How to Make an LED Dimmer

Updated June 07, 2017

LEDs cannot be dimmed like light bulbs, by simply varying their input voltages. Instead, the current running through them must be directly controlled either by a resistor placed in series with the LED or by more advanced methods like constant current regulators. Since LEDs lack a steady internal resistance, they tend to draw more current than would otherwise be sufficient for adequate lighting. As they glow brighter, their temperature increases and further decreases their internal resistance, potentially leading to a runaway thermal cycle which ultimately destroys the semiconductor. It is for this reason that a driver circuit is almost always necessary. Fortunately, solving the problem of regulating current to an LED will also provide a reliable dimming method.

Calculate the resistor values. If using a 9 volt battery and driving a 20mA LED the formula volts / amps provides the resistor value. 9 Volts / 0.020 Amps = 450 Ohms. To determine the power requirements for the resistor use the formula P = I^2_R, or in other words power equals the current, in amps, flowing through the circuit, squared, multiplied by the resistor value in ohms. In this particular case, P = 0.02A^2_450 Ohms, which equals 0.18 Watts. 1/4 Watt (0.25 Watts) resistors are a common type and will be more than adequate.

Strip back about 1/8 inch insulation from each of the two 9-volt battery terminal wires with the wire strippers. Also prepare several lengths of jumper wires, cutting them to an appropriate length and pre-stripping the insulation from the ends. This will come in handy later.

Solder one end of the resistor to the red wire on the 9-volt battery clip. Solder the other side of the resistor to one of the ends of the potentiometer.

Solder the short end of the LED to the black wire of the 9-volt battery clip. This is where polarity matters. The negative side of the battery must connect to the negative side of the LED, which is marked by the shorter lead.

Solder the free end of the LED to the free end of the potentiometer, completing the circuit.

Plug in the 9-volt battery to the battery clip. The LED will illuminate. If it doesn't, turn the potentiometer to one far end or the other. The 450 ohm resistor inline with the circuit prevents too much current from flowing through the LED regardless of turning on the potentiometer. To one extreme of the potentiometer the LED will be at maximum brightness, and to the other extreme, the LED will be maximally dimmed, or may be off entirely.


For high power LEDs the simple series current-limiting resistor is insufficient. A resistor in series with the circuit burns up the excess energy flowing through it and dissipates it in the form of heat. To drop from a high voltage source down to the low levels used for LEDs would produce huge power losses. For these reasons, switching current regulators must be used when driving medium- to high-power LEDs. Such circuits commonly employ PWM outputs and current sense feedback mechanisms, employing proportional integral derivative control.

Things You'll Need

  • LED (20mA)
  • Resistor (450 ohms, 1/4 watt)
  • Battery (9 volt)
  • Potentiometer (Variable resistor 0 to 1K Ohm)
  • Battery terminal (9 volt battery clip)
  • Jumper wire
  • Wire stripper tool
  • Soldering iron
  • Solder
Cite this Article A tool to create a citation to reference this article Cite this Article

About the Author

Jesse Randall studied mathematics and physics and works as an embedded electronics engineer, developing microcontroller firmware and digital interfaces. He writes about subjects including abiogenesis, electrochemistry and algorithm optimization. He has been writing on technology-related subjects since 2000.