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I'm a newbie in analogue circuits. I find it so confusing analyzing feedback amplifiers.

The typical feedback systems introduced in textbooks is shown below. The input signal \$X\$ is connected to a non-inverting port and the feedback signal is connected to the inverting port.

enter image description here

However, in some cases (like the two circuits hand drawn below), the input signal and the feedback signal (\$\beta \cdot Y\$, \$ Y = V_{\text{out}}\$) are connected to the inverting port at the same time. I'm wondering what the system block diagram would look like in this case.

enter image description here

I have seen it said that an "equivalent \$V_{\text{in}}\$" would be calculated as in Fig. 2, $$-V_{\text{in}}\cdot \frac{R_2}{R_1+R_2}$$ and will be the "equivalent \$V_{\text{in}}\$" as drawn in diagram above. If that's true, is it the case that to calculate the equivalent \$V_{\text{in}}\$, \$ V_{\text{out}}\$ needs to be grounded, and then the differential input (\$V_+ - V_-\$) would be the equivalent \$V_{\text{in}}\$? So the block diagram of the inverting amplifier would be as below?

enter image description here

In Fig. 1 (I think it's a charge amplifier), the input is a current. It makes things harder to understand from the perspective of feedback system (especially the block diagram). If in this case \$X\$ is a current, it doesn't make sense that \$Y\$ is a voltage, unless \$\beta\$ is a transconductance network, so that \$\beta Y\$ is a current, as only current and current can be summed. However, the input of the amplifier couldn't be a current. So, in this case what is fed in? What would the system block diagram look like?

If we just consider the feedback in a charge amplifier as a voltage-voltage feedback, then \$ \beta \$ should be:

$$\frac{V_x}{V_{\text{out}}} = \frac{C_\text{int}}{C_\text{int}+C_\text{pd}}$$

If that's true, the feedback factor \$ \beta\$ would be a constant when the frequency varies, but in the AC simulation, \$ \frac{V_x}{V_{\text{out}}}\$ is not a constant at all. I'm so confused about this.

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    \$\begingroup\$ I totally get the confusion. But you just need to learn and then work through a process a bunch of times. Then it starts to become more clear over time. So you won't get a magic wand that just makes things right in your mind. You need to work at it and solve a variety of situations. Then stuff starts to make sense. (Unless you are one of those super-geniuses one rarely meets who 'just get it'.) \$\endgroup\$
    – jonk
    Feb 22, 2022 at 11:12
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    \$\begingroup\$ Fig. 1: The feedback block, assuming \$A_{_\text{OL}}=\infty\$, is \$\beta=s\: C_{_\text{INT}}\$ in s-space. That's pretty much expected, from inspection. And it's obviously frequency-dependent. If you plug that into \$A_{_\text{CL}}=\frac{A_{_\text{OL}}}{1+A_{_\text{OL}}\:\beta}\$ then you find that \$A_{_\text{CL}}=\frac{A_{_\text{OL}}}{1+A_{_\text{OL}}\:\beta}=\frac1{\frac1{ A_{_\text{OL}}}+\beta}\approx\frac1{\beta}\$ which is an impedence. As the input is a current source, when multiplied by the \$A_{_\text{CL}}\$ impedance, you get a voltage. Which is also expected. \$\endgroup\$
    – jonk
    Feb 22, 2022 at 11:34
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    \$\begingroup\$ Fig. 2: Here, the feedback is \$\beta=-\frac{R_1}{R_2}\$. You can imagine that if \$R_2=\infty\$ or if \$R_1=0\$ then the feedback is zero. And if they are equal and non-zero then the feedback is equal to the input, or -100%. Here, \$A_{_\text{CL}}=\frac{A_{_\text{OL}}}{1+A_{_\text{OL}}\:\beta}=\frac1{\frac1{ A_{_\text{OL}}}+\beta}\approx -\frac{R_2}{R_1}\$. Also expected. \$\endgroup\$
    – jonk
    Feb 22, 2022 at 11:50
  • \$\begingroup\$ @jonk Thank you so much!!! It's much clearer now. Really appreciate it! \$\endgroup\$
    – Jack Black
    Feb 22, 2022 at 12:22
  • \$\begingroup\$ I'm just glad it helped. :) And yes, you do need to keep track of units. \$\endgroup\$
    – jonk
    Feb 22, 2022 at 12:24

