Question: What is voltage drop, and how can I calculate it for my application?

Answer: In many cases (especially with low voltage lights), the output voltage from the power supply must be much higher than the lamp voltage. The reason for this is that a significant amount of power is lost due to the electrical resistance of the power cable.

The voltage drop over a length of cable can be calculated by using the formula, V = IR, where V is the voltage drop, I is the current draw of the light in amps, and R is the total electrical resistance of the power cable in ohms. The current draw of a particular lamp can be calculated if the wattage and voltage of the lamp are known. The current draw is equal to the lamp wattage divided by the lamp voltage, or, amps = watts/volts.

For example, referring to the table of electrical resistances of various wire gauges listed below, we can calculate the voltage required to operate a 24 volt-300 watt light at 24 volts over 250 feet of 16 gauge cable. The current draw, I, of a 24 volt-300 watt lamp operating at 24 volts is 300 watts/24 volts = 12.5 amps. The resistance of 16 gauge wire is approximately 4 ohms/1000 feet. Since the total path of the circuit is from the power supply to the light and back to the power supply, the total resistance of the cable is twice the length of the cable times the linear resistance, or for this example, R = (2 x 250 ft) x (4 ohms/1000 ft) = 2.0 ohms. Since V = IR, the voltage drop, V is equal to 12.5 amps x 2.0 ohms = 25 volts. This means that 25 volts is lost due to resistance and so the power supply will need to provide at least 49 volts to power this 24 volt-300 watt light over a 250 foot cable!