You guys got me thinking about phase and polarity. I looked up this site, and it offers the best and clearest explanation I could find. Remember; I'm looking for clear and simple solutions.
Phase vs Polarity explained | JustMastering.com
jj
Phase vs Polarity explained | JustMastering.com
jj
@ GUNFU and benb
You are both correct, so please bin my offending statements!
Phase or polarity. It depends on whether you are technical or not, it seems to me. I know only one is correct, but it is rather like bulb or lamp (in the UK). To an electrician the round glowy thing is a lamp. To the consumer it is a bulb, and a lamp is the thing it is fitted to.
The hifi consumer is more likely to understand phase than polarity in this context.
Thanks, jimmyjoe. The mathematical explanation is particularly helpful and I've simplified it slightly for future reference as follows:
I read the article jimmyjoe mentioned, and I don't understand this last part, the mathematical analysis.
"So from a mathematical point of view, this is the difference between changing the phase and polarity of a complex waveform (A) relative to another (B) within a mix:
When you reverse the polarity of complex waveform A, you change the phase of all its frequency elements equally, by 180°. This changes the phases of the individual frequencies in waveform A relative to B in the mix. For example, if the relative phases in A and B of the 200Hz and 400Hz elements were 170° and 10° respectively, the new relative phase of the 200Hz element will be 180° + 170° = 350°, and the new relative phase of the 400Hz element will be 180°+10° = 190°. As a result, the 200Hz element in the mix will get much louder, and the 400Hz element will get much softer.
When you delay complex waveform A relative to B, you change the phase of each of its frequency elements by a different amount. This again changes the relative phases of the individual frequencies in the mix, but in a different way. If we take the example above but delay waveform A by enough to shift the 200Hz element by 180° instead of changing polarity, the new relative phase of the 200Hz element will again be 180° + 170° = 350°, but the new relative phase of the 400Hz element will be 360° + 10° = 370°. As a result, the 200Hz element in the mix will again get louder, but the 400Hz element will keep the same volume."
Not having much in the way of maths, I would have thought that every frequency would have nulled or doubled. Can someone explain it in words of less than one syllable?
"So from a mathematical point of view, this is the difference between changing the phase and polarity of a complex waveform (A) relative to another (B) within a mix:
When you reverse the polarity of complex waveform A, you change the phase of all its frequency elements equally, by 180°. This changes the phases of the individual frequencies in waveform A relative to B in the mix. For example, if the relative phases in A and B of the 200Hz and 400Hz elements were 170° and 10° respectively, the new relative phase of the 200Hz element will be 180° + 170° = 350°, and the new relative phase of the 400Hz element will be 180°+10° = 190°. As a result, the 200Hz element in the mix will get much louder, and the 400Hz element will get much softer.
When you delay complex waveform A relative to B, you change the phase of each of its frequency elements by a different amount. This again changes the relative phases of the individual frequencies in the mix, but in a different way. If we take the example above but delay waveform A by enough to shift the 200Hz element by 180° instead of changing polarity, the new relative phase of the 200Hz element will again be 180° + 170° = 350°, but the new relative phase of the 400Hz element will be 360° + 10° = 370°. As a result, the 200Hz element in the mix will again get louder, but the 400Hz element will keep the same volume."
Not having much in the way of maths, I would have thought that every frequency would have nulled or doubled. Can someone explain it in words of less than one syllable?
I'll give it a try! Consider when complex waveform A and complex waveform B each contain the same sine wave component.
A zero degree phase relationship between the two sine wave components would make them perfectly in phase, giving the highest combined signal level.
A 180 degree phase relationship puts them perfectly out of phase, resulting in total phase cancellation - and therefore silence as the combined output.
All the other possible phase relationships put the two sine wave components partially out of phase with each other, resulting in partial phase cancellation.
It is this phase cancellation between the constituent sine wave components to which the mathematics in jimmyjoe's article is referring - sometimes louder, sometimes softer and sometimes the same.
The attached diagrams are courtesy of Sound On Sound (SOS) magazine. Read the full article here: Phase Demystified
A zero degree phase relationship between the two sine wave components would make them perfectly in phase, giving the highest combined signal level.
A 180 degree phase relationship puts them perfectly out of phase, resulting in total phase cancellation - and therefore silence as the combined output.
All the other possible phase relationships put the two sine wave components partially out of phase with each other, resulting in partial phase cancellation.
It is this phase cancellation between the constituent sine wave components to which the mathematics in jimmyjoe's article is referring - sometimes louder, sometimes softer and sometimes the same.
The attached diagrams are courtesy of Sound On Sound (SOS) magazine. Read the full article here: Phase Demystified
Attachments
But the article I quoted refers to a 180 deg polarity switch changing the relative levels of different frequencies. That's what I don't understand. If that is so, how does the null test work?
Ah wait! I missed that it was adding to another wave of the same frequency but with a different phase angle. That makes more sense, as he is talking about mixing.
As you were.
Ah wait! I missed that it was adding to another wave of the same frequency but with a different phase angle. That makes more sense, as he is talking about mixing.
As you were.
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I don't know whether I should say this or not, but I'll take a chance:
I have no problem with words.
I have no problem with charts.
I have no problem with animations.
I have severe problems with mathematics. Don't know why; it's just a total mental block.
Due to the kindness and patience of the people in this thread, I've learned more about things I didn't understand in the last few weeks than I learned in my whole life. I can't thank you guys enough. jj
I have no problem with words.
