Most layouts I see show a 4.7k Resistor attached to the LED.
Why is this?
using the resistor calculator with a 9v supply a 3.3v LED (ratshack) with 25mA the resistor value comes out to 228R
Am I missing something?
You're not missing anything, you're just assuming you want to run your LED at 25ma. Depending on the LED that may be far too bright, and for a lot of stompboxes it means the LED will be using more power than the rest of the pedal.
Values of a few K Ohms work well for normal LED's for high brightness ones you can go higher.
+1 Slacker
As you probably know (I'm stating it for other reader's benefit) LEDs come in a variety of forward voltage specifications. The resistor limits the current that the LED sees and allows us to use the LED across a range of voltages without damaging it. The 4.7k resistor is a safe rule of thumb for almost any LED subjected to the voltages we normally use for our pedals.
Please see the discussion here http://www.diystompboxes.com/smfforum/index.php?topic=98455.msg860627#msg860627, especially PRR's explanation.
Thanks guys.
I suppose another question.
I have a set of "high Brightness" LED's from RadioShack 7000mcd, 25mA 3.3v (3.6v max)
SO the 4.7k would be just to dim it I suppose.
I ran two AA's in series measured at about 2.8v but the LED wasn't really all that bright. I would kinda expecting a blinded white light, but this wasn't the case. Any idea of the cause of that?
I make the LED as dim as possible as it's one of the biggest power hogs in a typical circuit. That is if I'm actually using a battery with my pedal(s). Rare, but I do do that sometimes.
It needs 3.3V to be fully switched on, with the caveat that you shouldn't try to power it directly from more than 3.6V. Your 2.8V is just starting to get it switched on, which is why it is very dim.
The other caveat is that you should not try to push more than 25mA through it. If you did supply it with more than 3.6V with no resistor in series, it would fizz in a heartbeat.
If you connect 9V with a resistor in series with it, the LED will drop 3.3V automatically (that's what it needs to be fully switched on, and that's exactly what it will take - it's just a special-purpose diode, really), leaving 5.7V dropped across the resistor. Say you decide to run it at 10mA, which is plenty for a high-brightness... probably too much, but you can play with the resistor values as you wish, as long as you don't exceed 25mA.
So, we need to find the resistor value that will pass 10mA of current when supplied with 5.7V... R = V/I = 5.7/0.01 = 570 ohms. Nearest common value is spookily close at 560 ohms.
560R, 1k, 2k2, 4k7... Whatever value you want - it's up to you how bright you want it and, as Paul Marossy says, how much battery power you're prepared to waste on a light.
Your 228 ohms calculation is spot on for 25mA for this LED, so don't go any lower than a 240 ohm resistor (just inside the limit).
Quote from: pappasmurfsharem on July 26, 2012, 04:16:14 PM
Thanks guys.
I suppose another question.
I have a set of "high Brightness" LED's from RadioShack 7000mcd, 25mA 3.3v (3.6v max)
SO the 4.7k would be just to dim it I suppose.
I ran two AA's in series measured at about 2.8v but the LED wasn't really all that bright. I would kinda expecting a blinded white light, but this wasn't the case. Any idea of the cause of that?
+1 RO. Here's what I was typing during your reply:
Looks like you might have these:
5mm High-Brightness White LED (2-Pack)
Catalog #: 276-017 Model: 276-017 |
This 5mm high-brightness white LED is ideal for hobbies and electronics projects.
Ideal for hobbies and electronics projects
High visibility, 5mm Round T-1 3/4
Intensity 7000mcd (typical);viewing angle 30°
FW current 25mA; FW supply 3.3 (typical), 3.6V (maximum)
RoHS compliant
But RS also has this:
NTE30045 White Super-bright LED Indicator
Catalog #: 55050633 Model: NTE30045 |
This is a 5mm all plastic mold type white super-bright LED indicator with a water clear lens.
Reverse voltage: 5V
Continuous forward current: 30mA
Peak forward current: 100mA
Power dissipation: 120mW
Operating temperature range: -25°C to 85°C
Storage temperature range: -25°C to 100°C
Lead temperature: 260°C
Forward voltage: 3.6V
Reverse current: 60 μA
Luminous intensity: 16,000 mcd
Peak emission wave length: CIE coordinates typically X: 0.3, Y: 0.31
Viewing angle: 22°
"High Brightness" and "Super Bright" are marketing terms (what's brighter than "Super Bright"? "Ludicrously Bright"?). The second LED listed above is more than 2x brighter than the first and the light is confined to a narrower viewing angle (I have to be careful here because there's a whole science dealing with how we measure light (candelas VS lumens), and perceive it.
