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
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.
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
^ 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.
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!
Quote from: Vallhagen on June 25, 2013, 07:05:45 PM
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! ;D Only I'm a bass player so I wrote it different. :P
Quote from: jimilee on June 25, 2013, 06:50:50 PM
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!
Quote from: jkokura on June 25, 2013, 06:53:01 PM
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!
Quote from: midwayfair on June 25, 2013, 07:02:10 PM
^ 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
Quote from: jimilee on June 25, 2013, 07:08:46 PM
Quote from: Vallhagen on June 25, 2013, 07:05:45 PM
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! ;D Only I'm a bass player so I wrote it different. :P
Well, that clears it up! Bengt to the rescue!!!!!! :D
I thought I was starting to understand this stuff, my apologies.i will continue to try to learn from you all.
Quote from: DutchMF on June 25, 2013, 07:22:12 PM
@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:)
Quote from: jimilee on June 25, 2013, 07:08:46 PM
Yeah that's I was thinking too! ;D Only I'm a bass player so I wrote it different. :P
Uhuh uhuh... Great minds think alike. ;D
...popular thread...:)
Quote from: jimilee on June 25, 2013, 07:31:08 PM
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
Quote from: jkokura on June 25, 2013, 07:42:10 PM
Quote from: jimilee on June 25, 2013, 07:31:08 PM
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!!!
;D Thanks Dutch
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.
Quote from: jkokura on June 25, 2013, 07:42:10 PM
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 :)
Quote from: Vallhagen on June 25, 2013, 07:05:45 PM
You can not, ever, place diods/LEDs parallell!
Huh? Of course you can. I really don't know where the rumor got started that the diode with the lowest Fv
completely takes over, but it's demonstrably false simply by USING LEDs so we don't have to use just out ears. If you put a green LED in parallel with a RED led on the same current limiting resistor, the green one will be dimmer, but it will still turn on. The red one will dim almost impercetably.
Now put them on two different LDRs -- they will work as normal. And it's a good thing, too, or we'd never be able to have parallel circuits with semiconductors -- including a rransitor-based pedal with an LED indicator in parallel.
I'm on a tablet and making dinner, or I'd go downstairs and put some LEDs on the breadboard for photo evidence. :)
It's possible I've misunderstood what you meant.
If you are using identical LED's, it works perfectly well too with both of them turning on just right.
I think maybe "the don't do this" comes from a couple of points. The first is that it's a bit weird to figure out what they are going to do since there is no parallel diode rule. The second and bigger point is that diodes don't stay the same over time and they don't age the same individually. If the characteristics change too much, then the circuit doesn't behave the same way and the failure isn't very predictable. If one gets too much out from the others, the brightness flops around drastically. If they each have their own resistor, then gradual changes will only cause small changes in brightness instead of major shifts.
Quote from: midwayfair on June 25, 2013, 10:28:15 PM
It's possible I've misunderstood what you meant.
To avoid possible misunderstanding; this is how i meant:
(https://dl.dropboxusercontent.com/u/10190945/LED_LDR_etc.png)
, where i would choose a Fig 2 design if i wanted to individually control broghtness/current flow through each LED. I would choose Fig 3 if the LEDs were "similar" or close to similar.
But i think we disagree on Fig 1?
I expressed myself a bit drastically; "you can not, ever"; which i maybe should change to: "if you are lucky, it might work". LEDs are - as we know them - components that we shall not really wear out at all, right. They work a lifetime and more. If you design as Fig 1, the risk is that you will wear out one of the LEDs early. The small difference (because there most likely is at least a small difference) in voltage drop cause big difference in current flowing through the LEDs. I'd say Fig 1 shows "bad design" and i put it on the "dont do this" list.
This link goes a bit deeper, and gives some calculated example as well (1 % voltage diff can cause 18% current diff): http://electronics.stackexchange.com/questions/22291/why-exactly-cant-a-single-resistor-be-used-for-many-parallel-leds (http://electronics.stackexchange.com/questions/22291/why-exactly-cant-a-single-resistor-be-used-for-many-parallel-leds)
Cheers
Yep, it's figure 1 that's the question. I think if you just change "Forbidden" to Bad Idea, there won't be any more question.
Quote from: RobA on June 26, 2013, 04:58:13 AM
Yep, it's figure 1 that's the question. I think if you just change "Forbidden" to Bad Idea, there won't be any more question.
You're right. Changes made. After all, it aint really illegal;). And pragmatically; in cases like this, where all currents are small and voltages low, we cant really cause any dangerous damage by designing things wrong or bad. But if we should setup parallell diodes in - say - a 230VAC rectifier, the word "forbidden" would be suitable. It could cause a fire.
Really good point there. I was only thinking in terms of LED's and low voltage. It could definitely be a dangerous failure mode in the situation you describe.