Negative Feedback (13.2.12)

Today in class we learned about negative feedback.  To understand this, we calculated the value of Vo for the following circuit:

Because the value of A is large, it leads to a more simplified model, saying that if:

V₊  > V_- then V_o = 12V

V₊  < V_- then V_o = -12V

When we add positive feedback to the OP Amp, like this:

Then it exhibits hysteresis and if:

V_- < 6V then V_o = 12V

V_-  > -6V then V_o = 12V

This might sound familiar, because it is an example of the Schmitt trigger we learned about last week.

So that was when we added positive feedback.  But let's say we have this:

Then V_- = V_o/2 and if

V₊ < 6V then V_o = 12V

V₊  > -6V then V_o = 12V

Which can be graphed like this:

When -12V < V_o < 12V, then V₊ = V_- .  We already established that V_- = V_o/2, but since V₊ = V_- we can replace V_- with V₊ and solve for V_o, so that V_o= 2V₊ .  When we graph this, we find a line like this:

With positive feedback, we don't see the vertical blue line , but with negative feedback, we do.  The vertical green line is the jump from -12V to +12V.