.
Then the whole picture gets very muddy and I can understand why it is hard to find hard formulas for calculating the power supply.
Cheers, Martin
Assuming a class AB amplifier without excessive idle current, the DC current from the PSU will be the average of the output current.
The average current, Idc (assuming a sineous signal) will equal Ipeak/pi.
Ipeak can be calculated as sqrt(Ppeak/R)
In your example Ipeak = sqrt(2*120W/8 Ohms) = 5.48A
Idc = 5.48A/pi = 1.74A and that will be the requirement for each power rail.
Sanity check: Total PSU power = 2 * 48V * 1.74A = 167.4W
Total power loss = 167.4W - 120W = 47.4W
Efficiency: 71.6% (120W out of 167,4W) - and the theoretical best is 78.5%.
However: I've already used the word "assuming" twice.
And when I then add that most simple power supplies will handle a temporary overload which in combination with the varying nature of music allows for an "under sized" power supply where in contrast several audio oriented people will opt for a power supply with plenty of reserve power.
Then the whole picture gets very muddy and I can understand why it is hard to find hard formulas for calculating the power supply.
Cheers, Martin
Hi Martin, thank you! Very interesting calculations. Where you typed pi, I assume it's not the number π but square root of 2 = 1.41?
I think 120 W is max. continuous power, not peak power, so dividing it by 1.41 isn't justified. So it's Idc that equals 5.48 A, but if we go along with your assumption that it's only the peak current, then RMS current is 3.88 A, just as in my original calculation (but for two channels, whilst I was only calculating for one, as if in dual mono configuration).
So the key thing is that indeed, each of the + and - rails must be able to produce this current, but only one is being under load at a given moment. In a stereo build, it would be great to connect the second power amp to an inverted signal so that its current draw is out of phase with the first one - that could produce better utilization of the + and - supplies so they're no longer 2x oversized. But I don't see a way to do that since both the line inputs and power outputs use shared ground connection.
It is an interesting observation that for dual mono you need 2x the total power supply current capacity, while for a shared left + right channel PSU you only need 1.41x. This seems odd, I wonder if there's a mistake in my reasoning.
Also, I don't quite understand how you were able to calculate efficiency of the power amp out of thin air, without knowing any implementation details. Isn't efficiency the attribute of the specific topology and implementation?
I think 120 W is max. continuous power, not peak power, so dividing it by 1.41 isn't justified. So it's Idc that equals 5.48 A, but if we go along with your assumption that it's only the peak current, then RMS current is 3.88 A, just as in my original calculation (but for two channels, whilst I was only calculating for one, as if in dual mono configuration).
So the key thing is that indeed, each of the + and - rails must be able to produce this current, but only one is being under load at a given moment. In a stereo build, it would be great to connect the second power amp to an inverted signal so that its current draw is out of phase with the first one - that could produce better utilization of the + and - supplies so they're no longer 2x oversized. But I don't see a way to do that since both the line inputs and power outputs use shared ground connection.
It is an interesting observation that for dual mono you need 2x the total power supply current capacity, while for a shared left + right channel PSU you only need 1.41x. This seems odd, I wonder if there's a mistake in my reasoning.
Also, I don't quite understand how you were able to calculate efficiency of the power amp out of thin air, without knowing any implementation details. Isn't efficiency the attribute of the specific topology and implementation?
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Hi Alexium
Allow me to elaborate:
When I wrote pi I meant π ideed - it comes from integrating a halfwave of a sinus over 2π which is what you do to calculate the average given the peak value (aka amplitude) - this is back to the math classes in college.
Do note that I talk about the DC current which usually is the standard for determining the load requirements for a power supply.
This is not the same as the RMS current in the speaker.
And you are right: 120W is the continous power - and therefore the peak power is 240W (no sqrt(2) in this).
With the peak power and load impedance known the peak current and then the averaged (DC) current supplied from any power supply rail can be calculated as I wrote.
Do note that the average is over 2π - implying a full period.
Do not mix a discussion on what rail is carrying the current into this as it does not make sense when averaging over a full period.
Efficiency: This is normally calculated as what percentage of the input power you get as output power - that is at least how I were taught it.
With a known DC current it is rather simple to calculate the input power (like in E * I).
Sorry, but I am not making this out of thin air - this is basic science class knowlegde.
Cheers, Martin
Allow me to elaborate:
When I wrote pi I meant π ideed - it comes from integrating a halfwave of a sinus over 2π which is what you do to calculate the average given the peak value (aka amplitude) - this is back to the math classes in college.
