# Regulate high current LED light with potentiometer (voltage regulator or voltage divider)

Can someone help me with designing a schema to regulate voltage to dim an LED light? I have 3 LEDs connected in parallel. One LED has following specification:

Forward voltage = 2.75 V (max. 3.4 V)

Forward current = 1000 mA

Power consumption = 3.4 W

Reverse voltage = 1 V

So, if I understand it correctly, I need output of 2.75 - 3.4 V and 3 Amps to power them.

My goal is to regulated voltage to these LEDs to be from 2 V to 3.4 V (or 0 V to 3.4V if not possible to limit the low voltage) and constand current of 3A. I also prefer the schema to consist only of necessary components and to make it as small as possible to fit it in the flashlight housing. I read about source voltage to be as close to the output voltage to minimize the heat dissipating. My 18650 battery, fully charged, gives me about 4.1V (I understand that the battery has to be capable of delivering 3A - in case I can connect more in parallel too), but some of my voltage regulators I have at home just won't provide as much as required 3.4 V. Then I read about low dropout voltage regulators (LDO) and did some research too, but have none at home to try it. I thought that you electrician geeks would probably help me out instead of me trying by errors to figure this out. Also, I have few 10K or 1K potentiometers laying around that looks like this, but I am afraid that they may not be suitable for my purpose, perhaps because of the high 3A current I need? Perhaps, it can be designed, so it could be used? It perfectly fits hole of my flashlight housing. Can you please provide schema, either for voltage regulator with some LDO or voltage divider schema, please?

I looked this forum for similar posts, but it doesn't answer all my questions, e.g. if my potentiometer will work or not as it is not heavy-duty.

• We can't determine if the potentiometer would work without at least the datasheet. Feb 6, 2017 at 10:19
• Unfortunately, I don't have datasheet for it. In other words, suitable potentiometer must withstand 3A of current for my project, is that correct? If so, this potentiometer is not suitable than as I believe it is good for some number of mA. Feb 6, 2017 at 11:10
• Assume a completely charged battery with the LEDs fully on. The battery voltage would be 4.1 V, the LED voltage 3.4 V and the current 3 A. To achieve this, the potentiometer needs to drop the remaining 0.7 V at 3 A, implying (ohms law) a resistance of 0.233 Ω, way too small to accurately set on a 10 kΩ or 1kΩ potentiometer. Even worse than that, potentiometers are usually rated for a few tens of milliwatts and yours would be dissipating 0.7 V * 3A = 2.1 W.
– jms
Feb 6, 2017 at 11:23

The simplest would be a MOSFET as a buffer plus your pot.

Fancier ones would use the pot to adjust thee regulators output to thee led, voltage or current.

Without knowing what you have, tough to be more specific.

• Hi. Lets say I have a flashlight that I want to modify. It has 3 LED that are currently soldered in parallel, but can change it to serial if it would make things easier. Same with the battery. I can use one or up to four 18650 (rated ~4,1V fully charged). And then, I want to, somehow, connect potentiometer to dim the LEDs. :) Feb 6, 2017 at 13:17
My goal is to regulated voltage to these LEDs to be from 2 V to 3.4 V (or 0 V to 3.4V if not possible to limit the low voltage) and constant current of 3A.

It doesn't work that way. You can regulate current or voltage, but the load decides what the other is.

In any case, what you want to do here is regulate current. The LED voltages then come out to whatever the LEDs decide, based on lot variation, temperature, age, and other factors.

Since you want to regulate current and have that be the same thru each LED, you should connect them in series, not parallel. So now you need a 1 A current supply that can output from 8.25 to 10.2 V.

Since the battery voltage is reliably lower than the lowest voltage you need to put out, you need a boost converter. Use a small low side current sense resistor to measure the current thru the LED string, then control the switcher to regulate the current to 1 A. For example, a 100 mΩ resistor will drop 100 mV at 1 A. That's enough to comfortably gain up to drive the A/D input of a microcontroller. However, it only dissipates 100 mW, which is a small fraction of the 8.3 W minimum going into the LEDs.

For dimming, connect the pot as a separate input to the microcontroller. The boost switcher control algorithm takes this input into account in deciding what the desired LED current should be. This has another advantage that the current doesn't have to be a linear function of the pot setting. Humans perceive light level logarithmically. You want a fixed rotation of the pot to cause a fixed ratio of current change. This can easily be done with a lookup table or simple formula run on the pot reading. Note that this can be done in foreground code and can be "slow" since it only has to respond in human time.

## Using more batteries

Yes, you can use multiple cells in series. Enough cells in series increases the voltage so that it wouldn't need to be raised, only dropped. Now you need a buck converter instead of a boost converter. The relative complexity of these two is similar, although a booster will be a little easier in this case (low side switch and low side current sense).

The worst case is when the battery voltage range and the LED voltage range overlaps. Then you sometimes need to increase the voltage, sometimes decrease it. That is more complex than always increasing it (boost converter), or always decreasing it (buck converter). I would stay away from this regime since it adds complexity and can be fairly easily avoided in this case.

Multiple cells in series present the additional problem of balancing during charging and detecting the lowest voltage of any cell while discharging. The latter is required to shut down to avoid cell damage.

You seem to be asking about how to use a pot directly in-line with the LEDs somehow to perform dimming, and seem to be wanting to use a linear regulator. Both these are a bad idea. This is a battery operated device, so presumably power efficiency is important. Both a pot in series with the LEDs or a linear regulator would dissipate significant power. Not only does that waste energy from the limited amount stored in the battery, but now you also have to get rid of the heat caused by that wasted power. A boost or buck converter will be smaller, cheaper, and lighter than something less efficient after considering the extra mechanisms for dealing with the heat.

• OK, I will re-solder my LED from parallel setup to serial. Also I can use two 18650 batteries to get ~8.2V (can use 3 if needed) to get enough of source voltage. This way, I don't need to use boost converter. Now, I understand that I cannot regulate both voltage and current, only one of it. But now, since I don't have to use boost converter, how do I connect my pot to dim my LED? I heard there are linear and logarithmic pots and when it comes to light I should consider the logarithmic ones? Feb 6, 2017 at 13:10
• @Paolo: See additions to the answer. Feb 6, 2017 at 14:10
• Thank you for your input. Therefore, I will try to use a boost or buck converter as suggested depending on the source voltage. Feb 6, 2017 at 21:21

High efficiency current regulator is required for battery powered equipment. See LT3474 for a possible circuit.

Rather put your LEDs in series as thermal runaway will destroy them in parallel.