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Discussion Starter #1
Does magical power exist in car audio? Some people think it does.


In a crossover with two speakers, a tweeter and a woofer, both @ 4ohms, with 100w of full range signal being fed into the crossover. Do both speakers get 100w?


In a subwoofer section with a 1000w mono amp pushing a single subwoofer at 1000w and we add a second subwoofer, is the second subwoofer now getting 1000w as well?

Enlighten me with your wizardry!!
 

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Think of it this way, while I dig up more reading for you.

If you send full 100 watts into a passive crossover, does the tweeter get quieter and quiter with the power power you send? The midwoofer will use a lot more power for low frequencies, so as you turn up the volume that would reduce the amount of power available to the tweeter. If your understanding of passive crossovers was correct, then the tweeter would end up muting as the volume was turned up.

That doesn't happen because the power is fullrange. The power that the mid uses doesn't reduce the amount available for the tweeter because they are playing different frequencies.
 

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Try this for me, it is a cheap experiment.

So if you send 100w of full range signal using white noise (so every frequency has equal energy).

You first just connect a mid that uses a single coil that cuts it at 2500hz to the system and measure its voltage that it is receiving.

You then just connect a tweeter that uses a single capacitor that cuts it at 2500hz to the system and measure its voltage that it is receiving.

You then split the signal at 2500hz to a woofer and tweeter using ZERO resistors in the circuit, just using a single coil on the mid and a single capacitor on the tweeter.

When you re-measure the mid, does its voltage change?

If your hypothesis holds, then the voltage going to each speaker should be around 20v when measured individually.

When connected together the voltage going to each one should be 14.14 so they each only get 50w.
 

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The first thing - I try to not think of 100W of power as a figure of reality. When we talk about "sending xx Watts" to a speaker, we're already not being truthful to ourselves unless we're listening to sinewaves or squarewaves. Our amp may deliver that on transients, of course power is logarithmic so it ramps up very quickly.


100W full range (pink noise?) https://en.wikipedia.org/wiki/Pink_noise is not linear. It falls off 10dB per decade.


White noise would be equal power per frequency, and it is dominantly higher-pitched.



When it comes to what the crossover is doing with that, the crossover (Assuming an "ideal" one) will allow pass-through of whatever frequencies are valid according to the passive or active crossover components, including digital components (DSP). No crossover has a "wall" for a slope, as that is a mathematical impossibility. So it too will have a fall-off in power and SPL at the cross frequency, at the slope determined by the DSP setting or physical electrical component. Power will fall with SPL, though not linearly.


Now, your second question is easier.


When you add a new sub, you are either loading in series, or parallel. Therefore, the load either halves or doubles, all else equal. Assuming you have an amp that can play ~1000W at both of those impedances, then you would have 500W per sub (all else equal.) And btw, that cone area would give you 3dB more volume. If you added equal power (so, 2000W total), you'd have another 3dB.
 

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Discussion Starter #5
Think of it this way, while I dig up more reading for you.

If you send full 100 watts into a passive crossover, does the tweeter get quieter and quiter with the power power you send? The midwoofer will use a lot more power for low frequencies, so as you turn up the volume that would reduce the amount of power available to the tweeter. If your understanding of passive crossovers was correct, then the tweeter would end up muting as the volume was turned up.

That doesn't happen because the power is fullrange. The power that the mid uses doesn't reduce the amount available for the tweeter because they are playing different frequencies.
Say the volume goes from 1 to 100, 100 being max volume.

Max power is set at 100 on the volume. Volume is increasing to the speakers at a constant rate from 1 to 100. The tweeter cannot mute. Again, you are using examples that lend no weight to the argument. We need facts, not hypothetical.

The second response... I have already used a multimeter at the outputs of a passive crossover and the voltage was lower for the tweeter.... is this what you mean?
 

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Power is always split when there's more drivers in the equation idk where the hell you got the idea that you get the same maximum amplifier power for each driver. Lay off the magic dust hahaha.

Not to mention impedance rise and how music is dynamic and recording levels directly affect output as well as voltage drops and soo many real world factors. Most people are never getting anywhere near as much "clean" power as they think.
 

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Found this on white noise, just neat info. Its really just random uncorrelated numbers, not really full power on all frequencies.
White noise is a signal made of uncorrelated samples, such as the numbers produced by a random generator. When such randomness occurs, the signal will contain all frequencies in equal proportion and its spectrum will turn flat.

