The simulation is giving you the correct output for your simulation schematic. Compare the two schematics. You have the 866k resistor in the wrong place. It needs to be on the other side of the 100k resistor and the 100pF cap.
Edit: Since you changed you post while I was doing mine, my answer doesn't make sense to others who read it. The regulator has a 1.24V reference at the FB pin 4. So for 5V output change the 866k resistor to a 309k resistor.
The simulation is giving you the correct output for your simulation schematic. Compare the two schematics. You have the 866k resistor in the wrong place. It needs to be on the other side of the 100k resistor and the 100pF cap.
Instead of editing you original post you should have just added another post to this thread. That way it doesn't confuse other people who read the thread.
Instead of editing you original post you should have just added another post to this thread. That way it doesn't confuse other people who read the thread.
You're right, but by the time i edited it didnt have any answers yet, and by the time i edited you were answering so you still got the old post. Anyway thanks for helping
Edit: Since you changed you post while I was doing mine, my answer doesn't make sense to others who read it. The regulator has a 1.24V reference at the FB pin 4. So for 5V output change the 866k resistor to a 309k resistor.
The LT1302 adjusts the output voltage until the voltage at the FB pin is 1.24V. Thus you add a voltage divider between the output and the FB pin to give 1.24V at FB when the output is at the desired voltage. Calculating the divider resistor values to get 5V at the output gives 309k and 100k for ≈1.24V at pin FB. 866k and 100k gives 12V at the output.