How to wire two open-drain logic outputs to a three pin bi-color LED

Question

I am very new to working with electronics; my main experience is in programming. I am working on a project and intend to use this MCP73833/4 Li-Ion Charger. It has two "Charge Status Output" pins in an open drain configuration. This table shows the state of the pins depending on the current state of charge.

The two states I care about are "Charge In Progress" and "Charge Complete". So for those two status, there will be one pin in low state, the other in high impedance. This is where my lack of knowledge creeps up.

I've been reading other answers about setting up LED's to open-drain pins, but none of it has fully made sense to me. I understand that when a pin is in a low state, it's connected to ground. So in the case of a three pin bi-color LED, I assume it'd be best to use a common anode configuration and the pins connected to their appropriate cathodes? Also, how do I set it up that the low state allows the flow of current through the LED without running too much current through the pin itself? I found this information in the datasheet, but I'm not sure I understand what each of it means:

The columns from left to right are "Parameters", "Symbol", "Min", "Typical", "Max", "Units", and "Conditions". I don't think I fully understand the significance of this information. Any help pointing me in the right direction to learn about this would be greatly appreciated.

Thanks for your time!

Answer 1

The output voltage is rated at a sink current of 4 mA, though it will typically sink 15 mA. Max sink current current is an odd specification in this part of the data sheet. Usually you'll have a min sink current specified in the electrical characteristics, and a max in absolute maximum ratings.

10 mA is a reasonable nominal current to run through a LED, most will handle 20 mA, and most will be pretty bright with much less. High accuracy is not needed for this application.

You can limit the current using a series resistor between the open drain output and the LED.

You need to know the nominal forward voltage drop of your LED, and the supply voltage you'll be running it from. Assume the open drain output will go down to 0 V, this will give you a conservative approximation. The voltage across the resistor will now be Vsupply - Vled.

Divide that voltage by your wanted current, ie multiply by 100 for 10 mA, and that will give you the resistor value in ohms.