Paradigm S4 signature measurement
I measured the voltages across the speaker terminals at the highest volume control I listen to.
120 Hz 3.0 v
220 H < 2.5V
Can I conclude that my amp needs to be 12 watts RMS. My present amp is b&K 7270 multi channel which is rated at 200 Watts.
This is an eye opener for me. Most people in the audio world brag about their amps and how powerful they are.
Also another thing to take into context is the room here.
Small rooms smaller voltages.
I was in plans for building Pass F5 with dual outputs. May be dual outputs are not needed now.
But anyway build the amps with heatsinking that later it can be used for dual outputs.
Also it is important to note about the first watt which is probably the most important. What is the point of having 200 watts of power when first watt is faulty.
I measured the voltages across the speaker terminals at the highest volume control I listen to.
120 Hz 3.0 v
220 H < 2.5V
Can I conclude that my amp needs to be 12 watts RMS. My present amp is b&K 7270 multi channel which is rated at 200 Watts.
This is an eye opener for me. Most people in the audio world brag about their amps and how powerful they are.
Also another thing to take into context is the room here.
Small rooms smaller voltages.
I was in plans for building Pass F5 with dual outputs. May be dual outputs are not needed now.
But anyway build the amps with heatsinking that later it can be used for dual outputs.
Also it is important to note about the first watt which is probably the most important. What is the point of having 200 watts of power when first watt is faulty.
paradigm S4 signature measurement
Following up with my previous post,
I connected the DMM's probes to the speaker terminals, and let the music run at my loudest levels.
I started observing the readings on the DMM. I saw the max voltage recorded at 32 Volts.
I would say less than 1 % percent of them exceeded 25 volt mark.
Less than 5 percent exceeded 20 volt mark.
Can we use watts = V*V/R here.
We dont have the value of R it could vary from min impedence to max impedence of the speaker. Let us say that 25 V is the maximum voltage at speaker terminals. 25* 25/8 = close 80 watts.
If we consider R as 4 ohms then 25* 25/4 = 160 watts.
There seems to be some thing wrong in my calculation.
Can we use the speaker terminal voltage when the music is running is my question.
Let me know.
The max watts used by the speaker can be between 80 to 160 depending on the impedence at that point of time.
I think I am missing something here.
Following up with my previous post,
I connected the DMM's probes to the speaker terminals, and let the music run at my loudest levels.
I started observing the readings on the DMM. I saw the max voltage recorded at 32 Volts.
I would say less than 1 % percent of them exceeded 25 volt mark.
Less than 5 percent exceeded 20 volt mark.
Can we use watts = V*V/R here.
We dont have the value of R it could vary from min impedence to max impedence of the speaker. Let us say that 25 V is the maximum voltage at speaker terminals. 25* 25/8 = close 80 watts.
If we consider R as 4 ohms then 25* 25/4 = 160 watts.
There seems to be some thing wrong in my calculation.
Can we use the speaker terminal voltage when the music is running is my question.
Let me know.
The max watts used by the speaker can be between 80 to 160 depending on the impedence at that point of time.
I think I am missing something here.
Thanks for reporting your results. You've shown why trying to read the musical signal with a multimeter is not a great idea. Reading the musical signal with an oscilloscope would be better. Most voltmeters are not very good at reading musical signals, which is why I provided the sine wave tones.
If you have done the test properly:
You might try measuring the pink noise signal I provided here: https://production.diyaudio.com/community/index.php?posts/2892860/
It should measure about half the value of the sine wave (it's 6dB lower - RMS). Please let us know what you find.
If you have done the test properly:
- set loudest level
- measure while playing -12dB test tone(s)
You might try measuring the pink noise signal I provided here: https://production.diyaudio.com/community/index.php?posts/2892860/
It should measure about half the value of the sine wave (it's 6dB lower - RMS). Please let us know what you find.
pink noise test
Pano,
surprisingly I got the 85db average value at the same level what I got 3.0 volts yesterday which is the max level I listen to.
