Simple LED question (definately leading to more)

angie1199 · Dec 31, 2005

audioguru said:
Good, we don't need to guess anymore which LED you are planning to use. Sorry about that. Didn't realise there was a datasheet there and there were so many different ones on the same sheet!

Is your supply 12V or 12.5V and does it have a 1.1V diode in series with it? Yes it's 12v and I have absolutely no idea if there's a 1.1v diode in series but if I knew what to look for I could take a peek.

These LEDs are the small 3mm size. They have 3-leads which makes switching them on and off separately very complicated. Erm, don't I just put the switch in between the LED and the incoming power so the switch decides which way the current is going to flow?

ljcox · Dec 31, 2005

angie1199 said:
ljcox said:

Its datasheet shows that it typically will operate at 20mA for four red in series and 16mA for four green in series with a 200 ohm current-limiting resistor and a 12V supply.

Click to expand...

I dont recall writing this and looking at the previous posts (on this page) it was from Audioguru. However, I think we have moved on anyway.
So it's safe to use a 200 ohm resistor for both. And if they're both of equal brightness or near enough then that's fine. The product information does say they have a uniform light output.

ljcox said:

..... So I suggest that you do some tests to determine what current (ie. the current needed for the required brightness) you really need and then do the calculations.

Click to expand...

Ok, I think I'm understading this. If I want them dimmer then I use a higher ohm resistor. Is that it?

Yes, a higher resistor will reduce the current and make the LEDs dimmer

ljcox · Dec 31, 2005

angie1199 said:
Nope, my maths is wrong... If I use 4 in series then it's 200 ohm resistor. And I can't put 5 in series in case they all decide to take 2.5v at the same time which would overload the power supply (12.5v) Wrong. You are looking at it the wrong way. The power supply is fixed at 12 Volt and will only reduce under load. So if it is a regulated supply (ie. has a regulator) the decrease in voltage with load (ie. as the load current increases) will be minimal. But if it is unregulated, there will be a greater fall off as the load current increases.

The voltage across the LEDs is a function of the current through them and vice versa. It is a bit complicated to explain, but if you have 4 LEDs in series with a resistor, the current is limited by the resistor and the voltage across the LEDs. So if you increase the resistor, the current will reduce and the voltage across the LEDs will also reduce. If you have a multimeter, you can check this by experiment. Also look at the graphs the Audio posted.
Angie

If you want to measure the current through the LEDs, the easiest way is to measure the voltage across the resistor and calculate the current using Ohm's law. If you state the resistance in k ohm, then the answer will be in mA.

eg. 4 Volt across a 2 k resistor means the current through it is 2 mA.

audioguru · Dec 31, 2005

You can wire them like this:

ljcox · Dec 31, 2005

audioguru said:
You can wire them like this:

Audio, That's essentially what I attached on a previous page.

audioguru · Dec 31, 2005

ljcox said:
Audio, That's essentially what I attached on a previous page.

I know, but since you are down-under you drew it upside-down. I made your pic right-side-up for us notherners and while I was at it I made it a little neater. I even coloured it with my crayons. :lol:

angie1199 · Dec 31, 2005

ljcox said:
I dont recall writing this and looking at the previous posts (on this page) it was from Audioguru. However, I think we have moved on anyway.

Sorry ljcox , that's what happens when I try to answer two post in one quote button click!

angie1199 · Dec 31, 2005

ljcox said:
angie1199 said:

Nope, my maths is wrong... If I use 4 in series then it's 200 <a href="#">ohm</a> resistor. And I can't put 5 in series in case they all decide to take 2.5v at the same time which would overload the power supply (12.5v) Wrong. You are looking at it the wrong way. The power supply is fixed at 12 Volt and will only reduce under load. So if it is a regulated supply (ie. has a regulator) the decrease in voltage with load (ie. as the load current increases) will be minimal. But if it is unregulated, there will be a greater fall off as the load current increases.

Click to expand...

Ok, here's where I get confused.

1. Power supply regulated 12v - which mine is , I checked
2. Run 5 of these LEDs from it in series, with the max. Vf of both red and green being 2.5v, which is a total of 12.5v

And this won't blow the power supply if all LEDs suddenly say 'hey I want 2.5v'.

