Lighting Bi-Color LED from Single Power supply

Maybe you could wire in an opto in place of one of the LEDs and drive a relay with that.

It's a bi-color LED (2 pin-2 LED), so that would not be an option.

How about parallel the LED with an opto. To do so I would have to...do what? Half the pull-up resistor and find an opto that could handle the same current? Sounds too easy. Sounds like I'm missing something. Would the opto and the LED have to have matched Vf ratings too?

With the existing circuit as drawn with the LEDs back to back so to speak the Vsat out of the comparator can never exceed the breakdown voltage of either LED, one of the two is always reversed biased so if the breakdown voltage of the LED is 5 Volts that is as good as it gets. The max current sink on a comparator like the LM393 is about 16 mA. Given a choice I would use the output of one comparator to drive a LED and the other to drive a transistor like a 2N2222 to drive the relay with a LED across it. This does away with the back to back LED configuration.

Ron

Use the relay coil to substitute for one of the pull-up resistors in your circuit. Put a resistor, or a zener diode, or a series string of ordinary diodes, in series with the LED that turns on in the state when the relay is not energized to limit the current to 20mA for that LED. It is not an ideal solution since some current will flow through the relay coil when it is supposed to be de-energized, but probably not enough current to switch the relay (pull-in voltage for a relay is typically half the coil operating voltage). Better, since you have to add a transistor to drive the relay anyway, use the input signal source to drive the added transistor and relay seperately from the rest of the circuit.


Edit: I forgot that your LEDs are package together so you don't have access to put parts in series with only one of the LEDs. So, in addition, lower the value of the other pull-up (to be equivalent to the relay coil resistance) and put a resistor in series with the prepackaged LEDs.

I already went down all those paths. Sorry.

ccurtis said:

Use the relay coil to substitute for one of the pull-up resistors in your circuit.

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That was the first thing I thought of and is cover in both design issues. A) LED (20mA) can't pass the relay current (77mA) and B) the relay turn on voltage plus the LED voltage is greater than the supply voltage, so neither will turn on.

ccurtis said:

Better, since you have to add a transistor to drive the relay anyway, use the input signal source to drive the added transistor and relay seperately from the rest of the circuit.

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C) The output of the comparator is not enough to drive either the relay nor the LED.

Reloadron said:

With the existing circuit as drawn with the LEDs back to back so to speak the Vsat out of the comparator can never exceed the breakdown voltage of either LED, one of the two is always reversed biased so if the breakdown voltage of the LED is 5 Volts that is as good as it gets. The max current sink on a comparator like the LM393 is about 16 mA. Given a choice I would use the output of one comparator to drive a LED and the other to drive a transistor like a 2N2222 to drive the relay with a LED across it. This does away with the back to back LED configuration.

Ron

Click to expand...

I figured out two LEDs would would, but I thought I'd like to try out the bicolor LEDs. I'm thinking it's not going to happen. I did figure out option three...add a second transistor to one comparator to drive the relay. But initially I thought I was giving up to easily and skipping some clever solution. I'm now getting the feeling there is no clever solution to this problem. (other than finding a comparator with more umph)

ADWSystems said:

I already went down all those paths. Sorry.



That was the first thing I thought of and is cover in both design issues. A) LED (20mA) can't pass the relay current (77mA) and B) the relay turn on voltage plus the LED voltage is greater than the supply voltage, so neither will turn on.



C) The output of the comparator is not enough to drive either the relay nor the LED.

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Yes, but for the sake of addressing both those design issues I assumed that a comparator could drive both the relay and the LED, as you would have me assume from one of your earlier replies to another poster pointing out that you could just add an open collector transistor buffer to each comparator output. Using that assumption, both those design issues are adequately addressed.

People talk about the TYPICAL current from the comparator as being 16mA. But really they should consider that its MINIMUM current is only 6mA which is not enough to drive anything except the low base current of a transistor.
The maximum saturation voltage loss is 1.5V when its current is only 6mA.

On one project I used an LM393 comparator in a wireless microphone base station to drive a tiny low power relay (6mA) that turned on a spotlight.

audioguru said:

People talk about the TYPICAL current from the comparator as being 16mA. But really they should consider that its MINIMUM current is only 6mA which is not enough to drive anything except the low base current of a transistor.
The maximum saturation voltage loss is 1.5V when its current is only 6mA.

On one project I used an LM393 comparator in a wireless microphone base station to drive a tiny low power relay (6mA) that turned on a spotlight.

Click to expand...

I found out imperically and then from the datasheet, the output transistor of the LM393 has a gentle climb on low mA and the has a "break point" and starts climbing rather steep. Even though it can handle 20mA, the Vsat loss is quite aweful.

The LM393 dual and LM339 quad comparators have a weak output because they are "low power" like the lousy LM324 and LM358 low power opamps.

It's a bi-color LED (2 pin-2 LED), so that would not be an option.

Click to expand...

Use two LEDs one being the opto.