Thanks all. Where does all this leave Bob Cordell's statement (as in post #1)? Is the consensus in agreement with AndrewT's comment in post #13 - "I believe Cordell should have used rms current at 50% duty cycle"?
Cordell is correct. To calculate power supplied from a constant DC supply you must use average current. If he wanted to calculate fuse ratings in the DC supply wires he should have used RMS current. This is because these two calculation use quite different voltages: one a DC value, the other a value from the voltage drop across the fuse.
As others have said, you need to use the correct 'averaging' method depending on what you are trying to do.
As others have said, you need to use the correct 'averaging' method depending on what you are trying to do.
I'll secong DF96 who posted while I was typing this:
Simple average current as per Cordell described in post #1 is correct because the power rails are constant.
RMS V or I is useful for power dissipated in a resistor in which voltage is proportional to current, thus average power is proportional to voltage squared or current squared: both V and I are changing and you can't pull one out of the integral you use for averaging, as you could if one were constant. So RMS is a convenient pseudo-average useful for power and thermal calculations: avg(P) = avg( V*I ) = avg( I*R * I ) = R * avg( I*I ) = R * ( rms( I ) )^2.
But for this question about _input_ power: The instantaneous power at any time is the (constant) rail voltage times the instaneous current. So the constant voltage can be pulled out of the integral use to determine average power. this leaves you with avg( P ) = Vrail * avg( I ).
Simple average current as per Cordell described in post #1 is correct because the power rails are constant.
RMS V or I is useful for power dissipated in a resistor in which voltage is proportional to current, thus average power is proportional to voltage squared or current squared: both V and I are changing and you can't pull one out of the integral you use for averaging, as you could if one were constant. So RMS is a convenient pseudo-average useful for power and thermal calculations: avg(P) = avg( V*I ) = avg( I*R * I ) = R * avg( I*I ) = R * ( rms( I ) )^2.
But for this question about _input_ power: The instantaneous power at any time is the (constant) rail voltage times the instaneous current. So the constant voltage can be pulled out of the integral use to determine average power. this leaves you with avg( P ) = Vrail * avg( I ).
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i thought that convention says, rms of the sine is the equivalent of dc voltage to produce the same heating effects in a resistor...i remember this was taught to us in college....some 40+ years ago...
Hi,
The point is we are not talking just P, V, I, W and resistors.
The power in the load is 100W but it is supplied by the output
stage, here +/- 40V. To supply 100W the output stage must
dissipate the difference between 40V peak (the supply) and
the actual output voltage x the actual current in the load.
40V into 8 ohms is 5A, that is the peak current, and 200W instantaneous.
No output dissipation (~), so the supply does 200W instantaneous.
20V into 8 ohms is 2.5A, 50W instantaneous, but the output stage also has
to dissipate 50W instantaneous, so the supply does 100W instantaneous.
30V into 8 ohms is 3.75A, 112.5W instantaneous, but the output stage
has to dissapate 37.5W instantaneous, so the supply does 150W instantaneous.
10V into 8 ohms is 1.25A, 12.5W instantaneous, but the output stage
has to dissapate 37.5W instantaneous, so the supply does 50W instantaneous.
Its clear the power from the supply is a linear function of output
voltage, 5W instantantaneous per instantantaneous output V.
So the power from the supply is the average V (not rms) of the output.
With a resitive load its also the average I (not rms) of the output.
rgds, sreten.
The point is we are not talking just P, V, I, W and resistors.
The power in the load is 100W but it is supplied by the output
stage, here +/- 40V. To supply 100W the output stage must
dissipate the difference between 40V peak (the supply) and
the actual output voltage x the actual current in the load.
40V into 8 ohms is 5A, that is the peak current, and 200W instantaneous.
No output dissipation (~), so the supply does 200W instantaneous.
20V into 8 ohms is 2.5A, 50W instantaneous, but the output stage also has
to dissipate 50W instantaneous, so the supply does 100W instantaneous.
30V into 8 ohms is 3.75A, 112.5W instantaneous, but the output stage
has to dissapate 37.5W instantaneous, so the supply does 150W instantaneous.
10V into 8 ohms is 1.25A, 12.5W instantaneous, but the output stage
has to dissapate 37.5W instantaneous, so the supply does 50W instantaneous.
Its clear the power from the supply is a linear function of output
voltage, 5W instantantaneous per instantantaneous output V.
So the power from the supply is the average V (not rms) of the output.
With a resitive load its also the average I (not rms) of the output.
rgds, sreten.
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Perhaps you are thinking of "RMS power," but that is NOT a real unit of measurement.
As covered by others in the thread, and regardless of the waveshape (and as I learned in AC circuits analysis), RMS voltage times RMS current is average power.
Circa 1973 the Institute for High Fidelity came out with some "standards" for measuring audio power amplifier capability and it mistakenly used the phrase "RMS power." The phrase that (perhaps) should have been used is average power.
There had been a lot of bogus phrases used in advertising, "peak power," "music power," "peak music power" and whatnot. The IHF had everyone standardize on ONE bogus phrase: "RMS power."
