Here I am again. There must be another problem in my build. I come back here because I know its a piece of cake for many. I tried my best.

 

Might try adding another 1N4148 diode in series.

 

The voltage through the temperature sensor is too big and it fails. Adding another diode didnt seem to help.

 

A major problem with your circuit is that as the diode junction voltage drop changes with temperature, there is a very similar change in the transistor base-emitter junction voltage drop, so the two 'cancel' each other.

 

The voltage through the temperature sensor is too big and it fails. Adding another diode didnt seem to help.

Define "fails".

 

What temperature change are you trying to detect?
You might want to use a thermistor instead of a diode and replace R2 with RV1.

 

This is just a schematic I found on the internet and I was curious to try it out. By fail I mean burn, because the voltage is ~7.29V and the minimum allowed is 5.5V. The resistor values are taken from the site where the picture is from and it says that the temperature should not go over 125C. I guess you are supposed to change its sensitivity with the potentiometer, so no defined values of temperature.

 

This is just a schematic I found on the internet and I was curious to try it out. By fail I mean burn, because the voltage is ~7.29V and the minimum allowed is 5.5V. The resistor values are taken from the site where the picture is from and it says that the temperature should not go over 125C. I guess you are supposed to change its sensitivity with the potentiometer, so no defined values of temperature.

What do you mean by "burn"?
What is there to burn in that circuit?
Your minimum resistance is 220Ω.
Even at 9V, I = V/R = 9V/220Ω = 41mA

The only thing that would be at risk is the LED.
Replace the 220Ω with 1kΩ and the LED would be safe.

 

This is what the simulator that I use says.

 

This is what the simulator that I use says.

Says what?
Post the simulator schematic and results.

 

A better result might be had if a stable V ref. was used like LM385Z, 1.235 V, 10uA to 20 mA & a comparator.
Pot. should be in V ref. leg .

 

At no point have you told us what it is you are trying to achieve. What do you want the LED to do, and when do you want it to do it? If the idea is that the LED comes on or goes off at a certain temperature, this is a very poor circuit for that task.

For this circuit to work, there must be a fairly large difference in temperature between the diode and the transistor. As mentioned above, if the transistor and sensing diode are in close proximity, there will be very little temperature difference between them. Once you get past that, the diode's forward voltage changes very little with temperature, so the transistor will turn on gradually as the diodes temperature changes by tens of degrees. A diode can be a reliable and low cost temperature sensor, but it takes more than this circuit can do to realize that.

The transistor's base-emitter junction characteristic is acting as a very low gain comparator. The circuit can be improved by adding an additional transistor and diode. For a real temperature sensor, use an actual comparator or opamp.

ak