Calculate Voltage Drop in Fire Alarm Systems

Calculate Voltage Drop for Fire Alarm Systems

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.

For example, let’s say we have a 300-foot NAC with 10 Model GE3-24 visible-audible signaling appliances made by Gentex.

Step 1: Find the total per unit current drawn by each device on the circuit. According to the specification sheet, each strobe draws 60mA (0.060A) and each horn (Temporal 3, high setting) draws 28mA (0.028A). Adding the two together we have a total of 88mA (0.088A) per unit.

Step 2: Total the current drawn by each device on the NAC to find total current. In this case, 10 units on a single NAC gives us a total current draw of 10 X 0.088A = 0.88A. Ideally we want 10 percent above this amount for headroom, or 0.88A X 0.10 = 0.088A + 0.88A = 0.968A total current.

Step 3: Determine the to-and-from distance of the circuit. In this case, we have a 300-foot run, or 300 feet X 2 (conductors) = 600 feet total.

Step 4: Utilize a conductor properties chart to determine the resistance of the cable at the end of the circuit. An 18 AWG conductor, for example, has a resistance of 7.77 Ohms per 1,000 feet, so divide 600 by 1,000 and multiply that by 7.77 to find the actual resistance: 600/1,000 = 0.6 X 7.77 = 4.662 Ohms. A 16 AWG conductor has a listed resistance of 5.08 Ohms at 1,000 feet. Again, do the math by multiplying 5.08 by 0.6 = 3.048 Ohms.

Step 5: Determine the voltage at the end of the NAC. To do this, we use simple Ohms Law: I X R = E. We know the total current in our circuit is 0.968A (see Step 2) and we know that the actual resistance of our circuit to be 4.662 Ohms using 18 AWG and 3.048 Ohms using 16 AWG wire. Doing the math: 0.968A X 4.662 Ohms = 4.513 Volts using 18 AWG, and 0.968 X 3.048 Ohms = 2.95 volts for 16 AWG.

Step 6: Determine the actual voltage drop at the end of the NAC. To do this, begin by subtracting the voltages in Step 5 from the operating voltage at the head of the NAC: 24V - 4.513V = 19.487V for 18 AWG, and 24V - 2.95V = 21.05V for 16 AWG.

Step 7: Determine which cable size to use based on actual voltage at the end of the circuit. Use the manufacturer’s specification sheet to find out the lowest and highest voltage allowable for proper operation. In this case, our GE3-24 visible-audible signaling appliances  have an operating voltage range of 16VDC to 33VDC. As you can see, using the voltages derived in Step 6, either gauge size will work — but clearly 16 AWG would be the better choice.