Okay, now I see your most recent post. You can find an old AC circuits analysis textbook online cheap, or maybe this will explain things:
www.ece.msstate.edu/~donohoe/ece3183ac_analysis.pdf
are we arguing about 10% difference?
to me it doesn't matter that much,
except if you are a scholar wanting to go about the math of things...
you can build practical amps and power supplies regardless...😉
to me it doesn't matter that much,
except if you are a scholar wanting to go about the math of things...
you can build practical amps and power supplies regardless...😉
No, we are teaching correct understanding. In this particular case the difference may only be 10% but in some other situation it could be greater. In any case, as getting it right is not difficult, why not get it right? Or is getting things right only a matter for "scholars"?AJT said:
I guess 10% doesn't make any difference, when you generally oversize your trafo by 50-100% compared to what they put in an "equivalent" store-bought amplifier. And if you're using a switcher it needs to be able to put out the PEAK current anyway. Unless you like it going into limit every time the bass hits.
Post # 11 says:
"This is correct. All DC current measurements are average by definition.
This gets a little confusing when the voltages and currents are DC with ripple, as they contain both DC and AC. The question is whether the AC component does useful work in the load, or merely heats up the wiring."
All the definitions of "alternating current" I've found describe it as current that changes direction every half-cycle (and some even talk about current-carriers reversing direction) - to do that it must cross to the negative side of zero on the "Y" axis. So if it doesn't do that then it's DC right? But suppose there is a sinusoidal variation from 0 volts to +10 volts - that is DC? So it can't be analysed with AC theory? What is the power in that waveform? Presumably it isn't the RMS value, since it's not AC. This has perplexed me for some time! Are we dealing with a sloppy use of the term "AC" when describing AC as a voltage current that changes direction/polarity every half-cycle - would it be more exact to describe AC as a current varying periodically about a mean (as someone hinted earlier)? Can anyone illuminate me? I've learnt a lot from this thread already, so it would be good to clarify that question.
"This is correct. All DC current measurements are average by definition.
This gets a little confusing when the voltages and currents are DC with ripple, as they contain both DC and AC. The question is whether the AC component does useful work in the load, or merely heats up the wiring."
All the definitions of "alternating current" I've found describe it as current that changes direction every half-cycle (and some even talk about current-carriers reversing direction) - to do that it must cross to the negative side of zero on the "Y" axis. So if it doesn't do that then it's DC right? But suppose there is a sinusoidal variation from 0 volts to +10 volts - that is DC? So it can't be analysed with AC theory? What is the power in that waveform? Presumably it isn't the RMS value, since it's not AC. This has perplexed me for some time! Are we dealing with a sloppy use of the term "AC" when describing AC as a voltage current that changes direction/polarity every half-cycle - would it be more exact to describe AC as a current varying periodically about a mean (as someone hinted earlier)? Can anyone illuminate me? I've learnt a lot from this thread already, so it would be good to clarify that question.
Your example is AC of 5Vtop with a DC offset of 5V=.MikeVou said:
Average voltage is 5V , the average of the AC part is zero.
Aveage of AC is allways zero unless there is a DC component.If one talks about the aveage of a sinewave they realy mean average of one half only.
The average current is to say how much charge is passed in a determing time.
If an amplifier draws current (fluctuated) there is a certain average and the supply must keep up with the demand with the same average=DC current.
Mona
Elementary introductions to electricity teach the difference between AC and DC. Like much early stuff, this is a simplification but not always flagged as such. Later on, people get confused (and argue in threads about - see the search function) about whether 'AC' has to cross the zero line and whether 'DC' can vary or not. The best option is to regard AC and DC as stories we tell newbies, and as useful descriptions in many contexts, but not definitive terms - Fourier theory is better.
A 5V (peak) AC wave added to a 5V DC level has an RMS value of about 8.5V. This is because the RMS value of the AC part is 3.535V (=0.707 x 5V). The RMS value of DC is equal to the DC value. Then, if there are two different frequencies (say, 0Hz=DC and 50Hz) their RMS values just add.
The average of a sinewave is always zero. The average of a rectified sinewave is not zero. People need to distinguish between these two quite different situations. If they mean the average of half a sinewave then they need to say so, and specify what are the starting and ending angles.
A 5V (peak) AC wave added to a 5V DC level has an RMS value of about 8.5V. This is because the RMS value of the AC part is 3.535V (=0.707 x 5V). The RMS value of DC is equal to the DC value. Then, if there are two different frequencies (say, 0Hz=DC and 50Hz) their RMS values just add.
The average of a sinewave is always zero. The average of a rectified sinewave is not zero. People need to distinguish between these two quite different situations. If they mean the average of half a sinewave then they need to say so, and specify what are the starting and ending angles.
That is a fair explanation. The reality is more like average and RMS, than "AC" and "DC".
I once designed drivers for Peltier cooling modules, little solid-state gadgets that pump heat from a hot side to a cold side. These require quite clean DC with no ripple. (Or should I say, current with a RMS value not much greater than its average. 🙂 ) The reason is that the DC component does the useful cooling work, while the AC just heats both sides of the device.
A similar explanation holds for battery charging: the DC component (average current) does the charging while the AC heats up the battery needlessly.
Don't ask me if this explanation holds for the supply to an audio amp output stage though. 🙂
I once designed drivers for Peltier cooling modules, little solid-state gadgets that pump heat from a hot side to a cold side. These require quite clean DC with no ripple. (Or should I say, current with a RMS value not much greater than its average. 🙂 ) The reason is that the DC component does the useful cooling work, while the AC just heats both sides of the device.
A similar explanation holds for battery charging: the DC component (average current) does the charging while the AC heats up the battery needlessly.
Don't ask me if this explanation holds for the supply to an audio amp output stage though. 🙂
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