Do note that I talk about the DC current which usually is the standard for determining the load requirements for a power supply.
This is not the same as the RMS current in the speaker.
And you are right: 120W is the continous power - and therefore the peak power is 240W (no sqrt(2) in this).
With the peak power and load impedance known the peak current and then the averaged (DC) current supplied from any power supply rail can be calculated as I wrote.
Do note that the average is over 2π - implying a full period.
Do not mix a discussion on what rail is carrying the current into this as it does not make sense when averaging over a full period.
Efficiency: This is normally calculated as what percentage of the input power you get as output power - that is at least how I were taught it.
With a known DC current it is rather simple to calculate the input power (like in E * I).
Sorry, but I am not making this out of thin air - this is basic science class knowlegde.
Cheers, Martin
I really think Pi shouldn't come up in these calculations, would love someone else to clarify which one of us is correct. There's effective current / voltage, also called RMS, and peak current / voltage, also called voltage amplitude or current amplitude. For sine waves RMS I or U = sqrt(2)*peak I or U. For DC RMS and amplitude are the same (the coefficient is 1.0).
Let's assume we do not have a burst power reserve in the form of large capacitors, all we have is the transformer alone, or an SMPS with negligible output capacitance.
But this is the key point that I don't understand - whether it's OK to divide the calculated required current by two because it's supplied by two independent rails, or not OK because each rail must meet this requirement.
Let's assume we do not have a burst power reserve in the form of large capacitors, all we have is the transformer alone, or an SMPS with negligible output capacitance.
is it not an oversimplification to assume that nothing but pure sines and their equivalent values are at play....and not that i'm an advocate of IP ratings (instantaneous peak ratings) even old school amps like a Crown DC300's supply is on the high side current wise for exactly that reason...if there's no reserve the distortion from current starvation will get gross fast.
and by the way what makes you think that all waveforms have equal energy +/-....
and by the way what makes you think that all waveforms have equal energy +/-....
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Maybe not an oversimplification - any band-limited signal is a sum of a finite number of sine waves. And we're not letting DC through.
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alright educate me how does a band limited signal equal a "finite" number of sine waves?
and please elucidate how it relates to RMS power?
and please elucidate how it relates to RMS power?
I'm not sure why you're asking me these questions, if you know the correct formula to calculate the PSU current - please do share. If you see a mistake in my reasoning, suggest the correction.
Raymas link shows that all signals are made up of sines. And if a waveform is not equal +/- then it contains a DC offset.
The math of the Fourier transform says that a repetitive waveform can be broken down
into an infinite (not finite) number of discrete frequency sine waves. If these are added together,
they will recreate the original waveform. There are some good examples of this in the article.
But the math only works if the original waveform is repetitive. In practice for transient waveforms,
a window is taken of the transient, and then the content of the window is repeated over and over
in time, to satisfy the requirements of the math so a spectrum can be calculated.
The original signal must be band limited because you can only sample a limited number of times
in each period, and two samples per period of the highest frequency present are needed
at a minimum to not lose data. Only when the signal is band limited does the number
of sine waves needed to recreate the original waveform become finite, and not infinite.
into an infinite (not finite) number of discrete frequency sine waves. If these are added together,
they will recreate the original waveform. There are some good examples of this in the article.
But the math only works if the original waveform is repetitive. In practice for transient waveforms,
a window is taken of the transient, and then the content of the window is repeated over and over
in time, to satisfy the requirements of the math so a spectrum can be calculated.
The original signal must be band limited because you can only sample a limited number of times
in each period, and two samples per period of the highest frequency present are needed
at a minimum to not lose data. Only when the signal is band limited does the number
of sine waves needed to recreate the original waveform become finite, and not infinite.
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The max current depends on many things. Where your speaker min z is at. (the lower the freq the more current, bass has more energy). The speakers reactance. (imagine a speaker with a capacitor across the input, the amp has to charge and discharge the cap. This takes current that does not contribute to SPL and burns extra power in the amp. What type of music. (EDM compared to vocal and acoustic guitar). How compressed the dynamics are. (less compression/limiting =more peak current.) And theres probably more.
Yes, but since the air will always go back to ambient pressure, air cannot support DC content in sounds.
Of course amplifiers do draw DC current from the power supply for biasing the output stage.
A small current for class AB, and a larger current for class A, plus the transient current for the audio.