I went otherway around, i increased the volume control by 6 db then measured the 120 and 220hz test tones you provided eariler.
The readings were around 5.0 volts.
May be my listening levels are not high because my room is small at 17ft X 12 ft X 10 ft.
Previous test 3.0v my normal listening levels
current test 5.0v 6 db above the normal listening levels.
By this way can we conclude that that wattage (rms) of my speakers could be 12 to 20 watts.
Thanks
Pano,
surprisingly I got the 85db average value at the same level what I got 3.0 volts yesterday which is the max level I listen to.
I went otherway around, i increased the volume control by 6 db then measured the 120 and 220hz test tones you provided eariler.
The readings were around 5.0 volts.
May be my listening levels are not high because my room is small at 17ft X 12 ft X 10 ft.
Previous test 3.0v my normal listening levels
current test 5.0v 6 db above the normal listening levels.
By this way can we conclude that that wattage (rms) of my speakers could be 12 to 20 watts.
Thanks
You're welcome.
From you measurements, I would say that you need an amplifier that can supply about 12V peak or 25V peak if you want that extra 6dB of headroom. A sine wave with its peaks at 12 or 25 volts would have an RMS value of 8.5 or 17.7 volts respectively. (not exactly, but close)
From that you can figure out power requirements.
From you measurements, I would say that you need an amplifier that can supply about 12V peak or 25V peak if you want that extra 6dB of headroom. A sine wave with its peaks at 12 or 25 volts would have an RMS value of 8.5 or 17.7 volts respectively. (not exactly, but close)
From that you can figure out power requirements.
Yes, extra headroom is good, no argument from me. However, let's look at this more closely.
As Tom Danley said a few pages back, you can clip a few peaks and not notice, unless you do a direct comparison with an unclipped signal.
That's why I've asked people doing the test to turn it up as loud as they ever will, then measure. The majority of listening will be done at a lower level, therefore gaining headroom. If you determine that you need a 13.6 watt amp not to clip, should you build yourself a 13.6W amp, or go with something bigger? Most guys here have enough sense to build in some margin, or headroom.
On the flip side, you may have your eye on a 10W amp and you will now know that you will be clipping a bit a loud settings. You might choose to live with that limitation because of other things you like about the amp. But at least now you know.
That's the point of the test.
- If you have an amp that is "clean" up to its rated power (voltage)
- You determine the max peak voltage you'll ever use
- Your peaks fall within the clean voltage range of the amp
- Your average loud level will be at least 12dB lower
- Only the peaks will be at the amp's limit
- You rarely actually play this loud
- How much more headroom do you need?
As Tom Danley said a few pages back, you can clip a few peaks and not notice, unless you do a direct comparison with an unclipped signal.
That's why I've asked people doing the test to turn it up as loud as they ever will, then measure. The majority of listening will be done at a lower level, therefore gaining headroom. If you determine that you need a 13.6 watt amp not to clip, should you build yourself a 13.6W amp, or go with something bigger? Most guys here have enough sense to build in some margin, or headroom.
On the flip side, you may have your eye on a 10W amp and you will now know that you will be clipping a bit a loud settings. You might choose to live with that limitation because of other things you like about the amp. But at least now you know.
That's the point of the test.
IIRC, the values you posted there are peak, not RMS. Perhaps that should be made clear.
Just for reference
8Ohms : (Vx4)
2v= 0.5w : 8w
5v= 3.125w : 50w
10v=12.5w : 200w
20v=50w : 800w
6Ohms : (Vx4)
2v=0.67w : 10.67w
5v=4.167w : 66.67w
10v=16.67w : 266.67
20v=66.67w : 1066.67w
4Ohms : (Vx4)
2v =1w : 16w
5v=6.25w : 100
10v=25w : 400w
20v=100w : 1600w
So worse case (not counting the under 4Ohm crowd)
36% need less than 16w peak
36% need less than 100w peak
17% need less than 400w peak
4% need less than 1600w peak
and 7% need new gear
Pano said:
?
