operational amplifier - How are positive and negative feedback of opamps so different? How to analyse a circuit where both are present?

In an opamp, feedback on the positive input places it in saturation mode and the output is of the same sign as V+ - V-; feedback on the negative input places it in "regulator mode" and ideally Vout is such that V+ = V-.

1. How does the opamp change its behaviour depending on the feedback? Is it part of a more general "behavioral law"? [Edit: Isn't it something in the lines of the voltage added increases the error instead of reducing it in the case of + feedback?]



2. How can we analyse circuits where both are present?

Whoever answers both at the same time in a coherent manner wins a pot of votes.

Answer

1. 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.


2. 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 $A_v\rightarrow\infty$


3. 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.