Great video, Professor. You are not wrong in your explanation, but I just want to point out that the main reason why the current in an inductor cannot change instantaneously is because the derivative must exist at every point. This requires continuity in the current function. Great work!
Just a few points: You should point out the importance of the resistance, as too low a resistance hold the output high long enough to significantly effect duty cycle - or prevent it from ever turning off. Using a Schottky and zener in series would have been a great addition to this video to see the dramatic changes forward voltage makes in switching time, ringing, and output voltage. And in higher power circuits, the power rating of the resistor is not trivial. It's also worth pointing out that this circuit is essentially the simplist and worst boost converter possible. I was going to say something about RC snubbers, but then I realized that's probably going to be your next video. Excellent video for >95% of people... I know I'm in the minority of people who learn every circuit by building one that does hundreds of volts and then can't figure out why everything explodes. 😁
Yes, there is a lot more that can be said on this topic, but it is an introductory video after all. I try to keep "the bites of information" small (pun intended). ;-)
With a motor being similar to a transformer than can physically rotate, is the inductive spike as coils are sequentially activated blunted because of physical rotation?
Fantastic explanation, is there any way to discharge the current loop in the snubber circuit so you don't overshoot the 10 volts (10.8V on the Vce waveform)?
10V is what's driving the circuit so I assume you're referring to the extra 0.8, yes? Generally it shouldn't be a problem, but you could get a diode with a lower forward voltage, e.g., schottky diode.
Thankyou. Presumably, if we put the diode astride the inductor only, the ringing/continuation of current is longer? In other words, how does the position of the load resistor, in this example, effect the outcomes?
@@simonyoungglostog It does make a huge effect, though. Because the spike gets limited to the forward voltage, you know the voltage and the resistance, you use ohm's law to get the amps, and then with the energy in the inductor you can calculate how long it will hold the voltage high. EEVlog has a video where a snubber diode on a relay holds it closed an extra half second or something crazy I never would have suspected.
Great video, Professor. You are not wrong in your explanation, but I just want to point out that the main reason why the current in an inductor cannot change instantaneously is because the derivative must exist at every point. This requires continuity in the current function. Great work!
True. I assume familiarity with RL transient response.
Just a few points:
You should point out the importance of the resistance, as too low a resistance hold the output high long enough to significantly effect duty cycle - or prevent it from ever turning off. Using a Schottky and zener in series would have been a great addition to this video to see the dramatic changes forward voltage makes in switching time, ringing, and output voltage. And in higher power circuits, the power rating of the resistor is not trivial.
It's also worth pointing out that this circuit is essentially the simplist and worst boost converter possible.
I was going to say something about RC snubbers, but then I realized that's probably going to be your next video.
Excellent video for >95% of people... I know I'm in the minority of people who learn every circuit by building one that does hundreds of volts and then can't figure out why everything explodes. 😁
Yes, there is a lot more that can be said on this topic, but it is an introductory video after all. I try to keep "the bites of information" small (pun intended). ;-)
Beautiful !
With a motor being similar to a transformer than can physically rotate, is the inductive spike as coils are sequentially activated blunted because of physical rotation?
Fantastic explanation, is there any way to discharge the current loop in the snubber circuit so you don't overshoot the 10 volts (10.8V on the Vce waveform)?
10V is what's driving the circuit so I assume you're referring to the extra 0.8, yes? Generally it shouldn't be a problem, but you could get a diode with a lower forward voltage, e.g., schottky diode.
What software are you using?
TINA-TI.
Thankyou. Presumably, if we put the diode astride the inductor only, the ringing/continuation of current is longer? In other words, how does the position of the load resistor, in this example, effect the outcomes?
In the real real world, you can't. Remember, the resistor/inductor combo is a simple model for a real world inductor. There isn't a separate resistor.
@@ElectronicswithProfessorFiore Yes, indeed. I did forget about every component having some resistance. Thankyou!
@@simonyoungglostog It does make a huge effect, though. Because the spike gets limited to the forward voltage, you know the voltage and the resistance, you use ohm's law to get the amps, and then with the energy in the inductor you can calculate how long it will hold the voltage high. EEVlog has a video where a snubber diode on a relay holds it closed an extra half second or something crazy I never would have suspected.