MAX999 draws current through negative input

Question

I have two versions of the input signal:

• 5V inactive level, 3.2V active;
• 5V inactive level, 0V active level.

Thinking on making universal circuit to support both signals, I decided to use comparator, having threshold voltage at about 4V. Here's the circuit:

I built a prototype, and it shows unexpected (for me) behavior:

Red is voltage on positive input pin, yellow is on negative. Capacitor C58 was removed.

Searching through internet I found this question. It is about operational amplifier, and not about comparator. But I took a closer look into the datasheet, and found the following:

Differential Input Clamp Voltage --- VCC = 5.5V, VIN- = 0V, IIN+ = 100µA --- 2.1 V

Is it the information I should be looking at? Does it mean that voltage between + and - inputs must be not larger than 2.1 V, otherwise there will be current path between the inputs? Is it naive to assume that if I have comparator powered from 5V, that it must be able to properly handle differential voltages on the whole power scale? And why datasheet does not state that there's such a limitation in its headline?

Update:

After considering answers below I decided to stick with MAX999 so far, and designed the following circuit:

I still think it is an overkill, but at least it is expected to work properly:

• supply current is about 8 mA. In my previous simulations MAX999 model was failing when IN+ was approaching IN- starting consuming about 60 mA, most probably because of issues in definition of the input stage circuitry.
• digital input was resized and re-leveled (green), with reference voltage placed into the exact center of the worst case for the input signal;
• circuit draws less than 1 mA from the positive input, and almost nothing through the negative input;
• power on current is much less than 10 mA, limited by the 470 Ohm resistor;
• differential voltage does never go out of 2.1 V range.

Answer 1

IIN+ = 100μA 2.1V means you may limit current the source voltage with a series R 2.1V/100 μA =21k is the threshold of conductance of the clamp.

Current into Input Pins .....................................................±20mA

Then consider driver Vol/Iol limits and (Vol(=0 @ 50 Ohms)) Thus divider divider is 4V (=Veq) / 1.8k (=Req) = 2.22 mA So you are OK. But with cap you need to add 330R ~1k in series = Rs on input.

Your driver Spec = TBD

Answer 2

If you look at the MAX999 data sheet again you will see this on page 2: -

Differential Input Impedance @ VCC = 5V 8 kΩ

And this is more likely giving you the effect you see although the clamping voltage at 100 uA is also contributing.