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I am trying to create a signal using two LEDs a Red and a Green. The problem is that I am replacing LEDs in a different circuit. However,  the Clear Red LED has a fwd Voltage of 2.2V and the Clear Green LED has a fwd V of 3.2 volts. 

So when I put them in the circuit the Red is bright and the Green is just a little green dot. 

DigiKey does not sell any Green LEDs, in the style I want, that have a lower Fwd V. of 2.2 Volts. Is there a way to add a component to equalize the voltages in  both LEDs so they are both BRIGHT?

It seems that a resistor on the Green LED side would just drop the voltage more.

I wish I had paid closer attention in the EE classes I was forced to take in Civil/ Structural  Engineering.

Last edited by AlanRail
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You can "tune" the resistor values to equalize the brightness of each LED.  LED's vary pretty widely in efficiency, so from a group of five different brand red or green LED's with the same current, you're liable to see a fairly significant difference in intensity.  This is the reason I buy my LED's in bulk and determine the relative brightness, makes it easier to build with them, I know what the performance is until I have to buy another batch.

Since you need current limiting resistors anyway, pretty much any DC supply will do the trick here.

NumberOne posted:

values, but a wide range of values will work. For example, using a 12VDC supply, use a 220 ohm resistor for green and 330 ohm for red.

Not really, those values are too low.  Standard LED's have a maximum current rating of 20ma, and the red LED's operate at about 2V, the green ones at around 2.4-2.5 volts.  So, you have to drop around 10 volts at 20ma from the 12V supply.  By rapid calculation, that's 500 ohms, using a 220 for the red results in 45ma of current, well over twice it's maximum rating, the 330 for the green is running at around 29ma, still well over it's maximum current rating.  Yes, it'll they'll bright, but not for long!

If you are running the external LEDs parallel to the existing LEDs on the board, that is the problem.      If not, you can ignore the rest of this.  

If the Green-Green pair are not very closely matched, all or most of the current goes to one LED leaving the other one dim or off.  Same thing for the Red-Red pair, although in your case they may be matched closely enough.  If this is what you are doing, the solution is to disconnect the LEDs on the board.  You could install the Green-Green pair in series, but then you may need to find and reduce the dropping resistor on the board.  This makes your task a bit more complex.

I have run LEDs from the same lot in parallel against conventional wisdom.  Some play well together and some do not.   I have even had pairs that shared current well for a while, then later one steals all the current.   Best practice is to run LEDs in series, not parallel.

Bob

So it appears you're trying to "tap into" and steal power from an existing red and green LED circuit.

You would be unbelievably lucky to find 5mm LEDs (red AND green) that would have identical characteristics to the existing red and green LEDs.  In other words, it just so happened that the reds were matched close enough that it seems they split/shared the available current in some fair/balanced ratio so both were suitably bright.  OTOH, the greens did not split/share in a suitable ratio.  Sure, you can fuss around with modifying the 2500C components, experimenting with other green LEDs, etc.

But as you're apparently willing to mess with components and perhaps do some simple soldering...

It appears it took some time to find 5mm LEDs that are suitable for whatever new signal head you are building.  Since this is a track-side (vs. rolling stock) application, you have plenty of power available.  I'm thinking you should simply "buffer" the DZ-2500 red and green signals using a 25 cent (or so) opto-coupler to drive your new red and green LEDs.  This would minimally load the DZ-2500 LEDs so they retain their original brightness.

Last edited by stan2004

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