When idle (not driven), 75W at mains' side seems okay.
Other readings are the max achievable output, probably only if that given channel is driven -> and nothing else of course.
The numbers, put together on a sheet of paper, under the specifications section, look impressive indeed, don't they... 🙂
In reality, this practice is very misleading/borderline criminal.
Do not buy anything from these manufacturers who indulge in such practices.
For the little heed paid to to it, I think the "correctness" amendments to advertising were all done and dusted in the 1990s and for some years, advertising claims did include industry standard definitions of what "music power" meant - albeit sanitized to make it a little easier to understand to the buying public. 'Trouble was, with cheap and popular gear, the public still only got to see the big power and lowest price numbers up front. The fine print had to be looked for but now, it's nowhere to be seen. Google "music power' and mostly you only get pop song titles.
In the 1960's and into the 1970's, wattage ratings for serious equipment were mainly accurate.
For instance, McIntosh amps gave a clear explanation of their wattages - posted in their service manuals in charts stating loaded impedences and voltage levels - the honest ratings.
And these products DID put out the watts - conservatively rated, and beyond.
Come the 1990's and today - I've seen more and more bizzare ratings/claims show up.
A blatent 1000 watts from a 2 1/2 inch high box? - I tested that thing in the shop and it fell way short of even 100 watts!
Yet predominently written on its front was that "1000 watts" crap.
Somehow, corrupt marketing and obvious "standards" by the Institute of High Fidelity are ignored.
The below chart can be used today with any amp, any wattage.
These are "honest watts at various voltages/impedences.
Simple to use - if checking a "supposed" 100w amp @ 8 ohms, merely divide the voltage readings accordingly, etc.
For instance, McIntosh amps gave a clear explanation of their wattages - posted in their service manuals in charts stating loaded impedences and voltage levels - the honest ratings.
And these products DID put out the watts - conservatively rated, and beyond.
Come the 1990's and today - I've seen more and more bizzare ratings/claims show up.
A blatent 1000 watts from a 2 1/2 inch high box? - I tested that thing in the shop and it fell way short of even 100 watts!
Yet predominently written on its front was that "1000 watts" crap.
Somehow, corrupt marketing and obvious "standards" by the Institute of High Fidelity are ignored.
The below chart can be used today with any amp, any wattage.
These are "honest watts at various voltages/impedences.
Simple to use - if checking a "supposed" 100w amp @ 8 ohms, merely divide the voltage readings accordingly, etc.
Attachments
Last edited:
For some time during the 90´s Peavey printed, say, "20 V RMS" on their user manuals, besides Watts rating.
I'll post a explanation of this "wattage" issue here for all to chew on.......
This is copied from an article that I found interesting, and may clear up some confusion.
"So... where does this bring us? As you have probably seen in amplifier spec sheets, manufacturers rate the output of their amplifiers in watts RMS. For example, let's consider the vintage amplifier Kenwood Model KA-9100. This amplifier is rated to put out 90 watts RMS per channel into an 8 ohm load. Technically speaking, the term "RMS" is not defined when referring to power (watts)! RMS is a valid term when referred to voltage or current, but not power! Watts are watts, period! Despite the term "watts RMS" being an incorrect term, it stuck with the community and has become the accepted way to rate an amplifier's output. The reason that audio amplifiers are rated in "watts RMS" is because they are rated to deliver that amount of power using a sine wave signal. Because amplifiers are rated this way, their peak power output will be twice the RMS rating. So, for the Kenwood KA-9100 (which is rated to deliver 90 watts RMS into 8 ohms), the peak power is 180 watts. Most amplifiers cannot sustain output at their peak capability for too long (and the characteristic of most music signals is such that this is not necessary anyway). So, despite "watts RMS" being a technically invalid term, it is used with audio amplifiers because of the sine wave signals that are used to determine their power output specifications.
Question: What are (at minimum) the rail voltages necessary for an amplifier to deliver 90 watts into an 8 ohm load? It is not too hard to figure this out. We simply look at the equations for power:
P=VI = I2R=V2/R
In this case, there are two known items: power (90 watts) and resistance (of the speaker, 8 ohms). Plugging these numbers into the equation (P=V2/R) yields a value for V of 26.83 volts. Is this the answer to the rail voltage question? No... REMEMBER, if we applied 26.83 volts of DC (note: DC) across an 8 ohm speaker the power to the speaker would in fact be 90 watts. However, amplifiers are rated using sine wave input signals, and (as described above) we need to apply more voltage to a load (for a sine wave) in order to get the same amount of power that would be delivered by a DC voltage. For sine waves, the multiplication factor is 1.414. So, if we take the voltage of 26.83 and multiply it by 1.414 (the square root of 2), we get a value of 37.94 volts. This value is the absolute minimum rail voltage needed for an amplifier to deliver 90 watts (with a sine wave signal) to a load! Had we mistakenly determined that 26.83 was the correct rail voltage, the amplifier would begin to clip as the output tried to exceed 45 watts of output. Note that 45 watts is exactly one half of the 90 watt value. This shows that by increasing the rail voltage by a factor of 1.414 results in the amplifier having twice as much output capability! This can also be deduced by looking at the equations for power.
