I am designing an LED light project using this Cree XLamp LED, (Forward voltage is 36V, forward current is 2400mA) and I am looking for a way to limit the output voltage of a boost converter by adding a potentiometer and a resistor. For reference, this was demonstrated in this video. However what I don't quite understand is how he chose to select a 6.8k ohm resistor to run in series with the added potentiometer.

What I believe is confusing me is the basic usage of Ohm's law. By my understanding, when the potentiometer is fully at the left and there is no resistance (through the pot), the total resistance is 6.8k. The output voltage of the boost converter is 6V. When fully at the right, the resistance is 9k ohms, and the output voltage is 12V. How can this be? What should go through my mind when trying to pick the resistor to use, since I'm wanting to use a different LED? Essentially, what is happening that lets the resistor determine the minimum voltage of the boost converter?

I know there's too many questions in my mind to ask them all, so I'll stick with the root of my confusion. Hopefully someone can understand my madness!


  • 2
    \$\begingroup\$ Most of us are not interested in watching a video. Please draw a schematic of the circuit you are talking about, using the built-in schematic editor. \$\endgroup\$ Aug 7, 2020 at 2:03

1 Answer 1


I watched enough of it to get what you're missing. It was obvious to me in a few seconds (but only because I'm already familiar with how those power supplies work)

That new resistor/pot combo is connected to an EXISTING potentiometer (the blue box). So what he's done is modify the resistance curve of the existing pot, not introduce a completely new element.

In the factory, a worker would tweak that blue pot until the output is right. But it would have a limited range - Maybe able to tweak the output +-0.5V (made up numbers). That's not enough range to do what the video guy wanted, so he modified it's total possible range.

Now as to HOW that affects the circuit....

The existing pot used to trim the output voltage of the supply taps into the output, and is tied to the controlling IC, giving the IC the ability to "read" the output voltage. The controlling IC monitors this tap and adjusts its output such that the tap reading is what it expects it to be.

Adding that potentiometer is basically tricking the IC into thinking the output is something other than what it is, and the IC responds by adjusting its output.

An analogy...

Consider the speedometer in your car. If you are going 50MPH, you only know this because the speedometer tells you that you are going 50MPH.

Inside the speedometer are a few gears that are designed such that when the car is going 50MPH, a needle on your dash points to the '50' mark.

Now suppose somebody opens up that gearbox and changes the gear ratio. They make it such that when the car is going 60MPH, your needle points at 50MPH.

So you think you're doing 50 but you're actually doing 60.

That's what this guy did to the pot. He manipulated it to report an erroneous reading back to the IC and trick it into boosting or cutting its output beyond the range that blue pot could do by itself.

He got lucky the power supply could support the new range. Just as if the guy messing with your speedometer adjusted it to report 470MPH as 50MPH, doesn't mean your engine could actually get the car going 470MPH. What would probably happen is it'd report you as going like 10MPH and you'd peak out there (while actually cruising at 90...)

However what I don't quite understand is how he chose to select a 6.8k ohm resistor to run in series with the added potentiometer.

To answer that, you'd need to know the value of that blue pot and any other resistors around it. The blue pot is not connected directly between output & ground. There's likely a resistor from output to the pot, and another from the pot to ground. You'd need to draw out all those resistors and analyze the circuit to figure out why 6.8k was the 'right' value.

A far easier way to accomplish what he did would be to modify those two extra resistors I described, rather than add a whole new pot. If those two resistors are made, say, 1/2 their current value, the range of voltages the blue pot could adjust to would increase significantly.

Why not just do that to the production pieces??? It's because they want to limit the possible range of output, and also give the ability to precisely control the output voltage. That blue pot is a '10 turn pot', meaning you turn the screw on the top 10 times to move the pot through its entire range. That makes very fine control of the resistance and therefore the output voltage possible - like to the millivolt level probably. The resistors set the total possible adjustment range - If they were not there, you would lose precision (because each turn of the pot now represents much more change in total resistance) and it would be really easy to accidentally fire up the power supply at a target output way out of it's possible range and smoke the device.

I might point out that it's highly unlikely you'll get a supply that's nominally designed to output in the 6-12V range to get up to 36v. Your best bet is just source a 36v power supply and forget about tweaking anything.

Maybe you can get a nominal 24 or 48V supply tweaked to hit 36V, but not a 12V supply - that's way too far. Especially at the current levels you're talking about. Don't get locked into thinking only about output voltage. The power supply can only deliver so much power (P=V*I remember). If you ask it for more "V", you'll get a reduction in the amount of "I" it can deliver.

  • \$\begingroup\$ Thank you so much! This is exactly what I wanted to know, I appreciate it so much! \$\endgroup\$ Aug 7, 2020 at 20:41
  • \$\begingroup\$ @Firedan1176 Sure thing brother! And thanks for saying "thanks" :) So many times I answer a question and get crickets in return..... When you're ready, if you wanna bounce your resistor values and the calculations off me, I'll try to give you a 'sanity check' before you power up a brand new supply and fry it. \$\endgroup\$
    – Kyle B
    Aug 7, 2020 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.