Bottom line is that you'll have to experiment with different LEDs and different resistors to figure out what YOU like.
Mark Hammer has a neat LED tester he built that allows him to quickly test an LED with different resistors:
http://www.diystompboxes.com/smfforum/index.php?topic=98233.msg858221#msg858221
(http://i1160.photobucket.com/albums/q485/jdansti/46f2433f.jpg)
Also, the resistor is not just there to dim the LED. We normally operate our pedals at voltages higher than the max voltage of the LED. To "get away" with this, we limit the
voltage [Edit]
current with the resistor to avoid burning up the LED. If you want your LEDs to operate at their brightest and you're not concerned about power consumption, then find the smallest resistor that protects the LED (as RO explained).
Actually, the explanation I gave above is not strictly true, from a purist's point of view, but it's more than good enough for government work.
For those that want the truth? (cue Jack Nicholson, "YOU CAN'T HANDLE THE TRUTH!" (http://growthnation.com/wp-content/uploads/2012/01/Jack-Nicholson.jpg)Sorry, couldn't resist...)
Open this spec sheet from Farnell for a Kingbright Blue (http://www.farnell.com/datasheets/1446130.pdf)
You'll see on page 2 the figures for Vf - Typ = 3.3V, Max = 4.0V and max If of 30mA.
Now scroll down to page 3 and look for the top left graph showing If vs Vf. It starts to switch on at about 2.4V, 2.8V gets you about 5mA, 3.5V gets you 30mA. What do we get from this? Well, the voltage drop across the LED is a function of the current you push through it.
Try this experiment on a breadboard - 9V battery, 470R and LED. Measure the voltage across the LED and note the value. Now add another 470R and repeat the measurement. You'll note that the voltage across the LED has reduced slightly. Have a look at that graph of Vf vs If again... it's not linear at all, which makes calculating precisely what the true voltages and currents really are very difficult. Add to that manufacturing tolerances (remember that 4.0V max figure?) and it all gets messy. As I said, the formula I provided above is good enough for government work...
For many of the high brightness LED's you should find that ½ma or less is plenty bright enough. I built the Mark HAmmer LED tester 'cause every LED's seems to run different and this makes it simple to ballpark an appropriate resistor value.
> The second LED listed above is more than 2x brighter than the first and the light is confined to a narrower viewing angle (I have to be careful here...
7,000mcd;viewing angle 30°
16,000 mcd; Viewing angle: 22°
2.2 times as bright with 0.733 times the angle.... hmmmm.....
0.707 times the angle is 0.5 times the spot-area at a distance. (For small angles and allowing some slop for soft edges and marketing protractors.)
Twice as bright in half the area means they make the same *total* light (pretty much) but one spreads it around and the other beams it.
If you know where your eye will be, the narrower one will put more light toward your eye, or can be run at lower current for adequate light.
If you jog around the stage, the wide-beam will cover more stage. (However even 30 degrees won't cover "most" of the stage.)
Far outside the nominal beam-width, these "way-bright" LEDs may seem _less_ bright than some frosted LED that spews "less" light in all forward directions.
Running these "stupor bright" LEDs a few feet away on the full 25mA is like standing next to a 200 Watt amplifier. If you get a good look at the LED you have spots in your eyes, makes it hard to read the crowd or a music sheet. Maybe your fans came to hear the 200 Watt amp; they didn't come for your LED. 1K to 5K is a better starting value. Try it and see. Accept that you may have to change that resistor after trial (in room light similar to where you play).
I use 1k for blue and white or 680ohm for violet LEDs
My pedals are stage lights!
I am back to regular LED's and using 4k7 resistors. With ultrabright leds I always ended up using 47k resistors (I build pedals, not flashlights :icon_lol:)
as a great fan of retina destruction it varies...i breadboard my leds first and use my resistor sub box to get the desired brightness for that colour... :icon_cool:
QuoteIf you get a good look at the LED you have spots in your eyes, makes it hard to read the crowd or a music sheet.
I learned this lesson the hard way. Man, nothing is more distracting and irritating than blinding yourself in the middle of an aria. ;) One more reason for using light plates.
Quote from: R O Tiree on July 26, 2012, 05:49:28 PM
Try this experiment on a breadboard - 9V battery, 470R and LED. Measure the voltage across the LED and note the value. Now add another 470R and repeat the measurement. You'll note that the voltage across the LED has reduced slightly. Have a look at that graph of Vf vs If again... it's not linear at all, which makes calculating precisely what the true voltages and currents really are very difficult. Add to that manufacturing tolerances (remember that 4.0V max figure?) and it all gets messy.
