Mr. Perreault, Thank you for your exceptional teaching ability. I enjoy electronics but you bring an enhancement to this enjoyment as well as a fascination.
This was a good one for learning the switching characteristics. I've seen similar waveforms in switchmode powersupplies for avionics equipment and now understand why it did what I saw.
An excellent lesson. Very comprehensibly presented - and by a very personable lecturer. Even for an amateur electronics hobbyist like me, every step is understandable. I learned al lot and I hope your students also appreciate the quality of your lectures.
I downloaded LTSpice for free and am simulating along with the lectures. I'm a Civil engineer and I can follow along quite well at this level, which is testament to the Professor's superb teaching style.
These few lectures have helped me start to appreciate how to work on circuits. I used to think that circuit analysis was just a collection of ad hoc tricks, but this professor makes a good point: these concepts are all very nonlinear, and a few solid principles can take you far. Very nice lectures, right on target.
Coming from being a mechanic and eventually an engineer this explains a lot to the “why”. In the alternator scenario explained by the professor, never disconnect the battery when a vehicle is running since it can cause a load dump or more appropriately it acts as a large capacitor handling any large load changes in the system.
When he gets to the end and asks what does this look like in the DC world, I was thinking from earlier that this looks like a battery, which can be modeled as a voltage source with a series resistance. As you pull more load, voltage sags, especially if you're using a lead acid battery.
I have built quite a few race cars and many with battery isolation\master cut off switches. The new style come with a giant resistor that is wired to ground and i always wondered why, now i know, to shed that spike from the sudden disconnection rather then sending it through the cars modules.
Some days I wish I had gone to MIT. I’ve never seen such a rigorous lead-in (pardon the pun) to Power Factor. I still hear in my sleep the phrase from military tech school, “ELI the ICE man”! I would have just shouted from the back, “add a capacitor!” :)
I'm an aerospace engineer, but I was thinking just add some capacitive reactance on the input. I think the next lecture is about power factor. So I think he will get there. Can I be an honorary EE? :-)
Don't we get a large current surge at turn on as well? Or are we just assuming components are already in place to absorb/dissipate this? Thank you for this series.
@@hugecannon yeah, inductor on output side has stored energy in form of magnetoc field because of te current, this energy is used to supply power to the load when D1 is off and D2 is on. you will see this very often in switched converters
Don´t think as "path to negative" you have a complete close circuit, then you can carie current, the one who provide the power is the inductor. Espero que se haya entendido. Suerte
because the integral of sine(x) is -cos(x) so for a bound [a,b] you would have - cos(a) -(-cos(b))= -cos(a) + cos(b) = cos(b) - cos(a) which looks like substracting the lower bound minus the upper bound
@michaelperkins3225 I am an EE, so I've taken all the classes for an undergraduate degree. I could not have easily passed this class in my sophomore year. I was still trying to wrap my head around electromagnetism.
It's a graduate course. See the course materials for more info at: ocw.mit.edu/courses/6-622-power-electronics-spring-2023/ Best wishes on your studies!
He changed the Integral from 1 period (time) to angular frequency (speed). The Period is the time taken for one complete cycle of the oscillation whereas angular frequency gives a signal oscillation's 'speed'. To switch between them, you use the formula w = 2*pi/T
if u integrate sin(wt) with respect to dt, you’ll have to change the integrated variables so they’ll match; do a simple u substitution (theta in this case) theta=wt;d(theta)/dt=w, so dt=d(theta)/w and now u sub these in the integral, and it becomes integral(sin(theta)d(theta)/w), w is a constant so you can pull it out the integral. hope it helps
No need for sarcasm. Everyone in the room of the lecture and the professor are smart enough to know he simply reversed the limits to account for the negative sign. This is MIT, not a community college.
Mr. Perreault, Thank you for your exceptional teaching ability. I enjoy electronics but you bring an enhancement to this enjoyment as well as a fascination.
Same as you
This was a good one for learning the switching characteristics. I've seen similar waveforms in switchmode powersupplies for avionics equipment and now understand why it did what I saw.
Hi
An excellent lesson. Very comprehensibly presented - and by a very personable lecturer. Even for an amateur electronics hobbyist like me, every step is understandable. I learned al lot and I hope your students also appreciate the quality of your lectures.
I downloaded LTSpice for free and am simulating along with the lectures. I'm a Civil engineer and I can follow along quite well at this level, which is testament to the Professor's superb teaching style.
These few lectures have helped me start to appreciate how to work on circuits. I used to think that circuit analysis was just a collection of ad hoc tricks, but this professor makes a good point: these concepts are all very nonlinear, and a few solid principles can take you far. Very nice lectures, right on target.