2 Answers 2

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Feedback Basics

Let's start with the two feedback situations, just to be pedantic. (Only the left side applies to the given opamp situation.)

enter image description here

Let's use the freely available sympy to work out the resulting closed-loop equations. Since I don't want to assume anything about the input and the output (whether voltage or current or whatever), I'll use p for process variable:

var('aol b pin pout')
solve(Eq(pout, (pin-b*pout)*aol),pout)[0]/pin
    aol/(aol*b + 1)
solve(Eq(pout, (pin+b*pout)*aol),pout)[0]/pin
    -aol/(aol*b - 1)

So the results for feedback that is applied as negative are \$A^{^\text{-}}_{_\text{CL}}=\frac{A_{_\text{OL}}}{1+A_{_\text{OL}}\,\beta}=\frac1{\frac1{A_{_\text{OL}}}+\beta}\$ and the results for feedback that is applied as positive are \$A^{^\text{+}}_{_\text{CL}}=\frac{A_{_\text{OL}}}{1-A_{_\text{OL}}\,\beta}=\frac1{\frac1{A_{_\text{OL}}}-\beta}\$.

As \$A_{_\text{OL}}\to\infty\$, \$A^{^\text{-}}_{_\text{CL}}=\frac1{\beta}\$ and \$A^{^\text{+}}_{_\text{CL}}=-\frac1{\beta}\$. (You can trivially solve for \$\beta\$, as well.)

Hopefully, that's clear.

Figure 1

Before diving into this one, it's worth stopping a moment to consider it. There's a current source/sink input (which may or may not be dependent upon time) at the negative opamp input node. Assuming an ideal opamp, whose negative input node neither sinks nor sources current, the current source/sink's instantaneous value must be matched by the sum of currents through the two capacitors. However, \$v_{_\text{X}}\$ (little-v used here for time-domain) must always be kept at ground (assuming an ideal opamp that responds instantly.) So it follows that there cannot be any current in \$C_{_\text{PD}}\$ and therefore all of the matching current must be coming solely through \$C_{_\text{INT}}\$.

Applying KCL, you know that:

$$i_{_\text{PD}}=C_{_\text{INT}}\,\frac{\text{d} }{\text{d} t}\:v_{_\text{OUT}}$$

If we stay in the time-domain, the time derivative of \$v_{_\text{OUT}}\$ multiplied by \$C_{_\text{INT}}\$ is that current. So we'd need to solve an integral to get back \$v_{_\text{OUT}}\$.

Usually, it's easier to just avoid such issues by staying in s-space where integral and derivative problems are reduced to simpler algebra problems. So the above equation is then (using capitals to represent the Laplace-equivalent variable names):

$$\begin{align*} I_{_\text{PD}}&=C_{_\text{INT}}\,s\:V_{_\text{OUT}} \\\\\therefore\\\\ V_{_\text{OUT}}&=I_{_\text{PD}}\cdot\frac1{s\,C_{_\text{INT}}} \\\\\therefore\\\\ A^{^\text{-}}_{_\text{CL}}=\frac{V_{_\text{OUT}}}{I_{_\text{PD}}}&=\frac1{s\:C_{_\text{INT}}} \end{align*}$$

But that's just the s-space impedance for \$C_{_\text{INT}}\$. All this means is that \$I_{_\text{PD}}\$ is pulled or pushed through the impedance of \$C_{_\text{INT}}\$ to get \$V_{_\text{OUT}}\$. Which makes a lot of sense.