White Noise
0 dBFS
Gaussian Distribution
Most white noise generators use uniformly distributed random numbers because they are easy to generate. Some more expensive generators rely on the Gaussian distribution, as it represents a better approximation of many real-world random processes. Both generators will sound the same though, and will exhibit the same flat spectrum. They will only differ by the distribution of their sample levels.
 

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You can easily think of a two-way speaker with a passive xover as a single single speaker. When the amp sends some music at a certain frequency, there will be a certain impedance of the signal driving the speaker(s):

If this is a single speaker the impedance will be a function of the single speaker properties and frequency of the signal.

If this is a two-way speaker, and the signal is a low frequency, the impedance will be a function of the woofer's properties and the frequency of the signal. If it's a high frequency tone, it will be a function of the tweeters properties and the frequency of the signal.

In other words, you can have a single full-range driver or you can have a crazy 5-way or 6-way design and the amp won't really know the difference. It will just know that the load is different at different frequencies which is true of all speakers - simple or complex. So there is nothing to figure out really.


Edit: These things are pretty close though: https://www.crutchfield.com/S-g5W8TgWIdQn/p_575PM100X1/Rockford-Fosgate-PM100X1.html

The magic there is that it is using the energy below the crossover point and storing that to power the rail voltage ... I guess? I wonder would it work at all if you set an active crossover in front of it with like 100Hz/24 and the freeload amps at 60 or 80 Hz.
 

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With passive crossovers, you can have impedance spikes and odd things, and that affects how the amplifier drives the load. That's really important, because a bad crossover can go essentially in the red on too low impedance, and blow the amp!


Been there, done that.
 

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You can easily think of a two-way speaker with a passive xover as a single single speaker. When the amp sends some music at a certain frequency, there will be a certain impedance of the signal driving the speaker(s):

If this is a single speaker the impedance will be a function of the single speaker properties and frequency of the signal.

If this is a two-way speaker, and the signal is a low frequency, the impedance will be a function of the woofer's properties and the frequency of the signal. If it's a high frequency tone, it will be a function of the tweeters properties and the frequency of the signal.

In other words, you can have a single full-range driver or you can have a crazy 5-way or 6-way design and the amp won't really know the difference. It will just know that the load is different at different frequencies which is true of all speakers - simple or complex. So there is nothing to figure out really.


Edit: These things are pretty close though: https://www.crutchfield.com/S-g5W8TgWIdQn/p_575PM100X1/Rockford-Fosgate-PM100X1.html

The magic there is that it is using the energy below the crossover point and storing that to power the rail voltage ... I guess? I wonder would it work at all if you set an active crossover in front of it with like 100Hz/24 and the freeload amps at 60 or 80 Hz.

Right, and having a wild 6 way setup is not going to reduce the output of the low frequency driver. Because the power is not split, the frequencies are.

You can split that bandwidth up into a bunch of sections and you will not take power away from the low frequency driver with each "way" you add. A 2, 3, 4, 5 way speaker system will not have progressively less bass response (or low frequency SPL) each time you split the frequencies. It's not the case that a 3 way will have (let's say) 60% power for the midbass, but each time you split the bandwidth you'll reduce the power to the midbass. A 5 way wouldn't drop the available power to that midbass from 60% to 40%.
 

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Do an FFT to make a spectrum like REW shows.
Each frequency bin has some amount of voltage or power... which trails off with freq for music.

Without any current flowing, there is only voltage that an amp can provide... so the amp is not sending power... it is driving voltage, until some load is attached to it.

If there is a 100w available "across the band", then there is a maximum voltage of 20v (or whatever it is)... but in each band, with some pink noise, that voltage will be trailing downwards as the frequency goes higher.

If one had a single 4-ohm speaker it would be consuming (or forced to work) at 100w.
If you had 2 speakers then they would be running at 100w each if the amp could supply the current... and most amp's supply around double the wattage into a two ohm load compared to a 4-ohm load.
Clearly if two 8-ohm speakers were in parallel, then they would be only able to consume 1\2 the current of a 4-ohm... so each should be consuming 50w.

One could actually just run pink noise and out a voltemeter on the tweeter and then the midbass and see that the RMS AC voltage is higher on the midbass than on the tweeter.
 

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One could actually just run pink noise and out a voltemeter on the tweeter and then the midbass and see that the RMS AC voltage is higher on the midbass than on the tweeter.
The problem with that is that the pink noise, by design, has less energy being sent to the tweeters.

Then, if it is a passive crossover, there are also resistors in the circuit that further decrease the voltage actually being seen by the tweeter (note: These resistors are not part of the crossover, they do nothing to split the frequencies, they are usually housed in the same container though)
 

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Discussion Starter #13
I think the nomenclature needs to be addressed. It is becoming obvious that what one person thinks a word means isn't what it means or what other know it to mean. If we are all to discuss this topic then our vocabulary needs to be the same. Otherwise there is going to be confusion.

