Current regulation - peak vs. average. - Page 1
Forum section: electronics / projects

author: PbFoot
author_rank: Contributor
timestamp: January 15, 2013, 05:34:32 pm
content: Hi,I am about to begin a project to make a very high power IR illuminator, for a special purpose (not CCTV). I plan to use around 250 Osram SHF4550 LEDs. These things have a peak current handling of 1 amp. I'll be modulating the LEDs at a few hundred Hz, and within the allowable pulse width (a few tens of microseconds) per the spec. sheet.I will be powering the device from a 12 volt source, like a car battery, and since each LED has a 3 volt drop (at 1 amp), I will have either two or three in series, and then have them all paralleled up to the power supply, with a MOSFET in between to do the modulation.My question is, since I will be running these at a very short duty cycle, the average current is pretty low, but the instantaneous current is very high. I was thinking about using an LM338 to do the current limiting. My question is, do I need to limit my current to the average or the peak? The peak would be 84 (!) amps, which seems really high. I expect the average would be maybe 10 percent of that, at most. Of course, I will have to have some big caps in there to provide all that juice.Can anyone advise? Am I going about this the right way?Thanks!-PbFoot

author: lewis
author_rank: Frequent Contributor
timestamp: January 15, 2013, 07:40:11 pm
content: You've answered your own question in a way; if the LEDs have a peak current of 1A for a few microseconds, you need to limit the peak current to 1A. But not quite...The datasheet says the IF, or forward current, is 100mA. This is a continuous rating. The surge current, IFSM, is 1A and this is the peak, but for one surge only (it says D=0 which is zero duty cycle). Have a look at the Permissible Pulse Handling Capability graphs on page 5 of the datasheet, this gives maximum current handling for some given duty cycles and on-times. You need to take into account frequency too.The power dissipation is the thing you need to limit on average, and it must bebelowthe maximum power dissipation of the LED,andthe forward current must be under the value you determined from the graph. You calculate that by the peak forward current, forward voltage and yourmaximumduty cycle: P = V x I x D. The power dissipation figure varies with ambient temperature (as does the VF) as well as thermal resistance, so ensure you design for worst case.Don't bother with a current source, a set of resistors in series with each string will do nicely assuming you have a stable 12V supply.I'm looking at this datasheet available here:http://catalog.osram-os.com/catalogue/catalogue.do?favOid=0000000200000190001c0023&act=showBookmark

author: PbFoot
author_rank: Contributor
timestamp: January 15, 2013, 07:52:49 pm
content: Hi,Thanks for putting me on the right track. I didn't realize what the D=0 meant. I will go with simple resistors in that case, taking into account the power dissipation. I am going to need some giant caps to source that much current. Basically, this thing is going to be a "beacon" to an edge detecting device that will sense the signal via a silicon photodiode. The pulse width will be as short as I can make it, without resorting to things like avalanche transistors. I need to get as much signal out of this because I need to detect it at long distance, and in suboptimal conditions.PbFoot

author: lewis
author_rank: Frequent Contributor
timestamp: January 15, 2013, 07:54:26 pm
content: Sounds like an interesting project, good luck with it!