Note: So why even bother specifying \$C_{_\text{PD}}\$? The reason has to do with noise analysis which is beyond your question here.

From the prior section, we know that \$\beta=\frac1{A^{^\text{-}}_{_\text{CL}}}=s\:C_{_\text{INT}}\$.

Let's just directly use KCL to find the transfer function:

var('cint cpd ipd vx vout iout s')
CINT = 1 / (s*cint)
CPD = 1 / (s*cpd)
eq1 = Eq( vx/CINT + vx/CPD + ipd, vout/CINT )     # KCL for Vx node
eq2 = Eq( vout/CINT, iout + vx/CINT )             # KCL for Vout node
eq3 = Eq( vx, 0 )                                 # opamp's goal
ans = solve( [ eq1, eq2, eq3 ], [ vout, vx, iout ] )
ans[vout]/ipd
    1/(cint*s)

That's nice to see.

(Also note that using KCL also causes \$C_{_\text{PD}}\$ to drop out of the solution. This is expected because of our prior analysis saying that there is no current through it.)

Figure 2

Let's just dive into the KCL for this one:

var('r1 r2 vin vx vout iout s')
eq1 = Eq( vx/r1 + vx/r2, vin/r1 + vout/r2 )     # KCL for Vx node
eq2 = Eq( vout/r2, iout + vx/r2 )               # KCL for Vout node
eq3 = Eq( vx, 0 )                               # opamp's goal
ans = solve( [ eq1, eq2, eq3 ], [ vout, vx, iout ] )
ans[vout]/vin
    -r2/r1

Again, from the first section we know that \$\beta=\frac1{A^{^\text{-}}_{_\text{CL}}}=-\frac{R_1}{R_2}\$.

That also makes sense. If you lower the input impedance (treating \$R_1\$ as the input impedance of the input voltage source) then the negative feedback factor is smaller as a result. Similarly, if you lower the output impedance (back to the input node and, again, treating \$R_2\$ as the output impedance of the output as seen by the input node), the negative feedback factor is larger as a result.

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  • \$\begingroup\$ Really appreciate your clear and patient explanation! It all makes sense now. However, more and more new questions come to my mind. I think I may have to endure a period of time full of doubts as a beginner. Thanks again! I will keep going. \$\endgroup\$
    – Jack Black
    Feb 23, 2022 at 21:41
  • \$\begingroup\$ @GeorgeGuo Thanks so much for letting me know that it helped! Appreciated! Do you need anything further for your question? If not, you can select an answer. This helps other contributors by saving their time in also adding an answer when you already feel you have one. If you don't feel that way, then just leave things as they are, of course. \$\endgroup\$
    – jonk
    Feb 24, 2022 at 0:21
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I'm wondering how the system block diagram would look like in this case.

It would look more complex than the actual circuit so, there's little point in trying to convert it.

Reason: the feedback block \$\beta\$ in your first diagram does not care about impedances; it's an immutable converter of output signal Y into a linearly modified version of Y i.e. if you tried to force a signal into it's output, it would not allow it. A resistor or capacitor on the other hand does allow this to happen so, inevitably, the symbol of a resistor (or a capacitor) is much more complex when treated as a symbol block diagrammatically.

For the op-amp your equivalent block diagram is a bit easier: -

enter image description here

If you wanted a block diagram of an inverting amplifier, the easiest and most straightforward method is simply this: -

enter image description here

Your formulaic interpretation is incorrect and this is down to how you look at the virtual ground point at the inverting op-amp node; you haven't considered that this node is effectively 0 volts i.e. Vx = 0 volts for an ideal op-amp.

But, even the block diagram of a simple inverting amplifier is a little "fragile" when it comes to a full analysis because it doesn't cater for the input impedance (R1). The circuit diagram is more expressive and more correct.

My advice; don't try and convert basic electronic components to individual blocks because it gets too messy. The actual circuit is simpler in the long run AND it gives you more information so, bite the bullet and recognize that the circuit is a better representation in most cases.

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