How it works is this.
Take a sine wave. If we take an average reading of its magnitude then we have to take an RMS value.
If we're working with digital, then the max level that any part of the sine wave will reach up to is 0dB. Obviously if we don't want to clip the sine wave then the peaks of the maximum amplitude sine wave will hit 0dBfs.
When we are discussing a sine wave in the context of this thread the -dB figure is with respect to an RMS sine wave value. If we take a sine wave that hits 0dB on the peaks then it has an RMS value of -3dB.
Lets take Pano's -12dB sine wave, as provided, this is an RMS value so the peaks are actually hitting -9dB.
Now when you measure the voltage with your volt meter you are also measuring an RMS value. This is handy because Pano's test is also using an RMS -dB value so we can directly compare the two.
When you measure 5Vrms from your volt meter we know that this is what's required to reproduce a signal with an average level of -12dB.
After we've measured and compared them, the fact that these are both RMS values isn't strictly important, what's important is that we now know that 5volts on the amplifier output = a signal level of -12dB on the digital scale.
-12dB is obviously 12dB short of the maximum digital signal level, so if the amplifier needs to output 5volts to reproduce -12dB, then we must multiply the voltage by a factor of 4 to scale it up 12dB.
This means that the 5V as measured that is necessary to produce a signal level of -12dB, ends up being 20V as necessary to reproduce a signal level of 0dB.
Now you cannot exceed 0dB, so for that volume position you have found the maximum voltage the amplifier needs to be able to output to reproduce the maximum signal level possible for the digital medium.
Lets look at this another way though.
Pano's -12dB test is an RMS value, but we know that the peaks here hit 3dB higher, that is -9dB.
We are also measuring an RMS value, 5 volts as per our volt meter.
From this we have also learnt something else.
First of all we know that to convert an RMS value to a peak value we multiply it by the square root of 2, or 1.41. In this case 5 * 1.41 = 7.05V.
We know that the peak values for the sine wave, as recorded digitally are hitting peaks at a digital signal level of -9dB, so for the amplifier to reproduce these peaks it must output 7.05V.
So for a signal level of -9dB the amplifier is called on to output 7.05V.
This is however 9dB short of the digital full scale output. To multiply a voltage by an amount that corresponds to a level change of 9dB we must multiply it by 2.83.
7.05 * 2.83 = 19.95 or rounded up 20V.
As you see, we ended up back at the same place with each method. 20 volts is required to reproduce full scale 0dBfs digital with the volume control set to that position.
Now how do we convert this to a watt figure so that we can compare it to our amplifier?
First of all this 20V is a peak figure. Amplifiers usually quote accurate RMS figures in their specifications so the first thing we need to do is divide 20 by 1.41. We therefore end up with 14.18 volts.
Power = Voltage squared/resistance.
so (14.18^2)/8 = you need a 25 watts RMS amplifier into an 8 ohm load.
Take a sine wave. If we take an average reading of its magnitude then we have to take an RMS value.
If we're working with digital, then the max level that any part of the sine wave will reach up to is 0dB. Obviously if we don't want to clip the sine wave then the peaks of the maximum amplitude sine wave will hit 0dBfs.
When we are discussing a sine wave in the context of this thread the -dB figure is with respect to an RMS sine wave value. If we take a sine wave that hits 0dB on the peaks then it has an RMS value of -3dB.
Lets take Pano's -12dB sine wave, as provided, this is an RMS value so the peaks are actually hitting -9dB.
Now when you measure the voltage with your volt meter you are also measuring an RMS value. This is handy because Pano's test is also using an RMS -dB value so we can directly compare the two.
When you measure 5Vrms from your volt meter we know that this is what's required to reproduce a signal with an average level of -12dB.
After we've measured and compared them, the fact that these are both RMS values isn't strictly important, what's important is that we now know that 5volts on the amplifier output = a signal level of -12dB on the digital scale.
-12dB is obviously 12dB short of the maximum digital signal level, so if the amplifier needs to output 5volts to reproduce -12dB, then we must multiply the voltage by a factor of 4 to scale it up 12dB.
