LED driver for multiple LEDs in series - Page 1
Forum section: electronics / beginners

author: danners430
author_rank: Regular Contributor
timestamp: October 09, 2018, 03:28:56 pm
content: Hey guys,I'm designing an LED matrix using PICAXE chips as an alternative to using proprietory LED matrix controllers, such as the LED171596A. Reason for doing this is that I have large groups, or clusters, of LEDs which will always be toggled on/off together, never individually, and so writing to so many registers over i2c would simply be wasteful in terms of programming overhead. That, and I'm a maker :-)The matrix will consist of MOSFETs switching on both the low and high sides, which means I could in theory use any sensible power supply voltage, independant of the control voltage. For current limiting, I was thinking of using the NSI45020AT1G LED driver, which provides a constant 20mA at up to 460mW to drive LEDs. So just to check my thinking here:If the driver is rated for 460mW, that would mean it can safely drop 23V at 20mA, right? Which means the maximum possible supply voltage would be about 24V, assuming a single 1V LED connected. If I were to connect multiple LEDs in series, the power drop across the driver would fall, and I could in theory connect 22x 1V LEDs to the driver, taking the 1.8V overhead into account.So in my particular scenario, where the supply voltage will be a maximum of 16V, it would be safe in theory to string any number of LEDs in series, for example up to 15x 1V LEDs, without impacting on also having single LEDs on other matrix nodes?I've attached a simplified diagram of what I mean - the left of the circuit represents the node on the matrix where you have multiple LEDs in series, and on the right represents the node where you only have a single LED connected. Obviously, I'll be driving this in the same way as any other LED matrix, by cycling rows/columns and switching the opposite side dependant on what nodes are activated.

author: sokoloff
author_rank: Super Contributor
timestamp: October 09, 2018, 04:25:10 pm
content: Note that the power ratings are less for thinner copper traces or less area for heat sinking on the PCB. (See notes 5-10 on the datasheet.) That works against you.What works for you is that the limit is most power limits like this are not an instantaneous power limit, but an average power limit, so for a lot of the cycle, your power being dumped into the device is effectively zero. (The hint that this particular power limit is an average power limit, meaning it's a thermal related limit, is that the power limit varies with heat sink ability on the PCB, per the notes above.)

author: danners430
author_rank: Regular Contributor
timestamp: October 09, 2018, 04:34:53 pm
content: Quote from: sokoloff on October 09, 2018, 04:25:10 pmOk, so if I were clever (dangerous, I know) and arranged all the drivers in such a way on the PCB that they were all neatly in line for a heatsink to be fitted in top, the max power would be achievable?Sent from my ONEPLUS A3003 using Tapatalk

author: Zero999
author_rank: Super Contributor
timestamp: October 09, 2018, 09:36:10 pm
content: How about using a BJT based current sink to both limit the current and switch the low side?R2 and R1 form a potential divider giving just under 1.25V to Q1's base, which drops about 0.7V, leaving 0.55V across R1, giving a current of 0.55/27 = 0.02A, which also flows through the collector and LEDs.If you don't have enough LEDs in one of the strings to dissipate the power, then how about adding a resistor as well?

author: danners430
author_rank: Regular Contributor
timestamp: October 10, 2018, 08:08:55 am
content: That sounds sensible, but wouldn't it be more complex in comparison? Also, I'm afraid I've very little clue how to work with transistors beyond using them as a simple switch

author: Zero999
author_rank: Super Contributor
timestamp: October 10, 2018, 08:29:51 am
content: Quote from: danners430 on October 10, 2018, 08:08:55 amYes, it would use more components, but might work out cheaper. If you want to simplify it, then R2 and R3 could be omitted, but it means you need 5V to spare, after all the LED voltage drops are factored in. The voltage on R1 would then be equal to 5-0.7 = 4.3V, so for 20mA, the value of R1 would need to be 4.3/0.02 = 215R, so use 220R.