Lecture 5: Intro to DC/DC, Part 1

[@carlosmarcelomathieu2712]

Pure gold, thank you so much for uploading these lectures!

[@minneso5424]

its just boring useless info, only because MIT people like this

[@عبدالعفوهقي]

​@@minneso5424 I watched these episodes and concluded that there are lessons or lectures that are linked to each other, so you will notice a deficiency or gap in these lectures. Although this university is the best university in the world, it seems that the matter is not normal, as there is a loophole or gap or deficiency of some kind. I am certain that there are lectures that were not announced. Or data from a previous year was studied.

[@minneso5424]

@@عبدالعفوهقيMIT is the best university 100%, thanks for explanation why all this mess

[@tajox]

@@عبدالعفوهقي what do you mean? could you please be more specific? thanks

[@عبدالعفوهقي]

@@tajox These lessons are good but there are other lessons we should learn. Most likely there are lessons that have not been published and they complement each other. Let's learn power electronics. There are other lessons with it and most likely they have not been published. I have a question. What country are you from. Are you a boy or a girl and how old are you? Thanks

[@edsonthackson8566]

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

[@hassegreiner9675]

The best known and perhaps oldest boost converter must be the gasoline ignition system where a coil is saturated with DC-current from the 12V battery and the rotating mechanical or electronic switch breaks the current at the right time in relation to each piston's phase to create a massive voltage drop over the sparkplug gap thus forming a spark that ignites the fuel/air kixture. Close followed by the electric fence which works sort of the same way and, of course, the defibrillator which resets a heart having lost it's internal chamber synchronization - a procedure known as cardioversion which I've personally benefitted from twice in my life. The two latter first charges a condenser to the energy (voltage) required which then feeds an 'ignition coil' to deliver the 'blow'. Thankfully you're in total coma, when they hit the defibrillator 'trigger', and you wake up fully restored and ready to go home for another MIT OpenCourseWare- session.

[@RA-ye3xl]

Thank you for the opportunity to watch your lectures. I learn a lot and really enjoy them.

[@SkyhawkSteve]

Switching power supplies are one of those areas of analog electronics that have continued to be interesting and challenging! Nice to have a review of the basics in this course.

[@ismailali1255]

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.

[@SamuelMolines]

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 ...

[@caleb7799]

I was going to ask the same question. The polarity should be reverse from what is written on the board, right?

[@nevis2769]

Spot on ​@@caleb7799

[@lukebennett5582]

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.

[@edsonthackson8566]

its due to the Counter Electromotive force generated by the Inductor

[@samad419]

people ask what i did during my vacation, I released electrical engineering for fun. The teacher, teaches a lot better than the ones I had in school.

[@johnpearcey]

Glad he talked about designs that blow up. I have a lot of those in my workshop lol!

[@TheKazMantic]

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.

[@TheKazMantic]

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.

[@ScottESchmidt]

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.

[@duhnboa5447]

Correct, if you consider

[@josephknapick5307]

For the boost converter, what is the practical limit on the amount of voltage boost achievable?

[@DavidLudlow]

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.

[@billargabright1610]

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.

[@TheKazMantic]

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.

[@daksheshgdvn7731]

Hii, to implement a power electronics design we need a schematic with ic, and which software can able to simulate with ic, which includes all ics ... I am really stuck at here...

[@JonitoFischer]

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...

[@dtung2008]

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?

[@pichitek8631]

it was described during the lecture, in the simplest scenario a MOSFET, IGBT or other power switch servers as a active switch, the diode is also "switching" just witouth the need of driving it. and the placement depends on the functionality you want to get. in buck converter you are essentially chopping input voltage and in boost converter you want to store energy in inductive element

[@sohamsuke]

The i looking like a lambda and the 0 looking like a D all the time is really stressful D;

[@wwgg1139]

It's Saul Goodman!

[@gungagalunga9040]

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?

[@aleckdzamara1738]

You get more voltage not power. The power is still conserved since output current is less

[@gungagalunga9040]

@aleckdzamara1738 oh yeah thanks

[@strajee]

The power in and power out are the same... (Vin * I in = Vout * Iout) The difference is you might boost 2v 10A to 20V 1A

[@POLMAZURKA]

where is lesson1,2,3,4,5,...............................as a sequence?

[@mitocw]

KZbin playlist: kzbin.info/aero/PLUl4u3cNGP62UTc77mJoubhDELSC8lfR0 View the complete couorse: ocw.mit.edu/courses/6-622-power-electronics-spring-2023/ Best wishes on your studies!

[@PROShineKITO]

boost and buck converters how damn tenacious the human being can be when we propose!

[@dimitararabadzhiev]

You can take buck converter and you can used as boost converter just change where you put power supply For example 24 V/12V buck converter can be use as 12V/24V boost converter 😂

[@jessstuart7495]

34:57 The moment you realize you have lost 85% of the class.

[@zzgsdcheng]

HELP: can someone explain to me: on the boost converter: when the switch is in the down position, or when the MOSFET is turned on, what is sustaining the voltage on the load side? is it only the capacitor sustaining the voltage? there is no discussion in the lecture about the function of this particular capacitor.

[@pichitek8631]

there is energy stored in the inductor, it was mentioned when current ripple was discussed. essentially you are pushing the current out of the inductor without any source connected.

[@tomcarr5307]

Yes, it is the voltage on the output capacitor that creates (sustains) the output current when the MOSFET is on (diode off). If this capacitor is too small the output voltage will droop as the charge is bled off.