This means that the 5V as measured that is necessary to produce a signal level of -12dB, ends up being 20V as necessary to reproduce a signal level of 0dB.
Now you cannot exceed 0dB, so for that volume position you have found the maximum voltage the amplifier needs to be able to output to reproduce the maximum signal level possible for the digital medium.
Lets look at this another way though.
Pano's -12dB test is an RMS value, but we know that the peaks here hit 3dB higher, that is -9dB.
We are also measuring an RMS value, 5 volts as per our volt meter.
From this we have also learnt something else.
First of all we know that to convert an RMS value to a peak value we multiply it by the square root of 2, or 1.41. In this case 5 * 1.41 = 7.05V.
We know that the peak values for the sine wave, as recorded digitally are hitting peaks at a digital signal level of -9dB, so for the amplifier to reproduce these peaks it must output 7.05V.
So for a signal level of -9dB the amplifier is called on to output 7.05V.
This is however 9dB short of the digital full scale output. To multiply a voltage by an amount that corresponds to a level change of 9dB we must multiply it by 2.83.
7.05 * 2.83 = 19.95 or rounded up 20V.
As you see, we ended up back at the same place with each method. 20 volts is required to reproduce full scale 0dBfs digital with the volume control set to that position.
Now how do we convert this to a watt figure so that we can compare it to our amplifier?
First of all this 20V is a peak figure. Amplifiers usually quote accurate RMS figures in their specifications so the first thing we need to do is divide 20 by 1.41. We therefore end up with 14.18 volts.
Power = Voltage squared/resistance.
so (14.18^2)/8 = you need a 25 watts RMS amplifier into an 8 ohm load.
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Indeed they are not, but they are limited by the absolute maximum of 0dBfs, the same maximum that the peaks of the sine wave reach with an average level of -3dBfs.
Yes. 5th E. is quite correct:
OR:
- The test signal is a sine wave.
- It has an RMS value of 12dB below digital maximum
- It has a peak value 9dB below digital maximum (all pure sines have an RMS value 3dB below their peak value.)
- Once you have measured the test signal, you need only multiply by 4 to find the digital maximum (peak).
- The loudest possible sine wave will have a value 3dB below that.
- Amplifiers are generally rated RMS power, not peak.
OR:
Measure the test tone- Multiply by 2.82 to find the highest sine wave possible on a digital recording
- That is the maximum sine wave RMS voltage at your volume setting.
- Use Ohms law to figure out how much power that is.
At least from what I've gathered weltersys drives things hard and uses lots of power. If he were to use a full scale sine wave then things might start exploding 😀
Bingo .....😀
They are all good 5th , one needs to watch the movie snatch 10 times , funny you should catch all of them seeing ur familiar with brick top's sarcasm ....
🙂
They are all good 5th , one needs to watch the movie snatch 10 times , funny you should catch all of them seeing ur familiar with brick top's sarcasm ....
🙂
I have (accidentally) used a 60 Hz 120 volt sine wave (probably more square than sine, the amp was clipped and limited) for distortion tests with a four ohm B&C18SW115-4 speaker with no "explosions".
A much lower level than that did cause a pair of AR voice coils to continue travel in to the magnet structure after separating from the cone when the spider bottomed against the frame, made surprisingly little noise.
Art
Peavey used to do the "plug it into the wall" demos with one of their speakers. Of course its impedance peak was right at 60Hz. 🙂 Still, an impressive trick.
Not 800 (or 1800) watts because of the impedance spike at 60Hz. It might have been 140 watts. The demo wasn't long, cause it was LOUD. A good trick.
At such voltages, wouldn't you need insulated binding posts? None of that fancy gold plated stuff here 🙂
Yes you do. There are some standards about that, tho I don't know what they are. Using Speakon connectors insulates you from that problem.
Peavy gets it right
I used to blame Peavy for driving me away from live shows. Chances are the entire audio chain was harsh in the 80's.
I used to blame Peavy for driving me away from live shows. Chances are the entire audio chain was harsh in the 80's.