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I managed to derive the "non-inverting op-amp" circuit using equations from the general formula of negative feedback. Unfortunately, I could not calculate the "inverting op-amp" circuit in the same way.

I couldn't find a solution on the Internet either, only for the "non-inverting" circuit. Could someone help me? Thanks.

Here are my calculations: (maybe they will be useful).

general feedback formula

non-inverting formula

inverting circuit

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3 Answers 3

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For an inverting amplifier you must consider the fact that the input signal to be amplified does NOT directly arrive at the opamp inputs.

Therefore, in your block diagram there must be an additional input block Hf (forward damping) in front of the summing junction having the transfer function Hf=-R2/(R1+R2). The corresponding feedback function in the return pass Hr (you call it beta) is Hr=R1/(R1+R2).

(These expressions result from superposition of both, input and feedback signal, at the inverting input.)

As the result - the closed-loop transfer function is

Acl=Hf * Ao / (1+Hr * Ao)

As you can see, for Ao approaching infinity this expression simplifies to

Acl=Hf/Hr=-R2/R1

Comment 1: (answer to Scott Seidmans question): The following block diagram is nothing else than a visualisation of the following derivation (following Kirchhoffs rules):

Voltage at the inverting opamp input:

Vn=Vin * Hf + Vout * Hr = -Vout/Ao

Vout=- Vin * Hf * 1/(1+Hr * Ao) = - Vin * Hf * [Ao / (1 + Hr * Ao)]

Comment 2: The reader may detect a slight inconsistency between the definitions in the text and the explanations in the above comment1. This "problem" can be solved very easily: In the text, I have assumed that Ao is positive and Hf is negative (due to the fact that the signal input is connected via Hf to the inverting opamp input node).

On the other hand, in comment1 the calculation (superposition) was based on a negative Ao and a positive function for Hf. Both assumptions are possible and interchangeable.

Diagram: In this diagram, according to the derivation (main text) the block Hf has a negative transfer function. However, a block diagram manipulation according to the derivation in comment1 is possible: Make Hf positive and Ao negative and change the "-" sign at the summing junction into a "+"

enter image description here

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  • \$\begingroup\$ A bit confused about the derivation and location of \$H_f\$. You state that the expression comes from a superposition of the input and output, but your \$H_f\$ block never "sees" the output. Obviously, your equations are right, but I'm just a bit hung up on that point. \$\endgroup\$ Commented Apr 10 at 14:56
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    \$\begingroup\$ @ Scott Seidman - perhaps my verbal explanation was not clear enough. Please, see the additional comment at the end of my contribution. \$\endgroup\$
    – LvW
    Commented Apr 10 at 17:34
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    \$\begingroup\$ @ Circuit fantasist - thank you for your comment. I understand your point. I think you have the same reservations as mentioned in Scott Seidmans comment regarding my term "input damping". In this context, please see the additional explanation I have provided above in "comment1". More than that, the corresponding block diagram explains per visualisation what I mean while saying "the input signal to be amplified does NOT directly arrive at the opamp inputs" (block Hf). \$\endgroup\$
    – LvW
    Commented Apr 11 at 13:41
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    \$\begingroup\$ @ circuit fantasist - we have the choice (as I have explained at he end of my contribution). Yes - we can assign a negative sign to the amplifier. However, in this case, the minus sign at the summing junction must also be changed (from minus to plus). \$\endgroup\$
    – LvW
    Commented Apr 14 at 14:59
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    \$\begingroup\$ @Circuit fantasist - OK, I agree. I know what you mean. It was my original intention to show a block diagram according to the classical model (with subtraction at the summing junction: comparison between "should" and "is"). Furthermore, thanks for the small info about the @ function. \$\endgroup\$
    – LvW
    Commented Apr 15 at 7:33
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Yes, the non-inverting configuration is easier to put into the standard closed loop form.

$$A_{?}(s)=\frac{V_{o}(s)}{U_{i}(s)}=\frac{A_{OL}(s)}{1+A_{OL}(s)\beta(s)}\tag{Equ 1}$$

So:

  • Include a finite op-amp input resistance.
  • Use a current summing junction instead of a voltage one.
  • Make a proper system block diagram.
  • Simplify using appropriate approximations. (Don't muck it up with infinity and zero) use insignificant instead.
  • The input to the canonical closed loop T.F. is the input to the summing junction, not any arbitrary input.

Whether \$U_{i}(s)\$ is a voltage or current is not clear at this point. The non-inverting amplifier can be analyzed in closed loop using a voltage summing junction. The inverting amplifier is easier to analyze using a current summing junction. This implies that \$U_{i}(s)\$ is a current.

The OP relaxes the constraints of infinite gain and zero input voltage, but leaves the ideal constraints of infinite input impedance and zero input current for the op-amp. Trying to implement a voltage subtraction is very difficult with these constraints.

