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An op-amp that drives a secondary gain stage might be given like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Where it is important to note that FB1 is some impedence (unlikely to be purely real) rather than a resistor, and the same for FB2_a and FB2_b.

However, I'm unsure how to describe such a circuit using a control theory diagram. The diagram that I naively believe to be correct is here:

schematic

simulate this circuit

However, The node between FB1 and FB2 doesn't seem correct (and I'm not sure how to deal with such a node). I have seen structures where FB1 and FB2 go through a summing block, or where they go through two difference blocks in series between In and G1:

schematic

simulate this circuit

However, this seems to ignore the fact that the voltages after FB1 and FB2 must be the same, they are not added together.

How does one properly represent the circuit given above using control theory diagrams?

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  • \$\begingroup\$ Your blocks labeled "FBx" I think are traditionally labeled "Hx". \$\endgroup\$ – Daniel Aug 5 '16 at 20:57
  • \$\begingroup\$ Orthodoxly, they are, @Daniel but I think that's for equations rather than notation (correct me if I'm wrong though, you're probably correct). \$\endgroup\$ – KingDuken Aug 5 '16 at 21:13
  • \$\begingroup\$ I have seen $H_i$, and $\beta$, though admittedly never FB. \$\endgroup\$ – Andrew Spott Aug 5 '16 at 21:19
  • \$\begingroup\$ Think voltage divider(s) \$\endgroup\$ – Chu Aug 5 '16 at 21:37
  • \$\begingroup\$ @Chu: care to elaborate a bit. I'm not sure what about voltage dividers I should be thinking about. \$\endgroup\$ – Andrew Spott Aug 5 '16 at 21:39
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I think this tells the story a little more accurately. It is almost identical to your first diagram, but you can't just connect outputs of blocks together. This creates an impossible condition -- two outputs defining one signal without any kind of defining operator. They need to be summed, and this diagram illustrates that.

schematic

simulate this circuit – Schematic created using CircuitLab

Another way to represent this would be to combine the op amp input summer and the negative input summing node into a (+ - -) summing operator, but circuit lab doesn't seem to have that.

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  • \$\begingroup\$ So, I believe this is equivalent to my second drawing, just rearranged -- Instead of summing then taking the difference, I took the difference then took the difference again. However, this still bothers me for the same reason: voltages between the two feedback blocks are NOT summed. To say that FB1 and FB2 are both connected so that their result is self-consistant might be more accurate. \$\endgroup\$ – Andrew Spott Aug 5 '16 at 21:15
  • \$\begingroup\$ Mathematically, yes, it is, but it rearranges things a bit so you don't have phantom nodes that don't really exist. \$\endgroup\$ – Daniel Aug 5 '16 at 21:19
  • \$\begingroup\$ But this is definitely a summation of feedback voltages at the negative input node. allaboutcircuits.com/textbook/semiconductors/chpt-8/… \$\endgroup\$ – Daniel Aug 5 '16 at 21:19
  • \$\begingroup\$ The resistors at each of those inputs is what allows the "adding". Two connections to the same node are not necessarily added. Though I must admit, I'm not sure how to show that using control system diagrams, it would appear that those would be represented as just an input summing node. \$\endgroup\$ – Andrew Spott Aug 5 '16 at 21:22
  • \$\begingroup\$ I think that FB1, FB2a, and FB2b are going to show up in both of the H blocks. \$\endgroup\$ – Daniel Aug 5 '16 at 21:28
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Here is my solution. Please note that in the given transfer functions the terms R1, RA and RB are identical to FB1, FB2a and FB2b, respectively.

Of course, you can replace one or all of these elements with other parts or any combination of two-pole elements (example: R||1/sC or R+sL or....).

Starting with this block diagram you can introduce several modifications according to the rules of block diagram manipulation. For example, you can divide the transfer function F2 by the factor G2 and - at the same time - connect the feedback path for F2 at the output of G2. In this case, the whole diagram is identical to Daniel`s schematic.

Proof: Setting G2=1 we arrive - afetr some mathematical manipulations - at a gain expression (out/in) which is equivalent to a non-inverting opamp with feedback resistors RA||R1 and RB. This is in accordance with the given original schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I'm not sure how you got the RB*R1/RA part. \$\endgroup\$ – Andrew Spott Aug 8 '16 at 16:35
  • \$\begingroup\$ Start with the output of G1 and calculate the feedback factor withnrespect to this output node. In this case, the fedback path is avoltage divider (G2+RA)||R1 in series with RB. Hence, we have two ways for the feedback signal. Then, calculate the current through R1 (two portions) - and you have the feedback voltage at the inv. terminal. With this, you have the whole feedback factor which consists of two parts. Hence, you can draw the main part of the block diagram with G1, F1 and F2. Finally, you have to add the output block G2 because the overall output is defined at the G2 output. \$\endgroup\$ – LvW Aug 8 '16 at 19:25
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Remove \$\small FB2a\$ to begin with, then you'll see a voltage divider feeding the inverting input. So you could have a feedback block around G1 with gain \$\small FB2b/(FB1 + FB2b)\$. Then do the same again for the outer feedback loop, and finally superposition gives the complete picture.

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  • \$\begingroup\$ You remove FB1 when calculating the outer loop transfer function? so FB2 is (FB2_a / (FB2_a + FB2_b)), and FB1 is (FB2b/(FB1 + FB2b)), and then they go through a summing node? \$\endgroup\$ – Andrew Spott Aug 5 '16 at 21:55
  • \$\begingroup\$ Yes that's correct \$\endgroup\$ – Chu Aug 5 '16 at 22:01
  • \$\begingroup\$ Except that FB2 should be FB2_b / (FB2_a + FB2_b)... but that was just a typo on my part. I'm still a little confused as to how this allows me to sum the two. The transfer functions of each of these are for the same node... their outputs physically can't have different values... \$\endgroup\$ – Andrew Spott Aug 5 '16 at 22:04
  • \$\begingroup\$ So \$\small H1=FB2b/(FB1+FB2b)\$ and \$\small H2=FB2a/(FB2a+FB2b)\$ \$\endgroup\$ – Chu Aug 5 '16 at 22:07
  • \$\begingroup\$ So, should these two be summed at the input to the difference amp? The outputs of H1 and H2 must be the same according to the circuit diagram, but I don't know how to deal with that in the control theory diagram. \$\endgroup\$ – Andrew Spott Aug 5 '16 at 22:12
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The (implementation of the) summing block for an op-amp is inside the op-amp, such that the non-inverting op amp input is connected to the "plus-sign" of the summing block and the inverting input is connected to its "minus-sign". Consequently, there is only one summing block available per op amp.

Also, as you observe, the output of the FB1 feedback network and that of the FB2 network are directly electrically connected. So there's no summing or differencing going on there, they share the same voltage (and other circuit node characteristics), and in that respect are part of a single circuit (having two "input ports" roughly speaking).

So I would be strongly inclined to prefer your initial diagram.

Perhaps a control-system expert can suggest a way to improve it, like possibly combining the two feedback paths into one or something, but I'd say it's safe to abandon the second diagram for a start :)

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  • \$\begingroup\$ Unfortunately, I'm pretty sure the initial diagram is not a "legal" control system diagram, however, I'm not sure how to fix it. \$\endgroup\$ – Andrew Spott Aug 5 '16 at 21:17

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