LM1084 Regulator for 1.25V - Page 1
Forum section: electronics / beginners

author: Genuigr
author_rank: Newbie
timestamp: July 13, 2013, 11:23:02 am
content: Hi guys,I want to use a LM1084 to drive IR-Leds. They need 1.25V and because I want to use a big bunch of them (about 50), I moved from the LM317 to the LM1084. Here is an example for a 5V Regulation:I used this calculation:Code:R2 = (Vout - Vref) * R1 / VrefR2 = (1.25V - 1.25V) * 100 / 1.25VWhich gives me a R2 value of 0.Now I am wondering. Should I connect ADJ to ground AND R1 (the 121 Ohm resister in the image), or just to R1?Should I use a heatsink? 50 Leds should use 2.5A (50mA each). I'm not so good at reading datasheets, I can't find hints on when to use a heatsink. I guess that's determined by the power usage?Code:1.25V * 2.5A = 6WHave a nice day,Fabian!

author: mariush
author_rank: Super Contributor
timestamp: July 13, 2013, 12:17:38 pm
content: IR LEDs are still LEDs - they're current devices, not voltage devices. The 1.25v is the minimum needed for them to actually work.You need to limit the current going through each led otherwise they'll go bad (burn up) and it's not a good idea to just rely on the fact that 50 leds would use 2.5A so you'd use a 2A regulator or something like that - they can do much higher currents for short periods of times.Look up in datasheets schematics on how to wire linear regulators as current regulators - LM317 / LM337 datasheets from National Instruments or Texas Instruments I think have quite a lot of schematics including current regulator examples.I'd recommend going with actual led drivers. You can either use very simple ones meant for individual leds such as this one which would limit current to 30mAhttp://uk.farnell.com/on-semiconductor/nsi45030at1g/ic-led-driver-45v-0-03a-sod123/dp/1794976or you could use something like this :http://uk.farnell.com/diodes-inc/al8807mp-13/driver-led-buck-36v-1a-8msop/dp/2138313(1A , 6-36v input )http://uk.farnell.com/diodes-inc/zxld13565ta/led-driver-550ma-60v-tsot23-5/dp/1672734RL(550mA , 6-60v)http://uk.farnell.com/micrel-semiconductor/mic2287cyd5-tr/pwm-wled-driver-1-2mhz-sot23-5/dp/1663103(500mA, 10v max input)and use one of these with at least 10 leds and you also get the possibility to adjust "brightness" by sending a pwm signal.In addition, the first one I linked also allows you to send a basic voltage between 0.5v and 2.5v to a pin to adjust power between 20% and 100% so you could use a plain voltage divider to adjust power if you'd so desire.

author: Genuigr
author_rank: Newbie
timestamp: July 13, 2013, 12:44:35 pm
content: Oh, wow. Totally forgot that.You just threw my whole planning in the trash binLooks like I have to rethink this.ThanksBut what's with the LM1084, just imagine I use it for something that works, just a simple resistor. I want 1.25V across that resistor. Would I have R1 and R2, or just R1?

author: mariush
author_rank: Super Contributor
timestamp: July 13, 2013, 12:58:12 pm
content: LM1084 has an internal voltage reference of about 1.25vBy using those 2 resistors , you form a potential divider that feeds 3.75v into the adjust pin instead of 1.25v .. that's how you get 5v out. If you leave the adjust pin unconnected, you'd probably get 1.25v out but the current is not limited so you'd burn the leds.You can use a resistor on the output and feed what's after the resistor to the adjust pin, this way you feed the voltage drop on the resistor to the adjust pin and limit the currentThis should work with LD1084 as well, but you have to keep in mind that as resistor heats up the current changes.. it's not a great, accurate current limiter. LED drivers are much better at this and have enough advantages to be worth buying them and they're not much more expensive.

author: MasterOfNone
author_rank: Regular Contributor
timestamp: July 13, 2013, 07:16:55 pm
content: Quote from: Genuigr on July 13, 2013, 12:44:35 pmI think if you really just wanted 1.25V, you would just connect ADJ to ground and forget R1 and R2. Think about what happens to the value of the Pot (R2) in an adjustable regulator when the power supply is set to the minimum voltage. The pot becomes zero ohms and the other resister R1 becomes a load resister (because it’s connected between 0V and the output).Also in your first post you seem to be calculating power dissipated by the LED rather than the Regulator. When calculating the power dissipated by the regulator you should use the voltage drop across the regulator and the current i.e. P = (Vin - Vout) * I.