The peak transient current in class AB is the same as the load current, for that polarity of signal.
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Fourier transforms for non periodic signals: https://www.siue.edu/~yadwang/ECE351_Lec11.pdf
Like a single pulse:
Rectangular Pulse and Its Fourier Transform - Wolfram Demonstrations Project
Which actually shows the other half of the information (usually missing in these diagrams) the phase
So any signal can be broken down into sines and then recreated. Isn't that how mp3s work?
Like a single pulse:
Rectangular Pulse and Its Fourier Transform - Wolfram Demonstrations Project
Which actually shows the other half of the information (usually missing in these diagrams) the phase
So any signal can be broken down into sines and then recreated. Isn't that how mp3s work?
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Alexium
I can not guess from your comments whether you do not like the concept of π or it is a simple matter of you not understanding the concept of DC current.
But, let me try once again:
The current you use for calculating the power rating of a power supply is the DC currect, which by definition is the averaged current.
Again assuming a sine signal and a class AB amplifier:
The the current from for example the positive rail will be a half wave sine with a peak current matching the peak current to the load.
Normalizing the mathematical expression for the signal it will then be defined as:
For
0 < X < π : sin(X)
π < X < 2π : 0 (zero)
To calculate the average over a period (0 - 2π) above expression must be integrated and then divided by the length of the period (witch is 2π):
The indefinite integral of sin(X) is -cos(X)
Thus the integral value for the signal over a full period (0 - 2π) is
For
0 < X < π : [(-cos(π)) - (-cos(0)] = 1 + 1 = 2
π < X < 2π : 0
Summed for 0 - 2π: 2
The average is thereby 2/2π = 1/π
Given that I used the normalized expression we have to multiply with the real peak value (which is the Ipeak).
This is how you arrive to the DC current being Ipeak/π.
So both rails must deliver the DC current.
And you are completely wrong about the assumption about the RMS current being the same as the peak current if is from a amplifier power supply.
Your assumption only works is you are talking about a constant, non-varying current. And I can assure you that the current from an amplifier power supply is not constant (it will follow either the positive or the negative half of the speaker current).
Above deals with the power requirements for the power supply.
However: the peak current will be significantly higher (actually π times bigger).
Simple unregulated power supplies will usually not have a problem with this - but for regulated power supplies and SMPS's this must be taken into consideration (and this is another area where things gets muddy, as I mentioned in post #2)
Cheers, Martin
I can not guess from your comments whether you do not like the concept of π or it is a simple matter of you not understanding the concept of DC current.
But, let me try once again:
The current you use for calculating the power rating of a power supply is the DC currect, which by definition is the averaged current.
Again assuming a sine signal and a class AB amplifier:
The the current from for example the positive rail will be a half wave sine with a peak current matching the peak current to the load.
Normalizing the mathematical expression for the signal it will then be defined as:
For
0 < X < π : sin(X)
π < X < 2π : 0 (zero)
To calculate the average over a period (0 - 2π) above expression must be integrated and then divided by the length of the period (witch is 2π):
The indefinite integral of sin(X) is -cos(X)
Thus the integral value for the signal over a full period (0 - 2π) is
For
0 < X < π : [(-cos(π)) - (-cos(0)] = 1 + 1 = 2
π < X < 2π : 0
Summed for 0 - 2π: 2
The average is thereby 2/2π = 1/π
Given that I used the normalized expression we have to multiply with the real peak value (which is the Ipeak).
This is how you arrive to the DC current being Ipeak/π.
No it is not OK - look at above calculation: it is the average over the full period.
So both rails must deliver the DC current.
And you are completely wrong about the assumption about the RMS current being the same as the peak current if is from a amplifier power supply.
Your assumption only works is you are talking about a constant, non-varying current. And I can assure you that the current from an amplifier power supply is not constant (it will follow either the positive or the negative half of the speaker current).
Above deals with the power requirements for the power supply.
However: the peak current will be significantly higher (actually π times bigger).
Simple unregulated power supplies will usually not have a problem with this - but for regulated power supplies and SMPS's this must be taken into consideration (and this is another area where things gets muddy, as I mentioned in post #2)
Cheers, Martin
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An interesting video that shows how complex signals (line drawings) can be broken down into sines.
But what is a Fourier series? From heat flow to drawing with circles | DE4 - YouTube
But what is a Fourier series? From heat flow to drawing with circles | DE4 - YouTube