Electronic Circuits and Projects Forum

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?

I already went down all those paths. Sorry.

Originally Posted by ccurtis

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

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.

Originally Posted by ccurtis

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.

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

Originally Posted by Reloadron

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

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)

Originally Posted by ADWSystems

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.

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.

Originally Posted by audioguru

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.

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.