LED quick and simple

[Papa_lazerous]

Ok here it is,

When using LED's on my stomp effects I generally buy 5mm 12Volts LED's

Mainly because 5mm fits te bezels I have and the voltage I never know what to buy 

I wire up my LED's with a 390 Ohm Resistor and a 5.1v Zener.  I do that as thats what I was told when I made my first effect and so that the LED will fade as the battery dies.

But what Voltage LED should I buy am I correct with 12V or is there another lower voltage one I should buy?

See I told you it was simple, oh and if you feel like tackling this one.....

Dual colour LED's what they all about and do you have any ideas on uses

[John Lyons]

I just use standard 5mm LEDs (2 volts or less) amd a 4.7K current limiter to +v.
The Dual color LEDs are nice for a box where you have A or B type selections with a switch, or where you would use two LEDs but only have space for one. They have three leads and are two LEDs in one package. I made a looper/recorderwith a dual color LED and when it was in play mode the green light was on, then in record mode the red light was on. Simple!

John

[Papa_lazerous]

I get a feeling here that Everyone is going to have a different way of doing it.

I didn't know you could get 2Volt LED's.  I'm hoping I'll be able to learn some more once some more people share what they use.  I don't really get it, as far as the voltage to use and the type as in high brightness etc and what value resistor to use.  It just escapes me

[The Tone God]

I don't really get it, as far as the voltage to use and the type as in high brightness etc and what value resistor to use.

You are thinking about voltage which is not that important with LEDs. As long as there is enough forward voltage the LED will light. You have to start thinking about current. The brightness of the LED is controlled by the amount of current going through the LED. You can find out how much current the LED can take and the brightness by looking at the LED's specs.

To find the resistor value use Ohm's Law.

Andrew

[Paul Perry (Frostwave)]

http://ourworld.compuserve.com/homepages/Bill_Bowden/led.htm for a led resistor calculator.

[amz-fx]

My LED resistor calculator (might be a bit simpler to use):

http://www.muzique.com/schem/led.htm

regards, Jack

[R.G.]

I get a feeling here that Everyone is going to have a different way of doing it.
I didn't know you could get 2Volt LED's.  I'm hoping I'll be able to learn some more once some more people share what they use.  I don't really get it, as far as the voltage to use and the type as in high brightness etc and what value resistor to use.  It just escapes me

Let me try.
An LED is a Light Emitting Diode, of course. The subtlety if there is one is that it is a diode, and acts like one; but it puts out light in rough proportion to the current through it.

So to get an LED to light, you have to have it forward biased, meaning that the anode (arrow) end must be more positive than the cathode (bar) end. Get it reversed with respect to the voltage and it will stay dark. And since diodes of any kind do not limit their own current, you have to have some external means of limiting the current through it.

This external means is usually a resistor. The typical circuit is an LED in series with a resistor. The resistor value is chosen to let just the right amount of current flow through the LED so the LED is bright enough but also does not burn out. LEDs can usually only withstand currents less than 20ma without burning up. There are special cases both bigger and smaller, but the normal kind of panel mounting LED indicator is rated for 0-20ma.

Being a diode at heart, the LED has a forward voltage drop. This is a kind of voltage penalty you must pay to get the thing to conduct at all. For LEDs, the material making up the LED diode junction is different for each color of LED. Different materials also make for different forward voltage drops. The old standard red LEDs had a voltage drop of about 1.2V. That means that you had to supply more than 1.2V in the correct direction or they would not light at all. When you supplied more than 1.2V, the LED voltage would come up to 1.2V and then start passing current - and emitting light - but no matter how much current you put through the LED, the forward voltage was stuck at about 1.2V, not increasing. So if you put a 9V battery across a red 1.2V LED, the LED would let enough current through to burn up.

The way LEDs are actually used is that we put a resistor in series with the LED and apply a voltage across both the resistor and LED in the correct direction to make the LED light. Let's continue to use the 9V battery as an example. The total voltage across both LED and the resistor is the battery voltage. The voltage across the LED can never be more than its forward voltage, which we'll assume is a red 1.2V one. So of the 9V, the LED needs 1.2V. That leaves 9V - 1.2V or 7.8V across the resistor.

Now comes the trick: Ohm's Law. We need to choose a resistor that will keep the LED current under 20ma (or whatever the rating current is for the LED we're using). So we  choose a resistor to let 20ma through it while there is 7.8V across it.

