Since NFPA doesn’t show their work, I can’t say exactly what line of reasoning they are following. Designing to “-0.5 dB loss” at the load is not a Code requirement, and, frankly, is kind of useless in the real world.
This is the formula they provide to determine the distance (in feet) a load can be from the amplifier for -0.5 dB loss on a 70.7 volt system:
Max length of cable = (59.25 x 4998)/(Wire resistance x Circuit load)
“Wire resistance” is resistance in ohms of two-conductor cable per 1000’.
“Circuit load” is “total watts required by the audio circuit.” This is confusing. Do they mean the power required at the load itself? Or the total power available to the circuit (loss due to wire + power at load)?
My students ask about the value 59.25. It’s obvious that it is 1000 x 0.05925, where 1000 is the conversion from resistance per 1000’ of cable to feet of cable. The 0.05925 is the adjustment for -0.5 dB loss. But this is the adjustment for +0.05 dB. IMHO, it SHOULD be 0.05594 for -0.5 dB.
I think the difference is in the formula in cell E5. As I mentioned earlier, NFPA is using the formula
dB = 20 * log10(Load impedance/(Load impedance + Circuit resistance)),
where
“Load impedance = Circuit voltage squared/Circuit wattage”.
This implies that they mean -0.5 dBV, calculated as 20 * log10(Voltage at load/Voltage at amp output) rather than -0.5 dB in actual power (watts).
If this is what they mean, then:
-0.5 dB = 20*log10(Vload/Vsource)
-0.0025 dB = log10(Vload/Vsource)
0.9943 = (Vload/Vsource)
0.9943 = (Vload/70.7 volts)
70.29 = Vload [voltage at load terminals equivalent to -0.5 dBV loss]
70.29 * 70.29/Load impedance = Power for load device
0.41 * 0.41/Wire resistance = Power dissipated by wire
I’m lousy at math, so there may be some weirdness here. But I think it shows what I suspect they are doing.
And I swear that I’m not making this up just to be difficult. This is in the Code (NFPA 72) and I’m just trying to make sense of it. For example, the formula they give for calculating -0.5 dB loss doesn’t work if you want -1.0 dB loss, so how do we get the right number for -1.0 dB? How about -1.5 dB?
OK, I’ll just jump in and solve one using NFPA’s approach (16 AWG wire, 70,7 volt amp, 20 watts):
Max length of cable = (59.25 * Vsource * Vsource)/(Wire resistance x Circuit load)
Max length of cable = (59.25 x 4998.5)/(8.032 ohms x 20 watts)
Max length of cable = 296131.5/160.64
Max length of cable = 1843.4 feet
Now let’s backtrack and use their formula for dB loss and the figures we just developed:
dB = 20 * log10(Load impedance/(Load impedance + Circuit resistance))
dB = 20 * log10((Vsource * Vsource/Ptotal)/((Vsource * Vsource/Ptotal)+(Rwire))
dB = 20 * log10((4998.5/20)/((4998.5/20)+(14.81))
dB = 20 * log10(249.925)/(249.925 + 14.81)
dB = 20 * log10(249.925/264.35)
dB = 20 * log10(0.9454)
dB = 20 * (-0.0244)
dB = -0.488
Hmmm…this is pretty darn close to our target of -0.5dB. I was very slap-dash with significant digits, too. Nobody would ever argue beyond one significant digit in a dB calculation like this.
Ahhhh, so wire resistance is the total resistance for both wires per 1000 feet? If so, I redid my calculations. In this spreadsheet I use correct formulas, and there’s a big difference between my final distance (cell C20) and NFPA’s distance (cell C22):
In the spreadsheet below, I use the incorrect formula of 20xLog(V1/V2) in cell C8, which makes my final distance (cell C20) fairly close to NFPA’s distance (cell C22):
In the above spreadsheet, if I change -0.5 dB to -0.53 dB in cell C5, the lengths are nearly equal: I calculate 1841.9 feet in cell C20, and NFPA calculates 1843.4 feet.
It’s much, much closer than we would ever care about in real life. Good job!
In one of NFPA’s sample formulae, they actually show the loss in dB calculated to 5 decimal places (-0.15738 dB). This is precision that is orders of magnitude beyond anything useful in system design work.
Ha ha! Yes, indeed. But NFPA also rounds off 0.707107 to just 0.7 when spacing spot-type detectors, so it’s in line with where they round off (and where they don’t).
They also use 25.2 volts for what most of us would call a “25 volt” speaker system, which makes a difference of less than 2% when calculating power.
And, of course, most designers pretend that power-limited fire alarm systems are a nominal 24 VDC, when they are usually closer to 25.2 VDC…and often much different from either value. Just depends on the manufacturer and design.
It is, I believe, from a manufacturer’s installation manual. I may be able to track it down. But NFPA is notorious for co-opting things manufacturers write and inserting them in the Codes.
Even aside from all the other issues, saying there is a “loss of -0.5 dB” is nonsense. A loss of a negative amount is… a gain. They should just say a “loss of 0.5 dB”, or maybe an “amplification of -0.5 dB” if they really want to go that way (“gain of -0.5 dB” is admittedly a bit confusing even if strictly correct).
True. But I think the intent is to be saying, “the voltage loss is equivalent to -0.5 dB” and it’s easier to say “loss of -0.5 dB.” In fact, nobody would care about -0.5 dB or even +0.5 dB. Both are absolutely insignificant when it come to the final installed system. The software most often used to calculate speaker circuit loss usually gives a warning when you reach -1.5 dB, but the Code does not specifically state that a certain loss is acceptable or unacceptable. The criteria are performance-based…make the alert tone 15 dBA above ambient in all locations and keep the voice-announcements intelligible and you’re good to go.
Not only is equation A.18.3.7.2.1a incorrectly applied, but their terminology is all f’d up.
In equation A.18.3.7.2.1a we have the (again, incorrectly applied) equation:
dB circuit loss = 20*log(load impedance/(load impedance + circuit resistance))
This can only make sense if load impedance is the resistance of the actual load (the speaker), and the “circuit resistance” is the total resistance of the wiring only.
In equation A.18.3.7.2.1b they say,
load impedance = circuit voltage²/circuit wattage
Well, since load impedance must be the resistance of the actual load (the speaker), then according to this equation “circuit voltage” must be the load (speaker) voltage, and “circuit wattage” must be the load (speaker) power. If so, why in the hell are they using the term “circuit voltage” for the load voltage, and “circuit wattage” for the load power? With the exception of current - which is the same everywhere - who in the hell would use the term “circuit” when talking about the load (speaker)?
And then they say the circuit voltage is 70.7 V. But isn’t that the source voltage?
Now some people might say, “Well, the source voltage and load voltage are roughly the same anyway.” No, their difference matters. If they were assumed to be the same, there wouldn’t be a dB loss!
I am still reviewing it, and it’s a complete train wreck. For now, two things are for certain: 1) whoever wrote that section should be taken out back and profusely beaten (j/k), and 2) it needs to be rewritten.
The only good thing you can say is that the error is conservative; the length limit is shorter than it needs to be.
