A question from a hobbyist, not a professional! -
In "Designing Audio Power Amplifiers" (page105), Bob Cordell writes in respect of a 100W amplifier feeding 8ohm load: "At full power into an 8ohm load, the output voltage is 40V peak and the output current is is 5A peak. The current drawn from each supply rail is a half-wave rectified waveform with a peak current of 5A. The average of a half-sine is about 63% and the duty cycle is 50%, so the average rail current is 0.5*0.63*5A = 1.58A. Average input for the two rails is then 126W."
Can someone explain for me why average current is used and not rms current for a (half) sine waveform?
Thanks🙂
Mike
In "Designing Audio Power Amplifiers" (page105), Bob Cordell writes in respect of a 100W amplifier feeding 8ohm load: "At full power into an 8ohm load, the output voltage is 40V peak and the output current is is 5A peak. The current drawn from each supply rail is a half-wave rectified waveform with a peak current of 5A. The average of a half-sine is about 63% and the duty cycle is 50%, so the average rail current is 0.5*0.63*5A = 1.58A. Average input for the two rails is then 126W."
Can someone explain for me why average current is used and not rms current for a (half) sine waveform?
Thanks🙂
Mike
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Hi, the average is rms, rgds, sreten.
The above implies an amplifier with 40V rails
will hit 100W into 8 ohms and in doing so will
dissipate 26W in the amplifier for 100W.
The interesting point is at what power does
the amplifier dissipation peak, around 63W,
hence 2/3 maximum power soak tests. *
* Not really applicable to audio amplifiers
but very applicable to industrial amplifiers.
The above implies an amplifier with 40V rails
will hit 100W into 8 ohms and in doing so will
dissipate 26W in the amplifier for 100W.
The interesting point is at what power does
the amplifier dissipation peak, around 63W,
hence 2/3 maximum power soak tests. *
* Not really applicable to audio amplifiers
but very applicable to industrial amplifiers.
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if we are talking sine waves, average is 0.636, while rms is 0.7071 of the peak values of current or voltage...Alternating Current
Thanks Sreten & AJT - what's puzzling me is why the calculation uses average current and not rms current. I'm wondering is it because you presumably can't have rms of a half sine wave?
Thanks, but that confuses me - if average is 0.636*peak and rms is 0.707*peak, how are they interchangeable. Something I'm not understanding...
Hi,
My fault for a misassumption and misunderstanding.
P= V/I, = Vsquared/R = IsquaredxR.
For the latter two you always use rms,
as its related to the square powers.
For the first with V fixed I must be a simple average, not rms, and
I admit I missed the point, but its crucial in the ratio of amplifier
output to amplifier dissipation to power drawn from the supply,
where the latter always equals the two former added up.
rgds, sreten.
My fault for a misassumption and misunderstanding.
P= V/I, = Vsquared/R = IsquaredxR.
For the latter two you always use rms,
as its related to the square powers.
For the first with V fixed I must be a simple average, not rms, and
I admit I missed the point, but its crucial in the ratio of amplifier
output to amplifier dissipation to power drawn from the supply,
where the latter always equals the two former added up.
rgds, sreten.
sine waves are used for convenience, music being what it is can not be used in calculations although we listen to music, sine waves are boring to listen to....😉
they are computed differently...Average Voltage of a Sinusoidal AC Waveform
MasteringElectronicsDesign.com : How to Derive the RMS Value of a Sine Wave with a DC Offset
they are computed differently...Average Voltage of a Sinusoidal AC Waveform
MasteringElectronicsDesign.com : How to Derive the RMS Value of a Sine Wave with a DC Offset
Thanks again both. Getting clearer, and looks like I'll have to do some work on AC theory!
Sreten - I had assumed that P= Vrms * Irms, but from what you said it should be P=Vave * Iave - have I got that right? Still a bit confused over that, and when I just peeked at the Nuffield Foundation website, it said: "...The rms values of current and voltage multiplied together give the actual power. This is a vital fraction when trying to do quantitative power and energy experiments such as specific thermal capacity...". Not trying to be obtuse or argumentative, but it is confusing... I'm sure I'll get there in the end!
Sreten - I had assumed that P= Vrms * Irms, but from what you said it should be P=Vave * Iave - have I got that right? Still a bit confused over that, and when I just peeked at the Nuffield Foundation website, it said: "...The rms values of current and voltage multiplied together give the actual power. This is a vital fraction when trying to do quantitative power and energy experiments such as specific thermal capacity...". Not trying to be obtuse or argumentative, but it is confusing... I'm sure I'll get there in the end!
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Dear Sreten, sorry but no.
AJT is right:
What's the difference and why, when dealing with the exact same sinewave 😱 , we choose to use either average or RMS?
