Timeline for Nodal analysis -> transfer function -> step response
Current License: CC BY-SA 4.0
16 events
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| Feb 13, 2019 at 12:55 | comment | added | LvW | And realize that it is the task of R1 only to provide a kind of frequency compensation (loop gain reduction). It does not influence the closed loop response. Such a loop gain reduction is often necessary when there are two opamps in a closed-loop (tend to instability). | |
| Feb 13, 2019 at 12:37 | vote | accept | 89f3a1c | ||
| Feb 13, 2019 at 12:37 | comment | added | 89f3a1c | I think I got it now! Thank you very much for your time!! Thanks to all the other users who contributed in some way or another, too! | |
| Feb 13, 2019 at 9:00 | comment | added | LvW | The rest is simple math: You have two equations: Vo=f(Vi,VE) and VE=f(Vo). You simply can introduce the 2nd equation into the first one - and the result is Vo=f(Vi), which allows you to isolate the wanted transfer function Vo/Vi. | |
| Feb 13, 2019 at 8:56 | comment | added | LvW | Yes, your assumptions for F1 and F2 are OK (but, you should not use the term Z1 for 1/sC2 as well as for R4). Regarding your last comment: A buffer with unity gain in the feedback loop of any amplifier does not change trhe closed-loop gain at all, right? In the present case, we have an inverting buffer (gain "-1") because we want to realize negative feedback (as always necessary) at the "+" terminal of the opamp (because we need a non-inv. integrator). Hence, R6 together with the most left opamp (having C3 and a "-1" gain stage in the feedback loop) form a non-inv. integrator (1/sR6C3). | |
| Feb 12, 2019 at 18:11 | comment | added | 89f3a1c | I do understand the -1 gain of the non inverting opamp. But I don't quite get the classical two-integrator loop part, and therefore the 2nd equation. I tried to look it up, and didn't find anything which would help me understand; would you mind pointing some source to help me understand that part please? | |
| Feb 12, 2019 at 18:11 | comment | added | 89f3a1c | Just to be sure, F1 would be -Z2/Z1 where Z2=the paralle of R5 and C1, and Z1=C2 (which would be 1/sC2). And F2 would be -Z2/Z1, where Z2 is the parallel between R5 and C1, and Z1=R4. Right? | |
| Feb 12, 2019 at 18:10 | comment | added | 89f3a1c | I see the damped integrator, but I get confused when the stages between 'classic opamp circuits' are not independent, because I don't really understand how to analyze the circuit in that case. For example, in this case if there would be no C2 and R6, the relation between Ve and Vo would be straightforward; but this way I just don't get it. That's why I got to nodal analysis (+ it's supposed to be the safest, since it should work with any circuit). Nevertheless, reading your answer I think I understand the first equation, although I'm not sure I'd be able to reproduce it in other circuits. | |
| Feb 12, 2019 at 17:29 | history | edited | LvW | CC BY-SA 4.0 | deleted 61 characters in body |
| Feb 12, 2019 at 17:23 | history | edited | LvW | CC BY-SA 4.0 | deleted 61 characters in body |
| Feb 12, 2019 at 17:14 | history | edited | LvW | CC BY-SA 4.0 | deleted 61 characters in body |
| Feb 12, 2019 at 16:34 | history | edited | LvW | CC BY-SA 4.0 | added 34 characters in body |
| Feb 12, 2019 at 16:14 | history | edited | LvW | CC BY-SA 4.0 | added 1 character in body |
| Feb 12, 2019 at 15:59 | history | edited | LvW | CC BY-SA 4.0 | added 4 characters in body |
| Feb 12, 2019 at 15:53 | history | edited | LvW | CC BY-SA 4.0 | added 179 characters in body |
| Feb 12, 2019 at 15:47 | history | answered | LvW | CC BY-SA 4.0 |