Voltage Drop For Fire Alarm Circuits

·     Code Requirement for Voltage Drop Calculations?

NFPA 72 2013 Edition Section 7.2.1 - "Where documentation is required by the enforcing authority, the following list shall represent the minimum documentation required for all fire alarm and emergency communication systems, including new systems and additions or alternations to existing systems.", Within this list, you will find #7, Battery Calculations  and #8, Voltage Drop Calculations.

Keep in mind almost all fire alarm control units are 24 volts DC.  Also note that there are a few fire alarm control panels that are 12 volt DC.  Now these panels are typically combination fire and burglar system.  Just remember that the calculations for NAC voltage drops are the same for these systems, however the cut off voltage for a 12 volt system will be roughly half that of a 24 volt system.

·        What is the reason for voltage drop calculations?

NAC voltage drop calculations are critical for determining if your notification appliances will in fact work with the provided head end equipment.  (This is of course based on the installation contractor installing the system per plans and noted wiring distances).  If you perform your NAC voltage drop calculations correctly at the time of design, you will know exactly how many remote power supplies and NAC circuits are needed as well as the wall space requirements and 120 VAC connections required by the electrical contractor.  Keep in mind that this is a requirement of NFPA 72.

·         Voltage Drop Calculation Methods

There are primarily two methods to perform a NAC voltage drop calculation.  These methods are better known as “Point to Point (PTP)” and “End of Line (EOL)”.

·       Point to Point Voltage Drop requires much more math than the “EOL” method.  However, the extra work pays off as this method is more accurate.

•          Designers generally use this method with a spreadsheet as the math can become tedious
•          This is the method typically used by panel manufacturers within their own calculation programs
•          Since it is less conservative of the “EOL” method, it allows for more devices on a circuit.
•          There are cases of a 30% difference between the PTP and EOL methods

End of Line method is the easiest, quickest calculation

•          Can be done easily by hand or with a calculator
•          Results are less accurate that provide lots of head room for future expansion

Did you know the 2021 International Fire Code now requires 120 VAC single and multiple station smoke alarms to produce a 520 Hz low frequency audible tone? 

Starting Voltage and Cut-Off Voltage

UL (Underwriters Laboratories) 864, 9^th Edition Standards for Fire Alarm Control Panels:

•          All panels must have a demonstrated 20.4 VDC panel cut off voltage.

You may be asking, “Where did they come up with 20.4 VDC on a 24 Volt system?”

It is actually quite simple.  20.4 VDC is 85% of 24 VDC. Or like we stated earlier, there are a few 12 VDC systems floating around.  In their case, the demonstrated voltage shall be 10.2 VDC.  Once again, 10.2 VDC is 85% of 12 VDC.

Now, above we mentioned a term “Cut-Off Voltage”.  All fire alarm control units (FACU) have and internal voltage drop.  The voltage at the actual NAC output terminal is always less than 20.4/10.2 volts at cut-off.

•          This amount of drop varies with every panel.  The variance can be as much as .5 VDC to 2.5 VDC.

Keep in mind that this value is not often found within the panel documentation.  I have found that the easiest way to obtain this value is to contact the panel manufacture and get it in writing.

You now may be asking yourself, “Why is it so critical that I get this value from the manufacturer and not just use the 20.4/10.2 value figured from the 85% set forth by UL 864 9^th Edition?”

In order for your NAC voltage drop calculations to be precise and as accurate as possible based on the facts and information provided, you must use the specific panel/power supply terminal cut-off value.

How to determine your NAC Voltage Drop using the End of Line method:

Step #1

Add up the total current draw for your entire notification appliance circuit.  This is based on the manufacture, type (strobe only, horn/strobe, mini horn, audible level, wall, ceiling, etc.)  Make sure to consult with the appliance documentation to get these figures.

Step #2

Add up the total wire length for the run and multiple it by 2 (if class B).  The 2 represents the number of conductors used in the run.

Step #3

Multiply the total wire length times the wire resistance value per foot for a total circuit wire resistance.  The wire resistance per foot can be found in Table 8 “Conductor Properties” in Chapter 9 of the National electrical Code.

Step #4

Using Ohm’s Law we know that Current (I) x Resistance (R) = Voltage (E).  Simply take the total current found in step #1 and multiply it by the resistance found in Step #3.  This will give you the volts dropped.

Step #5

Subtract the volts dropped from the panel/power terminal cut-off voltage to obtain the voltage that will be supplied to the last appliance on the circuit.  This value MUST exceed 16 volts.

Keep in mind that this method is not as accurate as the Point to Point method.  This method assumes that the voltage drop at each appliance will be the same when in reality, they are not.

