Try a full lesson: Power supply and loading

Why this task matters at Level II

At Level I you install what the drawings show. At Level II you are expected to check that the design actually works: that the batteries will carry the system through an outage, that each notification circuit is not overloaded, and that the last strobe on the circuit still gets enough voltage to flash. NICET puts this under task 2.3.4, inside the submittal and layout domain, which is 20 to 30 percent of the Level II exam.

You will not be asked to recite code text. You will be asked to run the numbers and pick the right result, often with the calculator built into the test software. This lesson walks through the three calculations you need, each with a worked example you can repeat with your own figures.

Keep your own copy of NFPA 72 open while you work. We cite sections by number so you can find them; we never reproduce the code's wording.

1. Secondary power: sizing the standby battery

When normal power fails, the batteries have to run the system in standby for a set time and then still power a full alarm. For a typical fire alarm system that means 24 hours of standby followed by 5 minutes of alarm. An emergency voice/alarm communications system needs 15 minutes of alarm at full load instead of 5. Find these in NFPA 72-2022 Chapter 10, under the secondary power supply requirements (section 10.6.7).

The method is the same every time:

1. Add up the standby current: the panel's own standby draw plus every device that draws current while the system is quiet (detectors, modules, annunciators).
2. Add up the alarm current: the panel's alarm draw plus everything that turns on in alarm, which is mostly your notification appliances.
3. Multiply each by its time in hours. Standby is 24 h. Five minutes of alarm is 5/60 = 0.0833 h; fifteen minutes is 0.25 h.
4. Add the two amp-hour figures.
5. Add the capacity margin. The margin comes from the code edition your AHJ enforces (Chapter 10) and the panel manufacturer's worksheet, so take it from those. This lesson uses 20 percent as an example input, not as the code figure.
6. Pick the next standard battery size up, never down.

Worked example

A small office building has a conventional panel and one NAC power supply. From the data sheets:

• Panel standby 0.120 A, panel alarm 0.250 A
• 28 smoke detectors at 0.00009 A each in standby (0.0025 A in total)
• Notification appliances draw nothing in standby and 1.80 A in alarm

Standby current: 0.120 + 0.0025 = 0.1225 A

Alarm current: 0.250 + 1.80 = 2.05 A

Standby amp-hours: 0.1225 A × 24 h = 2.94 Ah

Alarm amp-hours: 2.05 A × 0.0833 h = 0.171 Ah

Total: 2.94 + 0.171 = 3.11 Ah

With a 20 percent margin: 3.11 × 1.2 = 3.73 Ah

The next common size up is a pair of 12 V, 4.5 Ah batteries (wired in series for 24 V). Some designers would choose 7 Ah for future additions, which is fine; choosing 3.5 Ah is not.

What the exam tests here. The common wrong answers come from forgetting to convert 5 minutes to hours (using 5 instead of 0.0833 inflates the alarm term about 60 times), adding the margin to only one of the two terms, or rounding down to a battery that is "close enough".

2. Notification circuit loading

Each notification appliance circuit (NAC) has a current rating, and so does the power supply that feeds it. Before you add appliances, you need to know the circuit can carry them.

1. Take each appliance's current at the setting you will use. Strobe current changes with the candela setting, so a 75 cd strobe draws more than a 15 cd one. Use the manufacturer's figure for the voltage the circuit will run at; many data sheets list a higher current at the low end of the voltage range, and that is the one to use.
2. Add them up for the circuit.
3. Compare the total with the NAC's rating and with whatever is left on the power supply after its other circuits.

Worked example

A NAC is rated 2.5 A. The plan puts these appliances on it (example values, not from a real data sheet):

• 8 horn-strobes at 15 cd, 0.070 A each = 0.560 A
• 4 horn-strobes at 75 cd, 0.165 A each = 0.660 A
• 2 strobes at 110 cd, 0.210 A each = 0.420 A

Circuit total: 0.560 + 0.660 + 0.420 = 1.64 A

That is within the 2.5 A rating with 0.86 A to spare, about 34 percent headroom. If the power supply has two other NACs drawing 1.9 A and 2.1 A and is rated 6.0 A in total, the supply carries 1.64 + 1.9 + 2.1 = 5.64 A, which is within its rating but leaves only 0.36 A. Note that on the submittal: one more circuit of strobes will need another power supply.

3. Voltage drop on a notification circuit

Appliances are listed to work down to a minimum voltage, often 16 V for 24 V regulated appliances. The circuit starts at the lowest voltage the power supply delivers at the end of its battery life (the panel manual gives this; 20.4 V is common for 24 V systems), and the wire eats into that voltage on the way out and back.

1. Find the wire's resistance per 1,000 ft. For solid uncoated copper, NFPA 70 Chapter 9 Table 8 gives about 3.07 ohms per 1,000 ft for 14 AWG and about 1.93 ohms per 1,000 ft for 12 AWG.
2. Use the loop length: current goes out and comes back, so double the one-way distance.
3. Resistance of the run = (ohms per 1,000 ft) × (loop length ÷ 1,000).
4. Voltage drop = current × resistance (Ohm's law).
5. Voltage at the last appliance = starting voltage minus drop. It has to be at or above the appliance's listed minimum.

Treating the whole load as if it sat at the far end of the circuit (a "lumped" load) overstates the drop, so if the circuit passes that way it passes. Real designs often split the load into segments for a more accurate figure.

Worked example: does the circuit pass?

Take the 1.64 A circuit above, wired in 14 AWG solid copper, with the last appliance 260 ft from the power supply.

Loop length: 2 × 260 = 520 ft

Resistance: 3.07 × (520 ÷ 1,000) = 1.596 ohms

Voltage drop: 1.64 A × 1.596 ohms = 2.62 V

Voltage at the last appliance: 20.4 - 2.62 = 17.78 V

17.78 V is above the 16 V minimum, so the circuit passes, with 1.78 V to spare.

Worked example: how far can the circuit run?

Turn the question around. With the same 1.64 A load, how long can the 14 AWG circuit be?

Allowed drop: 20.4 - 16.0 = 4.4 V

Maximum resistance: 4.4 V ÷ 1.64 A = 2.68 ohms

Maximum loop length: 2.68 ÷ 3.07 × 1,000 = 873 ft

Maximum one-way length: 873 ÷ 2 = about 437 ft

Moving up to 12 AWG: 2.68 ÷ 1.93 × 1,000 = 1,390 ft of loop, about 695 ft one way. That is the usual fix when a long corridor fails the check, along with splitting the circuit or adding a NAC power supply closer to the load.

What the exam tests here. The classic mistakes are using the one-way length instead of the loop, starting from 24 V instead of the end-of-battery voltage, and using the 15 cd current for a strobe set to a higher candela.

Check yourself

The free practice set below has questions on these three calculations, written in the style of the exam, with a full explanation for every answer.

Where to look in your own books

• NFPA 72-2022 Chapter 10: power supplies, secondary power capacity (10.6.7) and battery requirements
• NFPA 72-2022 Chapter 18: notification appliances, including how appliances are listed and installed
• NFPA 70 (2020) Chapter 9, Table 8: conductor properties
• The panel and power supply installation manuals: end-of-battery voltage, NAC ratings and battery calculation worksheets