Computing for the dB power loss for a PAGA speaker Loop
Computing for the dB power loss for a PAGA speaker Loop
(OP)
Hello,
I'm currently making a technical report on the PAGA location for our project. But I got bogged down because I disagree with my KP on the subject of computing the line for a loop of speakers.
Our PAGA speaker loops/system, has the following parameters: 100v supply, 300w amplifier(but will only be loaded to 240w), maximum tap settings at 15w per speaker, maximum of 1.5dB loss
Our approach was to consider the speakers (there were 16 pcs) in the loop as a lump load as seen by the amplifier. So you would arrive at a simple diagram --- a 100v supplying series resistances (resistance of the line and the lump resistance of the speakers).
My KP's approach in getting the resistance of the speaker is to divide the line voltage by the max. tap of the speaker (100v*100v/15w) and then get the total R for the 16 speakers.
Mine is, first to get the total R (100*100/240W) of the system. Then get the resistance of the lump speakers by subtracting the resistance of the line (the line btw, stretched for 500m). So R of speakers(lump) = R total - R of the line.
Could you guys shed a light on this? Just felt that my brain needs the little oil and grease to remove some rust. Btw, we're using the formula dB loss = 10log(p1/p2) for computing the max. dB power loss.
I'm currently making a technical report on the PAGA location for our project. But I got bogged down because I disagree with my KP on the subject of computing the line for a loop of speakers.
Our PAGA speaker loops/system, has the following parameters: 100v supply, 300w amplifier(but will only be loaded to 240w), maximum tap settings at 15w per speaker, maximum of 1.5dB loss
Our approach was to consider the speakers (there were 16 pcs) in the loop as a lump load as seen by the amplifier. So you would arrive at a simple diagram --- a 100v supplying series resistances (resistance of the line and the lump resistance of the speakers).
My KP's approach in getting the resistance of the speaker is to divide the line voltage by the max. tap of the speaker (100v*100v/15w) and then get the total R for the 16 speakers.
Mine is, first to get the total R (100*100/240W) of the system. Then get the resistance of the lump speakers by subtracting the resistance of the line (the line btw, stretched for 500m). So R of speakers(lump) = R total - R of the line.
Could you guys shed a light on this? Just felt that my brain needs the little oil and grease to remove some rust. Btw, we're using the formula dB loss = 10log(p1/p2) for computing the max. dB power loss.





RE: Computing for the dB power loss for a PAGA speaker Loop
240 watts distributed to 16 speakers is 15 watts per speaker. That's a lot. If the speakers are ceiling mounted in an office area, you'll need to hand out ear protectors. If they're horn speakers in a warehouse, 15 watts into a horn is still a lot. Maybe if the warehouse was full of noisy fork-lifts...
The whole point of using such 100-volt or 70-volt systems (instead of 8-ohm) is that you don't need to worry too much about loss provided you're within the chart of wire length and distance provided by the equipment supplier.
The loss would be primarily caused by voltage drop in the long wires. The voltage drop depends on the current. The current depends on the 'power level tap' being used at the speakers. Also, if there are L-pads at each speaker, that would also affect the load/current/voltage drop/dB loss.
If you're worried about loss, then you could wire up each speaker on its own pair of wires back to the amp. If you want to save wire, then you'll have more loss due to the currents sharing one pair.
To calculate the loss, figure out the power in the wire pair (15 watts max per speaker). Knowing the voltage (100 volts RMS ?) calculate current (150mA). Calculate the voltage drop (V=I*R), don't forget it's two-way (x 2). At this point you'll probably see the reason high voltage is so useful (small number compared to 100-volts). Then calculate the power lost in the wire (P=I*V). And calculate the dB loss (probably a very small fraction of a dB).
If you use lower-power taps at the speakers, then the loss will be less due to lower current (even if the loss mattered at that point, which it wouldn't).
The advantage of high voltage is squared. Less current and more voltage both help with voltage drop.
RE: Computing for the dB power loss for a PAGA speaker Loop
16 loud speaker is actually the maximum number of speakers we set per loop. I'm designing a refinery plant in Doha Qatar, so all of our speakers are outdoor, Zone 1 certified. It seems that our speakers are many but the plant is actually big. We also have lots of motors and airfin coolers so our sound map software came up with lots of speakers just to maintain the required signal to noise ratio.
We arrived at similar dB ratio (0.769, you can view the attachment). If you have time you can take a look of the computation that my KP and I made (sorry for the hand writing, Ü ). He's computation is the one w/c has an encircled number 1 at the bottom of the page.
What he did was to compute for the resistance when the load consumes 240W at 100v supply. And then still used that value in series with the line resistances. Ü
Regards,
RE: Computing for the dB power loss for a PAGA speaker Loop
For example, is the amplifier putting out 240 watts, or are the 16 speakers each consuming 15 watts (15*16=240)? Do the speakers really consume 15 watts each if the applied voltage isn't really 100 volts due to the drop in the wire?
Since the amplifier is almost certainly a voltage source, his numbers (page 1) look good.
RE: Computing for the dB power loss for a PAGA speaker Loop
Actually he got a different answer from us. The reason for making all of the speakers consume 15w (the highest tap) is to simulate the largest loss possible. This translate to getting the largest current the lump load will draw since power is directly proportional to the square of the current (P=I*I*R, using your approach ).
What he did was to compute for the resistance using Rt=v*v/r (100v*100v/240w=41.67 ohms). There's nothing wrong with this, but what he did was to add the resistance of the cables (as if the cable resistance is not part of "Rt") in representing the equivalent circuit diagram. The right approach is to deduct the resistance of the line from "Rt". This Resistance "Rs" is the resistance of the lump speaker. That is why, if you'll check the attachment, he got a different db loss.
Anyway, my KP already agreed that he overlooked this one Ü,
Cheers!!
RE: Computing for the dB power loss for a PAGA speaker Loop
Of course, the 'calibration' of PA equipment is probably only accurate to maybe +/- 20% anyway. So the correct answer is "about a dB".
RE: Computing for the dB power loss for a PAGA speaker Loop
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By the way, what does KP stand for?
RE: Computing for the dB power loss for a PAGA speaker Loop
Oh, Im sorry about the abbreviation Ü. KP stands for "Key Person" in our corporate lingo. He's responsible for checking of the deliverables (drawings, calculation reports, tender documents, etc.) of our team.
Regards,