Below is an example of an End of Line voltage drop calculation:

Diagram notes:

• ·        We will assume that the terminal cut-off voltage is .5 volts below the 20.4 VDC giving us a voltage of 19.9.
• ·         Use the wire lengths shown in the diagram
• ·         V1=85mA / V2=75mA / V3=115mA / V4=100mA
• ·         The circuit is using #12 AWG wire
• ·         Use the Table 8 of the NEC (National Electric Code) provided previously in this document

Using the Diagram and notes above, can you provide the voltage drop for this circuit using the End of Line method?  Give it a try and when you are ready move on to the next page where we will break it down for you.

End of Line Voltage Drop Calculation Break Down:

Step #1

Add up the total current for all four notification appliances within the circuit.  We know that V1 = 85 mA, V2 = 75 mA, V3 = 115 mA and V4 = 100 mA.  The total of all four of these is 375 mA.

Step #2

Add up the total wire length and multiply it by 2.  We know based on the diagram that the first section is 200 feet, the second section is 150 feet, the third section is 25 feet and the final section is 70 feet.  This totals up to 445 feet x 2 = 890 Total Feet

Step #3

We know from the Table 8 “Conductor Properties” we have a value of 1.98 Ohms/1000 feet of #12 AWG stranded uncoated wire.  To find our resistance for our circuit simply take the total wire length (890 Feet) and divide it by 1000.  This gives us a value of .89.  Now take .89 and multiply it by the 1.98 value found in the NEC Table.  (.89 x 1.98 = 1.7622 Ohms of Resistance)

Step #4

Using Ohm’s Law we know that Current (I) x Resistance (R) = Voltage (E).  Take the total current found in Step #1 (.375) and multiply it by the total found in Step #3 (1.7622).  .375 x 1.7622 = .660825 volts dropped.

Step #5

Finally we need to subtract the .660825 volts from our terminal cut-off voltage.  We know from the previous diagram and notes that we have a terminal cut-off voltage of 19.9 volts.  19.9 volts - .660825 = 19.239 Volts at the last appliance.

Point to Point Voltage Drop Calculation Break Down:

Diagram notes:

• ·         We will assume that the terminal cut-off voltage is .5 volts below the 20.4 VDC giving us a voltage of 19.9.
• ·         Use the wire lengths shown in the diagram
• ·         V1=85mA / V2=75mA / V3=115mA / V4=100mA
• ·         The circuit is using #12 AWG wire
• ·         Use the Table 8 of the NEC provided previously in this document

Calculation Breakdown:

To perform a point to point voltage drop calculation is basically the same as the End of Line method however; we are going to do a breakdown of each path/appliance.

Calculation #1

• ·         First wire run section resistance multiplied by the total current for appliance V1, V2, V3 and V4
• ·         Subtract the total from the terminal cut-off voltage to get the voltage drop for V1.

Calculation #2

• ·         Second wire run section resistance multiplied by the total current for appliances V2, V3 and V4.
• ·         Subtract the total of V1 from the terminal cut-off voltage to get the voltage drop of V2

Calculation #3

• ·         Third wire run section resistance multiplied by the total current for appliances V3 and V4.
• ·         Subtract the total of V2 from the terminal cut-off voltage to get the voltage drop of V3

Calculation #4

• ·         Fourth wire run section resistance multiplied by the total current for appliances V4.
• ·         Subtract the total of V3 from the terminal cut-off voltage to get the voltage drop of V4

If this last value is greater than 16 volts, the circuit should work.

Using the Diagram and notes above, can you provide the voltage drop for this circuit using the Point to Point method?  Give it a try and when you are ready move on to the next page where we will break it down for you.

Calculation # 1

• ·         200 Feet x 2 = 400 Feet.  400 / 1000 = .4 x 1.98 = .792 Ohms (From the FACU to V1)
• ·         .792 x .375 (Current of all Appliances) = .297 volts dropped @ V1
• ·         19.9 (Terminal Cut-Off Voltage) - .297 = 19.603 VDC @ V1

Calculation #2

• ·         150 Feet x 2 = 300 Feet.  300 / 1000 = .3 x 1.98 = .594 Ohms (From the FACU to V1)
• ·         .594 x .290 (Current of Appliances V2-V4) = .17226 volts dropped @ V2
• ·         19.603 (Terminal Cut-Off Voltage) - .17226 = 19.43074 VDC @ V2

Calculation # 3

• ·         25 Feet x 2 = 50 Feet.  50 / 1000 = .05 x 1.98 = .099 Ohms (From the FACU to V1)
• ·         .099 x .215 (Current of Appliances V3-V4) = .021285 volts dropped @ V3
• ·         19.43074 (Terminal Cut-Off Voltage) - .021285 = 19.40946 VDC @ V3