To summarize, amplifiers use the technically incorrect term "watts RMS" in their output ratings because amplifiers (those used for audio anyway) are rated using sine wave signals. Music is not a simple sine wave, and it is not DC... however a sine wave is more representative of a music signal than DC is. So, the standard practice in use today is for manufacturers to rate amplifiers in "watts RMS". Be aware that watts are watts, there is technically no such thing as watts RMS."
This is copied from an article that I found interesting, and may clear up some confusion.
"So... where does this bring us? As you have probably seen in amplifier spec sheets, manufacturers rate the output of their amplifiers in watts RMS. For example, let's consider the vintage amplifier Kenwood Model KA-9100. This amplifier is rated to put out 90 watts RMS per channel into an 8 ohm load. Technically speaking, the term "RMS" is not defined when referring to power (watts)! RMS is a valid term when referred to voltage or current, but not power! Watts are watts, period! Despite the term "watts RMS" being an incorrect term, it stuck with the community and has become the accepted way to rate an amplifier's output. The reason that audio amplifiers are rated in "watts RMS" is because they are rated to deliver that amount of power using a sine wave signal. Because amplifiers are rated this way, their peak power output will be twice the RMS rating. So, for the Kenwood KA-9100 (which is rated to deliver 90 watts RMS into 8 ohms), the peak power is 180 watts. Most amplifiers cannot sustain output at their peak capability for too long (and the characteristic of most music signals is such that this is not necessary anyway). So, despite "watts RMS" being a technically invalid term, it is used with audio amplifiers because of the sine wave signals that are used to determine their power output specifications.
Question: What are (at minimum) the rail voltages necessary for an amplifier to deliver 90 watts into an 8 ohm load? It is not too hard to figure this out. We simply look at the equations for power:
P=VI = I2R=V2/R
In this case, there are two known items: power (90 watts) and resistance (of the speaker, 8 ohms). Plugging these numbers into the equation (P=V2/R) yields a value for V of 26.83 volts. Is this the answer to the rail voltage question? No... REMEMBER, if we applied 26.83 volts of DC (note: DC) across an 8 ohm speaker the power to the speaker would in fact be 90 watts. However, amplifiers are rated using sine wave input signals, and (as described above) we need to apply more voltage to a load (for a sine wave) in order to get the same amount of power that would be delivered by a DC voltage. For sine waves, the multiplication factor is 1.414. So, if we take the voltage of 26.83 and multiply it by 1.414 (the square root of 2), we get a value of 37.94 volts. This value is the absolute minimum rail voltage needed for an amplifier to deliver 90 watts (with a sine wave signal) to a load! Had we mistakenly determined that 26.83 was the correct rail voltage, the amplifier would begin to clip as the output tried to exceed 45 watts of output. Note that 45 watts is exactly one half of the 90 watt value. This shows that by increasing the rail voltage by a factor of 1.414 results in the amplifier having twice as much output capability! This can also be deduced by looking at the equations for power.
To summarize, amplifiers use the technically incorrect term "watts RMS" in their output ratings because amplifiers (those used for audio anyway) are rated using sine wave signals. Music is not a simple sine wave, and it is not DC... however a sine wave is more representative of a music signal than DC is. So, the standard practice in use today is for manufacturers to rate amplifiers in "watts RMS". Be aware that watts are watts, there is technically no such thing as watts RMS."
Doesn’t clear up all the confusion, even when it is an “RMS” rating. There seems to be three schools of thought. One is to measure the output power based on the full-load voltage, ie, what the rail voltages drop to under a steady state sine wave, including the voltage drops in the transistors and emitter resistors. This is like the (now obsolete) “FTC” rating. The most common approach these days is to rate “RMS” power based on the peak voltage the amp can deliver under a light average load. Basically, the nominal rail voltage minus transistor drop, divided by the square root of two, for the “RMS” voltage. That is more like the old “iHF” rating, and is a true peak power. The “EIA” rating is the same, but allows 1% distortion. Which is clipping. The third is more marketing driven, taking the maximum possible rail voltage, and calculating the maximum possible “RMS” power given the formulas in the above post.
