Simple LED question (definately leading to more)

[ljcox]

Hi Ron,

Ok, contact with the green annode (is that right) (no - too many Ns), ie anode will be disconnected when the door is open and at the same time contact to the red is made.

So as far as I can see this is the same as having a straight forward 2 way toggle switch with no off position (I've used them before)

Would this be easier with small (3mm dia 8.5mm long) bulbs instead of LEDs? Either works good for me.... It makes no difference ion principle

Agie

 

[ljcox]

I gather that you have 3 lead LEDs with a common cathode. If so, then it can be done as attached.

 

[angie1199]

Sorry that I haven't even the basic knowledge for this...

ljcox - I meant in using normal bulbs I wouldn't have the problem of resistors as the bulbs are 12v and they choose to draw the 65mA current.

What you've drawn is what I was thinking of but.... and there's always one of those.... The resistance of the red LED is different to that of the green LED and that's where I start to get confused. As the same resistor is being used wouldn't that cause a problem?

<note to self - wish I'd listened to Guy at school!>

Angie

 

[ljcox]

Sorry that I haven't even the basic knowledge for this... No problem, that's why we are here to help

ljcox - I meant in using normal bulbs I wouldn't have the problem of resistors as the bulbs are 12v and they choose to draw the 65mA current.
Agreed, that's your choice; but you will consume much more power and you would need one red and one green globe per door.

What you've drawn is what I was thinking of but.... and there's always one of those.... The resistance of the red LED is different to that of the green LED and that's where I start to get confused. As the same resistor is being used wouldn't that cause a problem? Possibly yes. But if there is noticable difference in brightness, you could insert a resistor in series with whichever LED is brighter (ie. insert it between its anode and the switch). I know that means you need an extra 20 resistors but again you have to decide.Angie

 

[angie1199]

ljcox - I meant in using normal bulbs I wouldn't have the problem of resistors as the bulbs are 12v and they choose to draw the 65mA current.
Agreed, that's your choice; but you will consume much more power and you would need one red and one green globe per door. I'm learning a lot here. Is there that much difference between the 2.2v 30mA LED and the 12v 62mA bulb?

What you've drawn is what I was thinking of but.... and there's always one of those.... The resistance of the red LED is different to that of the green LED and that's where I start to get confused. As the same resistor is being used wouldn't that cause a problem? Possibly yes. But if there is noticable difference in brightness, you could insert a resistor in series with whichever LED is brighter (ie. insert it between its anode and the switch). I know that means you need an extra 20 resistors but again you have to decide.Resistors at 1p each I can handle but the spec sheet for the LEDs say the typ mcd is 40 for the red ad 35mcd for the green. Is there much difference between those values.

The product page also says "which has uniform light output and a low power consumption" about the LED. does this mean it won't be noticeable or is it trial and error?

 

[ljcox]

typ mcd is 40 for the red ad 35mcd for the green. Is there much difference between those values.

I don't know, but someone else may. I would have to test view.

 

[checkmate]

Just some clarifications for you to get a clearer picture.
LEDs are a type of diode, and diodes dont have resistances unlike resistors. The function of an ideal diode is very simple, it allows current to flow only in one direction. The key thing is that it allows ANY amount of current to flow in one direction. Therefore, the main role of the resistor is to ensure that the right amount of current is fed into the LED. The bigger the resistor, the less current that flows. If the resistor is too small, or non-existant, the LED simply take in whatever current the power supply provides, generate heat and burn itself out.
The only other thing you have to take note is that diodes in the practical world are non-ideal. They have an exponential V-I curve. As such, it has a non-zero turn on voltage, and when turned on, is assumed to pass ALL currents fed into it at that voltage. Therefore, their specifications often come with a maximum forward current (max If). If you know these specs, the resistor provided can simply be calculated using simple ohm's law.

 

[angie1199]

........Therefore, their specifications often come with a maximum forward current (max If). If you know these specs, the resistor provided can simply be calculated using simple ohm's law.