Ohm's Law is something that is fundamental to electronics. Ohm's law states that the voltage across a resistor will be equal to the resistance times the current through it. So in this case, 7.8V = R * 20ma. We can do some high school algebra here and get R = 7.8V/20ma, or R = 390 ohms. And that is the "big secret" in the on-line LED calculators. You have to (1) know what voltage you have available (2) know the LED forward voltage (3) know what current you want. The on-line calculators do you the service of subtracting the LED voltage from the available voltage and then dividing that difference by the current you want.  And in return, they keep LED current and resistor calculation a mystery to you so you never learn what's really happening. As you can guess, I'm big on understanding and not a proponent of mysterious calculators.

So we know how to calculate resistors for one LED. How about all kinds?

Easy. Different color LEDs have different forward voltages. Red is generally lowest, then orange, yellow, green, and blue. It gets tricky with UV and white because there are several technologies for these and they are sometimes not simple LEDs. Red is generally under 2V, orange and yellow about 1.8-2V, green about 2-2.4V and blue 3-5V. The manufacturer tells you the forward voltage and it's usually listed in catalogs. If you use one of these, you just change the forward voltage in the calculation to match.

With that knowledge, you are now independent of the internet for working with LEDs and don't need an ethernet umbilical cord to set up your indicators.
As a final observation:
(1) There are no 12V LEDs. Did you mean 1.2V?
(2) The 5.1V zener eats up 5.1V out of the 7.8V we calculated for the LED; that leaves 9V -5.1V - 1.2V = 2.7V across the resistor. The current is then, by Ohm's Law: I = V/R = 2.7V/390 = 6.9ma.
(3) The LED will fade as the battery dies. In fact, the LED will not glow at all under about 6.3V on the battery.

You don't need the zener. You can use a resistor of  (9V-1.2V)/6.9ma = 1130 ohms and get the same brightness on a full battery. It just won't fall off the edge and quit glowing at 6.3V.

Did that help?

[GibsonGM]

FWIW, I usually find the 'suggested' resistor, and then up the value with a trim pot in series, adjusting the brightness by eye to an "acceptable" level.  This decreases current consumption...not a big deal in a Dist+ or anything, but can have an impact on battery life in a current-hungry application like a rate LED in an EasyVibe...

btw, starting off with something like a 1K will get you close and in a 'safe zone' if you don't know the forward voltage/current specs (think Radio Shack multipacks, lol)...with a 9v battery, that is 9v/1000ohms = .009, 9mA.  The majority of LEDs are set up to accept up to about 20mA IIRC so this gets you started.  Gotta burn a few out to get a feel for it!      

[96ecss]

I always get an education whenever I read a post or reply from R.G. I can't believe how much I've learned just by reading.

Thanks R.G. and everyone else who posts here.

Dave

[Papa_lazerous]

RG you are the man!!

I have printed this out and am going to read through a few times so it sticks in my head.

In reply to your 3 points at the end....

(1) As far as voltage goes I went to the counter in Maplins asked for some 5mm Red LED's the guy wanted to know what type and asked what voltage, I know he said 12Volts.  But then my experience of Maplins lately is that its staffed by people who know less than the customers.  Started buying from Farnell now   And small bear for harder to get stuff.  The 12V issue was the main part of confusion I think

(2) ok that makes sense about the Zener using 5.1V Thanks

(3)  I am glad that the LED will fade with a low battery that was the aim.  Maybe the voltage that it fades at could be tuned.

One more Question R.G while I have your attention.  I know now how to work out the current going through the LED but..  What about the Zener surely that must be consuming power can I calculate that using Ohms law? if so how do I apply the rule I am still not too confident on things.  I am just curious if they consume more power than they are worth

[Papa_lazerous]

Just searched LED's at maplins and these are being sold as 12Volt LED's now thats just thrown a bit stick into what I just learnt

http://www.maplin.co.uk/Module.aspx?ModuleNo=2054&criteria=LED%2012v&doy=9m1

[Minion]

Just searched LED's at maplins and these are being sold as 12Volt LED's now thats just thrown a bit stick into what I just learnt

http://www.maplin.co.uk/Module.aspx?ModuleNo=2054&criteria=LED%2012v&doy=9m1

Thats doesn"t throw a stick into it at all because if you read the Specs of the LED"s they are just regular 1v to 2.5v LED"s but they just have the Resistor built into the LED so the same principle would apply if you were for instance going to use a 12v LED with a 18V supply for example...

