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

How to Locate Ground Faults Fire Alarm System

What is the worst service call you can possibly receive?  In my mind it is the infamous ground fault on a fire alarm system.  These can either be a quick fix (pending it's locked in and you are familiar with the circuit pathways) or a very time consuming correction.  If the ground fault is minuscule due to compromised wire insulation, water or vibration causing the ground fault to swing in and out, you could potentially be looking at a long day.

Note there are two types of ground faults.  

• An absolute short to ground is caused by a portion of the conductor touching a solid earth ground.  This is the case with pinched our cut insulation, wires stripped back to far, strands of your conductor poking through tape, etc.  In all these cases, the actual copper of the conductor is touching physical ground.  

• The other form of ground fault is commonly known as a "soft ground fault".  These faults are typically caused by moisture or compromised insulation where the copper is not necessarily touching ground however the insulation/voltage threshold is lowered and can be the cause of a ground fault.  Water can also be the cause by penetrating the insulation through hairline fractures and bridging the gap between the copper and ground.

How ground faults are determined by control panels and multi-meters

A fire alarm control panel uses an internal 12-24 volts DC to seek out ground faults on all connected circuits including power, SLC, IDC and NAC.  Most ohmmeters used for troubleshooting fire alarms systems have an internal battery with an output voltage of 1.5 - 3 volts DC (In the picture below, you will notice the analog ohmmeter I am using for this article puts out 1.628 volts DC).  In cases where the insulation/voltage threshold is below 12 volts but above 3 volts, the ohmmeter will not detect the ground fault.  If the ground fault does register on the ohmmeter, it will show some conductance to ground where a clean circuit will show infinite ohms or an open circuit.  With a soft ground fault present on the conductor, the ohmmeter will show a reading in the range of high K-ohms (K) or Meg-ohms (M).  Note the fact that these readings can be unstable and hard to decipher if the technician is not completely seasoned and experienced with their multi-meter.

The solution to quickly locating a soft ground fault on a fire alarm system.

In order to locate a soft ground fault as easily as the fire alarm control panel, we will need a ohmmeter that has a slightly higher output voltage than the factory range of 1.5 - 3 volts DC.  This higher voltage output will allow you to locate the source of the ground fault more quickly and accurately saving you a great deal of time in the field.  Once you know what conductor the ground fault is on, you can use the industry standard of breaking circuits in half until the fault is isolate to its source.
   

How to build your own high voltage output ground fault detection meter. 

Below I have outlined step-by-step instructions on how to build your own ground fault meter using materials found at any hardware store for around or below $30.00.

Above is a picture of all the necessary equipment to build a very simple to operate fire alarm ground fault tester.  The equipment and links to purchase in its entirety is as follows:

• 1 - Analog ohmmeter (roughly $21.00)
• 4 - 9 volt Batteries (roughly $7.00)
• 4 - 9 volt Battery Connectors with Flying Leads ($.49)
• 1 - Set of Ohmmeter Test Leads ($8.00)
• 1 - Heat Shrink Tubing (< $1.00)
• 1 - Limiting Resistor (< $1.00)
• 1 - Set of Velcro Straps ($5.00)

Equipment you should already have in your arsenal

• Digital Multi-Meter
• Soldering Iron
• Electrical Tape
• Wire Strippers

Lets get started with the build!

Step #1:

Use your digital multi-meter (set to read current) to take a current draw of your analog ohmmeter.  To be more detailed, you need to find out the current used by the analog ohmmeter to read exactly zero Ohms or a dead short.  Move your positive test lead on the digital multi-meter to AMPS and turn the selection dial to DC Amps.  Now place the two test leads of the digital multi-meter into the test lead ports of the analog ohmmeter.  When done correctly as pictured below, your analog ohmmeter should show zero Ohms and your digital multi-meter will indicate the amps needed to show true zero.  In the picture below the digital multi-meter shows a reading of 0.016 Amps.  Remember this number for determining the limiting resistor value in step #3.

Step #2:

In order to achieve the higher output on our new analog ohmmeter, we will be using the four 9 volt batteries in series. Remember when batteries are connected in series the voltage doubles and the Ah remains the same.  So in the case of four 9 volt batteries in series, we will achieve 36 volts DC.  Start out by stripping and connecting the four 9 volt battery connectors with flying leads in series.  Once connected, I highly recommend a solid bond via solder finished off with a clean heat shrink tubing.  See the below pictures for examples.

Step #3:

Now that we know what amount of voltage we have with our four 9 volt batteries in series (36 volts) as well as the current used to read absolute zero on the analog ohmmeter (found in step #1), we can use Ohm's Law to determine our limiting resistor value.  Ohm's Law states that E (volts) / I (current) = R (Resistance).  In this case 36 volts / 0.016 amps = 2,250 or 2.25K.  For the purpose of this article, I will be using a 2.2K resistor.  When you attempt this project, you may need to slightly increase/decrease the value of your limiting resistor in order to read absolute zero.  Remember, lower value resistors will move your analog ohmmeter needle closer to the right (0 ohms). 

Step #4:

We will now need to insert this limiting resistor in series to the negative test lead of our analog ohmmeter.  I recommend cutting the negative test lead 5 - 7 inches away from its meter connection side.  Once the lead is cut, splice the resistor in series, solder the connections and finish off with heat shrink tubing.  IMPORTANT, make sure the limiting resistor is placed between the analog ohmmeter and the 36 volt battery connectors.  See below picture.

Step #5:

This next step is preference as you can clean up the wire connections anyway you see fit.  For this article, I just taped them up.  Once you get your ground fault meter built, you will want to play with some different ideas that will insure longevity and ease of use in the field.  With the negative test lead disconnected from the analog ohmmeter, connect all four of your 9 volt batteries to the connectors.  Once connected, use your digital multi-meter to verify 36 volts insuring your connections are correct.

Step #6:

Now that you have verified the correct output voltage, connect the negative test lead to the analog ohmmeter.  With the ohmmeter set to Ohms, short the test leads together.  You should get absolute zero.  In the picture below, I have the 36 volt battery pack Velcro strapped to the back of the analog ohmmeter.  You will see both test leads are shorted and the ohmmeter is reading absolute zero.

The picture below shows the new output voltage once your fire alarm ground fault tester is complete.  If you remember from earlier in the article the original analog ohmmeter put out 1.628 volts DC whereas the new setup is showing 39.53 volts DC.

This 39.53 volts is a combination of the 36 volt battery pack as well as the internal battery within the analog ohmmeter.  This higher voltage will make locating a ground fault a lot easier as well as increase your accuracy tenfold.  Also note the output voltage is still low enough to not damage devices and appliances on fire alarm circuits.  Make sure when using this tester, you disconnect all circuits from control boards.  To locate a ground fault, look for continuity to ground on each circuit.  This new analog ohmmeter will show infinite ohms when the conductor is not exposed to an earth ground.  If the insulation is compromised and/or the copper is directly connected to ground the ohmmeter will indicate 0 ohms.  If a soft ground fault is present, the meter will fluctuate from infinite (left) towards zero (right).  This fine movement should allow you to track down circuits with faults in half the time.

Good luck with your build and be sure to let us know if this new fire alarm circuit tester helps you in your daily troubleshooting.

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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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