# Can someone please explain how to add an external potentiometer to a voltage regulator with a trim pot? [closed]

I'm building an LED light panel and I'm setting up the electronics according to this video.

As you can see, he uses a 22K potentiometer and a 6.8K resistor. I would like to understand why he uses those values. As in, why a 22K potentiometer and would any other value work? And why would a 6.8K resistor result in a minimum voltage of 5V? And how do the two potentiometers work together?

• You need the schematic diagram of the voltage regulator board and/or the datasheet for any ICs used on the board, to answer these questions. – Peter Bennett Oct 19 at 1:44
• You can't expect us to watch a video to find out what you are doing. Even if we did, without a schematic any answers would not make sense to anyone else reading them. If the video becomes unavailable, then all is lost. If you scan other questions, you will see what is expected in the way of information. The idea of this forum is not only to answer individual questions but to provide information that could help others. – Barry Oct 19 at 2:34

In a typical DC/DC converter a proportion of the output voltage is fed back through a voltage divider into the regulator IC, which compares it to an internal reference voltage and raises or lowers the output voltage to make the feedback and reference voltages equal.

The block diagram looks something like this:-

simulate this circuit – Schematic created using CircuitLab

The effect of regulation is to maintain a constant voltage across R2 equal to the reference voltage (eg. 0.9 V), which Ohm's Law tells us must then have a constant current flowing through it (eg. 0.9 V / 1.2 kΩ = 0.75 mA). This current also flows through trimpot R1, so the voltage across it (and therefore the output voltage) will be directly proportional to its resistance (eg. Vout = 0.9 V + 0.75 V per kΩ of R1).

If we wire another pot in parallel across R1 the total resistance will be lowered and the output voltage will be reduced. The new pot can also have a resistor wired in series to set the minimum total resistance. The voltage divider circuit then looks like this:-

simulate this circuit

Everything inside the box is effectively a single resistor as far as the feedback circuit is concerned. Setting R4 to its maximum value and adjusting R1 sets the maximum resistance of the combination, thus setting the maximum regulated output voltage. Setting R4 to 0 Ω puts 6.8 kΩ across R1 which sets the minimum output voltage. Intermediate settings of R4 cause the output voltage to vary between these minimum and maximum voltages.

why a 22K potentiometer and would any other value work? And why would a 6.8K resistor result in a minimum voltage of 5V?

The values of 22k and 6.8k were chosen (possibly by experiment) to suit the particular DC/DC converter. Using different values would result in a different voltage range. If your converter is not identical to the one in the video then you may need to experiment to determine the correct values for your device.