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# Different Colour Leds in a Serie

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#### Budiman

##### Junior Member level 1
I want to put one red and one green led in serie.
if the leds are 2.5V 25mA and 3.5V 45mA ( PSU 9V).

Regards,
Budiman

In series there will be same current in both leds, so if one is designed for 25mA and other for 45mA, one of them will ligth briter. I guess better is to calculate two resistors and connect them in series with diodes, and then two sets of diodes and resistor parallel.

You can connect the two LED's in series. The voltages you are mentioning are really just ratings, possible maximum ones.

The current should be more like 5mA or so.
Therefore, just assume a 2V drop across each LED and calculate the resistor for about 5mA:
R=(9-2*2)/5=1kΩ

Hello there,

Correct me if I'm wrong but LED's are still 'diodes'. Meaning their voltage drop across them is quite deffinate (like 0.6v in silicon diodes). This may vary slightly with temperature, but essentially, a 3.5v LED, will not work with less than 3.5 volts. Its not really a 'maximum rating'. That infer's that is could work at lower voltages. Theoretically..you could connect a 40v supply to an LED...if the max current available was low enough, say 10ma. But of course, normally a current limiting resistor is used in series. 10v supply, with a 2v LED in series with a resistor. Because the total voltage from the supply is 10v, and the LED has a 2v voltage drop...the voltage across the resistor is 8v. For a 20ma supply V=IR. R=V/I = 8/0.02 = 400 Ohms.

I, of course, could be wrong...(university seemed like so long ago...) but LEDs in parallel doesn't work...because of the voltage dorp..each one requires its own 'resistor'. But series....sure, why not. Just add up all the voltage drops of each LED, use a supply thats higher than the total LED votage...and work out the appropriate resistor to use in series..to limit the current flowing through the circuit.

2.5V 25mA and 3.5V 45mA ( PSU 9V)

So..2.5v + 3.5v = 6v. With 9V supply...the v-drop across the resistor would be 9-6=3v. For 25mA (0.025A) thats R=V/I = 3/0.025 = 120 ohms <- a value commonly available.

Of course, it wouldn't hurt to measure the voltage drop of the LED's, it may be different to the spec. It works fine in theory though.

Good Luck.

BuriedCode.

Hello,
Use resistor for each LED.
Bye

Here are snippets from the datasheets of two ordinary 5mm LED's, produced by Kingbright.
First, look at the forward voltages: at the current I suggested, 5mA, the green LED needs about 2.0V, while the red one will have about 1.9V. Therefore, the calculation I suggested was close to reality.

Next, look at the typical brightness, given at 10mA. The green LED is about 25% less efficient, but in practice you may not even notice the difference. Plus, the intensity varies so widely from LED to LED, that in practice you may find brighter the LED that is supposed to be dimmer.

IF you wish, use separate resistors, one for each LED. The advantage of this method is that it allows you to adjust the brightness of each LED separately.
The disadvantage, however, is that you will need approximately twice the current to run the two LEDs.
Therefore, I suggest you connect them in series and place a resistor across the one that is brighter. By adjusting that resistor, you can dim the brighter LED, while keeping the total current low, since it is limited by the common ballast (series) resistor.

Buriedcode :
Theoretically..you could connect a 40v supply to an LED...if the max current
available was low enough, say 10ma.

If the Led with 25mA 2.5V connected to CMOS output ( 12V 12.5mA max),
what is the R value ? How can I calculate the R value.
Is this work :

LED need : P = V x I => 2.5V x 25mA = 62.5mW
CMOS output : P = V x I => 12V x 12.5mA = 150mW
R must hold : 150mW - 62.5mW = 87.5mW
R Value : P = I^2 x R => 87.5 = 12.5^2 x R
R = 560 ohm

Can this bring the optimal bright to led ?

Budiman

First of all, I have to repeat that the rating you mention is an ABSOLUTE MAXIMUM rating and so you should stay away from it.

If you look at the graphs I posted, you will see that even for 25mA, the voltages across each LED are still in the 2V range.
Therefore, use 2V and calculate the resistor as I suggested:

R=(12-4V)/12.5mA=640 ohm. Use 620 ohms.

hi
All red and green diodes do not have same I-V chart, there for it is better that use separate resistor for each of them.

hi

they are right
you have to separate the 2 leds
see this drawing may be usefull

bye
Ahmed

I totally disagree.

The CURRENT in the circuit will be the same. Yes, that means the LED's will have different voltage drops across them. But the LED light intensity is dependent on the CURRENT, not the voltage. Therefore, if the LED's current/ intensity characteristics are similar, they will light up with much the same intensity.

VVV said:
I totally disagree.

The CURRENT in the circuit will be the same. Yes, that means the LED's will have different voltage drops across them. But the LED light intensity is dependent on the CURRENT, not the voltage. Therefore, if the LED's current/ intensity characteristics are similar, they will light up with much the same intensity.

ok
but if current was sufficient and forward voltage was not sufficient , it will not turn on. some LEDs such as blue LEDs need more forward voltage.

Yep, thats back to my point. LED's, dispite their intensity being proportional to current, obviously have a 'minimum' voltage as well. VVV. you may well be right about the maximum rating voltage, but I've done a few little measurements, and the V-drop across LED's is pretty much fixed, depending only on the colour/material of the die.

I agree that the voltage specified in datasheets isn't spot-on, I measured several blue LED's, and they varied from 3.8 - 4.2v. That said, each individual LED gave a constant voltage drop throughout the test.

I still don't understand why everyone seems to suggest 'parallel' configuration, when the topic is quite clearly about 'series' connection. I agree that I would nearly always use parallel for powering multiple LED's, but of course they can be connected in series! After all, I'm sure we've seen a few silicone diodes connected in series for a specific voltage drop of multiples of 0.6. (ie: 4 diodes = 2.4v drop). Also, some LED arrays used in LCD backlights are configured in both series and parallel, although it must be said, those are generally all the SAME colour/material, and therefore have the same voltage drop.

Again, why not do some tests? Its all very well in theory, we could even start talking about energy bands, band gap etc.. but nothing is beter than good 'ol experimentation. I stand by my old post, make sure you have a PSU that can put out more voltage than the sum of the LEDs, then work out a series resistor.

I'm not going to talk about 'intensity' since that is subjective. Our eyes sensitivity to light depends on the wavelength (colour) and LED's are available in various viewing angles which determine its so called 'luminous intensity' (mcd). In series, multiple colours could be very hard to match brightness, try to get ones that match intensity vs current, since each LED will get the same current.

Buriedcode :
Theoretically..you could connect a 40v supply to an LED...if the max current
available was low enough, say 10ma.

Notice how I started with *theoretically* The maximum current that could flow through the LED would be 10ma. That was my point, it would still probably send the LED to 'Gallium Aluminum Arsenide heaven'.

Post was a tad too long, sorry.

BuriedCode

If the forward current is limited, the LED will not be destroyed. You can even power the LED from the AC mains with an appropriately sized resistor. The only catch is that the reverse voltage would be too high and destroy the LED on the negative portion of the cycle. But if you connect a regular diode in anti-parallel, you CAN actually power the LED from the mains. I have done it and it works just fine. DO NOT FORGET THE ANTI-PARALLEL DIODE, THOUGH!

I agree with Buriedcode, the original question was about connecting the LED's in series. And a little easy-to-do experiment can provide the answers in no time.

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