This discussion thus includes a finite \$R_{in}\$, allowing a small input current \$I_{in}\$ as shown in Figure 1. Then the voltage between the op-amp's inputs is: \$V_{in}=R_{in}I_{in}\$ and \$I_{in}=I_{i}-I_{f}\$ which is the summing junction, SUM2 in Figure 2.

schematic

simulate this circuit – Schematic created using CircuitLab

At this point \$V_{in}\$ cannot be considered insignificant and so must be included to calculate \$I_{i}\$ and \$I_{f}\$ . So by Ohm's law, SUM1 and F1 convert \$V_{i}\$ into \$I_{i}\$.

Continuing in this vein gives the control system block diagram in Figure 2.

schematic

simulate this circuit

This is not what was expected. The diagram shows not one but two linked closed loop systems, so some approximations must be made to find what the OP is looking for. So consider that \$V_{in}\$ is small enough to be insignificant to the outputs of SUM1 and SUM3. So the open loop gain must be large enough to achieve this. (Let's not use infinity. It will muck things up). So these summing junctions can be removed as shown in Figure 3. (The negative input of SUM3 must be applied to H1).

schematic

simulate this circuit

In order for the standard form to work, the \$U_{i}(s)\$ must be the input to the summing junction. It cannot be just any input. Clearly that is not the case for the inverting amplifier. So from Equ 1, $$A_{?}(s)=\frac{V_{o}(s)}{I_{i}(s)}=\frac{-R_{in}A_{o}}{1+\frac{R_{in}}{R_{f}}A_{o}}$$ where \$A_{OL}(s)=-R_{in}A_{o}\$ and \$\beta(s)=\frac{-1}{R_{f}}\$.

So finally, the overall gain of the closed loop inverting amplifier

$$A_{CL}=\frac{1}{R_{i}}\frac{-R_{in}A_{o}}{1+\frac{R_{in}}{R_{f}}A_{o}}=\frac{-R_{f}}{R_{i}}\frac{R_{in}A_{o}}{R_{f}+R_{in}A_{o}}$$

To obtain the traditional gain approximation, \$R_{f}<<R_{in}A_{o}\$.

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  • \$\begingroup\$ @ RussellH - I agree with your alternative approach. The inverting op amp is a circuit with voltage controlled current feedback. This is also shown by the input resistance which, according to system theory, should decrease with this negative feedback variant. Therefore, it is obvious to compare the input current with the feedback current. When deriving the block diagram in my article, I assumed that the current superposition at the input of the inverting opamp generates a voltage which acts as input voltage for the "naked" voltage amplifier (opamp). Both ideas are of course equivalent. \$\endgroup\$
    – LvW
    Commented Apr 11 at 13:33
  • \$\begingroup\$ @LvW: Thanks. 1) “This is also shown by the input resistance which, according to system theory, should decrease with this negative feedback”: Rin=Vin/Iin does not decrease by negative feedback. R’in=Vin/Ii does decrease by negative feedback. The former is used in the answer appearing as a constant multiplier to Ao in the forward path. 2) Superposing currents cannot generate a voltage unless the resulting current passes through a resistance. In this case the resistance is Rin. \$\endgroup\$
    – RussellH
    Commented Apr 11 at 15:56
  • \$\begingroup\$ @ RussellH - What I mean with "decreased input resistance" is the following: Without any feedback the input resistance of the opamp is 1Mohm (for example). For a non-inv. finite amp the relevant input resistance is higher than 1Mohm (non-inv. input) and for inverting amplifiers the input resistance of the whole amp becomes smaller (e.g. 1k). Regarding your 2nd. point: Imagine we have two resistors in series with a voltage V1 at the left side and V2 at the right side - and the voltage between the resistors is (V1R2+V2R1)/(R1+R2), right? This is the background of the "virtual ground" concept. \$\endgroup\$
    – LvW
    Commented Apr 11 at 16:50
  • \$\begingroup\$ @LvW: 1) No argument about the input resistance. I was just clarifying why the Rin in my answer does not decrease in closed loop. 2) Agreed, voltage superposition works well to answer the question requiring only that Iin is insignificant, which seems implied by the question.|| My preference is still Figure 2 in my answer. Perhaps others may like it as well. Cheers. \$\endgroup\$
    – RussellH
    Commented Apr 12 at 6:10
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Using this image from here: -

enter image description here

You've already done the hard work. Now you just find an expression for the voltage at the inverting input and equate it to zero (knowing that the op-amp gain can be assumed to be infinite for an ideal device.

The inverting input voltage is made from the R1 and R2 potential divider hence: -

$$0 = V_{IN} + \left(V_{OUT}-V_{IN}\right)\cdot\dfrac{R_1}{R_1+R_2}$$

$$V_{IN}\cdot\left(1 - \dfrac{R_1}{R_1+R_2}\right) = -V_{OUT}\cdot\dfrac{R_1}{R_1+R_2}$$

$$\dfrac{-V_{out}}{V_{IN}} = \left(1 - \dfrac{R_1}{R_1+R_2}\right)\cdot \dfrac{R_1+R_2}{R_1}$$

$$\dfrac{-V_{out}}{V_{IN}} = \dfrac{R_2}{R_1+R_2}\cdot \dfrac{R_1+R_2}{R_1}\hspace{2cm} = \hspace{2cm}\dfrac{R_2}{R_1} $$

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