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
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
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.
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.
myhrrhleine said:
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
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.
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.
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.
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.
From : http://www.zaphaudio.com/audio-speaker18.html
FWIW 4 drivers handle 4 times the power, which ever way you drive them.
/sreten.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?
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?
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!
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!
It did help.
But really, my question should read "if ...... what watts would the resulting speaker then be rated at"
So if I wired the two 15W 8 ohm in series, what would the resulting speaker be rated, in terms of 8 ohms (even though the resulting speaker is 16 ohms, I do believe you could convert it to 8 ohms and change the power rating, right?)
Thanks,
Matt
But really, my question should read "if ...... what watts would the resulting speaker then be rated at"
So if I wired the two 15W 8 ohm in series, what would the resulting speaker be rated, in terms of 8 ohms (even though the resulting speaker is 16 ohms, I do believe you could convert it to 8 ohms and change the power rating, right?)
Thanks,
Matt
Ok. bear with me. so if I wanted to match the amp wattage ballpark to the speaker wattage, and
a) wired them in parallel, I would need an amp that output 30 W at 4 ohm?
b) wired them in series, I would need an amp that output 30 W at 16 ohm?
I know those are odd 'ohmages', but I'm just trying to understand the math.
Thanks
a) wired them in parallel, I would need an amp that output 30 W at 4 ohm?
b) wired them in series, I would need an amp that output 30 W at 16 ohm?
I know those are odd 'ohmages', but I'm just trying to understand the math.
Thanks
There are 2 limitations to the power handling capability of a driver:
1. Thermal limit. This is how much power the voice coil can dissipate as heat before it overheats and begins to meltdown
2. Excursion limit. This varies with frequency. going down one octave requires 4 times the excursion to maintain the same volume. Your power handling reduces as you go down in frequency (why I can't explain - it just does).
So above a certain frequency, a driver will be thermal limited. Below a certain frequency the excursion limit starts to dominate and power handling reduces.
1. Find the excursion limits of the drivers
I presume you are planning to drive your HiVi drivers full range? If so - you want to model their excursion at given power input levels to ensure you are not exceeding their excursion limits.
For example - subwoofers might be thermally rated at 250 watts, but can be excursion limited to only say 40 watts at 30Hz.
2. Determine how loud (in dB) you want the speaker to play.
It takes double the amplifier power for every 3dB increase of volume. Sound pressure levels reduce by 6dB with every doubling of distance from a point source (or reduction of only 3dB from a true line array).
ie. At 2 metres distance from a standard speaker (ie. point source), the sound from a nominally 86dB speaker with 1 watt input will be 6dB down (removing any reinforcement you'd typically get in a room). Therefore at 2 metres, you ears receive 80dB. If you want to play at 86dB, you would need to increase the output by 6dB - which is 4 watts (double the power = 2 watts = 3dB increase, then double that again to get 6dB = 4 watts)
The above reduction is itself reduced depending on the room (reflections etc...)
You therefore need to work out how loud you want to play, and whether the amplifier power output required to play that loud is going to exceed BOTH the thermal AND excursion limits of your line array.
You are usually better to have a more powerful amplifier. There's an old saying that high power amps rattle woofers, under power amps blow tweeters (due to clipping induced distortion... it's more complicated than that but I can't explain).
I run a 185 watt/channel into 8 ohm load amp into my 100 watt mains speakers. this is a fine combo as long as you pay attention to the volume knob and listen for the first signs of either overdriving the speakers or amp (first more likely in my scenario), which is a loss of dynamics in the music (ie. sounds start to "compress").
I hope the above helps. I know others can elaborate / correct.
Cheers,
David.
1. Thermal limit. This is how much power the voice coil can dissipate as heat before it overheats and begins to meltdown
2. Excursion limit. This varies with frequency. going down one octave requires 4 times the excursion to maintain the same volume. Your power handling reduces as you go down in frequency (why I can't explain - it just does).
So above a certain frequency, a driver will be thermal limited. Below a certain frequency the excursion limit starts to dominate and power handling reduces.
1. Find the excursion limits of the drivers
I presume you are planning to drive your HiVi drivers full range? If so - you want to model their excursion at given power input levels to ensure you are not exceeding their excursion limits.
For example - subwoofers might be thermally rated at 250 watts, but can be excursion limited to only say 40 watts at 30Hz.
2. Determine how loud (in dB) you want the speaker to play.
It takes double the amplifier power for every 3dB increase of volume. Sound pressure levels reduce by 6dB with every doubling of distance from a point source (or reduction of only 3dB from a true line array).
ie. At 2 metres distance from a standard speaker (ie. point source), the sound from a nominally 86dB speaker with 1 watt input will be 6dB down (removing any reinforcement you'd typically get in a room). Therefore at 2 metres, you ears receive 80dB. If you want to play at 86dB, you would need to increase the output by 6dB - which is 4 watts (double the power = 2 watts = 3dB increase, then double that again to get 6dB = 4 watts)
The above reduction is itself reduced depending on the room (reflections etc...)
You therefore need to work out how loud you want to play, and whether the amplifier power output required to play that loud is going to exceed BOTH the thermal AND excursion limits of your line array.
You are usually better to have a more powerful amplifier. There's an old saying that high power amps rattle woofers, under power amps blow tweeters (due to clipping induced distortion... it's more complicated than that but I can't explain).
I run a 185 watt/channel into 8 ohm load amp into my 100 watt mains speakers. this is a fine combo as long as you pay attention to the volume knob and listen for the first signs of either overdriving the speakers or amp (first more likely in my scenario), which is a loss of dynamics in the music (ie. sounds start to "compress").
