The key is to think about what happens to the gain at DC when you couple with the capacitors (this is assuming the capacitors are large enough not to attentuate the signal passed through significantly - I'm assuming this probably isn't the point of the question since the values are not shown)
If you apply no signal to the input in each scenario, what is the output voltage?
- Op-amp always behaves as a differential amplifier and the behavior of circuit depends on the feedback network . If negative feedback dominates, the circuit works in linear region. Else if positive feedback dominates, then in saturation region.
- I think the condition \$V^+ = V^-\$, the virtual short principle, is valid only when the negative feedback dominates. So if you are not sure that negative feedback dominates, consider op-amp as a differential amplifier. To analyze the circuit, find \$V^+\$ and \$V^-\$ in terms of \$V_{in}\$ and \$V_{out}\$. Then substitute in the following formula,
$$V_{out} = A_v(V^+-V^-)$$ calculate \$V_{out}/V_{in}\$ and then apply the limit \$A_v\rightarrow\infty\$
- Now, net feedback is negative if \$V_{out}/V_{in}\$ is finite. Else if \$V_{out}/V_{in} \rightarrow \infty\$, then the net feedback is positive.
Example:
From the circuit given in the question,
$$V^+ = V_{in}\ \text{and}\ V^- = V_{out}/2$$
$$V_{out} = A_v(V_{in} - V_{out}/2)$$
$$\lim_{A_v\rightarrow\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\rightarrow\infty}\frac{A_v}{1+A_v/2} = 2$$
$$V_{out} = 2V_{in}$$
\$V_{out}/V_{in}\$ is finite and net feedback is negative.
\$\mathrm{\underline{Non-ideal\ source:}}\$
In the above analysis, \$V_{in}\$ is assumed to be an ideal voltage source. Considering the case when \$V_{in}\$ is not ideal and has an internal resistance \$R_s\$.
$$V^+ = V_{out}+(V_{in}-V_{out})f_1\ \text{ and }\ V^- = V_{out}/2$$
where, \$f_1 = \dfrac{R}{R+R_s}\$
$$V_{out} = A_v(V_{out}/2+(V_{in}-V_{out})f_1)$$
$$V_{out}(1-A_v/2+A_vf_1) = A_vf_1V_{in}$$
$$\lim_{A_v\rightarrow\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\rightarrow\infty}\frac{f_1}{\frac{1}{A_v}-\frac{1}{2}+f_1}$$
$$\frac{V_{out}}{V_{in}} = \frac{f_1}{f_1-\frac{1}{2}}$$
case1: \$R_s\rightarrow 0,\ f_1\rightarrow 1,\ V_{out}/V_{in}\rightarrow 2\$
case2: \$R_s\rightarrow R,\ f_1\rightarrow 0.5,\ V_{out}/V_{in}\rightarrow \infty\$
\$%case3: R_s \rightarrow \infty,\ f_1 \rightarrow 0,\ V_{out}/V_{in} \rightarrow 0\$
The output is finite in case1 and so net feedback is negative in these conditions (\$R_s < R\$). But at \$R_s = R\$, negative feedback fails to dominate.
\$\mathrm{\underline{Application:}}\$
Case1 is the normal working of this circuit but it is not used as an amplifier with gain 2. If we connect this circuit as a load to any circuit, this circuit can act as a negative load (releases power instead of absorbing).
Continuing with the analysis, the current through \$R\$ (from in to out) is,
$$I_{in}=\frac{V_{in}-V_{out}}{R}=\frac{-V_{in}}{R}$$ calculating the equivalent resistance \$ R_{eq}\$
$$R_{eq} = \frac{V_{in}}{I_{in}} = -R$$
This circuit can act as negative impedance load or it act as a negative impedance converter.
Best Answer
When you solve positive feedback circuits like this, you need some initial values.
We can say that \$V_{sat+}\$ as the upper limit to what the opamp can drive to and \$V_{sat-}\$ as the lower limit.
If we make an initial assumption that \$V_{out} = V_{sat+}\$ then you will get $$ V_+ = V_{sat+}\dfrac {R_1}{R_1+R_2} $$ $$ V_{out} = A_v(\dfrac{R_1}{R_1+R_2} V_{sat+} − V_{in})$$
When \$V_{in} < \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat+}\$ When \$V_{in} > \dfrac{R_1}{R_1+R_2} V_{sat+}\$ the output will be \$V_{sat-}\$
You would do the exact same procedure with an initial assumption that \$V_{out} = V_{sat-}\$ to see when you would see a transition going in the opposite direction.
Check out page 7 of Opamp circuits - Comparitors and Positive Feedback