Buck Converter

[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/

[@slushy6972]

Chemistry, Biology, Algebra, Calculus, Electronics, this guy does it all

[@kiche_allan]

Since my first Yr in 2018 enrolled in school of Engineering, this channel has been a great resource. Towards the finishing line and am still here. Awesome work guys

[@tjdlamini4062]

You are I both😁

[@photon2724]

its insane to me why lecturers dont explain like this. Soo good!

[@FaraiMutemeri]

12 minutes JUST 12 MINUTES!!!! I thank you!

[@eswing2153]

Great job!!! Nicely explained just how I learned it in school. It’s wonderful to get such explanations available on KZbin.

[@williamfolk3099]

Fantastic explanation with a very easy to understand steps during the tutorial. Then with the math added to demonstrate the outcome. Great Job! I have read a couple other descriptions ahead of this video and this video completed my understanding. I graduated college in 1981 with a Biomedical Engineering Degree. However, have forgotten most of the rudimentary electronics over the years. This was really easy to understand and explained every element of the circuit well! Thank you!

[@garethcopping3213]

This is one of the best explanations of buck converters I have come across. Keep up the good work!

[@nikolatesla399]

That's why I love youtube. You can learn the whole engineering concepts for free here that too from best teachers..

[@SohamChakraborty42069]

At 3:00 shouldn't it be VL = Vin - Vout considering the polarity of the voltage across the inductor?

[@mgakhtar1001]

VL = vin-vout. Not the way you showed, otherwise it would a negative voltage across the inductor. Good videos but this point got me confused so I just went back to KVL.

[@rastabanja6444]

Thanks for the correction ❤

[@bondamekur4837]

Holy mother fuck man 😰

[@d614gakadoug9]

On discharge of the inductor, the voltage will *always* be sufficiently high to deliver current to the capacitor and/or load until it is fully discharged. This is a fundamental property of an inductor. The equation Vout = Vin • d applies *only* if the inductor current is continuous, that is, it is never drops to zero. Once the inductor current becomes discontinuous (spends some time a zero each cycle) the relationship of Vout to Vin has to be calculated based on the energy stored and delivered each cycle.

[@Re-lx1md]

Can't believe an ochem tutor is helping me with my power electronics homework

[@NeoNeo96352]

Thank you very much for your great explanation, it helped a lot! Greetings from Germany. ✌

[@motivationalspark1681]

101%

[@puwartech7111]

Wow! great Explaination through this video i understood the vrm section of motherboard Thkkkku sir 😍

[@Anon-nv7bp]

Really great video, explained much better than my teachers managed.

[@felixcat4346]

The "switch" is actually more circuitry to get it to oscillate to produce the whole effect.

[@christopherhall8925]

You always give the best explanations.

[@adityakumar2649]

I can bet you're a good singer.

[@dawooduthaymeen]

because he almost knows everything.

[@ovalwingnut]

Good stuff. "Repetition Breeds Confidence" (I think I'll watch it again:). Cheers from So.Ca.USA 3rd house on the left

[@atexnik]

5:27 why are you adding them? These two currents do not flow at the same moment of time, so you can't just take and sum them up. Please, clarify.

[@patrickpaclibar9252]

I think its supposed to be averaging

[@kammanmok7294]

I really like your video. Writing and explanation are clear. It is easy to understand the concept s.

[@dpu3100]

If the duty cycle is responsible for determining the output voltage, what is the role of inductor in this circuit to reduce the output voltage as you explained in the beginning like "if 5v dropped across the inductor, remaining 7v will be there across the RL" Appreciated if anyone help me to figure out this doubt.

[@plinker439]

Without the inductor, the RL load would get 12V 'spikes' controlled by the 555 timer instead of the 7V. Let's say RL is a 1.4V CPU, 12V spikes would kill the CPU instantly. You just 'turn off' the COIL before it could charge up to the full potential of the battery (12V).

[@akimijohni8398]

you are the best one to understand keep going man and thanks to you i learn alot

[@EpicGamer-ux1tu]

woah, that video really helped me, easy, understandable... Thanks!

[@skampp]

Best explanation I've seen, thank you!

[@solapowsj25]

The inductor stores energy in the magnetic field that develops when current flows through the coil. Buck converters are useful in solar power supplies.

[@randhircheema1642]

Perfect!. Love it. Thank you heaps.

[@Jinjukei]

5:27 ... Didn't understand it well. When you add A and add B, then the sum of A and B is larger than A? Clearly yes... what's the point here?

[@jessef9041]

Great video! Simple and complex all in one.

[@kabandajamilu9036]

So nice and educative like how you explained logic gates so nice

[@yacinegourya9238]

Thanks for the video but why at 3:00 VL=Vout-Vin and not Vin-Vout :) ?

[@wilisonilagi1143]

Best explanation, m just curious with the period of the duty cycle

[@muhammadbilal2000]

At @2:59, shouldn't the VL = VIN - VOUT?

