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So my problem is simple: I have an IC that provides an output pin to alert you when some condition is met. The output pin is open-collector, high-impedance when active. I want to monitor this output on another IC (a microcontroller, specifically).
Is this as simple as having a pull-up on the output so when it's "on"... the level goes low and I just monitor for the high-to-low transition? I feel like I'm overthinking this, but it's not immediately setting off a light-bulb in my head.
Yes, all you have to do is to add a pull-up resistor to the open-collector output, provided that the other IC input doesn't need much current it will be OK.
The typical value for the pull-up resistor is 10 K.
Yes! It's that simple. That's part of the reason open collector is so popular.
You'll find that all the cheap comparators (e.g. LM393) have OC outputs.
If you attach a pull-up a resistor from the range of 1kΩ − 15kΩ everything should be O.K. If you want to do everything properly, consult the IC's datasheet for the OC current ratings and use an appropriate resistor.
Jonny and Bruno already confirmed that this is the right way to do it. Most microcontrollers have internal pull-up resistors on their I/O pins, so you could use that and save an external part. You may have to enable the pull-up resistor, since they're not always enabled by default.
Open collector outputs have two main uses:
For high speed the open collector has a disadvantage, however. While the transistor can sink enough current to discharge a capacitive load fast (think of a long wire), the pull-up resistor will make rising edges slower, depending on the RC time constant.
One point not yet mentioned is that while a resistor may be used to pull up an open-collector line, it will waste more current for a given level of performance than would some other methods. The problem is that while the open-collector output is simply sitting low, the resistor passes the most current even though nothing is happening and thus speed really doesn't matter, and only half as much current when the output is approaching half-rail and speed does matter. A constant-current source would pass the same amount of current in both scenarios. Some more sophisticated circuits can do even better, passing only a little bit of current when the output is simply sitting low, but increasing the current as soon as it starts to move. Such approaches can greatly reduce quiescent current without significantly sacrificing performance; the biggest danger is that if the output isn't being driven hard low, they may (depending upon the exact approach) draw excess current and/or oscillate. Still, active approaches may be helpful in some cases where one needs to e.g. detect quickly when a switch opens but doesn't want to burn much current through the switch while it's sitting idle.
