Why is \$V_o=-V_c\$ in this circuit?

Question

Why is \$V_o=-V_c\$ in this circuit? Is it a derivation from Kirchhoff's voltage law?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

The key is to recognize that this circuit has negative feedback, so for an ideal op amp we can assume that \$V_- = V_+\$. Since \$V_+ = 0\,\text{V}\$, \$V_-\$ is effectively connected to ground (a virtual ground). Having made that assumption, KVL can be used to get the result.

Answer 2

Reasoning: -

• Both the op-amp inputs are held at the same potential due to negative feedback.
• Therefore, because +Vin is at 0 volts, -Vin is forced to be 0 volts.
• You have placed plus and minus signs on the capacitor nodes and, the plus node is forced to be 0 volts (as per what I said above).
• Therefore, the op-amp output is intrinsically and indisputably the reverse polarity of the capacitor voltage.

Is it a derivation from Kirchhoff's voltage law

No, it's an implication of how an op-amp uses negative feedback and, how the polarity symbols you have applied force \$V_{O}\$ to be the negative of \$V_{C}\$.

Answer 3

That circuit is really an idealized integrator topology. So it's used for cases where you want to integrate. Not perform some KVL solution.

But given your question, the KVL is:

$$V_{_\text{OUT}}+V_{_\text{C}} +I_{R_1}\cdot R_1 - V_{_\text{IN}}= 0\:\text{V}$$

Since \$I_{R_1}=\frac{V_{_\text{IN}}-0\:\text{V}}{R_1}\$, we can re-write the above as:

$$\begin{align*} V_{_\text{OUT}}+V_{_\text{C}} +\frac{V_{_\text{IN}}-0\:\text{V}}{R_1}\cdot R_1 - V_{_\text{IN}}&= 0\:\text{V} \\\\ V_{_\text{OUT}}+V_{_\text{C}} +\frac{V_{_\text{IN}}}{R_1}\cdot R_1 - V_{_\text{IN}}&= 0\:\text{V} \\\\ V_{_\text{OUT}}+V_{_\text{C}} +V_{_\text{IN}} - V_{_\text{IN}}&= 0\:\text{V} \\\\ V_{_\text{OUT}}+V_{_\text{C}} &= 0\:\text{V} \\\\ V_{_\text{OUT}} &= -V_{_\text{C}} \end{align*}$$

In more practical systems the integrating capacitor will have a resistor in parallel to it or else there will be a switch across the capacitor -- perhaps a COTO relay, for example, or else a semiconductor switch -- and the input source will be a photodiode instead of a voltage source. There would also be other switches included, as well. That might be what you'd see for low-frequency photodiode integrator. You can see the basics of such a system illustrated by the Burr Brown ACF2101, for example.