This is the official RU-vid channel of CAN Education. CAN Education provides courses for students, self-learners and industry around the world.
Main subjects are the following. You will find numerous videos about these subjects on this channel.
➤ Electrical Engineering: Analog Electronics, Power Electronics, Electric Circuits, Fourier Series, Transient Response, and Electronic Noise Analysis and Design ➤ Control Systems: Controller Design, Steady-State Errors & Sensitivity, Stability, System Identification, Two-Degrees of Freedom Controller System, State-Space Design, and Laplace Transform ➤ Electrical Machines & Drives: Magnetic Circuits, DC Motors, AC Motors (Induction & Synchronous), and Transformers ➤ Mathematics: Calculus, Linear Algebra, and Dynamic Systems ➤ Semiconductor Device Physics ➤ Physics and Chemistry
⭐ For questions, collaboration or consulting 👇 📧 can.mehmet.tr@gmail.com ☎️ +31616179479 🌐 www.canbijles.nl
For the loop bode plot, why does the phase shift start at -90 degrees at low frequencies instead of -270 degrees or +90 degrees. Shouldn't the negative in the Type-III Compensator transfer function (due to the inverting opamp) cause an additional -180 degree phase shift on top of the -90 degree phase shift from the pole at the origin?
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?
4:08 how did you get the "normalized output with LC filter" graph? Is it derived using numerical simulation due to the non linear nature of the diodes?
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.
@@mohameddrissi1075 Yes, sure. Here is the playlist about inverters. Inverters (DC to AC Converter): ru-vid.com/group/PLuUNUe8EVqlkDI0dky7_JIMC4TyVSS99Q
Great to know! Here is the link for the feedback controller design for the DC-DC Buck Converter: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-p5q5jMvsjto.html
HI, CAN, could you explain why i shouldn't considered the sqrt(3) for 3-phase in this particular case leading to sqrt(6) for Vo,LN. is used of value of 6 define six-steps of switching?
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).
I also assumed that, (wt=a) and boundary is between (+ -a) in order not to zero due to symmetry but not that great, i found the magnitude to contain negative with is totally wrong too! if you can explain to me that can be great and thanks
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).
Hello CAN, sir i have question about integrating magnitude of harmonic amplitude formula. inside integral did you let differential element (dwt), be (da) hance VSin(nwt) into VSin(nat)?? how about boundary condition.
Would you consider doing an example for a Delta-Y, Unbalanced Load, with Line Impedances, NOT converting the source to a Y equivalent, calculating the source, line, and load currents, power delivered, and power consumed.
I don't understand why doing I_L=I_F-I_{2-} doesn't work. because this gives: I_L=I_F V_L/R_L = (V_{1-} - V_L)/R_F (1/R_L + 1/R_F) V_L = V_{1-}/R_F V_L = V_{1-}/(R_F/R_L + 1) and since V_{1-}=V_{1+}=0V, then V_L must be 0V too. I don't get where I am going wrong, yet it's clearly the wrong answer.
Thanks for your message. The equation you write for the output node as: I_L = I_F - I_{2-} is not correct, because you are not taking the current flowing out of op-amp 2. In your case, it would mean I_F = I_L, which is of course not correct. I hope this clarifies the situation.
Greetings!!but because in some texts the formula for mos saturation is Id=kn(Vgs-Vth)^2 and in other parts they report it as Id= 1/2*kn(Vgs-Vth)^2. I saw in Razavi's text that he reports that multiplied by 1/2.. thank you very much!!
How do you get the given transfer function, if you compare to the videos from Prof Marcos Alonso, he designs by choosing the inductor and output capacitor first from other design criteria , and then the transfer function is established from that, followed by the actual component values, Thank you for great Videos, Greetings, Petrus Bosman.
Hi Petrus, thanks for your mail. Great to know you liked the video! The transfer function F(s), which is taking into account the filter and load, is determined using the output voltage divided by the input voltage for F(s) circuit only. When you carry out the analysis, you will get the transfer function F(s). The calculations of the component values really depend on the design specifications and of course on the available components.
Can you explain in more detail the relation between the calculated output noise voltage and the simulated output noise voltage. I didn't quite understand how you picked your frequency for comparison.
For comparison with the simulation results, I used the integration bandwidth as the frequency for determining the total output RMS noise voltage. This is a rule of thumb, but is valid for most practical purposes. Also, the noise will not contribute much for larger frequencies.
Hi, Sir, in your expression of current max/min, why are you using Vdc/R instead of Vdc/Zp???........ Is it because of the RL response of the steady state where inductor become short-circuited. I see in your expression if am not wrong, you using force response. would you mind to derive the expression of the formulas of load current and voltage above in your calculation and thank
The V_DC/R is the steady-state value of the load current. At steady-state, the current in the inductor is zero. The derivation of this formula can be found in many book, for example: Power Electronics, Daniel Hart
intressing! this video will be useful for me after summer :D will check it deeper later, atm on e-plan and few days left before summer brake! thx you for uppload.
Great to know! You can watch different power converter types and feedback control for buck converter also in this playlist: ru-vid.com/group/PLuUNUe8EVqlmo8U7EEBS6W1NpMBkA0JjI Good luck!
i have a question if you don't have a problem. In a FV system, as source in a buck converter, the In capacitor is important to define a input current ripple isn't? Do you have some equation for this Capacitor? thanks, kind regards!
@@JoseGonzalezFernandez-bb6ow The capacitor, which is in parallel with the load, will determine the output voltage ripple. Increasing the value of this capacitor will decrease the output voltage ripple.
