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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