The dotted lines shows (in the previous post.) where the lm35 will be connected in a real circuit.
Sorry, a terrible idea. A couple of diodes in series is a much more sensible and reliable solution.
The LM35 requires 4mA. If ther is a resistor between it and Vcc it will make a voltage divider.
No That is not the way to use the LM35 (or any other IC, for that matter)!
Did you miss Figure. 18 in Reply # 30? If you want to offset the output of the LM35 up into a usable range you put something to offset the output UP from zero/ground. Putting anything in series with the supply pin does the OPPOSITE of what you need!
Be it 5.8 or 6volts does not matter.
Sorry, I don't know what that means?
In the image nothing say that the comparator inputs has to be higher than the output. It can be two sources of voltage, no? One to 7, 8 volts and the ouput can be connected to 12 volts going to the same ground as the 7, 8 volts? No?
The comparator simply compares the two inputs. It switches one way or the other depending on whether one pin is higher than the other or vice-versa.
If the + input sees 6V, then the comparator will turn OFF when the - input goes higher than 6V, and it will turn ON when the - input goes lower than 6V.
Connect the power input of the LM35 directly to the power bus (12V or whatever).
Then connect the ground pin of the LM35 through a couple of diodes to ground (as shown in Fig. 18 in Reply #30)
The diodes will OFFSET the LM35 voltage UP to a more manageable signal level of around 1.4V at zero degrees Celsius.
Then use a pot (like your "U3") to adjust the setpoint/threshold voltage into the + pin of the comparator.
You can set the output of the pot at around 1.5V so that your compartor switches the relay ON just above freezing.