Use of superposition principle for the inverting amplifier

Question

My current understanding of the inverting amplifier goes as far as assuming the second golden rule: no current flows through the OpAmp.

From here on, up to the end result of \$ H(\omega) = -\frac{R_2}{R_1} \$, I'm having a hard time understanding what's going on.

My first confusion arises when applying the superposition rule to obtain the reference voltage \$ U_{-} \$ at the inverted terminal (the non-inverted on being grounded).

I want to understand this crucial step so I find an expression for the gain voltage \$ G\ U_{-} \$ to which the OpAmp sets its output.

I stumble upon the same problem in calculating the reference voltage at the junction point at \$ V_{out} \$.

I'm not sure of the approach of converting this circuit into a Thevenin equivalent circuit, by treating the OpAmp as a current source, either.

For the problem at hand, I don't assume the first golden rule where \$ G(U_{+} - U_{-}) = 0 \$. This is what I actually want to derive. :)

Answer 1

I normally would just rush forward with the obvious voltage divider equation. But I'm going to assume, because you reference it, that you want each voltage source activated one at a time while all other voltage sources are "shorted." It's more busy work. Oh, well.

You only have two voltage sources, \$V_\text{IN}\$ and \$V_\text{OUT}\$. So this isn't too much work. With \$V_\text{IN}\$ active and \$V_\text{OUT}=0\:\text{V}\$. There is a current \$\frac{V_\text{IN}}{R_i+R_f}\$. With \$V_\text{OUT}\$ active and \$V_\text{IN}=0\:\text{V}\$. There is a current in the opposite direction, \$\frac{V_\text{OUT}}{R_i+R_f}\$. The sum of these two superimposed currents is then \$\frac{V_\text{IN}}{R_i+R_f}-\frac{V_\text{OUT}}{R_i+R_f}\$. To find the voltage at the inverting terminal, we start with \$V_\text{OUT}\$ and add the voltage drop across \$R_f\$, so we find that:

$$\begin{align*} V_-&=V_\text{OUT}+R_f\cdot\left(\frac{V_\text{IN}}{R_i+R_f}-\frac{V_\text{OUT}}{R_i+R_f}\right)\\\\ &=V_\text{OUT}+R_f\cdot\frac{V_\text{IN}-V_\text{OUT}}{R_i+R_f}\\\\ &=\frac{V_\text{OUT}\:R_i+V_\text{OUT}\:R_f+V_\text{IN}\:R_f-V_\text{OUT}\:R_f}{R_i+R_f}\\\\ &=\frac{V_\text{OUT}\:R_i+V_\text{IN}\:R_f}{R_i+R_f}\label{eq1}\tag{1} \end{align*}$$

But you know, given that the non-inverting terminal is grounded, that \$V_\text{OUT}=-G\cdot V_-\$. So:

$$\begin{align*} V_-&=\frac{V_\text{OUT}\:R_i+V_\text{IN}\:R_f}{R_i+R_f}\\\\ &=\frac{-G\: V_-\:R_i+V_\text{IN}\:R_f}{R_i+R_f}\\\\&\therefore\\\\ V_-&=V_\text{IN}\frac{R_f}{R_f+R_i\left(G+1\right)}=V_\text{IN}\frac{1}{1+\frac{R_i}{R_f}\left(G+1\right)} \end{align*}$$

Then,

$$\begin{align*} V_\text{OUT}&=-G\cdot V_-=-G\cdot V_\text{IN}\frac{R_f}{R_f+R_i\left(G+1\right)}\\\\&\therefore\\\\ \frac{V_\text{OUT}}{V_\text{IN}}&=-G \frac{R_f}{R_f+R_i\left(G+1\right)}\\\\ &=\frac{-R_f}{\frac{R_f}{G}+R_i\frac{G+1}{G}}=\frac{-R_f}{\frac{R_f+R_i}{G}+R_i} \end{align*}$$

As \$G\to\infty\$ then \$\frac{V_\text{OUT}}{V_\text{IN}}=\frac{-R_f}{R_i}\$.

Not sure if that's what you wanted, though.

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Nodal, done my way

I gather you worry about keeping the signs right. I also worry about that, too. I dislike, with some vigor, the way that nodal analysis is usually taught, where you have to consistently keep track (and continually re-inforce it mentally) of what is being subtracted from what. So here is my nodal approach that makes it a lot easier (for me) and I never worry about signs.

I set up the nodal equations such that outflowing currents are on the left and inflowing currents are on the right. Note that there are no minus signs anywhere.

