Great to know 👍

Type 2 or 3 or any other compensator can be used for current-mode and voltage-mode control. I have not seen a restriction for able to do both methods. Where did you read this?

@@CanBijles One can find the Type II/Type III compensator selections in practical designs and suggested in several books, such as in Basso's books and also in the TI-literature. Let me add one part of a TI Report SLVA662, listed as follows: "Demystifying Type II and Type III compensators". A Type II compensation amplifier adds an RC branch to flatten the gain, and improve the phase response in the mid-frequency range. The increased phase is achieved by increasing the separation of the pole and zero of the compensation. Note that this type of compensator always has a net negative phase, and it cannot be used to improve the phase of the power stage. For this reason, Type II compensators cannot be used for voltage-mode control in CCM where there is a large phase drop just after the resonant frequency. Type II compensators are usually reserved for current-mode control compensation, or for converters that always operate in the DCM region. The Type III amplifier is the one for the voltage-mode converters operating in CCM. ------- Regards Udo

@@udohuhn-rohrbacher1406 I see that is written in this paper of TI www.ti.com/lit/an/slva662/slva662.pdf, but for example in this paper, it does use the Type 2 in voltage-mode control: www.infineon.com/dgdl/an-1162.pdf?fileId=5546d462533600a40153559a8e17111a I have to check those in more detail to understand why TI is mentioning this in his paper.

@@CanBijles Check this from Basso's book: Basso book Switche-Mode Power Supplies 3.5.9 pages 297/298 Selecting the Right Amplifier Type Type I As it does not offer any phase boost, type I can be used in converters where the power stage phase shift is small at the desired crossover frequency. Type II: Most widely used and works fine for power stages lagging down to 90° and where the boost brought by the output capacitor ESR must be cancelled (to reduce the gain in high frequency). This the case for current-mode CCM and voltage-mode (direct duty ratio control) converters operated in DCM. Type IIa: This application field looks the same as for type 2, but when the output capacitor ESR effect can be neglected, for example, the zero is relegated to high-frequency domain, then you can use a type 2a Type IIb: By adding the proportional term, it can help reduce the under-or overshoots in several design conditions. We have seen that it prevents the output impedance from being too inductive, therefore offering superior transient response. Nevertheless, you pay for it by a reduction in the dc gain, hence a larger static error. Type III: You use this configuration where the phase lag brought by the power stage can reach 180°. This is the case for CCM voltage-mode buck or boost-derived types of converters.

@@udohuhn-rohrbacher1406 Thanks for the detailed feedback. I will check this also. Great to have this discussion.

The closed-loop gain of the TIA is given by -R_F with R_F being the feedback resistor only. In this expression, there is not feedback capacitor or any other component included.

@@CanBijlesThanks for your reply! But I would like to know how that conclusion was derived. Thanks!!!

You should use the same formula as for the inverting amplifier. Did you see my video?

@@CanBijlesYeah, that's what I thought. And I learned how to calculate the RMS noise from your video. Thanks for your patience and your video!!!

@@龙源-b1k Great to know. You are welcome. The derivation for the inverting amplifier and the TIA is very similar.

Thanks for your message. Great to know you liked the video! 👉 For more videos, see the playlist of Transistor Current Sources: kzbin.info/aero/PLuUNUe8EVqlk_MLKSjvb9oYSVelzHr-E-

Thanks for your message. This is a nice question and perhaps a video worthy. I will look at it in details later. After a fast search, I found this, but I think you can find more detailed further. electron6.phys.utk.edu/PhysicsProblems/E&M/5-AC%20circuits/ladder-circuits.html

@@CanBijles Thanks for the info, I'll look into that path!

​@@Alan96555 You're welcome.

I explained this in the video also. Check the steps after time 00:10:20 for the details.

This is a nice topic I still want to discuss in detail later. Here, you can read some info about this subject: www.mathworks.com/help/mcb/gs/open-loop-and-closed-loop-control.html

@@CanBijles thank you so much for your help , that will be so usefull .. Is there any design guideline , book that explains step by step how to design the loop and choose component .. because i really need these information for a project .

​@@mohameddrissi7215 There are some books that touch on the topics loop control of power converters, but I think the application notes of Analog Devices, Texas Instruments, etc. are more to the point and practical. See this video for more info and link to resources in the video description: kzbin.info/www/bejne/pmbUZp2Dq9idqtE

@@CanBijles thank you sir , i already saw this video and i found it so great by i m looking specifically for control loops of inverter to control motors

​@@mohameddrissi7215 Got it, I will look at it.

