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In chapter 3 of the book Designing Control Loops for Linear and Switching Power Supplies: A Tutorial Guide, there is a figure 3.59:

enter image description here

The expression of the divider ratio is as follows:

$$\alpha=\frac{R_{lower}}{R_{lower} + R_{upper}}$$

And the expression is derived from figure 3.58.

Why doesn't the divider ratio appear in the closed loop? I thought the divider ratio \$\frac{1}{\alpha}\$ should be placed inside the closed loop:

enter image description here

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  • \$\begingroup\$ The @Verbal Kint in the Autor of this book. \$\endgroup\$
    – G36
    Sep 26, 2022 at 13:32
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    \$\begingroup\$ Please post the section of this question under "The answer" as an answer to this question. You can and should answer your own question if you can, rather than post your answer as part of the question. \$\endgroup\$
    – Null
    Sep 27, 2022 at 13:59

2 Answers 2

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This is classic in loop control discussion and the answer varies depending on the selected architecture. Most of the block diagrams shown in the literature include the divider block as indicated in the book from Erickson and Maksimovic:

enter image description here

In this representation, you can see the sensor gain affecting the loop gain as it should. However, depending on the compensator implementation, there can be some subtle variations. If you take a classical type II compensator based on an op-amp (but a type I or III would also do), you can see that the op-amp operates in closed-loop conditions considering the network \$Z_f\$:

enter image description here

In this mode, the op-amp implements a so-called virtual ground and maintains both inputs (-) and (+) at the same potential, providing the open-loop gain \$A_{OL}\$ is very large. Practically speaking, it implies that if the reference voltage is set to 2.5 V on the (+) pin, then, when the converter regulates, you measure 2.5 V across \$R_{lower}\$ (static loop gain magnitude is large in dc). If you now modulate the output voltage \$V_{out}(s)\$ via a sinusoidal source, the voltage level at node (-) remains unchanged. This is because \$V_{ref}\$ is, as its name implies, a reference source and is insensitive to the on-going modulation: \$\frac{dV_{ref}(t)}{dt}=0\$. Therefore, if you have 0 V ac across \$R_{lower}\$, the resistance is naturally excluded from the gain calculation.

If you study the same compensator in dc conditions (\$s\$ = 0), for instance to derive the static loop gain, then after opening the capacitors (as SPICE does in a bias point calculation), then you can see that the virtual ground is gone and, in this case, the divider plays a role to determine the output voltage. Therefore, if I plot the transfer function of a closed-loop converter after changing the division ratio, you see the dc gain changing (with a slight effect on the output dc error) but crossover remains untouched:

enter image description here

Now, if you think of a compensator built around a transconductance operational amplifier (OTA), then there is no local feedback and the virtual ground does not exist: the low-side resistance plays a role in the equation.

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Finally, when you deal with a digital controller, the error voltage processing is a simple subtraction between the error voltage and the internal reference without a virtual ground of course:

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As shown in this SIMPLIS simulation schematic diagram of a PID (see my ready-made compensators here, there is no local feedback as with an op-amp and no virtual ground either: the division ratio needs to be factored in.

As a conclusion, engineering judgement is necessary to check whether the division ratio enters the picture or not.

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  • \$\begingroup\$ It is the local feedback that creates the virtual ground in the case of an op amp. But if there was an OTA instead, the application note ti.com/seclit/ml/slup386/slup386.pdf mentioned in electronics.stackexchange.com/questions/681970/… , which prohibits injection point 3 for op amps, would not be relevant for the OTA, right? I mean the OTA has no local feedback, so its input impedance would be much higher than the resistor divider impedance, so something like Bode 100 could use injection point 3 for an OTA. \$\endgroup\$
    – Hyp
    Feb 16 at 11:57
  • \$\begingroup\$ The output impedance of the resistive divider is usually quite high so I am not sure if injecting at point 3 makes sense, even with an OTA. Points 2 and 4 seem more common for measuring the loop anyhow. \$\endgroup\$ Feb 16 at 12:53
  • \$\begingroup\$ The feedback resistors would be tens of kOhm and the LT1534 e.g. has feedback input current 250 nA typ (900 nA max), so 1.25 V / 250 nA = 5 MOhm. And 5 MOhm is higher enough than the feedback resistance, isn't it? \$\endgroup\$
    – Hyp
    Feb 16 at 13:28
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Why doesn't the divider ratio appear in the closed loop?

Because it's not in the closed loop. The demand (the required set-point for any control system) cannot be inside the control loop.

Why should it be in the control loop when all it does is modify a reference voltage level. Maybe you are thinking that because the signal transfer diagram shows \$I_{OUT}\$ that somehow that current has a ripple effect back to the divider? It doesn't of course.

EDITED SECTION

After the op edited their question, a new picture appeared that shows part of a feedback system in what appears to be a voltage regulator. The potential divider (\$R_{UPPER}\$ and \$R_{LOWER}\$) IS within the closed-loop but, if you look closely, the reference input isn't within the closed-loop. The reference input is the control-loop demand and cannot be inside the loop.

enter image description here

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  • \$\begingroup\$ The divider ratio is the sensor gain H. See this fiugre: link.springer.com/chapter/10.1007/978-3-030-43881-4_9/figures/2 . It should be in the closed loop, isn't it? \$\endgroup\$
    – T L
    Sep 26, 2022 at 11:07
  • \$\begingroup\$ Hold on @TL you've altered your question since I answered it and your question now contradicts itself. The picture in your comment has sensor gain within the loop (and thus alters thTF because it is inside the loop. However, in your original question you are referring to the reference input (the demand input) which isn't at all inside the loop. First, what must be done is to is fix your question. I'll do that so that the new picture is clearly seen as an addition. \$\endgroup\$
    – Andy aka
    Sep 26, 2022 at 11:28
  • \$\begingroup\$ Then, when you are happy about that fix I'll add a few more lines to my answer that explains why one is inside and the other isn't inside the closed-loop. \$\endgroup\$
    – Andy aka
    Sep 26, 2022 at 11:28
  • \$\begingroup\$ @TL OK I've gone ahead and adjusted my answer but please please note that you should always take great care when amending any question once an answer is provided. \$\endgroup\$
    – Andy aka
    Sep 26, 2022 at 11:44
  • \$\begingroup\$ Sorry Andy. My unskilled expression makes you feel confused. But I haven't changed the question. The piture that I added is figure 3.58. The divider ratio (a=Rlower/(Rlower+Rupper)) is derived from figure 3.58. I add this pucture and the expression of divider ratio to make the helper know what is the divider ratio. And in the figure 3.59, the author doesn't put the divider ratio in the closed picture which makes me confused. \$\endgroup\$
    – T L
    Sep 26, 2022 at 11:52

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