From there it further degrades, calculating the power that it would be into 2 ohms if nothing current limited or had any losses, and perhaps a factor of two or four thrown in for good measure.
From there it further degrades, calculating the power that it would be into 2 ohms if nothing current limited or had any losses, and perhaps a factor of two or four thrown in for good measure.
Sorry but writer sounds somewhat confused himself.
90W "RMS" and 180W Peak are exactly the same.
When an amplifier is delivering 90W RMS at the same time it´s delivering 180W peak (always talking sinewaves, of course)
In your example it will be delivering 26.83V RMS *and* 37.95V peak at the same time, different ways to call the same voltage.
Actual power (water boiling power) is the same whether you apply 26.83V RMS or 37.95V peak to an 8 ohm resistive load.
And it will boil exact same water in a minute as 26.83 V DC
And that under load; if unregulated raw voltage will need to be , say, 15% or 20% higher, to compensate for voltage drop under load.
True, there are not actual "Watts RMS" but the term is in common use and as I see it (others may differ) it has come to mean: "Watts calculated *using* measured RMS Volts into a resistive load, just before clipping (or at a given distortion level) , continuously (or at least during an appreciable time, say 1 hour, minimum 15 minutes)"
This a repeatable and easy to reproduce method, and in principle will include power supply drop and losses due to overheating, specially resistance increase in transformer copper wire but also any other in the path.
Short term/transient power may also be measured, like anything else, but it will need a precisely defined method, say using a burst of sinewaves, so everybody reaches same measurements, or it will be a source of confusion, subject to manipulation.
As is, I don´t think there is a universally accepted method, but if available, please post it here.
Yes.
Yes.
?????????????????????
90W "RMS" and 180W Peak are exactly the same.
When an amplifier is delivering 90W RMS at the same time it´s delivering 180W peak (always talking sinewaves, of course)
In your example it will be delivering 26.83V RMS *and* 37.95V peak at the same time, different ways to call the same voltage.
Actual power (water boiling power) is the same whether you apply 26.83V RMS or 37.95V peak to an 8 ohm resistive load.
And it will boil exact same water in a minute as 26.83 V DC
Basically 37.95V DC + power transistors voltage drop (usually around 4V) + resistive losses (I peak into emitter/ballast resistors) + peak ripple voltage
And that under load; if unregulated raw voltage will need to be , say, 15% or 20% higher, to compensate for voltage drop under load.
True, there are not actual "Watts RMS" but the term is in common use and as I see it (others may differ) it has come to mean: "Watts calculated *using* measured RMS Volts into a resistive load, just before clipping (or at a given distortion level) , continuously (or at least during an appreciable time, say 1 hour, minimum 15 minutes)"
This a repeatable and easy to reproduce method, and in principle will include power supply drop and losses due to overheating, specially resistance increase in transformer copper wire but also any other in the path.
Short term/transient power may also be measured, like anything else, but it will need a precisely defined method, say using a burst of sinewaves, so everybody reaches same measurements, or it will be a source of confusion, subject to manipulation.
As is, I don´t think there is a universally accepted method, but if available, please post it here.
How LOUD an amplifier can play, without any clipping, or when slight clipping can be tolerated, is directly correlated to the old “IHF” rating. How loud it WILL play, regardless of the distortion that can be tolerated, is directly correlated to the old “FTC” rating.
The FTC ratings did change in recent years (2000- ish?) to go from requiring a 1/3 power preconditioning to a 1/8 power continuous without any thermal limit, before measuring and reporting RMS power fir the continuous sine wave test signal. Now I think they’ve dropped it altogether, because not even modern “Tour Grade” amplifiers can even do that anymore. Switch mode supplies at those power levels cannot deliver enough oomph fir a continuous sine wave - even for a 30 second test. They resort to a “burst mode” test, which is a bit more rigorous than the old “iHF” rating system, but not by much.
The FTC ratings did change in recent years (2000- ish?) to go from requiring a 1/3 power preconditioning to a 1/8 power continuous without any thermal limit, before measuring and reporting RMS power fir the continuous sine wave test signal. Now I think they’ve dropped it altogether, because not even modern “Tour Grade” amplifiers can even do that anymore. Switch mode supplies at those power levels cannot deliver enough oomph fir a continuous sine wave - even for a 30 second test. They resort to a “burst mode” test, which is a bit more rigorous than the old “iHF” rating system, but not by much.