This question came up before, so I will copy paste the answer I have given then: 1. Compensation network (Type 3 network) is a third-order system, so it can provide 270 degrees. 2. MATLAB and TINA-TI Spice use a different phase polarity for the inverting error amplifier. MATLAB considers the phase reversal of the error amplifier as 180 degrees and not as -180 degrees, which is in effect the same. At DC, the phase of the compensation network is -90 degrees and with 180 degrees from the error amplifier make it -90 degrees. 3. Before we use the formula to calculate the K Factor, we add a phase of 180 degrees to get the phi_comp to compensate for the phase inversion in the error amplifier's inverting amplification. This is also described in this video when we discuss the calculations and simulations. More details are given in this video or a similar video via this link: ⚡ DC-DC Buck Converter Design Part 2 ⚡ - Controller Design - Calculations & MATLAB & TINA-TI: kzbin.info/www/bejne/pmbUZp2Dq9idqtE

@@CanBijles Many Thanks 🙏

You are welcome.

I am busy with different topics at the moment. I hope to get back to power electronics topics soon.

@@CanBijles Many Thanks!

​@@biswajit681 You're welcome.

Thanks for your message and your appreciation! Great to know you are still eager to learn, that's the spirit. Good luck!

Maybe it will help just to focus on the procedure rather than why the formulas are correct. After all, I do not prove the formulas in this video and it is not necessary for the design.

Great to know you liked the playlist. It is a good point to know what value and type capacitors we can choose. If the frequency changes, the circuit will definitely respond different. Specifically about capacitors and types, I can advise this video of Alan Wolke: kzbin.info/www/bejne/gGG3m4qVZd2HgLc

@@CanBijles Thanks for the video suggestion. Regarding "If the frequency changes, the circuit will definitely respond different" - do you maybe have some resources (done by you or by someone else you can recommend) on how to calculate the right capacitance?

@@BentFunction I do not have a video going into details of the choices of capacitors, but I do discuss the effect of frequency on the circuit. See for example the playlist about Analog Filters: kzbin.info/aero/PLuUNUe8EVqll-7GNROQDvzXL9dEBa1oRp

Thanks for your message. You are welcome!

I use the simulation block from Simulink. Did you tried that also? What is the error?

Thanks for your message. Great to know that you liked the video 👍

Thanks!

Digital Controller Design: kzbin.info/aero/PLuUNUe8EVqln8g-yPt_2ZEtjH8Faegm3S

You are welcome!

Thanks for your message. Great to know that you liked the video 👍 You are right, there is a typo in the formula for the duty cycle. The right formula should be D = 1 - Vs/Vo.

Thanks for your message.I received your mail too. I will get back to this next week.

I do not use LTSpice.

The other ratios are also available. Not all ratios and orders are given in the tables. It depends on the application.

The rule of thumb is to use 5 times the time constant.

Great to know that 👍

This depends on the configuration of the motor and the load.

Thanks for your message. Great to know that you liked the video 👍

Thanks for your message. Great to know that you liked the video 👍

Thanks for your message. Great to know that you liked the video 👍

Which type of switch are you taking and in that switch input block name please

​@@sujathakemya7387 The switch is ideal. The names of the other blocks are mentioned in the video.

Gazlı ve benzeri sistemler üzerine bir çalışmam yok. Zaman müsait olursa, belki ileri vakitlerde bakarım inşaAllah.

@@CanBijles Çok teşekkürler hocam. İnternette simscape kütüphanesinin nasıl kullanılacağına dair düzgün bir tutorial yok hocam bu konuda öğrenilebilecek kaynaklar varsa paylaşabilir misiniz acaba. Hocam bir de bu sensörler modele seri mi paralel mi bağlanmalı acaba örneğin mass flow rate sensör, pressure temperature sensör vs. hangisi doğru sürekli hatalar alıyorum yardımcı olabilirseniz çok sevinirim

Buraya baktın mı? www.mathworks.com/help/simscape/getting-started-with-simscape.html

The system can have any poles and zeros, also complex poles and zeros. The definition for the phase margin frequency and phase margin is still the same. Phase of +90 degrees means an effective phase of -270 degrees and phase of +270 is equivalent to -90 degrees.

Because you do not take the fringe lines in the gap into account.

