Back to basics: Practical capacitor charging currents by LTspice demonstration
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- Опубліковано 7 лют 2024
- Note that an inductance at the input to a closed loop switch mode system may induce instability.
See:
• Instability and oscill...
• Correction to:Instabil...
Thank you very much. I need this to my MPPT DC-DC BUCK 2-phase.
👍😊
Very informative, Thank you very much.
Thanks
Thank you for the video. So, the maximum current will no be higher than Uin/sqrt(L/C). And the same effect one can see in gate drivers, where we need high gate current in hard switching aplication.
Yes, correct. 👍. Hold on to my next, eye opener video.
Very nice.
Keep in mind that LTSpice by defualt adds Rser=1mOhm to every inductor in the model to improve convergency. This can be changed to 0Ohmin Control Panel -> Hacks -> Always default inductors to Rser=0. Not sure though what is applied when L=0, whcih may be the rootcause of the difference in Energy you see :)
All this is known. But...you missed the REAL question: is the loss with L dependent on R?
@@sambenyaakov Interesting question, thank you. Intuitively, not. Battery provide energy E*q. And capacitor is charged finally to E*q/2 energy. So we always lose E*q/2 in heat, we assume charge save in the circuit.
@@zaikindenis1775 Hold on to my next video on losses. Coming soon😊
@@sambenyaakovI mean that the total energy is not measured only over R2 if Rser=1mOh, hence the difference between L=0 and L!=0 energies. L is irrelevant for the E 🙂
@@nikolaivic4480 OK, Got you now. You are of course correct.
Hi sir. I have a question about LTspice.
Is there a way that add TI models in to ltspice and use it?
Thank you
You may be lucky and the model might be compatible, but if it includes non generic sources or syntax it will take an expert to convert.
Great insight...Sir could you please make video on smps input filter design with respect to voltage mode and current mode control and input filter impact on loop characteristics
Will try
Thanks Dr Sam Ben-Yaakov. Is that also concluded that no matter how large the inductance is the energy dissipation is the same? Also seems it may takes a bit longer to dissipate the same amount of energy than no inductance at all.
Indeed. Hold on to my next video an eye opener.
🙏❤
👍🙏❤️
Numero uno😊
At 4:46 I calculated 28mJ not 38mJ?
75/2= 28? Or you used another method?