Imagining myself as standing in the middle of the \$V_-\$ node:

$$\begin{align*} \frac{V_-}{R_i}+\frac{V_-}{R_f}&=\frac{V_\text{IN}}{R_i}+\frac{V_\text{OUT}}{R_f}\\\\ V_-\left(\frac{1}{R_i}+\frac{1}{R_f}\right)&=\frac{V_\text{IN}}{R_i}+\frac{V_\text{OUT}}{R_f}\\\\ V_-\left(\frac{1}{R_i}+\frac{1}{R_f}\right)\frac{R_i\:R_f}{1}&=\left(\frac{V_\text{IN}}{R_i}+\frac{V_\text{OUT}}{R_f}\right)\frac{R_i\:R_f}{1}\\\\ V_-\cdot\left(R_f+R_i\right)&=\left(V_\text{IN}\:R_f+V_\text{OUT}\:R_i\right)\\\\ V_-&=\frac{V_\text{OUT}\:R_i+V_\text{IN}\:R_f}{R_i+R_f}\label{eq2}\tag{2} \end{align*}$$

You can see that eq. \$\ref{eq1}\$ matches eq. \$\ref{eq2}\$.

It's a lot easier, I think, to handle the signs consistently and flawlessly, time after time, this way. Mentally stand in the middle of each node and just put all the outflowing currents on the left; all the inflowing currents on the right; and the rest just follows trivially. I learned this technique by observing source code for a Spice program and seeing how it handled the equation setup. Made a lot of sense to me and I've never gone back to the book-learned way, which I now consider to be designed for failure.

(Now imagine that you had to include an input impedance for the opamp. Easy. Let's call that input impedance, \$R_\text{in}\$. On the left side, you'd simply add \$\frac{V_-}{R_\text{in}}\$. On the right side, you'd add \$\frac{V_+}{R_\text{in}}\$. That's it. The process is trivial to follow. You almost cannot mess up.)

Answer 2

Applying the superposition principle, is the most simple method for finding the closed-loop gain of the inverter circuit. As long as the opamp is operated within its linear region we can set the resulting voltage Vn between ground and the inverting terminal Vn=0 (in reality it is as small as some µVolts).

Superposition:

(1) Vout=0: Vn1=Vi*Rf/(Ri+Rf)

(2) Vi=0: Vn2=Vout*Ri/(Rf+Ri)

(3) Setting Vn=Vn1+Vn2=0 we arrive at Vout/Vi=-Rf/Ri

EDIT: Finite Aol.

For finite Aol we have Vn=Vn1+Vn2=-Vout/Aol

This expression leads to Vout/Vin=-Rf/[Ri+(Ri+Rf)/Aol]

For Aol infinite this expression reduces again to Vout/Vi=-Rf/Ri

Answer 3

If I understand your question correctly, the main goal you have is to show how the op-amp has a transfer function of \$A(s) = -R_2/R_1\$. If we're performing ideal op-amp analysis- which you seem to be starting to do- then there are 2 "golden" rules (Let's use them to get the transfer function and then we'll come back to why they are the way they are):

1. No current flows through the input terminals of the opamp.
2. The voltage at the input terminal nodes are the same (only in a closed loop).

For reference here is a figure of an ideal op-amp, from Sedra and Smith.

With these two rules we can start analyzing the circuit. The first thing I like to do is to try and list all the quantities I know. In this case, since the non-inverting terminal is grounded, by rule 2 we have that the voltage at the inverting terminal node is 0V. From here we can perform a nodal analysis to find \$A(s) = V_{out}(s)/V_{in}(s)\$. Performing a KCL at the inverting terminal node (and noting from rule 1 that no current flows to the inverting input terminal): $$\frac{V_{in}(s)-0}{R_1} + \frac{V_{out}(s)-0}{R_2} = 0$$ From here a little algebra gives the result.

Now going back to why the goldren rules are why they are: they are actually a consequence of several "ideal" assumptions that we make about the op-amp. Most textbooks will list the same assumptions that wikipedia does, but the ones that relate to rule 2 are mainly infinite open loop gain.

Consider the op-amp from the figure above. The op-amp is basically a difference amplifier, it takes the difference in voltage between the two terminals and multiplies it by its gain \$A\$. In a closed loop, if the op-amp is working- i.e. it is providing a finite output, then by its TF the relation \$A(v_2-v_1) = v_o\$ exists. Then, $$v_2-v_1 = \frac{v_0}{A}$$If \$A\$ is infinite, from our ideal assumption, then that means \$v_2-v_1=0\$ and thus rule 2 is established.