@@hiranthini1 Great to know 👍 You are welcome.

Hello, thanks for your message. You are right, the imaginary part should be +j300. The subsequent calculations will be also different because of this error, but the general idea is there I hope.

@@CanBijles Yes for sure, the general idea is more important. Thank you a lot for share your knowledge

​@@MarceloMoraesYou're welcome! Thanks again for your comment 👍

(*-*)

Thanks for your message. I will look at this in detail after the holiday break.

The resources for the videos will be placed in the description of the videos as soon as possible.

Okay I saw in a comment you mentioned the calculations are different between MATLAB and Tina-TI. Do I need to account for the 180-degree phase shift in MATLAB, because having a phase shift starting at 90 degrees is confusing when calculating stability margins.

​@@jadondewey1237 Thanks for your message. -90 degrees + -180 degrees is in totaal -270 degrees, but can be also written as +90 degrees by adding a 360 degrees.

@@CanBijles Thank you for your answer. Why does the phase bode plot for the loop at the end of the video start at -90 degrees instead of +90 degrees then?

​@@jadondewey1237 MATLAB and TINA-TI Spice use a different phase polarity for the inverting error amplifier. MATLAB considers the phase reversal of the error amplifier as 180 degrees and not as -180 degrees, which is in effect the same. At DC, the phase of the compensation network is -90 degrees and with 180 degrees from the error amplifier make it -90 degrees.

Great to know the videos are helpful. Share the knowledge :)

You can find them and more details in the following book: Feedback Control Systems, Charles L. Phillips & John M. Parr

@@CanBijles Thank you very much for your response. It is a pity that I could not find this book in the public domain.

​@@vovashv Here is the link: www.amazon.com/Feedback-Control-Phillips-Prentice-Hardcover/dp/B00LMTLLA6

The graph for "Normalized output with LC filter" is taken from a nonlinear equation. I left the details out, but it is somewhat work to get to the actual details. This is not due to the diode, but due to the discontinuous behavior of the inductor current.

@@CanBijles Do you have another video or source that covers these details? thanks

@@fablearchitect7645 You can look at Chapter 3 in Power Electronics, Muhammad Rashid for more details.

@@mohameddrissi1075 Yes, sure. Here is the playlist about inverters. Inverters (DC to AC Converter): kzbin.info/aero/PLuUNUe8EVqlkDI0dky7_JIMC4TyVSS99Q

Great to know! Here is the link for the feedback controller design for the DC-DC Buck Converter: kzbin.info/www/bejne/pmbUZp2Dq9idqtE

@@CanBijles thank you👌

@@sadeghmollaii9873 You're welcome!

Thanks for your message. I do not know if you mean the formula, but I answered this already for another question in this video. The formula for Vo,n,LN is correct, but when I moved to next slide to collect the formulas, I forgot the sqrt(2) in de denominator, so in total, it should be divided by sqrt(6). Actually, the formula should be written as Vo,n,LN where the harmonics order n is shown. Does this answer your question?

@@CanBijles yes, you re correct, you said sqrt(6) is the correct once below. I read it after asking you the question above, sorry for that. but moreover am asking why is it the case in the formula sqrt(6) also i see 6 appearing in Vo,LL in this divided by pie inside the cosine. the waveform across the loads no pie/6 but pie/3 of each phase being shifted. would you explain the concept behind it, if you don't mind it? and thank sir

yep, I see why it should be sqrt(6), you don't have to answer that part, its is clearer! that is from sqrt(2) multiplied by sqrt(3) at the denominator equal to sqrt(6) defining Vo,n,LN so nothing to do with six switches. however 30 degrees inside the cosine i have no idea since each phase is shifted by 60 degrees( pie/3).

You're welcome! Great to know that you liked the video!

Great you liked the video!

The boundaries are shown in the integration at 06:05. You can also see this from the plot of the output voltage.

yeah sir i did see that, the simplification if plugin boundaries into negative cosine after integration. of (negative COS(n(pie-a)) subtract way negative COS(n(a)) positive DC peak across load then multiply by 2 for full cycle. And end up with 2Vdc/npie (2COS(n(a))) also if am not wrong then i got an idea of how you got result. I try simplification using reduction identities of COS(pie + X)=negative COS(X) the only problem is insde COS(pie-a) is difference rather then sum! but it result into the solution you got if only i ignored the difference and X =+delay(a).

The integral at 06:05 has the variable wt, which has the unit of radians.