HE Red
Forward voltage at IF=20mA: 2.5V
Forward current max.: 30mA
Reverse voltage max.: 5V
Wavelength @ peak IF=20mA: 627nm
Power dissipation PT: 105mW
Light output min.@ 20mA: 12mcd
Light output typ.@ 20mA: 40mcd

Green
Forward voltage at IF=20mA: 2.5V
Forward current max.: 25mA
Reverse voltage max.: 5V
Wavelength @ peak IF=20mA: 565nm
Power dissipation PT: 105mW
Light output min.@ 20mA: 12mcd
Light output typ.@ 20mA: 35mcd

So using my newly aquired skill in working out the resistor required:

r=(12v-(2.5v*4))/25mA
so r = 80 Ohms which according to my chart I found for working out pretty colour codes means I need a 'black - grey - black - doesn't really matter' resistor for each starting point of each line in series.

Is that right?

Ah, they don't do a 80Ohm so need a 82 Ohm which is 'black grey red and whatever'

 

[Someone Electro]

Use rather 15 to 20 mA.

At 25mA the LED will blow.

 

[angie1199]

Use rather 15 to 20 mA.

At 25mA the LED will blow.So 100 Ohm for 20mA?

And what's the difference between an 'MF' resistor and a 'min res' one please?

 

[audioguru]

Angie,
You were looking at the max LED voltage rating which is 2.5V. You should also look at the typical LED voltage column which shows a red LED's voltage of only 1.85V (it could be less). With 4 red LEDs in series with an 82 ohm resistor from 12V, the current is 54mA and they will smoke.

 

[Styx]

...

Exactly.
angie1199, the thing with LED's (as well as other semiconductors if you do loss-calc) is you need to define the operating point and extract data.

So from this 20mA has been recomended, so check the datasheet (remember my original post...) and get a more accurate volt-drop. Then use this volt-drop to size yr burn-resistor

 

[angie1199]

Hang on guys!!! I didn't know what Ohms law was till yesterday now you're talking about loss-calc!

AudioGuru: I think you misread the datasheet. The datasheet covers three different bi-colour LEDs. The LED I want is the High Efficiency Red and Green (The first two listed under the 'device' section - L-3WEGW) which are 2.0 and 2.2 volts respectively. This is why I initially added up the highest possible voltage (2.5v) but was informed by Someone Electro that I should use the typical voltage so I added up the lower of the two being 2.0v which gave me .......

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)

Is this right now? 200 Ohm, 4 in series, typical voltage 2.0?

Angie

 

[audioguru]

The LED I want is the High Efficiency Red and Green (The first two listed under the 'device' section - L-3WEGW)

Good, we don't need to guess anymore which LED you are planning to use.
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.

Is your supply 12V or 12.5V and does it have a 1.1V diode in series with it?

These LEDs are the small 3mm size. They have 3-leads which makes switching them on and off separately very complicated.

 

[ljcox]

I'm sorry, but I made an error in my post yesterday. I wrote that if you need to reduce the brightness of one LED (say the red one for example) then you need a resistor in series with all of the reds.

Wrong. The resistors should be in parallel with their LEDs in order to bypass the extra current.

For example, say that, after testing, you decide that the green LEDs need 10 mA but the red only need 8 mA. Then you need to bypass 2 mA.

So the resistor required is R = 1.85/2 = 0.92 k. You can't buy a 920 , so use either a 910 (if you can buy them) or an 820 or a 1 k.

 

[ljcox]

Ah, they don't do a 80 Ohm so need a 82 ohm which is 'black grey red and whatever'

82 ohm is not 'black grey red' it is grey red black for a 3 band resistor.

 

[ljcox]

The LED current has been quoted as 15 mA by some and 20 mA or more by others.

Normal LEDs only need about 6 mA to give a reasonable glow.

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.

 

[audioguru]

Normal LEDs only need about 6 mA to give a reasonable glow.

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.

Hi Len,
You haven't seen my latest project yet, I sent it to another website but it isn't up yet.
I'm feeding a whopping 26mA to each ultra-bright LED, and my little 9V rechargable Ni-Cad battery nearly has a fit when they are all lit!

 

[audioguru]

It is a little on the bright side.

 

[angie1199]

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.

..... 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?