How wattage adds....HELP!

[Shpoop]

Hey guys. I'm building a kind-of line array project (but not really...only three or four drivers per side). And I was wondering how the wattage adds. I've studied this stuff, but I'm not quite sure how it applies to speakers. Below is what I've worked through in my head...tell me if I'm right.

Say for instance I used the HiVi B3N 3''. It recommends 15 watts RMS, and the impedance is 8 ohm. Now...it has never been clear to me how an amp can simply output "watts," which is a unit of power dissipation. Power dissipation is a function of current and resistance (impedance). For instance...the current going into my lamp is always the same, but I can plug in many different light bulbs to get different wattages of the circuit. Since an amp keeps current constant (for a given song, say), why won't different speakers produce different wattages for that circuit?

The answer I've come up with is that the speakers are all the same resistance...say 6 or 8 ohm. So even though they look and sound different, it's like switching out a 60 watt bulb for a 60 watt bulb. So I think that amps use wattage as a measurement due to familiarity, but what they really output is a constant CURRENT at that given 6 or 8 ohms.

So using the formula P = (I^2)*R for my HiVi, sqrt(15/8) = 1.37 amperes. So the speakers power dissipation (~15 W) will always be a function of the current running through it, but the REAL limitation of the speaker is that does not want current much higher than 1.37 amperes.


Now if I were to setup the speakers in two groups in parallel, with each group having two speakers in series, the total impedance is 8 ohms. This means I would need the same amp capable of 15 watts, same as if I were driving one speaker, because the impedance is the same.

However, what if I were to place them all in series. The impedance is then 32 ohms. Now, assuming the speakers want no more than 1.37 amperes, P = (I^2)*R = (15/8)*(32) = 60 W. So if I place them all in series, can I use an amp that provides 60 watts?

Is this logic correct?

_______

A couple things to mention before this is over. The math I did was for DC current. However, for AC current, you find the circuit's total impedance (in DC, you only use resistance R due to resistors, but in AC, impedance depends on resistors, inductors, and capacitors, I think) and then use that in the equation P = I^2*Z. Since the speaker is in AC, but gives you the whole apparatus' total impedance already, the math I did should work.

When you guys add crossovers and filters, the resistors and inductors and capacitors change the whole circuit's impedance, right? If I'm right in assuming that since an amp has a given wattage at a given impedance that means it outputs constant current, then won't these crossovers and filters change the current given to the speaker and therefore its power dissipation?

Do you guys take this into account when picking an amp? I do not see much of this type of math here....especially all the pain in the *** imaginary numbers it involves.

_______

Please let me know if the general assumptions made here as well as the specific conclusions are on point, or are way off the mark. If they are wrong, please help me clarify!

Thanks,

Matt

[myhrrhleine]

For the most part, amplifiers are providing a voltage out.
Current is then determined by the loudspeaker impedence.

[AndrewT]

Hi,
it's more useful to think in terms of voltage.
P=V^2/R but also P=V*I as well as your P=I^2*R.
When using gain and input voltage it makes it much easier to use voltage for comparing performance into different loads. In fact so much more useful I wonder why they ever thought of using power to measure amplifier output, particularly when our ears are logarithmic in response.

I would much rather buy an amp that does 30V into 8ohms and know that it does more work than 27.5V into 8ohms. And if both these amps manage 26V into 4ohms then it indicates to me that the 27.5V amp stands up better to lower load values.

[Sonusthree]

Originally posted by myhrrhleine
For the most part, amplifiers are providing a voltage out.
Current is then determined by the loudspeaker impedence.

I've often wondered about that!

So .... for example an amplifier pushing a tweeter a passive crossover (High pass filter) with a full range signal ........

The potential would be there at low frequencies but effectively zero current? This makes passive bi-amping seem more attractive if it's true.
Is this correct?

It's been bugging me for ages, please cure my insomnia!