 Calculation # 4

• ·         70 Feet x 2 = 140 Feet.  140 / 1000 = .14 x 1.98 = .2772 Ohms (From the FACU to V1)
• ·         .2772 x .200 (Current of Appliance V4) = .05544 volts dropped @ V4
• ·         19.40946 (Terminal Cut-Off Voltage) - .05544 = 19.35402 VDC @ V4

Both of these calculations are commonly used and widely accepted by your local AHJs.  As you can see the PTP method came up with a total voltage drop of 19.35402 while the EOL method came up with 19.239.  Remember both of these examples used the same parameters.  I personally recommend using the Point to Point method purely based on its accuracy.

Low Frequency Sounders for Sleeping Areas

What are Low Frequency Sounders for Sleeping Areas?

Anyone in the fire alarm design and installation industry has more than likely heard the term low frequency sounders.  These newer sounders first appeared in the NFPA 72 2010 edition under section 18.4.5.3 and stated an effective date of January 1, 2014.  In short these are re-designed audible appliances for sleeping areas only!  Multiple studies have been performed on sleeping parties to see how they react in different scenarios.  Some of the scenarios involved older individuals who have a difficulty with their hearing,  others were individuals who are under the influence of narcotics.  The original thought was that these people would have a difficult time awaking from a sleeping state via the standard 3 Hz sound put out by a typical mini-horn commonly found in apartment and hotel sleeping areas.  The newer low frequency sounders have a square wave signal with 520 Hz plus or minus 10 percent.  Below are two links to show you the difference between the two sound outputs:

Listen to a 520HZ Low Frequency Tone

Listen to a 3 KHz Standard Tone 

Now that it is 2024, most of the country is currently enforcing the NFPA 72 2022 edition.  If your AHJ is enforcing the 2013 version of this standard, you have been required to provide these low frequency appliances in all sleeping areas for nearly 4 years.

Are you Installing Low Frequency Sounders Correctly?  

Now the same installation factors found in the NFPA 72 standard apply to low frequency sounders. They shall have a sound level of at least 15 dB over average ambient, 5 dB or maximum sound level (duration of at least 60 seconds) or 75 dB whichever is highest.  Now seeing that NFPA 72 2013 has been nice enough to provide us with a chart depicting average ambient sound levels for different types of occupancies, this cannot be used in lieu of actual readings taken at the site.  However we can take away from this chart that a residential occupancy has an average ambient sound level of 35 dB.  If we use this as an example and tack on the additional 15 dB per NFPA 72 2022 18.4.5.1, we only come up with 50 dB.  This is 25 dB lower than the required 75 dB per the same standard section.  An example of the only time you would be going higher than 75 db is a sleeping area with an average ambient sound level of 61 dB.  61 dB plus 15 dB = 76 dB.  In short, 99 percent of the time, a designer will be aiming to achieve 75 dB.

NFPA 72 Occupancy Average Ambient Sound Levels

Tips to keep in mind:

• If you double the power to the sounder, you will gain 3 dB
• The dB from the appliance is measure at 10' away.  Every time you double the distance from the appliance, you loose 6 dB.  Read more here.

What Exactly is a Sleeping Area?

Seems like a pretty easy question to answer right?  Well I'll have you know this comes up a lot and more often than not, designers get it wrong!  A sleeping area is obviously associated with a bedroom and a bedroom is not classified as such unless it has a closet.  However, as defined by NFPA 72 2022 A.18.4.6.3, low frequency sounders are required for use in areas intended for sleeping (bedrooms) as well as areas that might be reasonably used for sleeping (Living Rooms).  Most apartments or hotel rooms are outfitted with a comfy couch in the living room.  It's not uncommon to find yourself falling asleep in this area.  This is what has prompted the requirement for low frequency sounders to be installed in not only the bedrooms but the living rooms as well. 

How Does this Affect my Design?  

It is pretty common knowledge that these 520 Hz low frequency sounders draw a lot more current than the standard 3 Hz mini-horns.  You will need to take this into effect when calculation voltage drop and battery calculations for the remote power supplies serving the areas in question.  Depending on the size of the project, this may mean you need additional notification appliance circuits and remote power supplies. 

Fire Alarm Calculations

If you are preparing to take the NICET exam for Fire Alarm Systems, there are numerous fire alarm calculations you must understand in order to properly design a code compliant system.  These calculations can break down exact requirements for sound pressure (dB) levels, voice intelligibility, voltage drop on a circuit, back up battery sizes, candela settings and dB line loss for speaker circuits.  There are additional calculations however these are some of the most common and important so this article will concentrate on the following:

You can also download our Fire Alarm Calculation Tool here.

