Need help selecting LED's - Page 1
Forum section: electronics / beginners

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 04, 2024, 01:40:57 pm
content: I have a small pcb that goes in a taillight it has 15 2-PLCC red LED's with 300ohm resistors for every 3 LED's running from ground to the LED's and I believe (it was partially chipped off) a 150ohm resistor between +12VDC and the LED's. I'm wanting to make this board work with 5V. I bypassed the 150ohm resistor and ran 5vdc directly to the LED's. They lit up, but they're dim so I tried a 100ohm resistors between ground and the LED's, but they didn't get brighter. The LED's are for an e-scooter, will be a solid on, and need to be as bright as they can for 5v. Any suggestions?

author: tunk
author_rank: Super Contributor
timestamp: June 04, 2024, 02:25:40 pm
content: Try replacing the 300ohm resistors with lower values.Or you could try to use a step-up/boost module.

author: Kim Christensen
author_rank: Super Contributor
timestamp: June 04, 2024, 03:38:08 pm
content: Quote from: Hootis Tigglebits on June 04, 2024, 01:40:57 pmIs this 8.7V measured across the LEDs?Be aware that many of these LEDs are multi chip devices, so they can have much higher voltages across them than you might expect.ie: White LEDs are typically 3.6V... So if you have a 3 chip LED, then that one would be dropping 10.8V ( 3.6 x 3 )

author: pcprogrammer
author_rank: Super Contributor
timestamp: June 04, 2024, 03:53:26 pm
content: Quote from: Hootis Tigglebits on June 04, 2024, 01:40:57 pmA red LED has a lower forward voltage then a white one and the OP stated them to be red. As such 8.7V might be high indeed, at normally ~1.9V would give 5.7V.Like tunk wrote lower the 300 Ohm resistors and see if the LEDs produce more light.

author: IanB
author_rank: Super Contributor
timestamp: June 04, 2024, 04:11:18 pm
content: Quote from: Hootis Tigglebits on June 04, 2024, 01:40:57 pmIn your diagram you have 450 Ω total resistance in the circuit with 22 mA through the LED and 2 V across the LED. To modify this for a 5 V supply, you need to drop 3 V through the series resistance at the same 22 mA, so your new series resistance would be 5 V / 0.022 A = 227 Ω.So in summary, to change from a 12 V supply to a 5 V supply, you would need to change the series resistance from 450 Ω to about 230 Ω.(Just as a check, 22 mA through 450 Ω would drop about 10 V, which is correct for a 12 V supply.)

author: jzx
author_rank: Regular Contributor
timestamp: June 04, 2024, 04:28:44 pm
content: Perhaps the leds are connected in 5 series of 3 leds each.In this case, to power with 5V you have to break these connections, or easier, use an DC/DV converter.

author: IanB
author_rank: Super Contributor
timestamp: June 04, 2024, 04:34:55 pm
content: Quote from: Hootis Tigglebits on June 04, 2024, 01:40:57 pmThey are not getting 8.7 V. According to your picture, they are getting 8.67 V Ω − 6.66 V = 2.01 V, which is reasonable for an LED.Remember that voltage is a difference between two points.Quote from: Hootis Tigglebits on June 04, 2024, 01:40:57 pmThis gives you a total series resistance of 300 Ω + 150 Ω = 450 Ω. If you have about 22 mA through the LEDs, then 0.022 A through 450 Ω gives a voltage drop of about 9.9 V, leaving 2.1 V across the LEDs, which is in line with expectations.

author: tunk
author_rank: Super Contributor
timestamp: June 04, 2024, 05:12:37 pm
content: Your schematic/simulation doesn't correspond to what you say, which issix resistors and 15 LEDs. I suggest that you measure the voltages.

author: pcprogrammer
author_rank: Super Contributor
timestamp: June 04, 2024, 05:27:20 pm
content: Quote from: tunk on June 04, 2024, 05:12:37 pmAnd draw up a proper schematic.

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 04, 2024, 07:18:09 pm
content: I added a schematic to my original post showing how I believe everything is connected.

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 04, 2024, 07:19:06 pm
content: I tried 100ohm, but that didn't change anything.

author: tunk
author_rank: Super Contributor
timestamp: June 04, 2024, 07:38:28 pm
content: As jzx suggested, it would make much more sense if the threeLEDs are in series. Connect it to 12V and measure the voltages.Edit: And it would probably explain what you're seeing.

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 04, 2024, 07:54:35 pm
content: The voltage is the same across the LED's.

author: IanB
author_rank: Super Contributor
timestamp: June 04, 2024, 08:07:55 pm
content: Quote from: Hootis Tigglebits on June 04, 2024, 07:54:35 pmThat's not a sufficient answer.When the assembly is running on 12 V, measure the individual voltages across each resistor, and across each LED, and let us know those numbers.

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 04, 2024, 10:46:40 pm
content: 300ohm resistors are:R1: 2.9VR2: 2.4VR3: 2.9VR4: 2.4V150ohm resistor is:R5: 8.5VAll led's are:1.86V

author: Kim Christensen
author_rank: Super Contributor
timestamp: June 04, 2024, 11:31:28 pm
content: If the circuit is how you drew it:Then simply shorting/bridging the 150should result in about the same brightness when running it at 5V.I really don't know why they designed it that way in the first place. Why bother with the 150at all when running on 12V? Then the 300resistors could have been 1K instead.Or better yet, have the 3 LEDs in series so they drop 5.6V and drop the remaining 6.4V across a 620resistor while operating at 12V. The result would have been a brighter light for approximately the same current draw.

