led's and resistors

Started by bsmcc2010, November 30, 2011, 04:25:02 AM

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bsmcc2010

this has probably been asked a million times-but how do i work out what value resistor i need for an LED?

I have a blue LED running of standard 9vdc so what value should the resistor be?

thanks

compuwade

Hey there!! I usually use a calculator like the one on this page: http://led.linear1.org/1led.wiz the worst thing that can happen is youll burn up your led. Not a major issue IMHO.

DavenPaget

Quote from: bsmcc2010 on November 30, 2011, 04:25:02 AM
this has probably been asked a million times-but how do i work out what value resistor i need for an LED?

I have a blue LED running of standard 9vdc so what value should the resistor be?

thanks
I either use 1K , 2K , 2.2K ( Some nasty shop recommended 100 ohm and the LED died easily  :icon_lol: )
Use http://ledcalc.com/
Hiatus

nocentelli

I used to use 4k7 for standard LEDs (e.g. standard red 5mm) and 15k for the super-bright clear I use on all my builds nowadays.
Quote from: kayceesqueeze on the back and never open it up again

R.G.

Light Emitting Diodes are diodes.They don't conduct much at all until the voltage across them reaches some minimal level, and then they conduct a current limited only by something outside them.The conduction voltage is a characteristic of the material. Silicon diodes, for instance, conduct at about 0.5 to 0.7V. LEDs conduct at a voltage determined by the stew of stuff they're made of. The different stews are to get different colors. So LEDs conduct at as little as 1.2V (for IR and deep red ones) through 2V (yellow/green/etc.) to as much as 4V (some blue LEDs).

Electrically they are just diodes with a big forward turn-on voltage.

The whole point of LEDs is that they emit light. They emit light proportional to the current through them. So no light at all until you hit Vf (the conduction voltage) and then light proportional to the current.

The easy way to control current is with a resistor. This is done by using Ohm's law: I = V/R. So to set up an LED you
1. Know how much voltage you have to work with. In pedals, this is usually 9V.
2. Know how much voltage the LED needs to turn on. Look at the datasheet. Yes, you can. That's what Google is for. Or put a 10K resistor in series with a battery across your LED and *measure it* with your meter.
3. Know how much current the LED can take. For 5mm/0.2"/ T1 3/4 "standard" LEDs this is usually 20ma, based mostly on how much heat can get out without damaging the LED. It's on the datasheet. Google or guess.
4. Subtract the LED voltage from the available power supply voltage you'll use. If you have a 9V supply and a 2.5V LED, then the available voltage for the current limiting resistor is 9V - 2.5V = 6.5V. (I have a stack of blocks 9 blocks high. I take away 2.5 blocks. How many blocks are left?)
5. Realize that your current limiting resistor will have this voltage across it, and you want a current somewhat less than you found in 3.
6. Calculate. The resistance is R = V/I, that Ohm's law thing again. So if you want 10ma in the hypothetical 2.5V LED, you want a resistor of R = 6.5V/0.01 = 650 ohms. If you want a current of 6ma, R = 6.5V/0.006 = 1083 ohms.
7. But how much current do you want? Time for introspection. LEDs vary in how bright they get per milliamp. Super-ultra-mega-hyper LEDs may be blindingly (literally!) brignt on 20ma. Old, weak first generation LEDs will be visible in dim light at 20ma. Know your LED. Try it with a pot. Guess, try, then adjust.

The basics are always (1) subtract the LED voltage from the available voltage (2) calculate a resistor equal to the available voltage minus LED voltage divided by the current you want.

