DC fire alarm voltage drop
DC fire alarm voltage drop
(OP)
Hey everyone,
I'm designing a fire alarm system in which i'll be using a parallel circuit with a few seperate legs. See attached.
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I prefer to do point-to-point voltage drop calculations with the following formula...
V - (R)(A)(W)
Where
V = Starting voltage
R = Resistance of wire (per foot)
W = Length of wire (both conductors)
A = Cummulative current
So for instance, lets say all devices are drawing 50mA each. Device one is 30' from the source, device two is 45' from device one, and device three is 45' from device two. We typically use a 14AWG wire which has .00307 ohms of resistance per foot (per Ugly's). Starting voltage is 24vdc. Here's how I calculate the voltage drop...
"Leg 1" voltage drop calculation
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Device 1... 24 - (.00307)(.150)(60) = 23.97
Device 2... 23.97 - (.00307)(.100)(90) = 23.94
Device 2... 23.94 - (.00307)(.050)(90) = 23.92
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For my circuit that has seperate legs (see attached), i want to calculate each leg point-to-point. I just calculated "Leg 1" above and can easily calculate "Leg 2" using the same method. My question is how do I calculate "Leg Main"? Would I use the cummulativ current for "Leg 1" and "Leg 2" like so?...
"Leg Main" voltage drop calculation
-------
Device 1 (Leg 1)... 24 - (.00307)(.500)(60) = 23.90
Device 2 (Leg 2)... 23.90 - (.00307)(.350)(120) = 23.77
Device 3... 23.77 - (.00307)(.200)(240) = 23.62
Device 4... 23.62 - (.00307)(.150)(90) = 23.58
Device 5... 23.58 - (.00307)(.100)(90) = 23.55
Device 5... 23.55 - (.00307)(.050)(90) = 23.54
-------
I'm designing a fire alarm system in which i'll be using a parallel circuit with a few seperate legs. See attached.
------------------------------------------------
I prefer to do point-to-point voltage drop calculations with the following formula...
V - (R)(A)(W)
Where
V = Starting voltage
R = Resistance of wire (per foot)
W = Length of wire (both conductors)
A = Cummulative current
So for instance, lets say all devices are drawing 50mA each. Device one is 30' from the source, device two is 45' from device one, and device three is 45' from device two. We typically use a 14AWG wire which has .00307 ohms of resistance per foot (per Ugly's). Starting voltage is 24vdc. Here's how I calculate the voltage drop...
"Leg 1" voltage drop calculation
-------
Device 1... 24 - (.00307)(.150)(60) = 23.97
Device 2... 23.97 - (.00307)(.100)(90) = 23.94
Device 2... 23.94 - (.00307)(.050)(90) = 23.92
-------
------------------------------------------------
For my circuit that has seperate legs (see attached), i want to calculate each leg point-to-point. I just calculated "Leg 1" above and can easily calculate "Leg 2" using the same method. My question is how do I calculate "Leg Main"? Would I use the cummulativ current for "Leg 1" and "Leg 2" like so?...
"Leg Main" voltage drop calculation
-------
Device 1 (Leg 1)... 24 - (.00307)(.500)(60) = 23.90
Device 2 (Leg 2)... 23.90 - (.00307)(.350)(120) = 23.77
Device 3... 23.77 - (.00307)(.200)(240) = 23.62
Device 4... 23.62 - (.00307)(.150)(90) = 23.58
Device 5... 23.58 - (.00307)(.100)(90) = 23.55
Device 5... 23.55 - (.00307)(.050)(90) = 23.54
-------
Nathan Scherneck, NICET II (fire alarm)
National Fire Protection
Honolulu, HI





RE: DC fire alarm voltage drop
Before you go too far, your wiring diagram does not meet NFPA or requirement for a supervised circuit. Your each "Leg" needs to be a separate notification circuit with its own end of line device. What you have is "T" tappings and is not permitted by NFPA for notification (singnalling) circuits.
Ugly (and many other basic electrical books) would have a formula for voltage drop calcs for a 2-wire circuit. Use them.
RE: DC fire alarm voltage drop
what we will be doing is running a 24VDC circuit from floor to floor in a high-rise building. we will tap this 24VDC circuit into an addressable control module (which will be the start of each leg) on each floor which will supervise the 24VDC.
Nathan Scherneck, NICET II (fire alarm)
National Fire Protection
Honolulu, HI
RE: DC fire alarm voltage drop
OK good, now that you have clarified supervision.
Your approach is correct, except that you need to account for the circuit length between the floors (legs) and from the source to the first leg (first addressable control module).