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?
angie1199 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.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.
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.
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
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.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
Audio, That's essentially what I attached on a previous page.audioguru said:You can wire them like this:
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:ljcox said:Audio, That's essentially what I attached on a previous page.
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.
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.
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?
audioguru said:You can wire them like this:
Well done, I did it with a finger nail dipped in tar.audioguru said: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:ljcox said:Audio, That's essentially what I attached on a previous page.
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.
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.
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.
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)
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.
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