I have no problem with charts.
I have no problem with animations.
I have severe problems with mathematics. Don't know why; it's just a total mental block.
Due to the kindness and patience of the people in this thread, I've learned more about things I didn't understand in the last few weeks than I learned in my whole life. I can't thank you guys enough. jj
I'm not an expert in mixing so you prompted me to check out what you mean by the null test - courtesy of the SOS article.
I'd just like to return to this as Jim did refer to medium priced solid state amplifiers.
I have, in my collection, a Rogers Ravensbrook Series III which was an affordable amplifier of the early 70s and is still sought after today. It was rated by the manufacturer at 20W into 8ohm.
Last night, I unearthed the Hi-Fi News magazine test results on a vintage Ravensbrook III that were carried out in 2016. Interestingly, the power output was measured as follows:
No figures for 16ohm are given, but I think it is safe to say that the maximum power is developed into 8ohm with this particular amplifier.
I wonder how a modern budget solid state amplifier (not one of those audiophile behemoths!) would fair under similar test conditions?
I thought I was finished, but I have another question.
There are many amplifiers that operate in what is called "Class A". It's been said that some are and most aren't ..... true class A, that is.
Here's what I've been told:
Class A devices pull constant power from the outlet. With no signal, a class A amp is simply producing heat. Signal sent to the speaker is simply less of that constant pull that is converted to heat. Whether the amp is driving a 4 ohm, 8 ohm or 16 ohm speaker makes no difference; the power drawn from the wall outlet is the same.
If that is true, wouldn't a simple amp meter around the power cord as several different signals are introduced to the amp tell the whole story?
Again .... thank you for your help. jj
There are many amplifiers that operate in what is called "Class A". It's been said that some are and most aren't ..... true class A, that is.
Here's what I've been told:
Class A devices pull constant power from the outlet. With no signal, a class A amp is simply producing heat. Signal sent to the speaker is simply less of that constant pull that is converted to heat. Whether the amp is driving a 4 ohm, 8 ohm or 16 ohm speaker makes no difference; the power drawn from the wall outlet is the same.
If that is true, wouldn't a simple amp meter around the power cord as several different signals are introduced to the amp tell the whole story?
Again .... thank you for your help. jj
There are very experienced amplifier designers on this forum, but here is my humble contribution which an expert may care to correct or expand upon.
In Class A, the output device (transistor or valve) is constantly carrying current, causing a continuous loss of power in the amplifier and generating heat.
The large amount of heat generated results in a class A amp having very low efficiency in converting DC supply power to RMS power in the loudspeaker - let's say around 30%
To produce 3W of RMS output power in the loudspeaker would therefore require 10W to be supplied by the DC power supply - the missing 7W generates heat in the amplifier.
To produce 6W in the loudspeaker would require 20W to be supplied by the power supply - the missing 14W generates heat.
So, the input power requirement increases as the output power increases and a wattmeter on the input cord would show that.
However, such disproportionate loss of power as heat requires a really beefy and expensive power supply, That's why Class A amplification is limited to low power applications.
In Class A, the output device (transistor or valve) is constantly carrying current, causing a continuous loss of power in the amplifier and generating heat.
The large amount of heat generated results in a class A amp having very low efficiency in converting DC supply power to RMS power in the loudspeaker - let's say around 30%
To produce 3W of RMS output power in the loudspeaker would therefore require 10W to be supplied by the DC power supply - the missing 7W generates heat in the amplifier.
To produce 6W in the loudspeaker would require 20W to be supplied by the power supply - the missing 14W generates heat.
So, the input power requirement increases as the output power increases and a wattmeter on the input cord would show that.
However, such disproportionate loss of power as heat requires a really beefy and expensive power supply, That's why Class A amplification is limited to low power applications.
I must say, people here are teaching me a good lesson regarding my poor ability to communicate clearly!
Let's try again.
A man told me that most amplifiers that are advertised as class "A" are actually not. In class "A", the amp has a constant pull from the wall (as measured by an amp meter) regardless of signal level or speaker impedance. This means that for all levels of output lower than clipping, the ampere draw of the amplifier will not change. If it does, the amplifier is NOT class "A".
True or not?
Thank you for your help. jj
Let's try again.
A man told me that most amplifiers that are advertised as class "A" are actually not. In class "A", the amp has a constant pull from the wall (as measured by an amp meter) regardless of signal level or speaker impedance. This means that for all levels of output lower than clipping, the ampere draw of the amplifier will not change. If it does, the amplifier is NOT class "A".
True or not?
Thank you for your help. jj
We both need an expert to chime in, Jim! 
However, this is what I now read!
"In a true class A amplifier, the power that is dissipated as heat in the output device at idle is passed on to the load at full power. The average current is the same as the idle value, so there is no change in the current from the supply."
Technical Q&A
However, this is what I now read!
"In a true class A amplifier, the power that is dissipated as heat in the output device at idle is passed on to the load at full power. The average current is the same as the idle value, so there is no change in the current from the supply."
Technical Q&A
We both need an expert to chime in, Jim!
However, this is what I now read!
"In a true class A amplifier, the power that is dissipated as heat in the output device at idle is passed on to the load at full power. The average current is the same as the idle value, so there is no change in the current from the supply."
Technical Q&A
Hmmmmmm. Seems to support what I heard. Thank you! jj
I have listened to a 120 watt class A amp, and its 100 kg or so. So yes PSU and heatsink gets massive. The amp was Dali Gravity.
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