# How to drop 28VDC for Voltage regulation between 14VDC and 16VDC

I want to convert 28VDC to run a LED map light that turns on at 14VDC and reaches its safe brightness at 16V (30mA). I used a 1k ohm potentiometer and a 1kohm resistor and dropped the voltage to where the potentiometer shows the max voltage of 16V.

Now the issue is, that the potentiometer rotates to about half the distance (0V to 14V) and then powers up the LED from 14V to 16V safely.

How do I modify the circuit such that I get the potentiometer to power the LED from range 14V to 16V and not have a dead zone as it is in the current setup?

EDIT: Thank you for the answers. I was able to run the LED safely using the circuit shown. But when I replace the 1k potentiometer with RHS1K0E, the circuit does not function in the same manner. The potentiometer stays off for half of its range, and then the voltage shoots up steeply and blows up the LED strip. Look at the pictures for details of the circuit that works. I do not have the detailed specs of the potentiometer in this circuit, but I have to replace it with the RHS1K0E for fitment purposes, that's all.

• Are you saying the LED in question has a forward voltage of 16V? That seems unlikely. Do you have specs on the LED? Can you include a schematic? Also why do you need a potentiometer as opposed to selecting a fixed resistor value? – JYelton Nov 8 '19 at 23:10
• Neil, I am gathering that you'd like to use a $1\:\text{k}\Omega$ potentiometer (or perhaps any-valued potentiometer) such that it sweeps a regulated voltage between $14\:\text{V}$ to $16\:\text{V}$ for the LED. But I think what you really want isn't voltage control, but instead current control. The LED module will take care of itself, given what you've already written about it, if you simply regulate the current up to a maximum of $30\:\text{mA}$. Could you measure the current when the voltage is at $14\:\text{V}$, for me? I'm curious. – jonk Nov 9 '19 at 5:45
• "But when I replace the 1k potentiometer with RHS1K0E, the circuit does not function in the same manner." if it was wired the same as in the schematic (with a 180 ohm resistor in series with the potentiometer) then it would not do this. Why are are you wiring it differently? – Bruce Abbott Nov 12 '19 at 0:19

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 – 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\$$)