? By using something other than the word "peak"? If you can think of a way to clarify my previous post please do.Sincerely,
revb
Last edited:
Hi Pano, & Rev,
Last question first.
No, I have not done the test yet.
In order to not prejudge what may come from the "test", I still await a consensus on what power of amplifier I should use for the test setup as described in my earlier post.
Now to the rubbish in Rev's post551.
2Vac as measured at the 8r0 load/speaker is by Pano's definition directly equivalent to a guaranteed non clipping output from the digital source when using a 4W amplifier.
When the 2Vac is measured when the load/speaker is 4r0, then Pano tells us to double the effective amplifier power to 8W. This again will give a non clipped signal from the digital source.
Similarly for the other measured output voltages all the "powers" stated are in error by +100%.
As a result the summary is also in error.
Last question first.
No, I have not done the test yet.
In order to not prejudge what may come from the "test", I still await a consensus on what power of amplifier I should use for the test setup as described in my earlier post.
Now to the rubbish in Rev's post551.
2Vac as measured at the 8r0 load/speaker is by Pano's definition directly equivalent to a guaranteed non clipping output from the digital source when using a 4W amplifier.
When the 2Vac is measured when the load/speaker is 4r0, then Pano tells us to double the effective amplifier power to 8W. This again will give a non clipped signal from the digital source.
Similarly for the other measured output voltages all the "powers" stated are in error by +100%.
As a result the summary is also in error.
Correct. Because the test signal is -12dBFS, the peak value of a full volume sinewave would be 12dB higher, or 4X the voltage.
4X2=8V. 8 volts into 8 ohms is 8 watts. (8,8,8 rule) That's peak, of course. Generally an amp rated at 4W RMS into 8 ohms can supply 8 volts peak without clipping.
Correct- but incomplete: If you have 4 ohm speakers, you should look at the 4 ohm rating of the amp. So the 8W RMS rating would be at 4 ohms.
How do you figure that? Did you read them as RMS values as I first did? He's talking about peak power. That is stated at the bottom.
No, it's not. Let's look at it again.
A slight overestimate for the 6 ohm crowd, but completely truthful. Did you miss the 6 Ohm reference?revboden said:
Needing 10.66 watts peaks (2V or less measured, 6 ohms) certainly fits within "less than 16W peak" Don't you think? 10.66 is normally considered less than 16. The others are the same.
Why? It's simple and fast. Don't you own a volt meter with a 20V (or less) AC scale?AndrewT said:
I can make the test even easier for you:
Once you have set your volume, the test tone I provided will be 9dB lower than the loudest possible sine wave from your digital source (non-clipped).
Just multiply your measured result by 2.83 to get the highest possible sine voltage. You can use Ohm's law to figure out how much power that is.
Easy, isn't it?
Andrew, I have deleted one of your posts. Please reread the Rules before commenting again.
Andrew. I hate to repeat myself but feel that I must.
Do you have anything new?
'Nuff said.Pano said:
An Easier Way
For those of you worried or confused about average levels, peak levels and decibels; here is an easy method for you.
You can use Ohm's law to figure out how much power that is, if you want to.
W = V2 / R
or
Voltage X Voltage divided by Impedance = Power.
Ohm's Law Calculations With Power
Please post your measurement results!
As a bonus, you might do this at several volume settings if you are curious about the difference in those volume settings. E.G., measure at your loudest setting, then measure at your normal settings. You may see a large difference.
For those of you worried or confused about average levels, peak levels and decibels; here is an easy method for you.
- Set your volume.
- Measure the sine wave tone (from post #2)
- Multiply your result by 2.83
You can use Ohm's law to figure out how much power that is, if you want to.
W = V2 / R
or
Voltage X Voltage divided by Impedance = Power.
Ohm's Law Calculations With Power
Please post your measurement results!
As a bonus, you might do this at several volume settings if you are curious about the difference in those volume settings. E.G., measure at your loudest setting, then measure at your normal settings. You may see a large difference.