BTW, as an observation on human nature, when you start using encapsulated calculators for things like LED resistors, you *stop trying to understand*, so you quit learning what's underneath. I think that's a mistake.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

LucifersTrip

The simplest way is to turn the pot until it is perfect for you:



warning: do not turn full counter-clockwise
always think outside the box

DavenPaget

I repeat the easiest way is to : 1K , 2K , 2.2K either one of those values are perfect .
Hiatus

deadastronaut

i use my retina's as a guide!... :icon_eek:  to  :icon_cool:.....that's probably why i need glasses nowdays too... :-\
https://www.youtube.com/user/100roberthenry
https://deadastronaut.wixsite.com/effects

chasm reverb/tremshifter/faze filter/abductor II delay/timestream reverb/dreamtime delay/skinwalker hi gain dist/black triangle OD/ nano drums/space patrol fuzz//

Gurner

#8
So the details are a 9V supply voltage & Blue LED which has a typical forward voltage of between 3.2V to 3.4V (let's say 3.2V for safety)

Since we know what voltage will be dropped across the lED (3.2V), we now need to find out what voltage needs to be dropped across theresistor - easy, just subtract the forward voltage from the supply voltage ......   so 9V minus 3.2V = 6.8V

Therefore 6.8V has to be dropped across the resistor ...so how are we going to ensure that?

Well, first, decide on what current you want through the LED - this is the bit where if you've not worked with LEDs before that it's difficult to know - if it's for a stompbox, I'd say go with the least current you can get away with - something like 6->8mA (less current = less potential for audible clicks) if it's a super bright LED type & you want max brightness, shoot for 25mA (I'd say this is  the maximum you should allow for a typical LED). Let's assume a super bright led.

Now we have all the input parameters we need to establish the correct voltage drop across the resistor (6.8V), and required LED current (25mA)

Ohms law V=IR   , therefore 6.8V=0.025A/Resistor   but we want 'R' so transposition of formula  ........ R = V/I       or 6.8/0.025     ...therefore the resistor needed is 272 Ohms. (so go with 270R, or if you want to err on the side of caution - for example a fresh battery puts out slightly higher than 9V, go with 300R)

LucifersTrip

#9
Quote from: DavenPaget on December 01, 2011, 03:10:38 AM
I repeat the easiest way is to : 1K , 2K , 2.2K either one of those values are perfect .

Those are good for some, but it is very common to use higher (5k-8K) with others. I use larger than 2K for 3mm and pinhole leds
always think outside the box

Ultrakd

Ive been using 820ohm, is that not enough?
Guitars: Ibanez S570DXQM
Amps:  Peavey ValveKing 112, Roland 15XL
Pedals: Big Muff w/ Tone & Wicker, Original Crybaby w/Modifications, BYOC Overdrive 2, Danelectro Cool Cat Chorus, Boss PH-3, Wave Breaker Tremolo,

R.G.

Quote from: Ultrakd on December 01, 2011, 11:35:00 PM
Ive been using 820ohm, is that not enough?
It depends on (1) the LED voltage (2) the supply voltage (3) the current you want.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Ultrakd

Quote from: R.G. on December 01, 2011, 11:49:09 PM
Quote from: Ultrakd on December 01, 2011, 11:35:00 PM
Ive been using 820ohm, is that not enough?
It depends on (1) the LED voltage (2) the supply voltage (3) the current you want.

The LED voltage is 3.3 V, supply voltage is 9v..

http://www.mouser.com/ProductDetail/Kingbright/WP7113QBC-D/?qs=sGAEpiMZZMt82OzCyDsLFDWEMTZfxH4OJWx7ztEMvOA%3D
Guitars: Ibanez S570DXQM
Amps:  Peavey ValveKing 112, Roland 15XL
Pedals: Big Muff w/ Tone & Wicker, Original Crybaby w/Modifications, BYOC Overdrive 2, Danelectro Cool Cat Chorus, Boss PH-3, Wave Breaker Tremolo,

R.G.

So that means the resistor gets 9V -3.3V, or 5.7V.

The correct resistor is then R = 5.7V /(how much current you want it to be) .

Using 820 give a current of  7ma. Is that what you want?

Put another way, is the LED too bright? If so, use a bigger resistor to lower the current. Is the LED too dim? If so, use a smaller resistance to raise the current.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Ultrakd

Quote from: R.G. on December 02, 2011, 09:50:00 AM
So that means the resistor gets 9V -3.3V, or 5.7V.

The correct resistor is then R = 5.7V /(how much current you want it to be) .

Using 820 give a current of  7ma. Is that what you want?

