Nice lecture and topic. It is very important to understand stored energy and therefore switching losses, and I hope many of my peers in the semiconductor industry would take care on learning those concepts. Talking of switching losses, I actually quite like the approach used by Matioli [Perera; Matioli - 2020 - Analysis of Output Capacitance Co-Energy and Discharge Losses in Hard-Switched FETs] and [Perera; Matioli - 2022 - Hard-Switching Losses in Power FETs_ The Role of Output Capacitance], where the turn on of one device has to dissipate its stored energy and the co-energy of the other device being charged. There is a large misconception, especially in industry, where the no load turn on and turn off losses are assumed to be Eoss, and the switching loss to be 2*Eoss*fsw. Firstly, the turn off loss should not include the Eoss stored energy, as it is not yet dissipated and therefore that energy should be removed from the turn off and considered in the turn on. In this case, at zero current, the Eon would be then 2*Eoss and Eoff=0. But that would not be right. Indeed, as definition of work and as explained by Matioli, the correct calculation should be Eon=Qoss V, which takes into account both energies. That's of course a first approximation. But maybe this can be a nice hint for a future lecture (unless I missed it).
One thing I didn't quite follow from this description is that it seems to imply that charging a capacitor is inherently 50% lossy, i.e. if you expend 1J of energy to charge a capacitor, you'll get at most 0.5J back when discharging. This doesn't seem to gel with my practical understanding of ripple current and dissipation factor, so I'm guessing I've misunderstood something along the way. Is the "E_loss" under the Q/V curve specific to contexts where the energy is being dissipated into a load, i.e. representing a symmetry between the energy you put in and the energy you get out?
@@sambenyaakov so if the series resistance is 1pΩ, and you charge the capacitor to some arbitrary voltage, then discharge it again, you lose 50% of the energy? That seems counterintuitive. The ohmic losses would be ~0. What's heating up?
@@gsuberland the ~1pOhm resistor is heating up. The intuition is that the current in that scenario is extremely high (say 1V/1pOhm is is tera-amp), so despite it being a low resistance, the power dissipation is huge!
The assumption here is that charging occurs via a perfectly stepped voltage source. This is not always the case, especially for switchmode cap chargers. If the output current is instead limited by a switched inductor instead of the voltage difference between Vin and Vcap dropped on a dissipative element then there is no loss implied by QV and efficiency can approach a theoretical 100%.
Great Video. Could you please point publications where the method you mentioned at the end for calculating losses, is used? I mean the graphical one in which two capacitors are connected
Thanks for the video sir, it would be great to have a video on the discharging of the inductor, specially in the boost converter, since I’ve always been confused as it seems the voltage across the inductor is constant during discharge. Thanks
Nice lecture and topic. It is very important to understand stored energy and therefore switching losses, and I hope many of my peers in the semiconductor industry would take care on learning those concepts.
Talking of switching losses, I actually quite like the approach used by Matioli [Perera; Matioli - 2020 - Analysis of Output Capacitance Co-Energy and Discharge Losses in Hard-Switched FETs] and [Perera; Matioli - 2022 - Hard-Switching Losses in Power FETs_ The Role of Output Capacitance], where the turn on of one device has to dissipate its stored energy and the co-energy of the other device being charged.
There is a large misconception, especially in industry, where the no load turn on and turn off losses are assumed to be Eoss, and the switching loss to be 2*Eoss*fsw. Firstly, the turn off loss should not include the Eoss stored energy, as it is not yet dissipated and therefore that energy should be removed from the turn off and considered in the turn on. In this case, at zero current, the Eon would be then 2*Eoss and Eoff=0. But that would not be right. Indeed, as definition of work and as explained by Matioli, the correct calculation should be Eon=Qoss V, which takes into account both energies. That's of course a first approximation. But maybe this can be a nice hint for a future lecture (unless I missed it).
Good points. Thanks. Perhaps.
One thing I didn't quite follow from this description is that it seems to imply that charging a capacitor is inherently 50% lossy, i.e. if you expend 1J of energy to charge a capacitor, you'll get at most 0.5J back when discharging. This doesn't seem to gel with my practical understanding of ripple current and dissipation factor, so I'm guessing I've misunderstood something along the way. Is the "E_loss" under the Q/V curve specific to contexts where the energy is being dissipated into a load, i.e. representing a symmetry between the energy you put in and the energy you get out?
Well, this is a physical fact. In charging of a cap from zero you loose 50% due to the rms current on series resistor
@@sambenyaakov so if the series resistance is 1pΩ, and you charge the capacitor to some arbitrary voltage, then discharge it again, you lose 50% of the energy? That seems counterintuitive. The ohmic losses would be ~0. What's heating up?
@@gsuberland the ~1pOhm resistor is heating up. The intuition is that the current in that scenario is extremely high (say 1V/1pOhm is is tera-amp), so despite it being a low resistance, the power dissipation is huge!
@@gsuberland I am talking about this in video.
The assumption here is that charging occurs via a perfectly stepped voltage source. This is not always the case, especially for switchmode cap chargers. If the output current is instead limited by a switched inductor instead of the voltage difference between Vin and Vcap dropped on a dissipative element then there is no loss implied by QV and efficiency can approach a theoretical 100%.
👍🙏❤
Thanks
Great Video.
Could you please point publications where the method you mentioned at the end for calculating losses, is used? I mean the graphical one in which two capacitors are connected
I don't remember off hand, but there are other videos in my channel that explain this approach.
Thanks for the video sir, it would be great to have a video on the discharging of the inductor, specially in the boost converter, since I’ve always been confused as it seems the voltage across the inductor is constant during discharge. Thanks
And also if the current does infact decrease, why is the voltage across the load resistance constant under discharge also?
Good suggestions. Will try.
Dear Sam,
Would you please address how to achieve a certain amount of leakage inductance in a transformer?
Thank you.
For small leakage separation of primary and secondary will do, For larger leakages you ferromagnetic plug in between.