Hi all:

I need to make a circuit with 3 LEDs in series, with a 220ohm resistor and 12V source. Can I put the resistor "in-between" any of the LEDs?

I know the typical layout is:
12V (+) power--------220ohm resistor---------LED1------LED2-----LED3------(-) ground

but can I do it like this:
12V (+) power---------LED1------LED2-------220ohm resistor------LED3------(-) ground

Will this result in any of the LEDs being overloaded in either voltage or current?

Thanks in advance!
-chris

 

hi,
That will be OK.
E

 

Just to check, replace each LED with a 100Ω resistor. Do a simple Ohms Law calculation applied to the two scenarios.
See if there is a difference in the voltage across each resistor and current through the resistor.

 

Hi all:

I need to make a circuit with 3 LEDs in series, with a 220ohm resistor and 12V source. Can I put the resistor "in-between" any of the LEDs?

I know the typical layout is:
12V (+) power--------220ohm resistor---------LED1------LED2-----LED3------(-) ground

but can I do it like this:
12V (+) power---------LED1------LED2-------220ohm resistor------LED3------(-) ground

Will this result in any of the LEDs being overloaded in either voltage or current?

Thanks in advance!
-chris

As long as everything is in series, it will work.

 

That's good to hear!

Second question: Will this (see pic below) arrangement work?

 

Yes...

 

What's the reason you want to put the resistor in series between the leds?

For a 3V/20mA/5mm led, using 80% of led rating current and it will be:
R = (12V - (3V*3) )/16mA
= 3V/16mA
= 187.5 Ω, choose 180 Ω

For a 2V/10mA/3mm led, using 80% of led rating current and it will be:
R = (12V - (2V*3) )/8 mA
= 6V/8 mA
= 750 Ω, choose 750 Ω

 

In general, it doesn't make any difference as to the order of the elements in a series circuit.

 

Guys, you're awesome!! It WORKS!

A project I'm working on requires me to have several clusters of 4 SMDs, in 2 x 2 squares. But i could only have 3 in series, not 4.

I first did this with 4 in series, but the source voltage wasn't enough, divided by 4.

Here was my design (after you all confirmed it for me):

And here is the proof of concept!

What do you think?

 

So that's four sets of two in parallel. Trace the power; you'll see what I mean.