Because both values explain different phenomena, we choose one or the other depending on what we want to know.
To compare , if we were talking DC, just one value is necessary, which we plain call current or voltage,no qualifiers, because it does not vary along time.
To be more precise, if we want to calculate two very different phenomena such as
a) charging/discharging a capacitor
b) heating a resistor (either to know power into a load or losses in conductors or active devices)
in both cases we use the same values.
But, on AC, the same value does not explain both.
So :
a) to design a power supply, capacitor size and ripple, we use average.
b) to calculate power dissipated, we use RMS
Dear mikevou: hope this clears things up for you.
See that it applies to all cases which look contradictory to you 🙂
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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.
In Cordell's example, he (reasonably) assumes that only the DC component is useful to the output stage, so he uses the average values. However, if he were sizing rail fuses, he should use the RMS current, 1.77A.
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.
In Cordell's example, he (reasonably) assumes that only the DC component is useful to the output stage, so he uses the average values. However, if he were sizing rail fuses, he should use the RMS current, 1.77A.
@J M Fahey -
"So :
a) to design a power supply, capacitor size and ripple, we use average.
b) to calculate power dissipated, we use RMS"
That's what is confusing. In the passage I quoted at top of thread, BC is specifically talking about "Estimating the Input Power" (as a step in calculating power dissipated in heat), not about sizing capacitors or the voltage/current minutiae of the power supply. He does seem to be talking precisely about your "power dissipated". That's why I was confused that he wasn't using rms values.
He is an expert, and I'm certainly not, so I'm not doubting his words - just trying to understand the theory here. It may also be true that any errors from using the wrong factor in calculations just gets swallowed up in the other working assumptions you have to make about transformer specifications/loading, uncertainty over heatsink thermal capacity, etc. The smallish (!) 10% difference between 0.636 and 0.707 will often probably get swamped by practical safety margins (if you're an amateur like me anyway). But it would be good to understand the theory BC is using!
"So :
a) to design a power supply, capacitor size and ripple, we use average.
b) to calculate power dissipated, we use RMS"
That's what is confusing. In the passage I quoted at top of thread, BC is specifically talking about "Estimating the Input Power" (as a step in calculating power dissipated in heat), not about sizing capacitors or the voltage/current minutiae of the power supply. He does seem to be talking precisely about your "power dissipated". That's why I was confused that he wasn't using rms values.
He is an expert, and I'm certainly not, so I'm not doubting his words - just trying to understand the theory here. It may also be true that any errors from using the wrong factor in calculations just gets swallowed up in the other working assumptions you have to make about transformer specifications/loading, uncertainty over heatsink thermal capacity, etc. The smallish (!) 10% difference between 0.636 and 0.707 will often probably get swamped by practical safety margins (if you're an amateur like me anyway). But it would be good to understand the theory BC is using!
I believe Cordell should have used rms current at 50% duty cycle.
The total current and thus total power fed into the amplifier comes out as a total of the internally dissipated heat and the load dissipated heat.
The load heat is based on rms currents.
The Total input - the load output = the internal dissipated heat.
{rms current input * rms supply voltage} - {rms load current * rms load voltage} = internal heat, when the load is a resistor.
I read the book and missed the anomaly.
The total current and thus total power fed into the amplifier comes out as a total of the internally dissipated heat and the load dissipated heat.
The load heat is based on rms currents.
The Total input - the load output = the internal dissipated heat.
{rms current input * rms supply voltage} - {rms load current * rms load voltage} = internal heat, when the load is a resistor.
I read the book and missed the anomaly.
A meter read averages. That applies to the analogue galvanometer type and the digital sampling type.
The meter measures the average and converts that to an equivalent for a sinewave without any distortion, i.e. it multiplies the AC average by ~1.12 and displays that bigger number on the scale/screen.
When measuring DC, they omit the x1.12 multiplication factor.
The meter measures the average and converts that to an equivalent for a sinewave without any distortion, i.e. it multiplies the AC average by ~1.12 and displays that bigger number on the scale/screen.
When measuring DC, they omit the x1.12 multiplication factor.
AndrewT - "I believe Cordell should have used rms current at 50% duty cycle".
That would make complete sense to me too. It's confusing because I just couldn't see why BC would have shifted from the obvious "rms" factor to the "average" in a tightly focused discussion of power unless he had a very good reason for it.
That would make complete sense to me too. It's confusing because I just couldn't see why BC would have shifted from the obvious "rms" factor to the "average" in a tightly focused discussion of power unless he had a very good reason for it.
Andrew - but I also wondered (in an earlier post) if it actually makes any sense to talk about the rms value of a half of the ac waveform (positive or negative parts of cycles only). Surely rms is by definition a factor of the whole-cycle waveform...? But my AC theory may be shaky here.