Power (watts) = V2 (volts) ÷ R (Ω)

20v squared = 400
400 divided by the resistance 4ohms = 100
100 is the wattage

An amplifier sends an amplified signal to the crossover. What is that signal?
It is A/C voltage correct? The amplitude of the voltage that is sent can be measured in wattage. So where there is voltage there is wattage. Send voltage in one direction and wattage will go with it. Send it another direction and wattage will go with it. Turn the gain down and the amount of voltage is going to be less... we know this because it can be measured.

The crossover is splitting the voltage and sending it to a specific speaker. The amount of voltage being sent to the speakers can be measured at the outputs of the crossover just like the amplifier outputs. The formula works the same here for wattage. Because we can measure voltage to get wattage and we know the crossover is splitting up the voltage, we can also find the amount of wattage that is split.
 

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I think the nomenclature needs to be addressed. It is becoming obvious that what one person thinks a word means isn't what it means or what other know it to mean. If we are all to discuss this topic then our vocabulary needs to be the same. Otherwise there is going to be confusion.

Power (watts) = V2 (volts) ÷ R (Ω)

20v squared = 400
400 divided by the resistance 4ohms = 100
100 is the wattage

An amplifier sends an amplified signal to the crossover. What is that signal?
It is A/C voltage correct? The amplitude of the voltage that is sent can be measured in wattage. So where there is voltage there is wattage. Send voltage in one direction and wattage will go with it. Send it another direction and wattage will go with it. Turn the gain down and the amount of voltage is going to be less... we know this because it can be measured.

The crossover is splitting the voltage and sending it to a specific speaker. The amount of voltage being sent to the speakers can be measured at the outputs of the crossover just like the amplifier outputs. The formula works the same here for wattage. Because we can measure voltage to get wattage and we know the crossover is splitting up the voltage, we can also find the amount of wattage that is split.
But crossovers are generally parallel circuits....voltage is equal on ALL branches of parallel circuits and that is equal to the supply voltage or 20v in this case....So by your own equations above would then prove that the voltage would be the same on each path, the resistance should be similar on each path, so therefore power should be the same.

So each speaker would see 20v, each speaker would be ~4ohms (I know this isn't right but lets go with it), so each speaker gets 100w.

In order to only see 50w each, that would have to break the laws of physics for parallel circuits. It would have to mean the input voltage of 20v somehow got "magically" reduced on a parallel circuit to 14.14v.
 

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Discussion Starter #15
But crossovers are generally parallel circuits....voltage is equal on ALL branches of parallel circuits and that is equal to the supply voltage or 20v in this case....So by your own equations above would then prove that the voltage would be the same on each path, the resistance should be similar on each path, so therefore power should be the same.

So each speaker would see 20v, each speaker would be ~4ohms (I know this isn't right but lets go with it), so each speaker gets 100w.

In order to only see 50w each, that would have to break the laws of physics for parallel circuits. It would have to mean the input voltage of 20v somehow got "magically" reduced on a parallel circuit to 14.14v.
They can not both get 100w. 100w is the TOTAL output. It will be less at each speaker. There is no way around it.

To touch on another example...

300hz and 6000hz are sent to the crossover... they can be sent at the same time at 100w. Now what? You think they will both play to the speakers at 100w???
 

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They can not both get 100w. 100w is the TOTAL output. It will be less at each speaker. There is no way around it.
I know that parallel circuit physics probably has you stumped on how this could be, but it is right, you wanted to use physics to explain this here and I just did.

So please explain how it is wrong using physics.

You can't have something get 20v, be 4ohms, and not get 100w.
 

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Discussion Starter #17
I know that parallel circuit physics probably has you stumped on how this could be, but it is right, you wanted to use physics to explain this here and I just did.

So please explain how it is wrong using physics.
Show me the magic amp that will now supply 200w from a 100w channel.
 

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Show me the magic amp that will now supply 200w from a 100w channel.
So we are done with the physics now? Prove me wrong.
 

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Discussion Starter #19
I know that parallel circuit physics probably has you stumped on how this could be, but it is right, you wanted to use physics to explain this here and I just did.

So please explain how it is wrong using physics.

You can't have something get 20v, be 4ohms, and not get 100w.
You can if there is more than one.

Let me restate that... You are correct. As soon as a second 4 ohm is added they will not get 20v
 

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