How to find the correct Candela Strobe to cover a given space

This is a very important measurement as it allows us to properly calculate the necessary candela power needed for a given space.  If you do not have NFPA 72 2013 edition tables 18.5.5.4.1(a) and 18.5.4.4.1(b) handy or memorized, this formula will save the day!

Take the selected candela (ex. 75 cd) and divide it by 0.0375
75 cd / 0.0375 = 2000
Now take the square root of 2000 to get a spacing of = 44.72136 feet.

If you consult NFPA 72 2013 tables 18.5.5.4.1(a) it shows a spacing of 45 x 45 feet and table 18.5.5.4.1(b) shows a spacing of 44 x 44 feet.

Voltage Drop Calculation

Step #1:  Find the total current from all of your field notification appliances.  If you consult the appliance's specification sheet, you will find the current draw for each setting.  For example, you have four appliances on the temporal high setting and they each draw 50mA or (0.050A).  If you add all four appliances together (4 x 0.050) you have a total current draw of   0.2A

Step #2:  Determine the to and from distance of the notification appliance circuit (NAC).  For this example, we will saw the NAC is 450 feet.  We have to double this distance to account for both conductors.  450 feet x 2  = 900 feet.

Step #3:  Now that we know the distance, we need to know what type of conductor we are using for the circuit.  For this example we will use a #12 AWG solid coated copper conductor.  Once this is determined, we will need to consult the Conductor Properties table 8 in the National Electrical Code or NEC 2011 (click the link to view a copy of the table).  This table can also be found in chapter 9 on page 721.   On the table, located the section at the top under coated.  Now follow the line down under ohm/kFT (ohms per 1000 feet).  Keep scrolling down until you reach the 12 AWG with a quantity of 1 since it is solid.  If you line the left and top rows up, you will see a resistance of 2.01 ohms for 1000 feet of conductor.

Step #4:  Since we do not have a distance of 1000 feet for a out circuit, we will need to break down this resistance according to our actual distance of 900 feet.  To do this simply divide 900 feet by 1000 feet sown as 900/1000 = 0.9.  Now multiply your resistance per 1000 feet (2.01) by your distance breakdown of 0.9.  2.01 X 0.9 = 1.809 ohms per 900 feet.

Step #5:  To determine the voltage at the end of the notification appliance circuit we need to use Ohm's Law.  Since we know know the total amps (0.2A) and the total resistance (1.809) we can now find the voltage.  I X R = E or Amps x Resistance = Voltage.  0.2 x 1.809 = 0.3618 volts.

Step #6:  To find the voltage drop subtract your answer found in step #5 (0.3618) from the starting voltage of 24 volts.  24 - 0.3618 = 23.6382 volts.

Step #7:  Sometimes you may be asked to know the voltage drop percentage.  To find this, take the voltage drop (0.3618 volts) divided by 24 volts and multiply it by 100.  This is shown as (0.3618/24) x 100 = 1.5075%

See more examples of voltage drop for NACs here

Resistor Calculations

Calculating Resistors in Series

If you come across multiple resistors in series with each other, simply add the resistor values.

Resistors in Series

Example:
R1 = 3.3k
R2 = 4.7k
R3 = 10k
Total Resistance = 18k

Calculating Resistors in Parallel 

Resistors in Parallel

1/Rt = 1/R1 + 1/R2 + 1/R3

R1 = 200
R2 = 400
R3 = 800

1/Rt = 1/200 + 1/400 + 1/800

If there is a common denominator for the bottom numbers use it by multiplying up both the top and bottom numbers in the fraction.

Example: the common denominator is 800.

Resistor R1 has a resistance of 200.  200 goes into 800 4 times.  Therefore R1 = 4/800
Resistor R2 has a resistance of 400.  400 goes into 800 2 times.  Therefore R2 = 2/800
Resistor R2 has a resistance of 800.  800 goes into 800 1 times.  Therefore R3 = 1/800

Now add the top numbers together (4 + 2 + 1 = 7) and place it on top of 800 like this 7/800.  Now take the reciprocal to make the fraction reverse to 800/7.  Divide 800 by 7 to get your answer of  = 114.286k.

Battery Calculations

Take the total standby current and multiply by 24 (hours for standby)
Take the total alarm current and multiply by (.083 for 5 minutes or .249 for 15 minutes of alarm)
Add the total of (standby current x 24) to (alarm current x .083 or .249)
Multiply the total of above by a safety factor of 1.2.  This gives you a 20% spare buffer.
Round up to required battery amp hour size.

dB Loss and Gain

Every time you double the distance from the audible appliance, you loose 6 dB.

