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I have a 3 pin Bi-LED, two anodes and one cathode. I am using it with
two transistors, each transistor controls a anode. The only way I can
sucesfully use this Bi-LED is if I attach 2 signal diodes to the
cathode (gnd), thus creating two seperate cathodes, connecting them to
the collector of the transistor.

How will this effect the formula in choosing a resistor on the Bi-LED's anodes? I would assume that a smaller resistance would be chosen.
It should be noted that one of the transistors are also powering a 150mA relay.

http://www.myfilehut.com/userfiles/155002/BiLED.JPG

2006-08-12 14:10:45 · 3 answers · asked by Anonymous in Science & Mathematics Engineering

Pancakes, could you offer some details on your solution, where on the transistor would you connect the annodes?

2006-08-12 14:39:44 · update #1

the two current sources are for simulation

2006-08-12 14:51:09 · update #2

3 answers

I still don't quite get the point of this circuitry with the "OR" logic (the 2 diodes) to light up both LED colours (green & red). I would have thought only Q1 -> D7 -> D2 or Q3 -> D6 -> D2.
But that's your business....
To answer your question, each diode will only "steal" about 0.5 - 0.7 Volt away, not worth to change the resistor value, the LEDs will just be almost unnoticeably dimmer.

2006-08-12 15:03:40 · answer #1 · answered by Marianna 6 · 0 0

The only way to control each section of the bi-l.e.d. separately from Q1 and Q3 is to add two pnp transistors. You can probably figure out how to do that?

Tell me, what purpose do the two current sources serve?

2006-08-12 21:44:59 · answer #2 · answered by dmb06851 7 · 0 0

If either trans turns on, both leds will light. I think you want one trans to one led anode, and the other trans to the other led anode. The signal diodes will not be needed then.

2006-08-12 21:17:23 · answer #3 · answered by Pancakes 7 · 0 0

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