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

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

• Hey , If I were you I would either go through some text or 10min video about virtual short. If you understand that , many such problems can be solved. The answers given below are perfect , but along with that strengthen your basic in this concept since its important. And practice few problems in Sedra or any good text book @JDoeDoe Commented Jul 24, 2021 at 11:55

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.

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}\$$.

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.

• However, to get there you do have to assume that $V_{\text{IN}} = 0\text{V}$ and that $I_{R_1} = 0\text{A}$. I don't think this is true, @jonk . Because, in this special case the op-amp input voltages are $V_+ = V_- = 0\text{V}$ which causes $V_{\text{IN}} = I_{R_1} \cdot R_1$ which can be seen by writing the KVL equation for that loop. This causes the term $I_{R_1} \cdot R_1 - V_{\text{IN}}$ to disappear in your first equation, regardless of what $V_{\text{IN}}$ is. Don't you agree?
– Carl
Commented Jul 27, 2021 at 8:15
• @Carl Yes. Thanks for the kick in the head. I was focused on the static viewpoint of the author and attempting to over-simplify the situation. But poorly. I can do better and will. Thanks!
– jonk
Commented Jul 27, 2021 at 8:55
• No worries, @jonk. +1 for a great edited answer.
– Carl
Commented Jul 27, 2021 at 9:02
• @Carl Thanks so much for the privilege you offered in setting this right. It's terrible to have poor material floating about. Too much of it, already. It's my own shame I may have ignorantly added to that pile without your offer. Very much appreciated.
– jonk
Commented Jul 27, 2021 at 9:05