Mag Lock on the Only Stairwell Door
You are retrofitting access control on the only stairwell door serving the fifth floor of a six-story office building. The owner wants a 1,200 lb mag lock controlled only by a card reader, with no request-to-exit device, to prevent tailgating; the door has rim panic hardware. The plan reviewer cites free egress and will not approve the permit without a listed release. The reader is a 12 VDC unit drawing 250 mA, wired with 22 AWG at 310 ft one-way; 22 AWG copper is 16.14 ohms per 1,000 ft. The power supply is a 12 VDC, 4 A unit at the IDF, and the reader's minimum operating voltage is 10.5 V. The GC wants to reuse the existing 22 AWG cable.
1. What egress features must be installed, and can the owner's card-reader-only design be approved?
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Free egress cannot be compromised. A mag lock on an egress door is acceptable only when it releases upon (a) activation of the fire alarm, (b) loss of power (fail-safe), and (c) a manual release, and it must be installed with a listed sensor release or request-to-exit device that detects a person approaching and releases the lock. The existing rim panic hardware must remain functional, and the lock must not rely on the card reader for egress. The plan reviewer is correct to withhold the permit.
I would not install card-only control. I would propose: fail-safe mag lock, listed REX/exit sensor on the egress side, door position switch, fire alarm relay that removes power, manual release, and local alarm if the door is held open. Because this is the only stairwell serving the floor, I would also confirm with the AHJ whether delayed egress is permitted; it is generally not the right solution for a stairwell, and I would not use it without written approval.
2. Calculate the reader voltage at the end of the existing 22 AWG cable and recommend a fix.
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Conductor length = 2 × 310 ft = 620 ft. R = 620 × 16.14/1,000 = 10.01 Ω. Drop = 0.250 A × 10.01 Ω = 2.50 V. Reader voltage = 12.0 − 2.50 = 9.50 V, below the 10.5 V minimum - the existing 22 AWG fails.
Remedies: use 18 AWG (6.385 Ω/1,000 ft): R = 3.96 Ω, drop = 0.99 V, reader sees 11.01 V - acceptable but with only about 0.5 V margin. 16 AWG gives 2.49 Ω, drop 0.62 V, 11.38 V. Alternatively, power the reader locally with a listed 12 V supply and use the 22 AWG only for data, or use a 24 V reader/PoE. I would also account for connector and splice losses and the supply's actual output under load, and choose 16 AWG or local power for margin rather than designing right at 11.0 V.
3. The fire alarm contractor wired its relay to the access-control panel with a normally open contact, but the access panel expects a normally closed supervised input. How do you resolve this and prove it works?
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The interface must be fail-safe and supervised. With a normally open contact, the access-control panel may see a "normal" condition until the alarm operates, and the circuit is not supervised as the access-control manufacturer requires. I would not fix this by inverting software only. The correct approach: use a listed fire alarm relay with the contact type required by the access-control manufacturer (typically NC, supervised with an end-of-line resistor), wire it so alarm activation removes power to the mag lock, and test end-to-end with the AHJ: alarm → lock releases → door opens; restore → lock re-secures; trouble → panel indicates. Document the interface, update as-builts, and have the fire alarm contractor and access-control programmer sign the same test sheet. If the correct relay was not in the original scope, this is also a change order.