Coming from being a mechanic and eventually an engineer this explains a lot to the “why”. In the alternator scenario explained by the professor, never disconnect the battery when a vehicle is running since it can cause a load dump or more appropriately it acts as a large capacitor handling any large load changes in the system.
Very nice,thanks
Gracias, le daré las gracias por cada clase.
When he gets to the end and asks what does this look like in the DC world, I was thinking from earlier that this looks like a battery, which can be modeled as a voltage source with a series resistance. As you pull more load, voltage sags, especially if you're using a lead acid battery.
I have built quite a few race cars and many with battery isolation\master cut off switches. The new style come with a giant resistor that is wired to ground and i always wondered why, now i know, to shed that spike from the sudden disconnection rather then sending it through the cars modules.
i had the same question as that guy at the end glad he asked it.
The inductance 'slows down' the output capability to supply ac loads. I have blown up lots of ECUs with a Schaffner during load dumps!
Some days I wish I had gone to MIT. I’ve never seen such a rigorous lead-in (pardon the pun) to Power Factor. I still hear in my sleep the phrase from military tech school, “ELI the ICE man”! I would have just shouted from the back, “add a capacitor!” :)
I'm an aerospace engineer, but I was thinking just add some capacitive reactance on the input. I think the next lecture is about power factor. So I think he will get there. Can I be an honorary EE? :-)
40V with centralised load dump protection. Most cars have this now.
Damn good
Don't we get a large current surge at turn on as well? Or are we just assuming components are already in place to absorb/dissipate this? Thank you for this series.
👏👏👏good lesson!
I'm not understanding how current can flow when D2 is closed (short), D1 open. There is no path to negative?
Its the inductor on the output side acting as the power source for the second half of the wave essentially?
@@hugecannon yeah, inductor on output side has stored energy in form of magnetoc field because of te current, this energy is used to supply power to the load when D1 is off and D2 is on. you will see this very often in switched converters
Don´t think as "path to negative" you have a complete close circuit, then you can carie current, the one who provide the power is the inductor. Espero que se haya entendido. Suerte
Why, when he does the integral is he subtracting the lower bound minus the upper bound?
because the integral of sine(x) is -cos(x) so for a bound [a,b] you would have - cos(a) -(-cos(b))= -cos(a) + cos(b) = cos(b) - cos(a) which looks like substracting the lower bound minus the upper bound
@@louistiticaramel6848 thank you!
people who didn't get engineering degrees have no idea, how hard this is. this is a 200 level class, needed for 300 and 400 classes
This is a graduate level course
Why would they not? They’re watching the same video you’re watching.
@@_soupnazithis seems like undergrad 200 level after you've taken an electronics and circuits class
@michaelperkins3225 I am an EE, so I've taken all the classes for an undergraduate degree. I could not have easily passed this class in my sophomore year. I was still trying to wrap my head around electromagnetism.
According to the MIT website, this is a graduate level course.
does this idea of load regulation somehow connects with the back emf?
Why do we employ inductor?
how can find the data ?
Ur mom knows best about load regulation
Didn't know at the start, but I guess this is a power systems introductory level undergrad course.
It's a graduate course. See the course materials for more info at: ocw.mit.edu/courses/6-622-power-electronics-spring-2023/
Best wishes on your studies!
Can someone explain how the professor changed terms from Vs/L to Vs/wL? @21:58
He changed the Integral from 1 period (time) to angular frequency (speed). The Period is the time taken for one complete cycle of the oscillation whereas angular frequency gives a signal oscillation's 'speed'. To switch between them, you use the formula w = 2*pi/T
Changing variable of integration from " t" to 'wt'..
if u integrate sin(wt) with respect to dt, you’ll have to change the integrated variables so they’ll match; do a simple u substitution (theta in this case) theta=wt;d(theta)/dt=w, so dt=d(theta)/w and now u sub these in the integral, and it becomes integral(sin(theta)d(theta)/w), w is a constant so you can pull it out the integral. hope it helps
veesonegatee..
Not once did I hear Ripple, Sag, or RMS. Be nice if he could connect the ideas of average voltage and reactance to these.
I am assuming they are in later lectures.
Don’t like chalk 😟
please stop slowing down development
wdym
integral is -cos
The integral it's correct. He switched the terms in the substraction to compensate the -1
@@tomasvidal424 you are correct. I always plug the limits of integration in the reverse order when evaluating... lol
No need for sarcasm. Everyone in the room of the lecture and the professor are smart enough to know he simply reversed the limits to account for the negative sign. This is MIT, not a community college.
Limits were substituted straightaway...
@@donmoore7785 thanks for being an elitist douche