Thanks for your message! Glad you liked the video. Here are some playlists: Operational Amplifier Circuits: ru-vid.com/group/PLuUNUe8EVqlmhmvCuxr326mnNKHB5jwJx
@@CanBijles an example in which you cascade a common source stage to the differential stage to amplify the signal, did you do it by chance? I really mean the internal structure of an op amp (differential stage/common source stage/studio buffer) for example
@@eduardmihailoiu7609 I have not done a video about a full transistor design of an op-amp circuit. Sounds interesting to make one in the coming future, will be in list.
If you have only repeated real poles, like (s+5)^2, you will not have the term with A/(s+2). So, you only have the last two terms in the partial fraction expansion I have given.
Hello, what happens if I have repeated poles at the origin, one from the plant and one from the controller. Does it change the computation for the angular criterion in any way?
The calculations are the same. You only need to take into account the magnitude and phase contributions of each pole and zero as I discuss in the video.
dear professor, i am trying to learn controller design for power converters. for the power converters generally we consider gain margin parameter while designing a controller. in your video series i have seen that you did not mention about gain margin. as far as i know gain margin has an important role for transient responses such as line or load transients of power converters. could you please share your ideas about this and explain why you did not mention gain margin while designing pid controller?
Thanks for your message. This is a valid and good question. Indeed, for the stability analysis, we should check both the gain margin and phase margin. In control theory, we also look at the modulus margin, which is actually a better measure how safe we are from the unstable point in the Nyquist plot. Usually (more often, but not always), the phase margin is somewhat more important than gain margin. If the phase margin is sufficient, than the gain margin is probably sufficient too, but of course there is no guarantee. In the videos I discussed about compensator design for buck converter, the gain margin was already large. In the links shown below, they discuss transient response of power converter and the discussion is also based on phase margin only. I do not claim you should only look at phase margin for power converters, but I do not know how strict and important the gain margin is for power converters. Maybe you have a good reference about it, love to read it. www.ti.com/lit/an/snoa507/snoa507.pdf?ts=1717893510339 pdfserv.maximintegrated.com/en/an/AN3453.pdf
dear professor, while calculating the arg of loop transfer function at min 8.10 you did not consider the phase contribution of zpi. since it is a zero i think it should have a positive phase contribution to the system. also in denomintator you have taken the phase of w^2 as -90 degrees. it is real number and i think that it should have zero degrees of phase contribution. could you please explain these calculations? if i think wrong i may improve my insight for this example and pi controller design.
Thanks for your message. Very good questions. Firstly: The PI controller has a pole at the origin (s = 0) and a zero left to this pole. The total phase contribution of the PI controller will be negative. The location of the pole is set, but not the location of the zero of the PI controller. I use the rule of thumb to place the PI controller zero one decade below the phase margin frequency wpm. Secondly: The expression of the real part is -w^2, so the phase is -180 degrees and this is in the denominator of the loop transfer function. The phase of the PI controller zero at wpm is close to +90 degrees and this is in the numerator of the loop transfer function. So, the first part of the phase of the loop transfer function at wpm is 90 - - 180 = 270 degrees or -90 degrees. That is the reason for the -90 degrees.
@@CanBijles thank you so much for your really fast return and explanation. I have understood very well why you have made the calculations as in video. I appreciate for your help.
You can find more information about controller tuning and plotting here: nl.mathworks.com/help/control/ug/getting-started-with-the-control-system-designer.html
@@CanBijles yes. But the phi looks like it makes Kirchhoffs law loop for voltage and not current I am sorry I learned all this in a different language I have no idea how you call all this.
Capacitor in combination with a resistor (RC circuit) will generate noise voltage for the capacitor, but not a noise current. More details can be found here: Thermal noise on capacitors en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise
We determine the effect of the internal noise sources of the op-amp and focus om de op-amp first and then the output lowpass filter. For thermal noise calculations, you may combine the resistors and redo the calculations, but the final results will be the same. In fact, there are other parameters not included in this example, like the input and output impedance of the op-amp, which will have an effect on the actual noise performance of the circuit. You can create an accurate model of the op-amp having the input and output independences included.
Hier zijn voorbeelden over maximum power transfer in DC en AC (in het Engels): Maximum power transfer in DC: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-abPGjW_BQQg.html ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-vKEsUdvBAk8.html Maximum power transfer in AC: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-tJP9Cp0quTA.html ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bnOfxnhEbTE.html
Thank you very much. I made an adjustable power supply unit for TL494. 0-27 V; 0-12A.I calculated the phase compensation for the first voltage error amplifier according to your methodology. It works perfectly! The second error amplifier operates on current limitation and short circuit protection. The shunt is in the circuit on the underside, after the load resistance. A small voltage proportional to the flowing current goes from it to the input of the operational amplifier. Please tell me, can I calculate compensation for the second error amplifier using this technique? What should be taken into account, because in this case we take the feedback signal not from the capacitor, but from the lower side of the load resistance? I'm sorry, I have to work with a translator.
Thanks for your message. Great to know that the method works! I will need some time for this. Maybe you can send me some details via mail can.mehmet.tr@gmail.com so I can give a better answer. I will let you know coming week.
Wow! Talking about circuit A and B first then telling us about the opamp is just smart. In university or other youtube videos they only talk about the characteristics or conditions. which isn’t wrong but they never really show us whats the effect or why this configuration exists. Well done good sir. I appreciate op-amp now thanks to you.
Thanks for your message 😊 Great to know you have got better insight about the op-amp. Objective was to compare the two situations and thereby observing the effect of having an operational amplifier (op-amp) in the circuit. Good luck👍
Thank you sir 😊 We like to use a dual voltage source (-VDD) And how to design the R value when M1 and M2 w/L are different, and then find the Vov on this basic current mirror circuit...
