Led constant current source - Page 1
Forum section: electronics / beginners

author: amateur_25
author_rank: Regular Contributor
timestamp: April 27, 2012, 02:03:22 pm
content: HiI'm trying to build an uv led matrix using an Lm338 as a constant current driver. I'm a little confused as to how the voltage drop across the led is obtained. This is important because the uv leds I've ordered from China need 3.2 volts according to the datasheet. Using the following schematic how can ensure I met the 3v on voltage?http://www.ebay.co.uk/itm/ws/eBayISAPI.dll?ViewItem&item=280834124496&ssPageName=ADME:L:OU:GB:1123http://users.telenet.be/davshomepage/current-source.htm

author: deephaven
author_rank: Frequent Contributor
timestamp: April 27, 2012, 02:46:09 pm
content: You should set your constant current source to deliver the current you want to pass through the LED according to it's specs. The voltage drop across the LED is just a characteristic of that LED, it's what you get when you pass a certain current through the LED. So ignore the voltage, just make sure you set the right current.

author: amateur_25
author_rank: Regular Contributor
timestamp: May 23, 2012, 02:08:01 pm
content: My leds have FINALLY arrived. The supplier says the maximum continuous forward current is 30ma. So if I decided I going for 20ma, all I need to do is set the LM338 am using to constantly deliver 20ma? I've decided on using 200 leds because it's nice round number and my case is rectangular. I'm doing this in 20 rows of 10 leds. Each row is wired in parallel with other rows. Each row has 10 leds in series. Am I doing this correctly?My logic is if one goes then only one row is affected.

author: IanB
author_rank: Super Contributor
timestamp: May 23, 2012, 02:19:46 pm
content: Note that 30 mA is an absolute maximum. You might try even less than 20 mA, say 10 mA.At such low currents an LM338 is rather unnecessary. You may as well just use a current limiting resistor.Putting a string of 10 in series is fine as long as you have sufficient voltage available. Putting strings of 10 in parallel is also fine. (Putting series strings in parallel is OK, but putting individual LEDs in series is not good.)

author: Psi
author_rank: Super Contributor
timestamp: May 23, 2012, 11:47:40 pm
content: Na, 20mA is perfectly fine for 99% of leds.And if it says 30mA on the datasheet then you can be sure 20mA is ok.

author: codeboy2k
author_rank: Super Contributor
timestamp: May 24, 2012, 12:47:54 am
content: Quote from: deephaven on April 27, 2012, 02:46:09 pmdeephaven is basically correct, don't worry about the voltage, focus on the current, but he left out some details. So here are the details: you can't totally ignore that voltage. From the datasheet there will be a graph of voltage drop vs current. If not a graph, then at least there will be some table of typical data. Whatever the LED voltage drop is at the current you've chosen, you should multiply that voltage drop by the number of LEDs in series in the string, and then you should make sure your power supply can provide that much voltage, at the constant current you've chosen.Since you also have 20 strings in parallel, you need to multiply the current by the number of parallel rows. So 20 rows, at 20ma per row, means you need to set your constant current source to 400ma. The voltage output capability of the power supply must be able to reach the needed voltage output level to supply the total voltage drop of 10 LEDs in series. For your UV LEDs, the data says typical forward voltage drop is 3.2, but you should think about maximum, not typical. That is 3.8V. So the power supply should be capable of 3.8V x 10 LEDs in series , which is 38V. The actual voltage output by the power supply will be reduced or raised to meet the current source needs of the LED string you have, to maintain the constant current at 400ma.Finally, consider what happens if one LED in one string goes out; yes you lose one string. Then you will have the 400ma across 19 strings, or 21.05ma per string. This is still safe, and you will still be safe until you lose about 7 strings at once. At that point you will have 400ma across only 13 strings, or 30.77ma. If you raise the constant current to 500ma (500ma/20 = 25ma per string), then you can still lose one string safely (500ma/19 = 26.32ma per string). At 500ma, you approach the maximum current per string when you lose 3 strings (500ma/17 = 29.41ma)Summary.. your power supply should be capable of a minimum 38V and 400ma. Current regulate it to 400ma and you will be fine. Add more current capability if you ever think you will want to increase the LED current to 25ma. Operate safely below the maximum current so you can survive multiple string outages.

author: Zero999
author_rank: Super Contributor
timestamp: May 24, 2012, 09:23:33 am
content: The LM338 is a very inefficient solution. The minimum voltage loss is between 3.25V to 4.25V and any excess voltage is just dissipated as heat. A switching PSU would be a better solution. Unfortunately it's more complicated. Another solution is an off the shelf 18V SMPS powering a constant current source driving the LEDs connected in groups of five in series.