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This is a simple question and I just want confirmation before I buy the wrong power supply.

I am trying to test some LEDs which need about 2-20mA of current at 3.3v. I am gonna be testing a bunch so I need a variable power supply. The problem is that most power supplies have a current regulation of about 0.3% + 5mA. Looking at it I'm guessing that would cause some fluctuations but then I realized it also has a voltage regulation of about 0.3% + 5mV. If I use the voltage regulation at 3.3v and use a fixed resistor that should mean technically my current should only change at a value of VariableVoltage / FixedResistor and thereby essentially giving me a near constant current supply of around 2-20mA depending on resistors used right?

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  • \$\begingroup\$ You can either regulate voltage or regulate current. Never both at the same time – that's not how the physics work. So, for LED you need a regulated current source. \$\endgroup\$ – Marcus Müller Jun 7 '18 at 22:39
  • \$\begingroup\$ so using a regulated voltage source and a resistor will not accomplish what I am trying to do? I can't seem to find any power supplies that have a current source with good regulation in the 2-20mA range \$\endgroup\$ – phivms Jun 7 '18 at 22:44
  • \$\begingroup\$ A regulated voltage source plus a resistor will work fine. Just set the voltage source a bit higher than the LED needs, and calculate the resistor to let through the right current. \$\endgroup\$ – Simon B Jun 7 '18 at 22:49
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You are testing a 'bunch' of LEDs, so what are you testing:

  1. Go-NoGo
  2. LED voltage at a constant current
  3. LED current at a constant voltage
  4. Optical output at 2,3 above

As Marcus pointed out you can regulate (make constant) either Voltage or Current but not both.

I'd suggest you pick the parameters you want, and build either a well regulated voltage source or a well regulated current source.

Since you have a reasonably well regulated power supply (one assumes it's a lab supply, you could simply create a small fixture and use an external INA226 board to capture the LED current and voltage, that would probably be accurate enough for rough batching. Here you would alter the voltage to achieve whatever current level you require and calculate the LED Vf

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  • \$\begingroup\$ Essentially I am testing the Brightness of different leds at different current levels to find the best fit for the job. Once I determine the led with the correct current required for that brightness I will be putting it through a constant current driver TLC59284. I then would select the appropriate resistor value to achieve the current required. \$\endgroup\$ – phivms Jun 7 '18 at 22:54
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Just use a common inexpensive constant voltage (fixed) 5V supply with a 1/8 W (or greater) resistor for each of the various currents desired.

It is important to use the actual measured Vf for each type of LED when calculating the resistor value.

Calculator Source LED Series Resistor Calculator

For 2 mA a 850Ω 1/8 W:

enter image description here

For 20 mA 85Ω 1/8 W:

enter image description here



I made a PCB to do something similar.

enter image description here


I used a Vishay Dale SR3R0100FE66 0.01Ω 1% shunt resistor to measure the current with a volt meter.

enter image description here


Not shown in the schematic I have 8 LED outputs.
Originally I used a 2 row header rather than a switch.
The 0.01Ω shunt resistors go in the center row of the PCB (right of dime).
Just got the boards yesterday.

enter image description here

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You can make this work with an adjustable voltage supply and a fixed resistor of 1 K 1 W 1% or better, in series with the LED. Now you know already that you need to source a voltage above 3.3 V for the LED to even come on dim.

The resistor limits the current to 1 mA per volt above 3.3 volts. So 4.3 volts will put 1.0 mA into the LED. 13.3 volts is 10.0 mA, 23.3 volts is 20 mA = .466 W, which is why I suggested you use a 1 watt resistor. If you have a DVM you can measure the Vdrop across the resistor, which gives you the current in mA.

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  • \$\begingroup\$ okay that makes perfect sense and what I was thinking! Thanks! \$\endgroup\$ – phivms Jun 7 '18 at 23:00
  • \$\begingroup\$ Except that the Vf of the LED will vary so you will have to set the power supply for each individual LED ...not so accurate. \$\endgroup\$ – Jack Creasey Jun 7 '18 at 23:12
  • \$\begingroup\$ @JackCreasey could you follow up on what you mean concerning setting the power supply for each LED? Would not increasing the voltage to increase the current give you a good idea of the brightness of that LED at that current level? \$\endgroup\$ – phivms Jun 8 '18 at 0:23
  • \$\begingroup\$ If you have a fixed supply voltage and a fixed series resistor, then the current for every LED you insert will be different (since Vf will vary). If you simply have say 10 LEDs and you want the brightest with this supply and this resistor ...just test them all and find the brightest. The current will however be different for each LED with that resistor. \$\endgroup\$ – Jack Creasey Jun 8 '18 at 1:38
  • \$\begingroup\$ @JackCreasey. I mentioned several voltage settings and the given current for a 3.3 V LED through a fixed 1 K resistor. At NO time did I imply the source voltage was fixed. Use your eyes and read ALL of what I write please. \$\endgroup\$ – Sparky256 Jun 8 '18 at 1:57

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