Regards, Mike
Regards, Mike
He is taking the RMS of the whole cycle. It just so happens that at the point in the circuit under discussion, the current is zero for half of the cycle.
rms is Root Mean Squared
It takes the value of I² at every instant and totals them to give a value for each half cycle. Then takes the square root to give an effective I value.
The squaring and square rooting also performs an arithmetic manipulation that measures the -ve (below the zero voltage value) without having to invert all the the -ve currents (currents flowing in the opposite direction).
It's the summing of the squares of I (rms) that leads to a different value from the summing of the I values of the current used in average current.
It takes the value of I² at every instant and totals them to give a value for each half cycle. Then takes the square root to give an effective I value.
The squaring and square rooting also performs an arithmetic manipulation that measures the -ve (below the zero voltage value) without having to invert all the the -ve currents (currents flowing in the opposite direction).
It's the summing of the squares of I (rms) that leads to a different value from the summing of the I values of the current used in average current.
In the general case, you integrate the volt-ampere product over a complete cycle to get "average power". In the case of a DC supply (which does not change) and a sinusiodally varying current, the volt-ampere product is proportional to the average current. There is no "V-squared" in the equation. In reality, voltage *droop* and ripple on an unregulated supply are related to the instantaneous current draw, and to get a more accurate value, you integrate the true volt-ampere product.
The voltage across the fuse itself is not constant - V is proportional to I at all points in the cycle because it is a resistor. The average power which translates to heating and when it blows is proportional to RMS current because of the "V-squared" relationship.
The average power in the *output transistor* is calculated the same way - integrate the volt-ampere product over the full cycle. In this case, Vce = rail voltage minus instantaneous output voltage. For calculating a heat sink, this is sufficient. Whether to stay inside the DC SOA at all points in the cycle is a judgement call. Commercial amplifier makers don't.
Excitation does not have to be limited to a single cycle of a sine wave. To approximate music, you can discretize a pink noise signal over some finite length of time, and do the the integration - in a circuit simulator, computer program, or even Excel spreadsheet.
The voltage across the fuse itself is not constant - V is proportional to I at all points in the cycle because it is a resistor. The average power which translates to heating and when it blows is proportional to RMS current because of the "V-squared" relationship.
The average power in the *output transistor* is calculated the same way - integrate the volt-ampere product over the full cycle. In this case, Vce = rail voltage minus instantaneous output voltage. For calculating a heat sink, this is sufficient. Whether to stay inside the DC SOA at all points in the cycle is a judgement call. Commercial amplifier makers don't.
Excitation does not have to be limited to a single cycle of a sine wave. To approximate music, you can discretize a pink noise signal over some finite length of time, and do the the integration - in a circuit simulator, computer program, or even Excel spreadsheet.
Average current and RMS current are not the same thing. Consider the following:
1 Amp DC is flowing through a 1 Ohm resistor. The power in the resistor is 1 Watt.
Now pulse the current to 2 Amps for half the time and shut the current off for half the time. The Average current is still 1 Amp, but what is the RMS current?
When 2 Amps flows through a 1 Ohm resistor, the power during that time is 4 Watts. Now average in the time when the current is zero and the power is 2 Watts. What amount of DC current would cause 2 Watts to be dissipated by a 1 Ohm resistor? The Square root of 2 or 1.414 Amps.
Pulse the current at 4 Amps for 25% of the time. The Average current is still 1 Amp but the power during the pulse is 16 Watts. The Average power is 4 Watts and the RMS current is 2 Amps.
Pulse the current at 10 Amps for 10% of the time. The Average current is still 1 Amp but the power during the pulse is 100 Watts. The Average power is 10 Watts and the RMS current is 3.162 Amps.
You can see from the above that you can not use the Average current to compute power in the load, you must use the RMS current.
1 Amp DC is flowing through a 1 Ohm resistor. The power in the resistor is 1 Watt.
Now pulse the current to 2 Amps for half the time and shut the current off for half the time. The Average current is still 1 Amp, but what is the RMS current?
When 2 Amps flows through a 1 Ohm resistor, the power during that time is 4 Watts. Now average in the time when the current is zero and the power is 2 Watts. What amount of DC current would cause 2 Watts to be dissipated by a 1 Ohm resistor? The Square root of 2 or 1.414 Amps.
Pulse the current at 4 Amps for 25% of the time. The Average current is still 1 Amp but the power during the pulse is 16 Watts. The Average power is 4 Watts and the RMS current is 2 Amps.
Pulse the current at 10 Amps for 10% of the time. The Average current is still 1 Amp but the power during the pulse is 100 Watts. The Average power is 10 Watts and the RMS current is 3.162 Amps.
You can see from the above that you can not use the Average current to compute power in the load, you must use the RMS current.
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