Thank you for replying. Yes .45 is a digital low. It worked fine as is. I was wondering this more from an "electronics" POW. Why does a 10k pullup causes a 0.45V low while a 100k pullup causes a 0.006V low. According to the LM393N datasheet, when I apply:
R-pullup (minimum) = Vout/I_output_lo = 3.3V/6mA = 550 ohms.
wouldn't a 10k be more than sufficient?
The LM393's outputs are designed to pull "fairly low". Often they're used with digital logic, where low is anything between 0V and 0.8V, often more.
There will be some variation from one manufacturer to another. I'm working from the data sheet from Texas Instruments for the device that was originally made by National Semiconductor.
The relevant specifications are VOL (low-level output voltage), which is typically 0.15V, maximum 0.4V, at room temperature when the output is sinking 4 mA, and IOL (output current with output low), which is specified as typically 6 mA when VOL is 1.5V. No minimum value for IOL is specified.
The output pin comes from the collector of a transistor whose emitter is connected to the negative supply. According to the equivalent diagram, the base is driven with 80 µA to turn it ON.
With 80 µA of base current into a common emitter stage, with a current gain around 100, the transistor's collector voltage depends on the current that it is sinking. For a relatively low current, such as 50 µA for example, the transistor will be saturated and will pull its collector pretty close to 0V. Typically less than 50 mV.
But when the collector current starts to approach the base current multiplied by the current gain of the transistor, the transistor is no longer saturated and the collector voltage will rise.
80 µA of base current is not necessarily enough to saturate the transistor when it is sinking the maximum current (typically 6 mA). That's why this figure is specified when VOL = 1.5V. In other words, if you force the output pin to sink that maximum current figure, you can expect the collector voltage to be more than 0.4V. The 0.4V figure is specified at a collector current of 4 mA.
So I would conclude from this that the collector voltage will change from being quite low (transistor saturated, or nearly) at a collector current less than 4 mA, to a significantly higher voltage (transistor out of saturation) at a collector current of 6 mA or higher.