This is a TTL logic level checker circuit. The highlighted blue area is a checker pin. Voltage at the checker pin is 0.65 V, so this must be logic 0, the LOW LED will on.
But the input of the first 7404 NOT gate in the first row is 1.3 V and the 2nd 7404 output is LOW, logic 0.
Why does the first NOT gate in the first row think 1.3 V is a logic 0 (LOW)?
Isn't a TTL voltage in the 0.8-2 V range invalid?
I assume since your circuit says it is using 7404, and not for example 74LS04 etc then it is really classic 7404 TTL.
So you are right, according to the datasheet the range for a low input is 0v to 0.8v, and a high from 2v to Vcc or 5v.
So, any input up to 0.8v is guaranteed to be a 0, and any input 2v and above is guaranteed to be a 1.
As you point out, the region in-between is a no-man's land. But that doesn't mean there is no output, it's just undefined. The input could be considered a 0 just as well as 1. But it has to be one or the other.
In your case, the gate considers the input of 1.3v to be a 0. This is not unexpected; the midway point between 0v and 2v is 1.4v, and this is slightly below that. Since this is a simulation, it is probably programmed to act that way. But in real-life, even if the input was 1.9v, it could still be considered a 0, just unlikely.
BTW your two voltmeters are not connected the same way, your top one (1.323v) shows the voltage going into the top 7404, but your bottom voltmeter (0.652v) shows the voltage at the junction of the two diodes, so the voltage is a diode drop above the voltage going into the input to the bottom 7404.
That circuit just doesn't work.
For instance, if the voltage divider R2 D1 D2 D4 is set up properly there'll be 0.8 volts at the junction of D2 and R4, but I don't think it's possible to set it up properly.
Then, if U1C is connected as shown, and there's 0.8 volts on its input, its output will go high, and LED1 will not light.
The same problem exists, in reverse, for the "high" sensor, with the sensing level set to Vih, which is 2 volts.
The schematic below shows one way to fix the problem.
Vcc is 5 volts and is common on both circuits, as is GND.
The voltage divider sets the inputs to the unity-gain buffers at 2.0 volts and 0.8 volts, the buffers being used to prevent loading on the divider by the circuit on the right.
When its input is connected to "HI", the HI LED lights while the LO LED doesn't, and when its input is connected to "LO" the LO LED will light and the HI LED will not.
A double comparator in an 8-pin DIP package would be a single-chip solution:
R1,R2,R3 divide the input voltage approximately by half. With input open, the R1,2,3 central point is at 1.4V. That’s halfway between 0.8V and 2.0V. R4,R5,R6 establish the scaled versions of the 0.8V and 2V threshold. The comparators have open-collector outputs and can directly drive LEDs as shown. No LEDs are ON when the input is between 0.8V and 2V or open.
We can also discriminate between open input and an invalid input state by detecting presence of input current via R1. There are many ways to do it. One easy way to do it is using another pair of comparators. The four could be two doubles or one quad in a 14-pin DIP.