I hope the above helps. I know others can elaborate / correct.
Cheers,
David.
everyone is giving me very valuable information that I will no doubt use many times in the future. For that, I'm grateful.
but for this thread, forget I said anything about amps.
If my individual drivers are rated at 15 W and 8 ohms, and I connect four (in two groups in parallel with two drivers connected in series per group) to result in a total of 8 ohms for the entire "speaker", what will this resulting 4 driver 8 ohm speaker be rated at?
I'm assuming 60 W and 8 ohm.
That's my presumed plan of action. But also there is the chance I may want to only use two per speaker, which results in either 4 or 16 ohms. If someone asked you what the resulting speaker was "rated", what would you say? Is there a way to give the answer in terms of 8 or 6 ohms?
but for this thread, forget I said anything about amps.
If my individual drivers are rated at 15 W and 8 ohms, and I connect four (in two groups in parallel with two drivers connected in series per group) to result in a total of 8 ohms for the entire "speaker", what will this resulting 4 driver 8 ohm speaker be rated at?
I'm assuming 60 W and 8 ohm.
That's my presumed plan of action. But also there is the chance I may want to only use two per speaker, which results in either 4 or 16 ohms. If someone asked you what the resulting speaker was "rated", what would you say? Is there a way to give the answer in terms of 8 or 6 ohms?
Hi,
two drivers in parallel are rated at 30W into 4ohms. An amp that is rated at 30W into 4ohm will drive this adequately. If the amp is rated for 8ohms AND is capable of driving 4ohm speakers it would be described as 15W into 8ohms.
Two drivers in series are rated at 30W into 16ohms.
An amp rated at 30W into 16 ohms will drive this adequately.
This is an unusual spec and you will more likely see 60W into 8ohms and this will drive the 16ohms to their maximum.
Most music has a high average to peak ratio. Compressed and rock is likely to be around +10db from average to peak.
Classical & Jazz and acoustic could be as high as +30db from average to peak.
If we assume for system building purposes that our most common sources are about +20db from average to peak then it makes the following arithmetic nice and simple.
We tend to listen to average levels that sound comfortable often at conversation level of about 70db. This requires most average speakers to playback at about 0.01W to 0.1W to give listening position levels of about equal to normal conversation.
If the amplifier and speakers are not to distort or clip at that level then thay must be capable of that +20db increase to allow the peak levels to get through to our seated locations.
+20db = 100times the power. So our 0.01W becomes 1W and 0.1W becomes 10W. Your 15W speaker and 15W amp could easily cope with 70db average level and +20db peak level.
Now you decide you want loud music and turn the average volume up to 80db. This requires all the previous powers to be increased by a factor of 10. To cope with peak levels we now need amps and speakers that reproduce short term transients at 10W to 100W. See the problem we are coming up against.
Now you want to party for a room half full of dancers (or you son with some head banging friends in). This situation may require average levels of 90db and peak powers in the range 100W to 1000W. Yes the levels get that big as the volume goes up.
That's why auditorium often use horns loaded drivers. They convert the electricity more efficiently into sound and can require between 10% and 1% of the power for the same volume in the audience.
two drivers in parallel are rated at 30W into 4ohms. An amp that is rated at 30W into 4ohm will drive this adequately. If the amp is rated for 8ohms AND is capable of driving 4ohm speakers it would be described as 15W into 8ohms.
Two drivers in series are rated at 30W into 16ohms.
An amp rated at 30W into 16 ohms will drive this adequately.
This is an unusual spec and you will more likely see 60W into 8ohms and this will drive the 16ohms to their maximum.
Most music has a high average to peak ratio. Compressed and rock is likely to be around +10db from average to peak.
Classical & Jazz and acoustic could be as high as +30db from average to peak.
If we assume for system building purposes that our most common sources are about +20db from average to peak then it makes the following arithmetic nice and simple.
We tend to listen to average levels that sound comfortable often at conversation level of about 70db. This requires most average speakers to playback at about 0.01W to 0.1W to give listening position levels of about equal to normal conversation.
If the amplifier and speakers are not to distort or clip at that level then thay must be capable of that +20db increase to allow the peak levels to get through to our seated locations.
+20db = 100times the power. So our 0.01W becomes 1W and 0.1W becomes 10W. Your 15W speaker and 15W amp could easily cope with 70db average level and +20db peak level.
Now you decide you want loud music and turn the average volume up to 80db. This requires all the previous powers to be increased by a factor of 10. To cope with peak levels we now need amps and speakers that reproduce short term transients at 10W to 100W. See the problem we are coming up against.
Now you want to party for a room half full of dancers (or you son with some head banging friends in). This situation may require average levels of 90db and peak powers in the range 100W to 1000W. Yes the levels get that big as the volume goes up.
That's why auditorium often use horns loaded drivers. They convert the electricity more efficiently into sound and can require between 10% and 1% of the power for the same volume in the audience.
There may be a problem here... I have seen far more damage to drivers with an amplifier that is too small in size. Sooner or later the system gets pushed and the amplifiers clip causing problems. When I was younger I purchased a pair of Altec Model 9 speakers in 1976 that I drove with a 60 watt amplifier. I caused the voice coils in the woofers to be overheated by clipping the amp. I took out two woofers the first year I owned them. I then purchased a 200 watt amplifier and never had anymore problems.
If I were to error in amplifier size I would sooner have an amplifier larger to the load size than one too small.
Just my .02
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