Oh please, calculating it is not
that difficult, you're thinking about this too hard.
Yes, diodes are non-linear up to and just past their threshold voltage. Then the curve is close enough as makes no difference to straight and you can call it a linear device (i.e. Ohm's law applies) with low-to-no resistance.
Edit because I can't help myself: If you have to go to that level of silliness with LED V/I curves, then you should really be driving them with a current source.
Quote from: DiscoVlad on July 27, 2012, 08:47:53 PM
Quote from: R O Tiree on July 26, 2012, 05:49:28 PM
Try this experiment on a breadboard - 9V battery, 470R and LED. Measure the voltage across the LED and note the value. Now add another 470R and repeat the measurement. You'll note that the voltage across the LED has reduced slightly. Have a look at that graph of Vf vs If again... it's not linear at all, which makes calculating precisely what the true voltages and currents really are very difficult. Add to that manufacturing tolerances (remember that 4.0V max figure?) and it all gets messy.
Oh please, calculating it is not that difficult, you're thinking about this too hard.
Yes, diodes are non-linear up to and just past their threshold voltage. Then the curve is close enough as makes no difference to straight and you can call it a linear device (i.e. Ohm's law applies) with low-to-no resistance.
Edit because I can't help myself: If you have to go to that level of silliness with LED V/I curves, then you should really be driving them with a current source.
Oh please, don't spoil a teaching moment. :)
While it's not necessary to calculate the perfect resistor for each LED, it's a good idea to understand the underlying math ("maths" for y'all across the pond), especially when the person asking the question appears to be new to the topic of choosing LEDs.
Quote from: Jdansti on July 27, 2012, 11:14:59 PM
While it's not necessary to calculate the perfect resistor for each LED, it's a good idea to understand the underlying math ("maths" for y'all across the pond), especially when the person asking the question appears to be new to the topic of choosing LEDs.
Indeed, knowing the underlying maths is a good thing.
But words to the effect of "It's not actually linear + manufacturing tolerances = therefore the calculation is difficult", while true (Shockley's equation, EEK!), are to my mind irrelevant, and serve to only confuse someone new, especially when the simple approximation of:
Resistor Value = (V[supply] - V[led]) ÷ I[led]works perfectly well for >99% of cases.
Maybe things need to be updated for modern parts...
Vled is between 1.5 and 4V depending on colour
Iled can be much lower for the same brightness with the more efficient modern "high-brightness" LEDs compared to older less bright ones.
Quote from: DiscoVlad on July 28, 2012, 12:33:08 AM
Quote from: Jdansti on July 27, 2012, 11:14:59 PM
While it's not necessary to calculate the perfect resistor for each LED, it's a good idea to understand the underlying math ("maths" for y'all across the pond), especially when the person asking the question appears to be new to the topic of choosing LEDs.
Indeed, knowing the underlying maths is a good thing.
But words to the effect of "It's not actually linear + manufacturing tolerances = therefore the calculation is difficult", while true (Shockley's equation, EEK!), are to my mind irrelevant, and serve to only confuse someone new, especially when the simple approximation of:
Resistor Value = (V[supply] - V[led]) ÷ I[led]
works perfectly well for >99% of cases.
Maybe things need to be updated for modern parts...
Vled is between 1.5 and 4V depending on colour
Iled can be much lower for the same brightness with the more efficient modern "high-brightness" LEDs compared to older less bright ones.
Much better stated than you're previous reply. Thanks.
Quote from: Paul Marossy on July 26, 2012, 04:22:52 PM
I make the LED as dim as possible as it's one of the biggest power hogs in a typical circuit. That is if I'm actually using a battery with my pedal(s). Rare, but I do do that sometimes.
This is it... as dim as you can get it while still being able to see it on a bright stage will give you a functional indicator with minimum power consumption.
> Shockley's equation, EEK!
It is quite simple. Double the current is 20mV more voltage.
OK, the full equation has a term for Temperature. For room-temp work, sometimes you can ignore this. For reference, voltage drops 2mV per degree C.
And it has charge on an electron, but since that never changes it is just a "magic number", no different than "12" for figuring inches from feet.
Just to add to discussion.
As mentioned, the LED is a special purpose kind of regular diode. If your circuit always gets the same voltage then you could use another special diode, the Zener diode, to shave some further voltage off and make it more efficient:
http://www.dazatronyx.com/tech/low-battery-led-fading/ (http://www.dazatronyx.com/tech/low-battery-led-fading/)