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 05, 2024, 02:16:00 am
content: Actually after getting home from work and doing some probing I discovered they are in series of 3. I'll post up a new schematic here soon.

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 05, 2024, 03:44:53 am
content: Original post updated with proper schematic. Also attached to this post.

author: Kim Christensen
author_rank: Super Contributor
timestamp: June 05, 2024, 03:55:24 am
content: Well, that explains why you can't operate it from 5V... 3 LEDs in series is 5.58V drop (1.86V * 3) so they'll be dim even with zero ohm resistors (never a good idea anyway).So you'd need to use a boost converter, rewire them, or short/bridge one of the 3 in each series string and change the resistors.Or maybe connect them to a higher voltage somewhere else in the circuit ahead of the 5V regulator. Where does the 5V come from?

author: Hootis Tigglebits
author_rank: Contributor
timestamp: June 05, 2024, 04:13:19 am
content: The 5v comes from the controller on the scooter. It's either 5v or 36V unless I run a booster or stepdown. I've not done the math yet, but according to easyeda the last led drops the voltage down to 0.

author: IanB
author_rank: Super Contributor
timestamp: June 05, 2024, 04:18:46 am
content: Probably easiest to use the 36 V supply. Just change the 150 Ω resistor for a 1.8 kΩ resistor and it should work fine.

author: mariush
author_rank: Super Contributor
timestamp: June 05, 2024, 04:24:48 am
content: The formula for calculating the resistor that limits a string of leds is :Input voltage - (number of leds in series x forward voltage of single led) = Current x ResistorYou don't want to power the whole light with the output of a microcontroller, because that output can only supply a limited amount of current, like in the 10-25mA range.The way it's normally done is you use the output of a microcontroller as an on/off signal , to power an on/off switch.Most often, the positive ends of the lights (leds, incandescent etc ) is always connected to the battery power supply (6v, 12v, 24v, whatever) or a led driver's output (that would limit the current) and the negative is connected to ground through a on/off switch which can be a mechanical relay, a solid state relay, a npn transistor , a mosfet etc etcWatch this video that shows how one can control a high power device (in the video a solenoid that consumes 12v 100mA approximately) using a npn transistor and a 5v powered microcontroller :So you really should power the leds directly from your scooter's battery voltage (12v or 36v or whatever) and use that 5v only as a control voltage.First determine how many leds are in series, usually red leds have a forward voltage below 2v, and also check if there isn't some step-down regulator to convert that 36v down to something like 10-12v - if the light is powered with 36v it wouldn't surprise me to see up to around 15 leds in series.Then measure the input voltage going into this light module , and the voltage before the first led to see the actual voltage usedThen put the numbers in the formula above and see if you have to tweak things.

author: ArdWar
author_rank: Frequent Contributor
timestamp: June 05, 2024, 05:27:42 am
content: Tricky isn't it? 3 series LED just happen to be slightly above 5V supply, but it would be horribly inefficient for the 36V supply.If you plan to power it off 36V probably better to change the 300R resistors into 2k7 each. Changing the 150R into 1k8 sounds like bad idea, with 50mA it would be 5W dissipation!!Unconventional idea: just buy 3 modules and put them in series into 36V supply.

author: pcprogrammer
author_rank: Super Contributor
timestamp: June 05, 2024, 06:10:14 am
content: Quote from: ArdWar on June 05, 2024, 05:27:42 amOr modify the circuit so that all the LEDs are in series and use a small resistor to control the current to run it of the 36V.15 * 1.9V = 28.5V36 - 28.5 = 7.5VSo the resistor needs to drop 7.5V at ~10mA leads to ~775 Ohm. Not a normal value so go up to the next one 820 Ohm. It will dissipate ~75mW.

author: mariush
author_rank: Super Contributor
timestamp: June 05, 2024, 07:15:43 am
content: Yeah, if you can make cuts between the columns to separate them you could do this :remove first resistor, remove all but the last resistor use a wire to connect the bottom of previous column to top of next column to make 15 leds in serieschange last resistor to have your desired current through the seriesIf you want to control using 5v, add a npn transistor between the resistor and ground wire (for <100mA current through the series pretty much any common npn transistor would do, ex BC84*, BC85*, MMBT3904 , MMBT2222A, etc )... add a reasonable resistor on the base of the transistor ex 10-100 ohm otherwise you'd blow the transistor

author: tunk
author_rank: Super Contributor
timestamp: June 05, 2024, 10:10:50 am
content: If you assume that each LED has a 2.0V forward voltage, then at 12Vthe total current is almost 30mA, and each LED has 6mA. If you assumea 14.0V charging voltage, then it's around 40 and 8mA.

author: pcprogrammer
author_rank: Super Contributor
timestamp: June 05, 2024, 10:20:35 am
content: Quote from: Hootis Tigglebits on June 04, 2024, 10:46:40 pmThis does not add up at all.Each LED 1.86V means times 3 is 5.58V plus the voltage across a 300 Ohm resistor of 2.4V makes it 7.98V plus the 8.5V over the 150 Ohm resistor makes 16.48V. Way above the 12V stated and still much more than the ~14V max in a car.Based on the voltage across the 300 Ohm, assuming that is correct the current is 8mA is some strands or 9.66mA in others.Did you do your measurements in reference to ground or actually probing across each part specific.