I have the following problem:
Consider the circuit below
Assume that Vo = 0 V, and that Vin slowly increases. At what value of Vin does Vo switch from 0V to 10V?
I'm not sure how to tackle this. I know that:
\$V_o=10V\$ if \$V_{id}>0\$
\$V_o=0V\$ if \$V_{id}<0\$
But I don't know how you would determine the exact point that this happens? Can node equations help me? Or maybe mesh currents?
Best Answer
I hadn't observed this question, earlier. But I thought I'd summarize the discussion and provide a general approach to answering this kind of question.
Let's take the schematic and draw it up using the editor provided (you should have done this in order to get part numbering, though I understand you were quoting a problem that included a diagram.)
simulate this circuit – Schematic created using CircuitLab
(We'll assume a magically ideal opamp here that only has access to ground and \$+10\:\text{V}\$. It's open-loop gain will be large so that we don't have to worry about it, too.)
So, the nodal equation for \$V_\text{X}\$ is:
$$\frac{V_\text{X}}{R_1=R}+\frac{V_\text{X}}{R_2=2\cdot R}=\frac{V_\text{IN}}{R_1=R}+\frac{V_\text{OUT}}{R_2=2\cdot R}$$
The opamp is arranged with the \$V_\text{X}\$ node connected to the (+) input. This means that when \$V_\text{X}\$ rises over \$4\:\text{V}\$ the output, \$V_\text{OUT}\$ will move in the same direction (upward, positively) towards its \$10\:\text{V}\$ rail. And that means that \$V_\text{X}\$ will move even further above \$4\:\text{V}\$, leading to still more upward, positively directed motion of \$V_\text{OUT}\$. In short, this is positive feedback. So the opamp will "rail." Put another way, this is a comparator. Of course, we already knew that. But it helps to be certain.
Okay, with that out of the way, this means that if \$V_\text{X}\$ rises over \$4\:\text{V}\$ then the output will rail at \$V_\text{OUT}=10\:\text{V}\$ and if \$V_\text{X}\$ falls below \$4\:\text{V}\$ then the output will rail at \$V_\text{OUT}=0\:\text{V}\$. Simple enough.
Given that \$V_\text{X}=4\:\text{V}\$ right exactly at the cross-over mark, we can solve the above equation for the cross-over point, as follows:
$$3\cdot V_\text{X}=2\cdot V_\text{IN}+V_\text{OUT}=12\:\text{V}$$
Since \$V_\text{OUT}\$ can only have two values (except for a very short period during its transition between the two values), namely \$V_\text{OUT}=0\:\text{V}\$ and \$V_\text{OUT}=10\:\text{V}\$, we can find the following two solutions for \$V_\text{IN}\$:
$$\begin{align*}V_\text{IN}&=6\:\text{V}\tag{$V_\text{IN}$ rising, $V_\text{OUT}=0\:\text{V}$}\\\\V_\text{IN}&=1\:\text{V}\tag{$V_\text{IN}$ falling, $V_\text{OUT}=10\:\text{V}$}\end{align*}$$
So there is substantial hysteresis, here.