6.2 Standby and Alarm Load Logic
Key Takeaways
- Standby (quiescent) current answers a different question than alarm current and is multiplied by 24 hours.
- Alarm current is multiplied by the alarm duration (5 min = 0.0833 h, or 15 min = 0.25 h).
- Only loads served by the supply under review belong in its calculation.
- Mixing milliamps and amps is the most common error and shifts the answer by a factor of 1000.
Standby and Alarm Load Logic
The most common fire alarm power calculation has a simple concept and a subtle execution: the system must run for a long standby period after losing primary power, then still deliver a short burst of full alarm. The challenge is assigning each load to the correct condition and keeping units consistent.
Standby (quiescent) current is what the system draws while it monitors inputs, supervises pathways, and sits in normal condition. Alarm current is what it draws when alarm functions operate — notification appliances energized, relays driven, control modules active. Some devices draw the same current in both states, some draw more in alarm, and some operate only during alarm. The NFPA 72 secondary-power rule pairs the standby current with 24 hours and the alarm current with the alarm duration (5 minutes general, 15 minutes voice/ECS).
| Load type | Question to ask | Typical examples |
|---|---|---|
| Standby current | What draws current while normal on battery? | FACU CPU, SLC modules, supervised detectors, comm path |
| Alarm current | What energizes during the alarm condition? | Horns, strobes, horn/strobes, relay/control modules |
| Auxiliary current | What non-alarm load is on the FA supply? | Listed door holders, interface modules |
| Excluded current | What is NOT on this supply? | Loads on a separate listed booster supply |
Convert durations to hours before multiplying: 5 minutes = 5 / 60 = 0.0833 h, and 15 minutes = 0.25 h. The standby leg is always 24 h. So for any supply, standby amp-hours = (total standby amps) × 24, and alarm amp-hours = (total alarm amps) × 0.0833 (or × 0.25 for voice/ECS).
Building the Two-Column Schedule (Worked Setup)
Before touching the calculator, build two columns. Place each device's standby current in the standby column if it is powered in normal monitoring, and its alarm current in the alarm column if it operates during the described alarm. Multiply by quantity only after the device is in the right column.
Worked setup (numbers carried into 6.3):
| Load | Qty | Standby (A) | Alarm (A) | Standby total | Alarm total |
|---|---|---|---|---|---|
| FACU control unit | 1 | 0.250 | 0.350 | 0.250 | 0.350 |
| SLC / addressable modules | 1 | 0.090 | 0.120 | 0.090 | 0.120 |
| Smoke detectors (0.0003 A ea.) | 50 | 0.0003 | 0.0005 | 0.015 | 0.025 |
| Horn/strobes @ 75 cd (0.090 A ea.) | 20 | 0.000 | 0.090 | 0.000 | 1.800 |
| Totals | 0.355 A | 2.295 A |
Notice the horn/strobes contribute zero standby current (they are silent in normal condition) but dominate the alarm column. That asymmetry is the whole point of separating the legs.
Unit discipline: a detector standby of 0.3 mA is 0.0003 A, and a horn/strobe at 90 mA is 0.090 A. Keep everything in amps, or keep everything in milliamps and convert once at the end. A single decimal shift makes a battery look ten times too big or too small.
Exam trap: adding standby and alarm current together first, then multiplying the combined number by 24 hours. That charges the system for full alarm across the entire standby day and overstates the battery badly. Second trap: placing notification appliances in the standby column. Unless the device data says a strobe draws standby current, it belongs only in the alarm column.
NICET's Level II outline places power supply and loading requirements in the submittal/layout domain, so a question may ask what is missing from a load schedule — standby current, alarm current, quantity, supply capacity, battery size, or voltage-drop basis — not only what the final number is. Treat a complete load schedule as a deliverable, and learn to spot the gap.
Where the Current Values Come From
Device currents are never guessed — they come from the manufacturer's installation/data sheet for the listed model, and an open-book exam expects you to read the right value from a provided cut sheet. Three rules govern which number to pull:
- Match the operating condition. A data sheet usually lists a standby (supervisory) current and an alarm current. Detectors and modules have a small standby draw; horns and strobes typically show 0 standby and a defined alarm current.
- Match the setting. Strobe current is published per candela tap, and speaker current depends on the wattage tap. Reading the wrong row is the single most common loading error.
- Match the source voltage range. Some appliances list different currents for regulated 24 V versus full-wave-rectified/filtered operation; use the row that matches the panel/booster output type.
A worked unit sanity check: if a horn/strobe data sheet lists alarm current as 90 mA and you have 20 of them, the alarm contribution is 20 × 0.090 A = 1.80 A, not 20 × 90 = 1800 (a milliamp/amp slip) and not 20 × 0.90 = 18 A (a decimal slip). When answer choices differ by a factor of 10 or 1000, suspect a unit error before anything else.
Finally, remember the 24-hour standby leg almost always dominates the battery result because it runs 288 times longer than a 5-minute alarm. A large alarm current (amps of strobes) multiplied by 0.0833 h often adds only a fraction of an amp-hour, while a modest standby current multiplied by 24 h drives the total. Knowing which leg dominates lets you sanity-check a battery answer at a glance: if the standby leg and the final number are wildly different, recheck the durations.
How is a 5-minute alarm duration expressed in hours for an amp-hour calculation?
A smoke detector draws 0.3 mA in standby. What is that in amps?
Which approach is the classic standby/alarm load trap?