Access-Control Strike on a Long 18 AWG Class 2 Circuit
An access-control retrofit in an Albuquerque office building uses a listed 24 VDC Class 2 power supply at the IDF to feed a 2.0 A electric strike at the main entrance. The one-way circuit length is 190 ft of 18 AWG stranded copper, run through ceiling space and down a wall to the door. The strike datasheet lists an operating range of 21–28 VDC and a 2.0 A holding current; the installer measured 23.9 V at the supply terminals under load with a true-RMS meter. At the door, the strike chatters and sometimes fails to release. The helper suggests turning up the supply, but it is a fixed 24 V listed Class 2 unit. The building manager wants the door working before Monday and asks whether larger conductor is really necessary. The circuit is not bundled with other current-carrying cables, and the ambient temperature at the ceiling is 32°C.
1. Calculate the voltage at the strike using the copper circular-mil formula and state whether the installation will operate reliably. Show each step.
Read the worked answer
Use the standard copper formula:
VD = 2 × K × I × L ÷ CM
- K = 12.9 for copper
- I = 2.0 A
- L = 190 ft (one-way)
- CM for 18 AWG = 1,620 circular mils
The factor 2 accounts for the outgoing and return conductors.
VD = 2 × 12.9 × 2.0 × 190 ÷ 1,620 VD = 9,804 ÷ 1,620 = 6.05 V
At the supply measured 23.9 V under load, the strike sees approximately:
23.9 − 6.05 = 17.85 V
That is below the 21 V minimum on the datasheet, so the chatter is explained by voltage drop, not by a defective strike or a weak supply. The supply is already delivering nearly 24 V at its terminals; raising it is not an option with a fixed listed unit. The circuit is functionally inadequate even though Article 725 does not impose a general voltage-drop limit on Class 2 circuits. The installation must be corrected to meet the listed product's operating range.
2. What is the minimum standard copper conductor size that restores at least 21 V at the door, and what non-code factors should influence the final choice?
Read the worked answer
Test the next standard sizes with the same formula, holding current and length constant:
- 16 AWG (2,580 CM): VD = 9,804 ÷ 2,580 = 3.80 V; 24 − 3.80 = 20.2 V - still below 21 V, so 16 AWG is not enough.
- 14 AWG (4,110 CM): VD = 9,804 ÷ 4,110 = 2.39 V; 24 − 2.39 = 21.6 V - meets the minimum, but with only about 0.6 V of margin.
- 12 AWG (6,530 CM): VD = 9,804 ÷ 6,530 = 1.50 V; 24 − 1.50 = 22.5 V - comfortable margin.
14 AWG is the calculated minimum; 12 AWG is the better design choice because the supply measured 23.9 V rather than 24 V, terminals and splices add resistance, and strike inrush and aging reduce margin. Non-code factors include terminal landing size at the strike and supply, cable fill in the raceway, cost and difficulty of repulling, and whether the door will later receive a second lock or reader. Confirm the fixed supply's Class 2 output rating and listing remain valid; changing the conductor does not change the Class 2 classification. If the circuit were reclassified as Class 1, Article 725 Part II installation rules would apply instead.
3. The helper proposes paralleling the existing 18 AWG conductor with a second 18 AWG conductor to avoid repulling. Evaluate that proposal technically and under the NEC.
Read the worked answer
Technically, two 18 AWG conductors in parallel give 3,240 circular mils, equivalent to roughly 15 AWG. VD = 9,804 ÷ 3,240 = 3.03 V, so the strike would see about 20.9 V at 24 V supply, or about 20.8 V at the measured 23.9 V. That is still at or below the 21 V minimum, so the proposal does not reliably solve the problem even before code is considered.
Under the NEC, 310.10(H) permits conductors to be paralleled only in sizes 1/0 AWG and larger, with limited exceptions that do not apply to a 2 A Class 2 circuit. Paralleling small Class 2 conductors also creates termination and identification problems, and the second conductor is unlikely to be listed and identified as part of the original circuit. The correct fix is to replace the run with a single 14 AWG or 12 AWG conductor pair, or to relocate the power supply closer to the door if that is practical. Document the calculation and the as-built conductor size for the owner.