[@балаж98]

At 3:00 is probably a small mistake just put the values in: 5V = 7V - 12V is not correct. It should be VL = Vin - Vout

[@Elrussoendosruedas]

Nice job man, thank you for this explanation

[@joseluiscanalesgalvez9041]

Thank you for the classes, they are very educational, I have a question: Where did the 80% efficiency come from? Could you teach how to calculate?

[@namdoyle]

So in this example the Vout can never be more than 12v?

[@ragazzotedesco4216]

fantastic Video! thank you!

[@naessenskjell]

Thanks so much! Way better than my teacher!

[@lasithaamarasinghe9251]

simple great explanation

[@MuhammadBilal-im7yt]

Excellent explanation

[@Tyfreaky1]

Thank you for all your help and for providing great content Julio. Question, do you ever respond to comments or make posts on your Community page? I'm sure there's a lot of viewers who would love to connect more and learn more about you. We definitely appreciate your work.

[@sirgalantoe6325]

If the output voltage is only dependent on the input voltage and the duty cycle as explained in the final part, does that mean a random inductor and capacitor can be chosen for the cycle irrespective of their respective inductance and capacitance?

[@lucaspham7360]

Thanks for this video mate! Great explanation

[@mac_uk5464]

So most step down bucks converters use LM2596 as it's regulator, with a max iput of 40V. If I wanted a down converter that can handle 6ov input, I could use a ? as the regulator. Very well explained brill.

[@d614gakadoug9]

Once you get into voltages and/or currents higher than the integrated switcher ICs can handle you would use a controller IC and a separate switch, typically a power MOSFET. Properly driving the "high side" switch for a buck converter is a little tricky. Even powering the controller IC requires some attention to detail.

[@Zikeyboy101]

are you saying the output voltage isn't affected by the value of the inductor and the switching frequency?

[@vasiqshair]

At the end of the video, you say that if the converter were 100% efficient, the voltage and current are both modified by a factor by 2. Is it always the case that the factor is the same?

[@alans2416]

3:01 On wikipedia, they say VL = Vin - Vout, which makes sense. Node voltage method. Are you sure you didn't make a typo? You showed VL = Vout - Vin

[@NIFUable]

Yes it is Vin - Vout, when the inductor is seen as a source, which he did.

[@alans2416]

@@NIFUable No he didn't? I gave the timestamp too. When the switch is closed, VL = Vin - Vout

[@NIFUable]

@@alans2416 I meant it like this way. He has seen the coil as a source, so his voltage directions are the way he intended them to be, which means the voltage must be vin-vout

[@haydendoan7691]

Great explanations! Thank you.

[@TheLevano22]

At 3:03 minute, it says that the voltage of an inductor is the difference between the voltage of the load minus the energy source. Shouldn’t it be vice-versa?

[@sRockL]

I believe this explanation is incorrect. inductor will not impose any voltage when a static current flows through. It will generate voltage only when the current changes, which is well explained in wikipedia.

[@zachdavis5834]

@@sRockL The current through the inductor is always changing, thus there is always a voltage drop across it. This is the reason why buck regulators are horrible for EMI purposes and a poorly designed buck regulator usually will not pass FCC standards for radiated emissions.

[@Anon-nv7bp]

Yh it should be vice versa. VL = Vi - Vo. He just got them the wrong way round, probably a typo.

[@successthruknowledge]

So are Buck converter circuits very common? Because you don't want waste heat damaging the device?

[@sleddy01]

Phenomenal explanation. Thank you.

[@waterfuel]

What about circuit with switch in series with coil, (buck type), and needing the 2nd flyback pulse to go forward also- as additive?The diode anode is facing the source plus voltage , and cathode is at end of other diode in series with coil L. In other words 3 diodes, diode ,coil, diode and diode across coil and diode - all cathodes are facing forward, and NO capacitor is present in my reference circuit found from 2008.

[@rickpontificates3406]

I agree with everything you said EXCEPT; if S1 is static on (100% duty cycle), how do you figure you’ll get 7V out of that 12V supply at RL? With no PWM on S1, RL will see 12V after the lag time of your inductor and the R*C time constant of your cap and your load.

[@Ham549]

So why do you need an inductor? If you just get a raw PWM signal and put it through a capacitor the capacitor just smooth out the voltage. I built a 555 circuit that is adjustable and the output of the connector to a capacitor and by adjusting the oscillation of 555 timer I can lower the output voltage.

[@sirgalantoe6325]

Whats a 555 circuit tho?

[@Ham549]

@@sirgalantoe6325 a circuit that uses a 555 timer. 555 Timers are very handy but I am not even Shure what they do.

[@PranavMaru]

Awesome explaination 🔥

[@nosferatu6633]

How is there a voltage drop VL when the current is constant?

[@jpechacek80]

How do you calculate the inductor & cap values?

[@JTKhopefuls]

What’s the formulae to find Vin ,when given Vout,Iout and Pout?

