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Is it "circuit safe" to use signal diodes on the cathode of a Bi-Led?

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. 150R x 2 Bi-LED V+ ---------/\\/\\/\\/\\/---------==|\\___ Diode1 V+ ---------/\\/\\/\\/\\/---------==|/ \\---|\\______ TO COLLECTOR OF TRANSISTOR 1 \\ \\ \\ | |/ | | Diode2 |--|\\______ TO COLLECTOR OF TRANSISTOR 2 |/
Reply to
JedOs86
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Heres a readable schematic:

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Reply to
JedOs86

DOH,, NO

Both LEDs will light at all times.

Just use two LEDs.

Reply to
Donald

Yeah, youre right LOL. any BiLEDS with 4 leads?

Reply to
Mr.J Dizzle

Notwithstanding your "ASCII art" is practically incomprehensible, from your description, what you need is high-side switches. If each anode has either an NPN emitter follower or PNP high-side switch, then you should be able to just ground the LED cathode. And of course, give each anode its own dropping resistor and transistor.

Good Luck! Rich

Reply to
Rich Grise

Yeah - see my previous answer - the 1N914's are an OR gate, so both sides of the LED will light either way. Either ground the LED cathode and connect the collectors to the anod/resistor junction, which will waste power because the transistor will be conducting when the LED isn't, or put two PNP's for high-side drivers on top of the 150R's, and lose the 1N914's, and just ground the LED cathode. Then you'd have to use a couple of, say

4K7 resistors:

VCC VCC | | [4K7] | | E +-----------| PNP | C [4K7] | | [150R] signal C | -----| NPN [LED] E | K | | GND GND

Twice.

Cheers! Rich

Reply to
Rich Grise

Thanks Rich, I think I will just stick with 2 LEDs

Reply to
Mr.J Dizzle

How about two cathodes ??

You found an LED with two anodes, I am sure on the same web site will be one with two cathodes.

this in not rocket science.

Reply to
Donald

Drive each anode with an emitter follower connected back to S1 and S2.

Reply to
John Popelish

Oh, and drive the base of Q2 with the LED current coming through the common cathode, and eliminate D3,4 and R2.

Reply to
John Popelish

If you can find me a 4 pin Bi-LED, or a two cathode one anode Bi-LED, I will bow down to you. I have never seen such a thing

Reply to
Mr.J Dizzle

Surely you are just stuck in one mindset.

As someone pointed out, you can simply drive the anodes with separate drivers, which really isn't a big leap from the concept of a two color LED with two cathodes.

Or, one used to (I've not looked recently) be able to get two LEDs in a package arranged back to back, so there were only two leads. That was the original form of two-color LEDs. To use them, you simply had to put one lead high and the other low, and reverse the process for the other color. In other words, you drove lead directly, and the other via an inverter.

Michael

Reply to
Michael Black

HELL, LMAO!! I actually found a Bi-Led with common anode!

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Reply to
Mr.J Dizzle

that circuit is seriously wacky, what's Q2 for?

what is the purpose/function of the LED.

can s1,s2 be operated in any combination?

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Bye.
   Jasen
Reply to
jasen

You don't have near enough base drive on Q2, it is off by almost a factor of ten.

Reply to
Fred Bloggs

How do you figure that there is not enough base drive on Q2? thats a 1K resistor... Nevermind the BiLed, that was real stupid, dont even know why i added that.

Reply to
Mr.J Dizzle

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