20 volts RMS (28 volts peak) equals around 50 "watts" into an 8 ohm load.
Of course, the "variables" have to include such things as frequency, and the speaker's impedence, both of which can vary "the load" greatly.
So those published watts ratings of amps could be all over the place in real life.
With the majority of speakers designed for home use, and living rooms of typical homes, 5 to 10 watts per speaker is "room filling" and would certainly be difficult to have a conversation nearby.
These are sensible conditions.
Yeah yeah, I've heard plenty of "but you need at least 100 watts/per for those crescendos!"
Indeed I suppose, if your hearing is pretty much ruined.
Of course, the "variables" have to include such things as frequency, and the speaker's impedence, both of which can vary "the load" greatly.
So those published watts ratings of amps could be all over the place in real life.
With the majority of speakers designed for home use, and living rooms of typical homes, 5 to 10 watts per speaker is "room filling" and would certainly be difficult to have a conversation nearby.
These are sensible conditions.
Yeah yeah, I've heard plenty of "but you need at least 100 watts/per for those crescendos!"
Indeed I suppose, if your hearing is pretty much ruined.
Car decks in particular have been touting all kinds of numbers for years, and they still will all max out at 12w or so.
I had to explain that to someone recently, they seemed to grasp it quickly, having a strong background in thermodynamics. I probably would’ve just let it go, but with the expectation of reasonable clarity in a convertible auto, I felt I should bring it up.
My first introduction to fake power numbers had been the Kraco auto equalizers that were not much larger than a pack of cigarettes, using 20 AWG wires, and a “400 WATTS” silkscreened on the case. This is while the home system was my Dad’s 40w Scott 299.
I had to explain that to someone recently, they seemed to grasp it quickly, having a strong background in thermodynamics. I probably would’ve just let it go, but with the expectation of reasonable clarity in a convertible auto, I felt I should bring it up.
My first introduction to fake power numbers had been the Kraco auto equalizers that were not much larger than a pack of cigarettes, using 20 AWG wires, and a “400 WATTS” silkscreened on the case. This is while the home system was my Dad’s 40w Scott 299.
Speakers turned up to “room filling volume” use a handful of watts of average power. But it takes all 100 of your receiver’s “watts” to execute. You would see 40 volt peaks on your oscilloscope. Try playing at the same SPL on a 10 watt amplifier, and it would sound like a mistuned radio station. A very loud mistuned radio station. A 5 or 10 watt amplifier can produce a couple hundred milliwatts cleanly. The dishwasher or AC unit will pretty much drown it out unless you turn it up until there is at least some clipping.
My one stereo tube amp, an EL84 ultralinear amp (of my own design), puts out a clean 15 to 17 watts/ch before clipping, if cranked up.
This drives a pair of hungry floorstanding air-suspension speakers (Advent Maestros).
And if cranked, my neighbors will surely complain.
Usually, the level that I need is around 5 watts - with that extra 10w headroom standing by.
This drives a pair of hungry floorstanding air-suspension speakers (Advent Maestros).
And if cranked, my neighbors will surely complain.
Usually, the level that I need is around 5 watts - with that extra 10w headroom standing by.
Yes indeed, we want real watts - the ones written with a 'W'! 
Physics note: When a unit is named after a scientist it is written in lower case, but its abbreviation is written in upper case.
e.g. watt (W); joule (J); ampere (A); coulomb (C) etc. etc.
Physics note: When a unit is named after a scientist it is written in lower case, but its abbreviation is written in upper case.
e.g. watt (W); joule (J); ampere (A); coulomb (C) etc. etc.
Thanks Galu - I'd been wondering about that.
Those MacIntosh 2300's were sure something, eh wiseoldtech? I humped one of those around for a year and a half doing sound for a band playing one-nighters. None of the band members offered to help carry it more than once! That was fine by me, as I trusted myself more (not to drop it) anyway -- and was in considerably better shape back then.
Cheers
Those MacIntosh 2300's were sure something, eh wiseoldtech? I humped one of those around for a year and a half doing sound for a band playing one-nighters. None of the band members offered to help carry it more than once! That was fine by me, as I trusted myself more (not to drop it) anyway -- and was in considerably better shape back then.
Cheers
Indeed, those 2300's are heavy suckers.
Quite difficult to maneuver on the bench for servicing!
As I remember, they're over 100 lbs each, all the weight is in those potted transformers.
I had a pair of Acoustic Research LST speakers that needed serious repairing too - those monsters weighed 128lbs each.
One handy thing I have in the shop is a roll-around table with a motor-driven elevation level.
I see they now sell for around $1,100.
