driving LEDs with combination of series and parallel resistors

Thread Starter

mikewax

Joined Apr 11, 2016
138
hey guys so i got some 3mm white LEDs and no datasheet.
my first circuit was like this:

5V---82ohm--->|---G

i added it up like so: (5 - 3(?))V / 82 = 25mA, and it worked well enough, i couldn't feel any heat after about 5 minutes.
now i'm using 4 of the same LEDs like this:

circuit.png
my calculation goes: (5 - 3(?))V / (22 + 3.75) = 78mA
and 78 / 4 = 19.5mA
so it's kosher, right? but is my arithmatic correct? i don't care if all 4 currents are exactly the same. a variance of 2mA or so is fine.
 

dendad

Joined Feb 20, 2016
3,332
It is not generally recommended to run LEDs in parallel. You are best to run them in series with 1 current setting resistor, or have each one with its own resistor.
Running in parallel is a good way to pop then all as you cannot ensure current sharing. And it will change with temperature. LEDs are current devices, not voltage like "normal" light globes.
 

Thread Starter

mikewax

Joined Apr 11, 2016
138
It is not generally recommended to run LEDs in parallel. You are bets to run them in series with 1 current setting resistor, or have each one with its own resistor.
Running in parallel is a good way to pop then all as you cannot ensure current sharing. And it will change with temperature. LEDs are current devices, not voltage like "normal" light globes.
well maybe my reasoning isn't right. i figure that since Vf increases with rising temperature, then whoever has the lowest Vf and gets the most current, that one will heat up more and so the Vfs will balance.
i don't know what Imax is, but i know from my first circuit that it's more than 25mA.
so i figured that even if the current isn't balanced, the Vfs will be so it will be in equilibrium.
 
Last edited:

ericgibbs

Joined Jan 29, 2010
9,580
hi mike,
Consider the effect if one of the LED's failed open circuit, the next 'weaker' LED would most likely fail and so on , you would then get a cascade failure of all the LED's

E
 

Thread Starter

mikewax

Joined Apr 11, 2016
138
hi mike,
Consider the effect if one of the LED's failed open circuit, the next 'weaker' LED would most likely fail and so on , you would then get a cascade failure of all the LED's
hmmm...... yeah. the average current rises to 25mA, then 34mA, then 54mA. an unstable equilibrium.
i'd have to have LEDs with Imax = 55mA, and i'm pretty sure i don't.
i see what you mean..... THANX fellas
 

Thread Starter

mikewax

Joined Apr 11, 2016
138
You can buy 5V to say 15V booster modules, then connect the 4 LED's in series with a single resistor.
DOH!!! Holy %&*! WHY DIDN'T I THINK OF THAT? and i even have some boosters right here in my drawer!
Dammit, man, you just showed me how dumb i am. i'm so lucky i didn't wire it up yet.
much obliged, Eric. job. done.
 
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