Example:  If you have a speaker with 75dB at 10 feet, you will have 69dB at 20 feet and 63dB at 40 feet and so on at 80 feet, 160 feet......  Please note, these are not multiples of 10 feet!!!  These are broken down by doubling the distance from the last measurement.
Correct: 10 feet - 20 feet - 40 feet - 80 feet - 160 feet
Incorrect:  10 feet - 20 feet - 30 feet - 40 feet - 50 feet - 60 feet

If you double the power output of the appliance, you gain 3dB.

Example:  If you have a speaker tapped at a 1/4 watt with 75dB and you double the wattage to 1/2 watt, you will then have 78dB.

dB Line Loss Calculation

TLR = Total Load Resistance
TWR = Total Wire Resistance
TWR = Ohms/Foot X (Distance X 2)

12 AWG Ohm/FT is .00193
14 AWG Ohm/FT is .00307
16 AWG Ohm/FT is .00489
18 AWG Ohm/FT is .00777

TLR = (Voltage X Voltage) / Power
20 X Log (1- (TWR / TWR + TLR))

You cannot go over -1.5 dB

Voltage Drop for Fire Alarm Systems

Fire Alarm Voltage Drop Calculations

All electrical conductors include a small amount of resistance.  This resistance increases if the length of the conductor increases or or the conductor size decreases.  Think of blowing air through a hose. If the hose diameter decreases and or the length increases it would be harder to blow through.  You can also think of freeway traffic as resistance.  The freeway is the conductor.  The wider the freeway, the faster and smoother you travel.

As electrical current flows through the conductor it will experience a decrease in voltage between the source (starting point) and at various points along the conductor path.  Another example to look at is the voltage drop in a 1000 foot run of 16 AWG wire would be greater than that of a 1000 foot run of 12 AWG.  This is simply because a 16 AWG conductor is smaller in diameter than a 12 AWG conductor.

Fire alarm equipment LISTED to the standards of the National Fire Protection Association and Underwriters Laboratories (U.L.) is tested to determine if can operate properly at 85% of the rated nameplate voltage.  This limit was set in place to make sure the circuit can deal with a "brownout" condition or a possible voltage drop which might result from excessive resistance in the system wiring.

As required in the CFC (California Fire Code), fire alarm designers are required to prepare voltage drop calculations for the notification appliance circuits (NAC) as part of the design.  These voltage drop calculations must be included in the submittal plans and specifications.  This is to assure that the devices on the system are supplied with electrical power within the operating voltage range.

You as a designer can use several different methods to calculate voltage drop on a fire alarm circuit. One method calculates the actual voltage drop for each length of cable and device within the circuit and the other calculates the overall voltage drop.  Either method will have slightly different results but should be acceptable by your local AHJ (authority having jurisdiction)

The suggested maximum allowable voltage drop on a fire alarm circuit is 10% or the voltage drop included in the fire alarm control panel installation guide, whichever is less.

"Lump Sum Method"

Step #1)  Take the total current of the circuit.  You can achieve this figure by adding up the current draw of each device on the circuit.  This will represent "A"

Step #2)  Measure out the length of the circuit in feet.  Do not double the distance of the circuit for 2 wire loops unless you want to use a multiplying factor of 10.8 versus 21.6 (see step #3).  This will represent "L"

Step #3)  Use a multiplying factor of 21.6.  This number represents the resistivity of copper conductors.  This is a constant used in the formula.

Step #4)  Find the Circular Mils for the particular gauge wire you are using.  This can be found in the National Electrical Code (NEC) chapter 9 table 8.  #14 AWG is 4110 and #12 AWG is 6530.  This will represent "C.M".

A x L x 21.6
-------------  =   VD
C.M.

Example

.356 x 450' x 21.6
-------------------   =   0.530 Volts Dropped
6530

To find the percentage of voltage dropped do the following:
0.530 / 24 = 0.022
0.022 x 100 = 2.2
= 2.2% Voltage Drop

Now remember you can also perform this calculation for each individual length of wire and device on the circuit.  This is known as the "point-to-point" method.  This is a better way to perform the calculation as it gives you a chance to really break down the circuit and pin point exactly where a circuit must end do to voltage drop.  Simply use the above formula for each wire run and add the voltage drop totals for each circuit section together for the total voltage drop.  Then divide by the source voltage (in this example we will use 24VDC) and then multiply by 100 to come to a total voltage drop percentage.

I will be adding info in the near future on the calculations for the "point-to-point" method using Ohm's Law

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