Put another way, is the LED too bright? If so, use a bigger resistor to lower the current. Is the LED too dim? If so, use a smaller resistance to raise the current.

No the led isnt to bright...I was using it cause it let the led be bright enough with battery and its a little brighter when its plugged into a onespot. But it is still the perfect brightness imo  :icon_razz:
Guitars: Ibanez S570DXQM
Amps:  Peavey ValveKing 112, Roland 15XL
Pedals: Big Muff w/ Tone & Wicker, Original Crybaby w/Modifications, BYOC Overdrive 2, Danelectro Cool Cat Chorus, Boss PH-3, Wave Breaker Tremolo,

Mark Hammer

LEDs vary in efficiency...widely.  Consequently, the same brightness, using the same supply voltage, can take much more current to achieve...or much much less.  Ideally, one would like to conserve as much battery juice as possible for as long as possible (yes, yes, I know many of you don't use batteries at all).  As well, heavy current draw, when powering an LED via stompswitch, can generate audible pops, a topic that has emerged and re-emerged on this forum over the years.

So, for a variety of reasons, one would aim for the balance between the lowest current draw providing usable visibility.  After all, conserving current for something you can't even see is not exactly wise.

While many of the formulae provided by members here to calculate current requirements are accurate, what they can't predict is how visible the indicator LED is for YOU, within the context you will use it.  A purple LED against a similar purple chassis will require more current to be visible than will a white or blue against a solid black chassis, and so on.  If you only use your pedals in daylight or room light,then you'll need greater brightness for visibility than someone who tends to play in dark clubs.

This is why I like Luciferstrip's idea of using a pot (though I'd use a 1k resistor n series with a 25k pot myself) to identify optimal brightness.  The other thing is to make yourself a simple LED tester/current-predicter box/thingie with a 12-position rotary switch, socket for seating the LED, and known resistances in each switch position: 1k, 2k2, 2k7, 3k3, 4k7, 6k8, 8k2, 10k, 12k, 15k, 18k.  Plunk the LED in the socket, start the switch at 18k, hit the juice, and work your way down a step at a time until you hit what you feel will be satisfactory illumination.

The advantages are that you don't need to know anything about the LED in advance, it's empirical and based on what you need to see, it uses nearest standard values that don't require measuring anything (because you know in advance what they are), and covers pretty much the entire range of what contemporary high and low-efficiency LEDS might require.

Gurner

#16
Quote from: R.G. on December 02, 2011, 09:50:00 AM
So that means the resistor gets 9V -3.3V, or 5.7V.

The correct resistor is then R = 5.7V /(how much current you want it to be) .

Using 820 give a current of  7ma. Is that what you want?


bada bing...

Quote from: Gurner on December 01, 2011, 03:46:04 AM
if it's for a stompbox, I'd say go with the least current you can get away with - something like 6->8mA


the thrust of my earlier post, whilst there are a lot of variables wrt LEDS, using 6->8mA as a starting figure will get you very much in the ballpark wrt calculating the correct resistor before you even go anywhere near your breadboard if you want low brightness     .......& so will using 25mA as a 'finger in the air' if you want high brightness.

DavenPaget

Gurner ... 25ma is too much ...
Hiatus

Gurner

#18
Quote from: DavenPaget on December 02, 2011, 11:33:10 AM
Gurner ... 25ma is too much ...

Too much suggests you know the final purpose?

It's not too much if you're after a high brightness LED to really light up something ....that's the whole point of what some here are trying to say ....it depends very much on the final deployment...but you can certainly run a modern high brightness led at 25mA without any trouble. As ever the datasheet is your friend, here's the first one I found typing in "high brightness 3mm led" into google

http://www.farnell.com/datasheets/67038.pdf        .........."Absolute Maximum Ratings rating = 30mA" (continuous, and even then the figure normally cuts you a little slack)

....25mA is therefore allows a healthy bit of headroom below max (just under 20%)....I wasn't actually recommending 25mA for a stompbox, but put forward the  figure for someone wanting to go for maximum retina challenging LED brightness.

Tacoboy

Hmm, I use clear bright leds in my stompboxes with a 10k resistor...
Let's have phun!