Small LED Tester




Introduction: Small LED Tester

YALT, YALT, YALT..., or "Yet Another LED Tester". There are already several different types of LED testers on Instructables. So why another?

Well, it started after I accidentally bumped over one of my component drawers. The drawer was filled with four different color LEDs. And to make matters even worse, all the LEDs were high-bright transparent LEDs.

With about 400 LEDS to resort, I decided to build a tester to assist me.

I wanted the following features:

- Automatic power control

- Quick testing (no button to press to test)

- No test leads or wires

- Battery operated

- No exposed battery leads

Step 1: The Circuit

The circuit is very simple, and uses two CR2032 coin batteries.

The LED under test is inserted into the testpoints, and if the LED works correct, battery power will flow through the LED, and into IC1. IC1 is a LM317L voltage regulator, configured as a constant current source. R1 limits the LED current to around 18mA through the LED under test.

LED1 and Switch1 is optional, and is only used to test the circuit and batteries.

How it Works

The LM317 has a minimum output voltage of 1.25V, when ADJ is connected to 0V. This difference of 1.25V between the ADJ and OUTPUT pins will always be present, and constant

I used this 1.25V as the basis for current limiting. Connecting ADJ with OUTPUT via a resistor, will create a constant current source across the resistor.

Calculate the resistor value as follow:

Rout = Vout / Iout, or
Rout = 1.25V / 20mA, or

Rout = 62.5 ohm

This is not a standard resistor, so use the next available, or 68 ohm.

Now, Iout can be calculated using the correct resistor:

Iout = Vout / Rout, or

Iout = 1.25V / 68, or

Iout = 18mA

Dont Have a LM317? No problem.

You can also use a 78L05 voltage regulator. In this case, the difference between GND and OUTPUT is 5V.

So just calculate the new value for Rout:

Rout = Vout / Iout, or

Rout = 5V / 20mA, or

Rout = 250 ohm

This is not a standard resistor, so use the next available value, or 270 ohm.

Now, Iout can be calculated using the correct resistor:

Iout = Vout / Rout, or

Iout = 5V / 270, or

Iout = 19mA

Step 2: Construction

Not much to show here.

I used a small piece of strip board to mount the batteries, and other components.

For the test points, I used two standard 1 x 10 headers mounted next to each other on another piece of strip board. I then cut a hole in the lid, and secured the test points on the cover of a 70mm x 50mm project box.

The battery test LED and button was also mounted on the lid.

1 Person Made This Project!


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5 years ago

You can use 68||820 to get 19.9mA or 47+15 to get 20.16mA. :-)

Eric Brouwer
Eric Brouwer

Reply 5 years ago

Thanks for your comment, I agree with your resistor values.


6 years ago

id like to see the under neath of the circuit board to see where the bits go and what the metal wire is on the board want to make myself one

Eric Brouwer
Eric Brouwer

Reply 6 years ago

See pictures in Step 3


Reply 6 years ago

great will be making one soon