This is confusing. 5 LEDs need 12.5v at peak which is more than 12v from the supply so where does it get the other 0.5v from?

OR does the series only require 2.5v total but then we play with the resistance?

angie1199 · Dec 31, 2005

audioguru said:
You can wire them like this:

Thanks AudioGuru, I got that when ljcox posted the drawing and believe it or not I understood it

angie1199 · Dec 31, 2005

By the way guys, happy new year and I do appreciate your help here.

ljcox · Dec 31, 2005

The 2 graphs attached are based on the curve of a red LED that I had drawn in Powerpoint. I have used it to show how the operating point can be estimated.
Graph 1 is for a single LED in series with a 100 Ohm resistor connected across a 3 Volt supply. I did not bother to start the x axis at 0. Had I done so, the load line would have intersected the y axis at 3/0.1 = 30 mA. The operating point is where this line and the LED curve intersect.
Graph 2 is for 4 LEDs in series with a 220 Ohm resistor connected across a 12 Volt supply. Again, I did not bother to start the x axis at 0. Had I done so, the load line would have intersected the y axis at 12/0.22 = 54.5 mA. In order to represent the graph of 4 LEDs in series, I multiplied the voltage scale by 4.
This the graphical solution for finding the operating point of the LEDs.

ljcox · Dec 31, 2005

audioguru said:
ljcox said:

Audio, That's essentially what I attached on a previous page.

Click to expand...

I know, but since you are down-under you drew it upside-down. I made your pic right-side-up for us notherners and while I was at it I made it a little neater. I even coloured it with my crayons. :lol:

Well done, I did it with a finger nail dipped in tar.

angie1199 · Dec 31, 2005

ljcox said:
The 2 graphs attached are based on the curve of a red LED that I had drawn in Powerpoint. I have used it to show how the operating point can be estimated.

Ok, here's what I don't understand.

Graph 1:
To draw the line you've divided 3v by 0.1 which I assume is the resistance of the resistor in mA. But, as that's what I am trying to find out, how would I know that?

Graph 2:
I have absolutely no idea why the left end of the red line would point at 54.5mA. Again there's the division of 12v by 0.22 which is again the resistance of the resistor, but that's the bit I don't know.

Or ... and this could be Angie's Eureka moment ... is graph 2 being used to work out how many LEDs could be used in series if they all used a lower resistance...?

I know why I gave up physics 26 years ago but wish I'd had a teacher who I could understand.... then I wouldn't have given up and might have understood this.

ljcox · Jan 1, 2006

angie1199 said:
ljcox said:

The 2 graphs attached are based on the curve of a red LED that I had drawn in Powerpoint. I have used it to show how the operating point can be estimated.

Click to expand...

Ok, here's what I don't understand. I wrote earlier that it is not easy to explain

Graph 1:
To draw the line you've divided 3v by 0.1 which I assume is the resistance of the resistor in mA. The resistor is 100 Ohm, ie. 0.1 k. But, as that's what I am trying to find out, how would I know that? I chose the resistor to be 100 Ohm as an example.

Graph 2:
I have absolutely no idea why the left end of the red line would point at 54.5mA. Again there's the division of 12v by 0.22 which is again the resistance of the resistor, but that's the bit I don't know. Again, I chose a 220 Ohm resistor. So it crosses the y axis at 54.5 mA. 12 Volt across 220 Ohm results in a current of 54.5 mA.

Or ... and this could be Angie's Eureka moment ... is graph 2 being used to work out how many LEDs could be used in series if they all used a lower resistance...? No Graph 2 is showing the solution for the case where there are 4 LEDs in series with a 220 Ohm across 12 Volt. It shows what the voltage across the LEDs will be if the LEDs conform to the V/I characteristic shown in Graph 1 and the series resistor is 220 Ohm.

The procedure (assuming you want to do it graphically) is to:-
1. determine by testing one of the LEDs, what current you need to give you the brightness you want.
2. put a dot on the LED curve (Graph 2') at this current - call it point P.
3. draw a straight line from the point (12 V, 0 mA) through P and find the point where it crosses the y axis - call it Ix.
4. the series resistor is then R = 12/Ix.
5. choose the nearest preferred resistor value, eg. if R = 350, then use either 330, 360 or 390.

