I got an alert with an Instructable question (pardon the poor txtng-like spelling):
How would i used a joule theif and power 10 to 50 led on a circuit if i used a 5v power supply or a battery.?
I wuld like to make a banner out of leds for my wife and i am wondering if it would b possible to do it using the joule theif method but instead of a battery use a 5v power supply. wat resistor wuld i need and is it even possible?
My answer was:
It’s very simple: you put a 100 ohm (brown, black, brown, gold) 1/2 watt or 1/4 watt resistor from the +5V to the lead on the LED that does not have the flat spot next to it. The other LED lead with the flat spot goes to the negative 5V.
The white or blue LED will have a little more than 3 volts across it, and the resistor will have a little less than 2 volts across it, which will be about 18 to 20 milliamps, just about the max current for the LED. If it is a red or orange LED, the voltage across it will be less, and the current will be more, around 30 milliamps, which is about the maximum for red LEDs.
The resistor and LED use 1/10 watt of power and will get warm. If you put a lot of these in a confined space, they will get a lot warmer, so make sure they have good air circulation. In all cases above, the LED will be a standard 5 mm diameter LED.
Each resistor and LED combination goes across the 5V wires, so they are all in parallel, no Joule Thief is needed, nothing else is needed (other than wire and a power supply). For fifty white or blue LEDs the total current is .02 A times 50 or 1 Amp, at 5 volts that is 5 watts. Make sure your power supply can handle more than 1 amp, I would say at least 1.5 amps to be safe. Get 50 LEDs and fifty 100 ohm resistors and get to work!
USB, Etc. One thing I should have added is that this could be powered from a USB port or USB power adapter, but USB ports are rated for 1/2 amp, so you could use no more than 25 LEDs on one port. But that’s maximum, so it would be safer to use less than 25.
A PC power supply could supply 5V at dozens of amps, but it requires that the pin that turns on the PS be grounded. More can be found here about how to do this. By the way, I cannot be responsible for, and I cannot vouch for the safety or accuracy of this or any other website.
Higher Power LEDs As I said, the above applies to the standard 5mm LEDs. The higher power LEDs can be used, but the resistor will have to be changed to match the power of the LED. For example, using a 1 watt LED, the current would be approximately 350 milliamps or 0.35 amp. For 3.2 volts across the LED and 1.8V across the resistor, the resistance would have to be 5.14 ohms, but that is not a common standard value. Instead, use a 5.6 ohm resistor. At 1.8V and 0.35 amps, the resistor would dissipate 0.63 watts, so a 1 watt resistor would be needed. Other LED powers could be accommodated with different resistor values.
Important Do’s and Dont’s
Do use adequate heat sinking and ventilation for your LED project. LEDs make heat and LEDs can have their life shortened when they are too hot.
Do not put LEDs in parallel. Why? Because the LEDs have different forward voltages. Say for instance, we put two LEDs in parallel, one with a forward voltage of 3.3V, the other with a forward voltage of 3.2V. The 3.2V LED will hog more of the current than the 3.3V LED. The important point is that the 3.2V LED will get warmer, and the higher temperature makes the forward voltage drop. With a lower forward voltage, it will hog even more of the current, and get even hotter, and this can continue until the LED either gets damaged from excessive current or its life is shortened.
Think of the two LEDs as if they were the edges of a brim of a coffee mug in the sink. Say one edge of the mug’s brim is higher than the other (we put a knife blade under it). When we pour water into the mug, the water will spill over the lower edge before it spills over the higher edge. The current will flow through the LED with the lower forward voltage before it flows through the higher one.
Exceptions?? You’re probably thinking, “I bought one of those cheapo 9 LED flashlights and took it apart, and found that all 9 LEDs were connected in parallel on the round PC board.” You are right. How do they do it without burning up the LEDs? I have some explanations but I have not verified them to see if they are valid. First, the LEDs have forward voltages that are very close or the same. This is helped by having all of the LEDs being from the same maker and the same batch.
Also, the thought occurred to me that they may have tested the LEDs to match their forward voltages, but I don’t think they would do this because of the expense and labor involved, unless their automated assembly equipment tested them during assembly. Another factor is that these are in a flashlight that is used only briefly, for a few minutes at most. The LEDs don’t have enough time to get hot, and the flashlight is used only a few hours per year, so any LED light degradation would not be noticed for years.
Efficiency Some have expressed their concern about the low efficiency of this resistor and LED combination. I calculate the efficiency at about the same or a bit more than a conventional Joule Thief. The LED dissipates 3.2V times .018 amps or .0576 watts. The resistor dissipates 1.8V times .018 amps or .0324 watts. Total dissipation is .0576W + .0324W or .09 watts. The LED dissipates .0564W / .09W or .64 of the total, which is 64 percent. A typical conventional Joule Thief has an efficiency of 50 to 60 percent. So the LED and resistor is a bit better and simpler to build.