LED question

[DutchMF]

So, I've got my head around the whole CLR thing, have some preferred values for LED's from different suppliers etc. But how about 2 LED's with one CLR? The logical side of my brain says to half the value to get the same kind of brightness, but is this correct? Anybody have some experience with this? Thanks in advance,

Paul

[jimilee]

No, but simple circuitry tells you that have the value=twice the brightness. Now the led can only get so bright, so you will only notice a difference up until a point. Why 2, why not use like a 2.2k instead of 2 4.7k? And then of course 2 4.7k in a series will be 8.4 and half the brightness.

[jkokura]

breadboard....

Whenever you're playing with resistor values and an LED, a breadboard is an awesome opportunity. It's actually the most fun you can have because it's so simple. Set up LED networks with resistors and then try adding LEDs or changing the value of resistors and see what happens.

Also, you can use a 10K or 25K trim pot, dial in the 'brightness' you like, then measure the resistance to get your perfect CLR value.

Jacob

[midwayfair]

^ What Jacob said. But also, you need to know that it doesn't scale linearly.

Here's a calculator.
http://ledcalc.com/

Here's why it doesn't scale linearly:
http://en.wikipedia.org/wiki/File:Diode-IV-Curve.svg

It's probably better to wire LEDs in parallel because the results will be more predictable. But there's plenty of reading out there for current limiting.

[Vallhagen]

You can not, ever, place diods/LEDs parallell! So, if you have two diods and SWITCH between them, all ok with one CLR, but if you want them both to light up simultaneusly, nope.

***

But sure you can put (two or more) diodes in serie with one CLR.

you get the CLR value by some turns with Ohms law; U=RI.

Example:

System voltage: 9V
CLR resistor: 4.7kOhm
Voltage drop over your diod (LED): 1.5V
...gives voltage drop over your CLR=9-1.5=7.5V

U=RI<=>I=U/R=7.5/4700=1.6mA

So you want 1.6mA through your LED(s). Two LEDs in serie gives a voltage drop of 3V, so you want a resistor that make 9-3=6V drive 1.6mA

U=RI<=>R=U/I=6/0.0016=3.76kOhm

...choose... 3.9kOhm from the E12 series of resistors.

Note from the example: 3.9 is NOT half the value of 4.7.

Cheers!

Cheers!

[jimilee]

You can not, ever, place diods/LEDs parallell! So, if you have two diods and SWITCH between them, all ok with one CLR, but if you want them both to light up simultaneusly, nope.

***

But sure you can put (two or more) diodes in serie with one CLR.

you get the CLR value by some turns with Ohms law; U=RI.

Example:

System voltage: 9V
CLR resistor: 4.7kOhm
Voltage drop over your diod (LED): 1.5V
...gives voltage drop over your CLR=9-1.5=7.5V

U=RI<=>I=U/R=7.5/4700=1.6mA

So you want 1.6mA through your LED(s). Two LEDs in serie gives a voltage drop of 3V, so you want a resistor that make 9-3=6V drive 1.6mA

U=RI<=>R=U/I=6/0.0016=3.76kOhm

...choose... 3.9kOhm from the E12 series of resistors.

Note from the example: 3.9 is NOT half the value of 4.7.

Cheers!

Cheers!

Yeah that's I was thinking too!  Only I'm a bass player so I wrote it different. 

[DutchMF]

No, but simple circuitry tells you that have the value=twice the brightness. Now the led can only get so bright, so you will only notice a difference up until a point. Why 2, why not use like a 2.2k instead of 2 4.7k? And then of course 2 4.7k in a series will be 8.4 and half the brightness.

Jimi, I love you and the great builds you post, but I have no idea what you are talking about!

breadboard....

Whenever you're playing with resistor values and an LED, a breadboard is an awesome opportunity. It's actually the most fun you can have because it's so simple. Set up LED networks with resistors and then try adding LEDs or changing the value of resistors and see what happens.

Also, you can use a 10K or 25K trim pot, dial in the 'brightness' you like, then measure the resistance to get your perfect CLR value.

