Lecture 5: Intro to DC/DC, Part 1
ฝัง
- เผยแพร่เมื่อ 5 ต.ค. 2024
- MIT 6.622 Power Electronics, Spring 2023
Instructor: David Perreault
View the complete course (or resource): ocw.mit.edu/co...
TH-cam Playlist: • MIT 6.622 Power Electr...
This class introduces DC-DC power conversion, including methods for analyzing DC-DC converters in periodic steady state. Buck and boost converters are introduced as examples.
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
Support OCW at ow.ly/a1If50zVRlQ
We encourage constructive comments and discussion on OCW’s TH-cam and other social media channels. Personal attacks, hate speech, trolling, and inappropriate comments are not allowed and may be removed. More details at ocw.mit.edu/co....
![[ TEASER ] เงาะป่า - วงL.กฮ. | TMG RECORD](http://i.ytimg.com/vi/M4ObdFfcPww/mqdefault.jpg)
Thank you for the opportunity to watch your lectures. I learn a lot and really enjoy them.
Pure gold, thank you so much for uploading these lectures!
its just boring useless info, only because MIT people like this
it´s a great honor for me have the oportuniti to learn from one of the best Institutions of the whole word! Thaks for this content
Excellent lecture. At 39:40, the inductor will generate a positive voltage that will aid the supply voltage to try to keep the current flowing.
Yes .. the boost comes from the inductive kick (opposite polarity when the inductor was in the charging phase) that rides on top of the supply voltage thereby BOOSTING the voltage to a higher voltage ... by applying feedback control to change the duty cycle one can regulate the output voltage with respect to a reference voltage .. but i suppose this will be covered in the next videos .. this is an excellent lecture series .. will finish the entire lecture series .. thank you Sir David for the clear step by step explanation ...
I was going to ask the same question. The polarity should be reverse from what is written on the board, right?
Spot on @@caleb7799
The polarity marks on the board (in orange) describe the reference direction for the voltage across the inductor, not the instantaneous value of that voltage.
its due to the Counter Electromotive force generated by the Inductor
It is helpful to look at the starting waveform of the inductor current in Buck and Boost converters to further understand the boost action. Start with zero voltage at the output caps and zero current in the inductor. If you correctly apply v_L = L di_L/dt, you'll find that right from the start, the buck inductor current ramps up from zero in [0, DT] due to a +ve voltage (Vin - Vout) applied across inductor and then ramps down in [DT, T], due to -ve Vout immersed across inductor. However it doesn't return back to zero because the Vout has not built up large enough to cancel the Volt-sec of the previous ramp.
The boost inductor on other hand experiences ramp up in both [0, DT] and [DT, T] with +ve slopes Vin/L and (Vin - Vout)/L respectively. The ramp in [DT, T] keeps happening until Vout > Vin, making the slope -ve.
Once you've slopes of opposite signs in the two intervals, and if Vin & D stays constant, then both the converter naturally end up in steady state eventually.
Also interesting to note is that the voltage and current conversion ratios are independent of load, implying that regardless of load, the output voltage will follow for instance Vout = D Vin for an ideal buck converter.
But what if I put a dead short as the load !!! The expression won't hold.
Similarly for the boost converter if the output is an open circuit, the formula won't work.
The reason simply is, that in both cases you can never achieve steady state. Inductor current will run away to infinity for a buck with shorted output and the output capacitor voltage will run away to infinity for a boost with open circuited output. This you can work out yourself by drawing the waveforms for inductor current and capacitor voltage from the beginning when the converter is turned on.
In practice, what is the principle behind the placement of the diode and MOSFET? Why do the MOSFETs in buck and boost converters seem to operate in opposite configurations?
Getting more voltage out than the voltage inputted - a boost convertor... doesn't that break the law of conservation of energy??? where does the extra energy come from?
I like the way he slams at your face unconventional ways to draw circuits or think about a problem.
A voltage undivider would be a voltage multiplier for everyone else...
I think that this lecture pretty much answered a longstanding question of mine: If you leave your laptop charger plugged in to the wall but not to your computer, will it be using electricity/costing you money. I think that the answer is pretty much no.
Correct, if you consider
It's Saul Goodman!
For the boost converter, what is the practical limit on the amount of voltage boost achievable?
That depends on what your practical constraints are. As voltage goes up, components get physically bigger, which eventually impacts efficiency and cost. A quick DigiKey search finds a LT8361 that is rated to boost 2.8VDC to 100VDC. If you're talking about that kind of voltage, or higher, then you have other issues, such as safety, that make "practical limit" pretty muddy.
If you're looking for specifics, it's the reverse voltage rating of the diode, the Vds rating of the FET, and the voltage rating of the output capacitor. These things drive physical size (and reduced efficiency), as @DavidLudlow said.
Even if you've components rated for higher voltage, the achievable voltage can be limited by the parasitics such as inductors ESR. Also if your inductor current ripple is bigger than its average value (due to low switching frequency, low inductance, light load or intended by design), then you'll be operating in another regime called Discontinuous Conduction Mode, this can also effect your conversion ratio.
34:57 The moment you realize you have lost 85% of the class.