[JonFrum]

Another thread gets saved to the "Study and Learn" file - thanks R.G. 

[Papa_lazerous]

Just searched LED's at maplins and these are being sold as 12Volt LED's now thats just thrown a bit stick into what I just learnt

http://www.maplin.co.uk/Module.aspx?ModuleNo=2054&criteria=LED%2012v&doy=9m1

Thats doesn"t throw a stick into it at all because if you read the Specs of the LED"s they are just regular 1v to 2.5v LED"s but they just have the Resistor built into the LED so the same principle would apply if you were for instance going to use a 12v LED with a 18V supply for example...

Ohhhhh I see they put resistors in there already thats a clever little idea me be quiet now 

[R.G.]

I know now how to work out the current going through the LED but..  What about the Zener surely that must be consuming power can I calculate that using Ohms law? if so how do I apply the rule I am still not too confident on things.  I am just curious if they consume more power than they are worth

It's one of Mother Nature's laws that anything electrical generates heat equal to the voltage across it times the current through it.

So if we have an LED, a zener and a resistor in series across a fixed voltage, for instance 9V, then the current through them is identical. The sum of the powers generated by all three is and must be equal to the power put out by the battery. That is, the power out of the battery is 9V times whatever current the three draw. Each part generates heat equal to the voltage across it times the current through it. The LED will have a fixed voltage of whatever its specific forward voltage is. The zener is also a constant-voltage device, since that's what zener diodes do - they let any amount of current through, but will never increase beyond their zener voltage (until they burn out...)

So the only device in the string that is limiting current is the resistor. If you have a 5.1V zener and a 1.5 LED, then there is only 9 -5.1 - 1.5 = 2.4V left over to be across the resistor. And since the resistance must be the ratio of the voltage across the resistor divided by the current through it by Ohm's Law, we can pick a resistor value to set the current to be whatever we want.

The battery, the LED, and the zener all determine how much voltage the resistor has left. The resistance determines how much current flows in all three. If we wanted, say, 10ma to flow, then the resistor must be 2.4V / 0.01A = 240 ohms.

The total power coming out of the battery is 9V * 0.01A = .09W, or 90mW. The LED disspates 1.5V*.01 = 0.015W or 15mW, and the zener disspates 5.1V* 0.01A = 51mW. The resistor has 2.4V across it and 0.01A through it, so it dissipates 24mW. When we add 15mW, 51mW and 24mW, we get 90mW, which is exactly the same amount of power the battery puts out. Mother Nature is satisfied that everyone is playing by the Rules.

If we cut the resistor in half to 120 ohms, the current increases to 2.4V/120 ohms or 0.02A. The power out of the battery, and the power dissipated in the LED, the zener, and the resistor all double, because the voltages have remained the same while the current has doubled.

But what happens if the battery voltage goes down?

Let's say we have a 9V battery, a 1.5V LED, a 5.1V zener and a 240 ohm resistor. When the battery is 9.0V, the current is 0.01A, as we calculated. But when the battery sags to 8.0V, the LED and zener demand the same voltage, leaving only 8.0 -1.5 -5.1 = 1.4V for the resistor. The current must then drop to 1.4V/240 ohms = 0.0058A, or 5.8ma. The current is only 58% of what it was before. The LED puts out roughly half the light.

If the battery sags down to 7.0V, as it will if we keep using it, the LED and zener still get their fixed voltages but the resistor only gets 0.4V, and the current decreases to 0.4V/240 = 1.7ma, only 17% of what it was at 9V. The light from the LED is down by 4/5.

Finally if the battery sags down to 6.6V, the voltage left for the resistor is zero. Well, it's not exactly zero since the "fixed" voltages on the LED and zener are not perfectly fixed, but it's so near zero that almost no current flows. And the LED goes dark.

If we did not have a zener, the LED would not change brightness as quickly. So the zener is indeed making the LED go out about the same time the battery is no longer very good. The circuit does as represented to you.

I personally would rather have the LED in series with just a resistor and not have the "falling off a cliff" action from the LED. But that's a matter of personal preference.

[Papa_lazerous]

Thanks allot R.G. that really helped