Graph 2' is a graph equivalent to my Graph 2, but drawn for the LEDs you intend to use, extracted from the graph Audioguru posted)

The alternative procedure (and this is the way I would do it) would be:-
1. as above
2. put a dot on the LED curve (the one Audioguru posted) at this current (call it Iled) and look at the voltage across the LED at this point. call it Vx.
3. Now calculate R. R = (12 - 4 * Vx)/Iled. If Iled is in mA, R will be in k ohm. The 4 in the formula is because you have 4 LEDs in series.

angie1199 · Jan 1, 2006

ljcox said:
The procedure (assuming you want to do it graphically) is to:-
1. determine by testing one of the LEDs, what current you need to give you the brightness you want.
2. put a dot on the LED curve (Graph 2') at this current - call it point P.
3. draw a straight line from the point (12 V, 0 mA) through P and find the point where it crosses the y axis - call it Ix.
4. the series resistor is then R = 12/Ix.
5. choose the nearest preferred resistor value, eg. if R = 350, then use either 330, 360 or 390.

Graph 2' is a graph equivalent to my Graph 2, but drawn for the LEDs you intend to use, extracted from the graph Audioguru posted)

Ya-hey by jove I think I've got it. I did say it could be a Eureka moment.

Ok to test I pick any point of the curve that is below the Typical value. Draw the line from 12v through point on curve to y-axis. Read what y-axis says, translate that to Ohms, get resistor and test. If too dim, get lower resistance resistor, if too bright get higher resistance resistor.

ljcox · Jan 1, 2006

angie1199 said:
Ok to test I pick any point of the curve that is below the Typical value. Draw the line from 12v through point on curve to y-axis. Read what y-axis says, translate that to Ohms, get resistor and test. If too dim, get lower resistance resistor, if too bright get higher resistance resistor.

One point of clarification, all of the points on the LED curve are typical. If you read the the data sheet of a LED, it will give the min, typ and max values of the LED current at a specified voltage. On the graph below, this would be where the green line intersects the curves (provided they chose 1.77 Volt as the point to define these parameters).

I have attached the original graph showing the min, typ and max values. I drew it to explain to someone why LEDs should not be connected in parallel. So you can ignore those aspects if you wish.

Mike - K8LH · Jan 1, 2006

Hey Guys and Gals,

Please forgive me, but, has anyone considered the resistance of all the wire strung out to some 20 doors? What kind of distances are we talking about?

Happy new year... Regards, Mike

ljcox · Jan 1, 2006

Happy new year to you also Mike.

Yes that is a good point, but so far, Angie has been learning the basics, so this is an issue for later when he is ready.

angie1199 · Jan 1, 2006

Ok, so graph 1 and 2 are from the datasheet.

Graph one shows Fv at 15mA. So resistor required is (12 – 4 * 1.92) / 0.015 for red which is 288 Ohm. Resistor for green is (12 – 4 * 2.1) / 0.015 which is 240 Ohm.

So in the circuit there would be a common resistor of 240 Ohm which would be used by all LEDs and each red would require an additional 12 Ohm resistor run in parallel to the red LED so that particular red LED is then using 15mA when lit.

This would then give the Luminous intensity of each LED colour as shown in Graph 2 which, at 15mA, are virtually the same.

Then, as was said earlier, another resistor is placed in parallel to the red LEDs as the voltage is different therefore it has to be allowed for. This is the circuit diagram I did in visio with the extra resistors in parallel. I think this is where you meant them to go.

audioguru · Jan 1, 2006

A 12 ohm resistor across an LED is like a dead short and will turn-off the LED. 330 ohms across the red LED would reduce its current 5mA and reduce the green current 6.5mA.

If the LEDs have a typical voltage drop of 1.92V for red and 2.12V for green, then with a 12V power supply a 240 ohm current-limiting resistor in series with four LEDs in series results in 18mA for red and 14.7mA for green.

I don't think that Luminous Intensity is the same as Visual Intensity because your vision is more sensitive to green. Therefore even though the green operates with less current it might appear brighter than the red.

Simple LED question (definately leading to more)

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