Thank you so much professor. Clear and informative as always. It would be interesting to explore the effect inter-winding capacitance has on CM currents in HF application, coping mechanisms and the trade-offs between low leakage design and low parasitic capacitance design. It would be interesting to hear the opinion of an expert such as yourself.
Thank you very much Professor! Please, could you provide a lesson explaining the effects of increasing the switching frequency in a ferrite transformer using the full-bridge topology, and the phenomenon that allows for power gain, potentially resulting in a physical reduction in core size with frequency increase? An example would be increasing the frequency from 100 kHz to 500 kHz.
I get it now. Thanks once again for the paper. So, even though Lt inductance would be physically more appropriate (as an analog to mass in Newton's law with a special relativity correction, i.e. F=d(m(v)∙v)/dt being an analogy to V=d(Lt(I)∙I)/dt) it is actually more practical to define a different inductance Ld (I) so that we could use a simple formula V=Ld(t)∙dI/dt to obtain the same results. Very clever trick! Probably also more numerically stable as it eliminates the addition that could in some circumstances lose significant digits.
Thanks so Professor Sam, I do learn something new each time I watch you. I do have questions 1- the dc current across inductor will make it saturate? Is like you have DC voltage across zero ohm , so this will saturate core , and the average voltage across inductors should be zero 2- how can I calculate temperature rise due to dc resistance of the winding Thanks so much
1, DC current moves B up the BH curve an may cause saturation 2. The power is I^2*Rdc and there is an experimental equation for temp rise as function of the dissipated power and total surface area.
Thank you for the video, and your insights. I am curious, why when you have talked about AC inductors you stated that to achieve high currents you must air gap the core? In most application, the designing constraints are current and inductance -> same stored energy in different solutions. By using un gap core we can achieve the desired inductance by much lower number of turns (in comparison to gapped core) and which results in low magnetic field with a high current.
In most applications the requirement is NOT energy but L and I. The enegy is the consequence. Furthermore, energy density is B^2/mu, so air is a better magnetic energy storage than high mu material, contra-intuition .
Hello Sir great Video, If I am designing an inverter and I want to put an LC filter at the output of my inverter would you recommend a AC inductor or a DC inductor?
As I pointed out in the video, there is really no differece between the two. If the LC filter has to pass a high current, there will be a need for an air gap. Not so in a common mode filter in which the in and out currents cancel each other.
“Thanks for comment. Please indicate to what minute or slide of the presentation you are referring to.” ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-AfTTO4m1ti8.html Here, you say that everything here is constant, which allows you to factor it out of the derivative, but later you admit that μ depends on H and therefore on I, which depends on t. Therefore, it should have been differentiated with that in mind, in my opinion. PS: I'm re-posting to the main thread as it seems to me that youtube does not generate a notification when a reply is made to a reply.
By extension, should we gap the core of a transformer that is excited purely by AC (ie no DC component present). If I understand you correctly, the magnetising inductance will be enhanced for a given core size by having a gap.
Not so. A gap will allow higher stored energy (which a tranformer does not need, input=output at any instance) while reducing inductance and hence increasing magnetization current.
IF i have to design an inductor for a SMPS should i use stranded wire or 1 more wire? Asking since 1 end will be at DC while the other one will be switching all the time so it is effectively in AC. Skin effect is the thing that is my fear
The two ends have the same current which matters. You need Litz wire if the ripple component is large and of high frequency that skin and proximity effects are substantial.
Dear professor, thank you for another great video. However, I'm puzzled why the differentiation of the Faraday's law was not carried out like this, considering that μ is a function of H and consequently of I: V(t) = n∙dΦ/dt = n∙d(μ(I)∙A∙n∙I)/l)/dt = A∙n^2/l ∙ d(μ(I)∙I)/dt = (A∙n^2/l)∙( dμ/dI∣_I(t)∙dI/dt∙I(t) + μ(I(t))∙dI/dt) = (A∙n^2/l)∙( I(t) ∙ dμ/dI∣_I(t) + μ(I(t)) )∙dI/dt writing μ' for dμ/dI: = (A∙n^2/l)∙( I(t)∙μ'(I(t)) + μ(I(t)) )∙dI/dt = L(I(t))∙dI/dt (*) the inductance L(I) would depend on the instantaneous current and would be equal to L(I) = (A∙n^2/l)∙(I∙μ'(I) + μ(I)) if we wanted to write V(t) = L(I(t)) ∙ dI(t)/dt. If, on the other hand, we would adopt that V(t) = d(LL(I(t))∙I(t))/dt the LL inductance would be defined differently. Namely, by carrying out the derivative, we get V(t) = LL'(I(t))∙I'(t)∙I(t) + LL(I(t))∙I'(t) = (LL'(I(t))∙I(t) + LL(I(t))) ∙ I'(t) Comparing it with (*) we see that L(I) = LL'(I)∙I + LL(I) which hold if we defined LL(I) = (A∙n^2/l)∙μ(I). So, which of the inductance definition do you like more? L(I) or LL(I)? I'd go for LL(I) for its simplicity despite it requires us to modify the formula for voltage across the inductor to V=d(LL(I)∙I)/dt.
@@sambenyaakov ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-AfTTO4m1ti8.html Here you say that everything here is constant, which allows you to factor it out of the derivative, but later you admit that μ depends on H and therefore on I, which depends on t. Therefore, it should have been differentiated with that in mind, in my opinion.
