Ohm's Law and Voltage Drop in a Long Access Control Circuit
An access control contractor is installing a new card reader and electric strike on a remote door. The access control panel is located 350 feet away from the door in a telecom room. The panel provides a 12 VDC output for the reader and a separate 12 VDC output for the strike. The reader requires 150 mA at 12 VDC, and the strike requires 350 mA at 12 VDC. The contractor plans to use 18 AWG copper conductors for both circuits. The resistance of 18 AWG copper is approximately 6.5 ohms per 1000 feet. The panel's output voltage is regulated at 12 VDC, but it can tolerate a voltage drop of up to 10% at the device. The contractor has a copy of Ohm's law and the voltage drop formula: Vdrop = (2 × L × R × I) / 1000, where L is one-way length in feet, R is resistance in ohms per 1000 feet, and I is current in amperes. The contractor is also considering using a 24 VDC power supply and a voltage converter at the door, but the panel only outputs 12 VDC. The contractor needs to ensure reliable operation and avoid callbacks.
1. Calculate the voltage drop for the reader circuit and the strike circuit using 18 AWG wire. Will either device receive at least 90% of its rated voltage (10.8 V)? Show your calculations.
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For the reader: I = 0.150 A, L = 350 ft, R = 6.5 Ω/1000 ft. Vdrop = (2 × 350 × 6.5 × 0.150) / 1000 = (700 × 6.5 × 0.150) / 1000 = (4,550 × 0.150) / 1000 = 682.5 / 1000 = 0.6825 V. Voltage at reader = 12 − 0.6825 = 11.3175 V, which is above 10.8 V (90% of 12 V). So the reader is fine. For the strike: I = 0.350 A. Vdrop = (2 × 350 × 6.5 × 0.350) / 1000 = (700 × 6.5 × 0.350) / 1000 = (4,550 × 0.350) / 1000 = 1,592.5 / 1000 = 1.5925 V. Voltage at strike = 12 − 1.5925 = 10.4075 V, which is below 10.8 V (90% of 12 V). Therefore, the strike will not receive sufficient voltage and may not operate reliably. The contractor should use a larger wire gauge for the strike circuit or a local power supply. For example, using 16 AWG (R ≈ 4.1 Ω/1000 ft) would give Vdrop = (2 × 350 × 4.1 × 0.350)/1000 = (700 × 4.1 × 0.350)/1000 = (2,870 × 0.350)/1000 = 1,004.5/1000 = 1.0045 V, giving 10.9955 V, which is above 10.8 V. So 16 AWG would work for the strike. Alternatively, a 24 V system with a converter would reduce current and voltage drop.
2. If the contractor decides to use a 24 VDC power supply for the strike and a 24-to-12 V converter at the door, what would be the voltage drop on the 18 AWG wire? Assume the converter draws 350 mA at 24 V? Wait, the strike is 12 V, 350 mA, so the power is 4.2 W. At 24 V, the current would be 4.2 W / 24 V = 0.175 A (assuming 100% efficiency). Calculate the voltage drop and the voltage at the converter input. Would this be acceptable?
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If the strike is powered via a 24 V supply and a converter at the door, the current on the 18 AWG wire would be lower. The strike consumes 12 V × 0.350 A = 4.2 W. At 24 V, the current is 4.2 W / 24 V = 0.175 A (assuming ideal converter). Using the same length (350 ft) and 18 AWG (6.5 Ω/1000 ft), Vdrop = (2 × 350 × 6.5 × 0.175) / 1000 = (700 × 6.5 × 0.175) / 1000 = (4,550 × 0.175) / 1000 = 796.25 / 1000 = 0.79625 V. Voltage at converter input = 24 − 0.796 = 23.204 V. The converter then steps down to 12 V for the strike. This is well within the input range of most converters (often 9-30 V). The voltage drop is only about 3.3%, which is acceptable. The converter must be rated for the strike current and located at the door. This approach reduces voltage drop and allows the use of existing 18 AWG wire. However, the contractor must ensure the converter is listed and installed in a suitable enclosure. Also, the 24 V supply must be capable of providing the total current. This is a good solution.