How to drop 28VDC for Voltage regulation between 14VDC and 16VDC
Light-emitting diodes do not have linear resistance. Because of this, they don't have a "safe voltage" per se. Instead, they operate based on current. Once the forward voltage of the diode is reached, they start to conduct exponentially. Small increments in voltage thereafter result in disproportionally large current increases. Because of tiny variations in real world applications (temperature, materials, etc.), LEDs should be current-limited to prevent runaway conditions that would destroy them. The most common ways are to use a current-limiting resistor or a constant-current source/driver.

I'm guessing by the context of an "LED map light" that you're referring to an automotive light, and based on 14-16V it is designed for a "12 volt" automotive system that actually operates in the 14-16V range (otherwise it wouldn't effectively recharge the battery).

LEDs generally have a forward voltage between 1.8 and 3.2 volts; with red and green being the lower values and white and blue being higher. The likely case is that your map light already includes a current-limiting resistor designed for 14-16V operation.

Let's assume that you have the following arrangement, with a white LED that operates on \$V_f\$ of 3.2V and 30mA:







simulate this circuit &ndash; Schematic created using CircuitLab

I drew the schematic somewhat horizontally to avoid having it appear overly large in this post.

The diode drops 3.2V which leaves 12.8 across the resistor. If 30mA is desired, the resistor needs to be 427Ω. A close common value resistor is 470Ω.

Now you have a 28 volt source that you want to operate this device with. Assuming you're leaving the existing resistor in place, you essentially are just adding another resistance in series and therefore need to calculate the new value for 28 volt usage:

What voltage is left for the resistor(s)? \$28 - 3.2 = 24.8 V\$

What total resistance is needed for 30mA? \$\frac{24.8}{0.03} = 827 Ω\$

What resistance needs to be added if 470 Ω is already present? \$827 - 470 = 357 Ω\$

Based on this, you just need to add a 330Ω or another 470Ω resistor (common values) and you should be OK.







simulate this circuit

Now, if you're trying to variably dim the LED using a potentiometer, I refer back to the first sentence: Light-emitting diodes do not have linear resistance. If they did, this could work. For more information, see: Using a variable resistor to dim an LED

Caution: You will want to use 1/2 watt (or larger) resistors. 30mA through a 470Ω resistor means it will have to dissipate 420mW of heat. (\$P=I^2R\$)