Most access control specifications carry two power requirements. The supply shall be sized for the connected load. The system shall operate on battery for a stated standby period, often four hours.
Both read like one line in the bill of materials: pick a supply with enough amps and a battery that fits the cabinet. Then a scheduled unlock trips the supply at 7:00 a.m., or a power outage leaves the magnetic locks at the main entrance unlocked in the second hour of a four-hour requirement.
A supply and its battery are answering two different questions, and the answers come from different loads.
1. Sort every device by when it draws
Every device on the supply belongs in one of two groups.
Continuous loads draw current all the time: card readers, request-to-exit sensors, the controller itself, and fail-safe locks. A magnetic lock holds the door with current, so it draws for as long as the door is locked, which is most of the day.
Energized loads draw only when they operate. A fail-secure electric strike stays locked with no power and pulls current only while it is unlocked.
The split matters because the two groups drive different numbers. The supply has to carry both at once. The battery, in an outage, mostly carries the first group.
2. Find the worst minute
The supply rating is set by the moment when everything continuous is drawing and the largest number of energized locks are unlocked together. That moment is rarely one person badging through one door. It is the scheduled unlock at the start of the day, a global unlock command, or a lockdown override that moves every door at once.
If any of those can happen, plan for every energized lock at the same time. Then add headroom. A supply run near its full rating has nothing left for inrush when solenoid strikes operate, ages faster, and leaves no room for the door someone adds next year. Keeping the worst case at or below about 80% of the rating, roughly 25% headroom over the load, is ordinary practice.
Check one more line on the datasheet: many supplies share their rating between the load and the battery charger. If the stated output is before charging current, the load gets less than the number on the label.
3. Size the battery for the load that stays on
During an outage the strikes are usually locked and drawing nothing. The magnetic locks, readers, sensors and controller keep drawing for the whole standby period. That continuous load, multiplied by the required hours, is the battery.
Then derate. A sealed lead-acid battery delivers its rated capacity when it is new, warm and discharged slowly. It will not deliver it at four years old in an unconditioned closet. A factor of 1.20 to 1.25 is standard practice, and a heavy four-hour discharge pulls less from a lead-acid battery than its 20-hour rating suggests, so treat the result as a floor, not a target.
Round up to the next standard battery size, never down, and avoid more than two strings in parallel. Parallel strings age unevenly, and one weak battery drags the bank down without telling anyone.
4. Check the voltage at the farthest door
A supply with enough current can still deliver too few volts. Current flows out and back on the pair, so the resistance that counts is twice the run length.
A 500 mA magnetic lock 150 feet from a 12 VDC supply, on 18 AWG copper at about 6.4 Ω per 1,000 feet, loses about 0.96 V on the cable and sees about 11.0 V at the lock. That is before the battery takes over. Near the end of a long discharge a 12 V lead-acid battery sits well below its float voltage, and the lock at the end of the longest run sees the least of what is left.
Compare the result with the minimum operating voltage on each device’s datasheet, at end-of-discharge, not at float. Where it is tight, the fixes are heavier wire, a shorter run, or 24 VDC. Most locks draw about half the current at 24 VDC, which reduces voltage drop and lowers the current the supply has to carry.
Putting it together
Take a ten-door set at 12 VDC: four fail-safe magnetic locks at 500 mA, six fail-secure strikes at 450 mA, and ten readers and ten request-to-exit sensors at 150 mA and 30 mA.
The continuous load is 3.8 A: 2.0 A of magnetic locks, 1.5 A of readers and 0.3 A of sensors. The strikes add 2.7 A, but only while they are unlocked. If a schedule can release all six together, the worst minute is 6.5 A. With 25% headroom the supply needs 8.1 A, so a 10 A supply, which carries the worst case at 65% of its rating.
The battery sees only the 3.8 A continuous load. Four hours at 3.8 A, derated by 1.25, is 19 Ah, and the next standard size up is 26 Ah. These device currents are plausible examples, not values for any product; replace them with the datasheet figures for the hardware you are installing. Add the controller’s own draw too, which is the load most often left off.
All of these numbers come from three tools in our ESS Toolbox: the lock power budget calculator for the supply, the battery standby calculator for the battery, and the voltage drop calculator for the farthest door. Each shows its formula and every assumption it makes.
Then prove it
Life safety comes first. Fail-safe locking on egress doors, fire alarm release and delayed egress are governed by NFPA 101, the building code and the authority having jurisdiction, not by a power calculation, so settle what each door is allowed to do before you size anything.
At commissioning, test both requirements the way they will fail. Run the scheduled unlock and read the supply’s output current while every strike is open. Then pull AC power, let the system run on battery, and read the voltage at the farthest lock at the start and near the end of the standby period. Record both readings. The battery test takes an afternoon and is the only evidence that the four-hour line in the specification is true.