Regards,
Martin

[AndrewT]

Hi,
the passive crossover is usually a nominal impedance across the full audioband.
Split the crossover and measure both halves.
You will find that the passband impedance is still near the nominal value for the whole crossover. The big difference is the stop band impedance. It rises to near infinity. This draws very low current and considerably reduces the load seen by the amplfier that is driving just one half.

[Sonusthree]

Thanks Andrew (yet again).

This helps me to understand my system a little better.

Many thanks,
Martin

[Petriej]

So I think that amps use wattage as a measurement due to familiarity, but what they really output is a constant CURRENT at that given 6 or 8 ohms.

I think that amplifiers take in a low-level (in some cases high-level) signal and boost the voltage of the signal, up to a maximum related to the voltage supplied to the op amps inside (+ and - rails). This voltage is related to AC voltage, in the sense that it alternates from positive to negative according to the signal (in this case, a song).

The current supplied to the system is related to the voltage at a specific moment in the signal. Say the song has a loud bang from a bass drum; at that moment, the signal will "spike" since the amplitude of the voltage is related to the amplitude of the sound (signal). If the gain of the amplifier is too high, the signal may "clip" in which case the output signal will max out since the system tries to amplify it past the capabilities of the circuit (related to the input voltage to the op amps, i think).

Since V = I * R (or V = I * Z in AC circuits), manufacturers could tell you how much voltage the system is capable of, but since speakers come in various impedances, AND crossovers may affect the total impedance, it would not be as helpful. Thus they list the max, and RMS power ratings of their device (since P = I * V, or P = I^2 * R) because everyone's system could have many different impedances, and thus currents.

It recommends 15 watts RMS, and the impedance is 8 ohm.

In your example, since your driver has an RMS rating of 15W, the RMS of the voltage output of your amplifier should be around:

15W = V^2 / 8 (P = V^2 / R);
V = sqrt(15 * 8) = sqrt(120) = 10.95...


You can use the same formula to find the max voltage if you know the max your driver can handle.


What does this all mean?
Well, if you're looking for an amplifier to power your driver rated at 15W RMS, then try to find one capable of a little bit higher, that way when you want to play the driver loudly, the amplifier won't clip too soon and send a distorted mess to your driver. However, this also means you can't turn your amplifier all the way up or you risk damaging the driver.


I hope this helps, I am still wondering about a few of your questions as well, so I hope some of the experts reply to this, too.

[sreten]

From : http://www.zaphaudio.com/audio-speaker18.html

FWIW 4 drivers handle 4 times the power, which ever way you drive them.

/sreten.

[Shpoop]

ok thanks...

so if I have four drivers each rated at 15 W 8 ohm, and I have the four in a setup whose resulting impedance is 8 ohm, I want an amp that 60 W at 8 ohm. right?

What if I have two drivers in series so that the resulting impedance is 16 ohm. (I know they dont rate at 16 ohm, but the math can be done, right?) And parallel?

[Pano]

You don't have to match speaker watts to amp watts.

In other words, if your driver(s) are rated at 60W, that does not mean you need a 60W amp. You could use a 6W or 500W amp. They would both sound pretty much the same running at 1 watt or 2.

The difference being that the 500W amp has enough power to blow the cones right off your speakers. Or throw sparks (guess how I know).

The 6W amp will run into clipping long before it reaches the 60W max of your drivers. Clipping is not good for the drivers, for the amp - or your ears.

What is much more important than how much power the amp has is "how much do you need?" For every 3dB you go up in speaker sensitivity, you need 1/2 as much power for the same SPL.

In general people try to get an amp that is in the ballpark of the power rating of the speakers. You'll sometimes see it stated that you should have an amp capable of 2X the power rating of your speakers, so that you don't run into clipping. Remember, clipping is bad, at least heavy clipping is.

My 15" woofers are pro units rated at 250W. I run them with 6WPC no problem. But I can throw more power at them if I choose to.

OK, that came out much longer than it should have, but I hope it helped!