Jacob

You're right, I really should get me a breadboard and experiment more, damn dayjob! I'm actually avoiding getting one, because Mrs. Dutch thinks I'm spending waaay to much time on this already, and a nice breadboard setup would make that even worse! It's a tricky balancing act, but at least I know what you mean. Thanks!

^ What Jacob said. But also, you need to know that it doesn't scale linearly.

Here's a calculator.
http://ledcalc.com/

Here's why it doesn't scale linearly:
http://en.wikipedia.org/wiki/File:Diode-IV-Curve.svg

It's probably better to wire LEDs in parallel because the results will be more predictable. But there's plenty of reading out there for current limiting.

Thanks for the links Jon, there's probably an answer for my question in there. You do know your stuff (or, at least where to get it! )

@Vallhagen: Ohm's law has been singed in my brain ever since I learned it in high school, but sometimes there's a situation I just can't get my head around, as electronics is not my main subject. Your calculation cleared it up a bit, thanks!

Paul

[DutchMF]

You can not, ever, place diods/LEDs parallell! So, if you have two diods and SWITCH between them, all ok with one CLR, but if you want them both to light up simultaneusly, nope.

***

But sure you can put (two or more) diodes in serie with one CLR.

you get the CLR value by some turns with Ohms law; U=RI.

Example:

System voltage: 9V
CLR resistor: 4.7kOhm
Voltage drop over your diod (LED): 1.5V
...gives voltage drop over your CLR=9-1.5=7.5V

U=RI<=>I=U/R=7.5/4700=1.6mA

So you want 1.6mA through your LED(s). Two LEDs in serie gives a voltage drop of 3V, so you want a resistor that make 9-3=6V drive 1.6mA

U=RI<=>R=U/I=6/0.0016=3.76kOhm

...choose... 3.9kOhm from the E12 series of resistors.

Note from the example: 3.9 is NOT half the value of 4.7.

Cheers!

Cheers!

Yeah that's I was thinking too!  Only I'm a bass player so I wrote it different. 

Well, that clears it up! Bengt to the rescue!!!!!! 

[jimilee]

I thought I was starting to understand this stuff, my apologies.i will continue to try to learn from you all.

[Vallhagen]

@Vallhagen: Ohm's law has been singed in my brain ever since I learned it in high school, but sometimes there's a situation I just can't get my head around, as electronics is not my main subject. Your calculation cleared it up a bit, thanks!

Paul

Cool. Take my example and use it in Jons link; ledcalc.com. It might get even more clear:)

Yeah that's I was thinking too!   Only I'm a bass player so I wrote it different.  

Uhuh uhuh... Great minds think alike.  

...popular thread...

[jkokura]

I thought I was starting to understand this stuff, my apologies.i will continue to try to learn from you all.

No worries Jimi! Our logical minds want things to be linear. 2+2=4, right? However, in audio and electronics, things are rarely linear.

Jacob

[DutchMF]

I thought I was starting to understand this stuff, my apologies.i will continue to try to learn from you all.

No worries Jimi! Our logical minds want things to be linear. 2+2=4, right? However, in audio and electronics, things are rarely linear.

Jacob

Yeah linear would be nice...... Thanks for all the help dudes, I have a lot of stuff to learn and experiment with. And Jimi, please don't use the Bass Player thing as an excuse, you (and all the other Bass Players) are  the reason us Guitarists stay sane.... Heck, I even had a Bass Player as my best man when I married Mrs Dutch! Without you guys, music just wouldn't groove!!!

[jimilee]

  Thanks Dutch

[RobA]

If you do need to parallel two diodes from the same voltage, like say in an LFO where you want to run one to a vactrol LED and one to an indicator, as long as your source has enough current drive, you can do this by using two parallel circuits of a diode and resistor each tied to the same voltage. You just don't want to tie them both to the same resistor.

[Vallhagen]

No worries Jimi! Our logical minds want things to be linear. 2+2=4, right? However, in audio and electronics, things are rarely linear.

Jacob

But Jimi is correct, even if i think he answers another question (the question "what happens if i change the LDR value with the same LED?") than Dutch's, which were; "what happens if i double the amount of LED-s?". (That is, if i understood things right. maybe im not...). Fact: When it comes to LDR-s, half the value gives double current; double the resistance gives half the current. etc; given that you keep the same LED.

Cheers