Joined Mar 22, 2011
All you need to do is add a potential divider to V+ and another potential divider to v- pins on the trigger. From your values, your output voltage would be:
\(\frac{v_0}{v_i} = (\frac{100}{100+10k}) \)

What you need is different values for the two resistor, in series, forming the potential divider. Use 100 ohm connected to +12v supply and 500 ohm as the resistor connected to ground. Take the output between the two resistors and that will give you the 10v output. Just put in these new values into the above equation to see why you get a 10v output.

Hopefully that helped!


Joined Mar 24, 2008
The only thing that makes Schmitt Triggers hard is the output of the op amp is not rail to rail. You must know what the voltage of the op amp is to accurately design and predict what a Schmitt Trigger will do.

Being digital a Schmitt Trigger only has two conditions. They are set up by the output of the op amp and the resistors as voltage dividers as shown thusly...