[@jimmytorreon9615]

Does astable multi vibrator fit for this as a switch which is square wave ?

[@basiljey2902]

So when the switch is open the current will flow from negative towards positive of the inductor and so confusing how exactly a current flows through ground and back to the load?

[@iancanal1675]

Thanks for your explained... I need your help to undesrtand about solar panel control, so i can build for myself.. I need to fully understand everything about solar panel control.. Thank you sir..

[@andrealfiansah9432]

What if i remove diode and place capacitor across switch?

[@016632]

it would be nice if at the start of the video you tell what a buck converter does

[@sarvanasadov6685]

Hey thanks for a video, but are you sure that VL=Vout-Vinput? maybe that should be opposite like VL= Vinput-Vout?

[@lucasianii8178]

Thanks for nice explanation sir.

[@jdmccorful]

Good lesson, answers alot of questions as well as creating questions.

[@mandcali7613]

Hi, this video can confuse us learners. Coz there was no parr of the ckt drawn that would provide the duty cycle effect. This cld be devastating to learning if the context is not proper. But I want to thank you for the effort.

[@shivamchauhan3461]

Explain very nicely 👌

[@AlpakaAntifa]

Why is the voltage when turning off the switch double the amount of when turning on?

[@aelejaga3780]

shouldnt you have to calculate de RMS voltage in order to calculate the power on the load? which is Vin*sqrt(Duty cicle)

[@tientien0077]

This lesson is wonderful.

[@mohanramakrishnan3287]

Very nicely explained

[@midwestheatmechanicalinc.5148]

Great explanation! very detailed.

[@youssofasiry7233]

Thanks for your information Can you tell as witch program do you use ?

[@Theo005-k5e]

0 voltage means negative voltage?

[@varundesai7566]

How the voltage across Inductor is 5V when switch is closed. Please clarify.

[@ahoorapersia4698]

Hi sir. Tnx for your good job. Please explain minus or negative voltage ⚡ and its applications. 🙏🙏🙏

[@chibuzoeze4225]

Please make a video of full circuit diagram. Thanks

[@douggale5962]

It doesn't mention parasitics or ringing, or switching transients, or snubbers, but yeah, that would be overwhelming too early. There is a whole bunch of fun stuff to learn about switching converters. Good video.

[@kedharguhan]

Why is the diode required in the buck circuit?

[@d614gakadoug9]

That diode, sometimes called a "freewheeling diode" creates the path through which current flows as the inductor discharges. When the switch opens, the current through the inductor will be exactly the same as it was at the instant before the switch opened. It will make that magnitude of current flow through something. With the diode in the circuit, the current flows through the diode and the current ramps down linearly with respect to time. The current flows through the load and the capacitor. Without the diode the voltage across the inductor would rise to a level that would let that initial current flow somewhere. The voltage could reach hundreds of volts (in theory there is no upper limit with ideal components, but with real components you rarely get more than several hundred volts - the switch opening takes a bit of time and there is capacitance between the turns of wire on the inductor). The high voltage typically would destroy the switching transistor. If it were a mechanical switch it would make a arc across the opening contacts and most of the energy would be lost as heat.

[@arnokalayc2054]

Hi, I'm not an electrician, neither don't have any science base. I'd like to ask a question about "LM2596S Constant Current" buck converter. Let's say, we have 12V 1A adapter, power supply. Normally when we step down this to 3V, the current should increase in %100 efficiency to 4. I know this converter has 3A limit. So, this much decreasing the V can harm the converter? Additionally, when I regulate the converter to provide constant 0.5 or 1A, where the difference goes? Is this dangerous? I'm interested in modelling and going to use some LEDs. Therefore I'm trying to be cautious not to burn the house :) Thanks!

[@christopherowens2635]

My daughter’s 12v electric Razor scooter is too fast for her. I like to drop the voltage down to 9v. Do you know a buck converter that can handle the scooter without overheating? It’s a 10 mph E90 Razor scooter. Send me a link if you would. Thanks.

[@davesdigitaldomain]

can anybody tell me what program he's using For drawing his diagrams?

[@romulopartoza5598]

8 amp is a large value for output current....where it comes from? Show some computations, sir! From the Philippines... What are the values of the inductor, capacitor,diode and resistor?

[@serg3y]

Question, if the input voltage is changing the output will also change or will it regulate?

[@d614gakadoug9]

Without a feedback circuit to adjust the duty cycle the output voltage would rise.

[@hackthis1767]

Hello 👋 can someone help me out so I have a 24volt 1amp intel stick pc hypothetical if i we're to apply the convetors for a tft panel that runs off of 12 volts 2 amps along with this pc can I run it via 24 volt 5 amp plug 5.5mm to extension to two split 5.5mm males to two separate step downs buck convetor one for 24 volt 1 amp and the other to a buck converter that is rated for 2amp should this follow laws correctly in theory since the power rail is two separate DC lanes in series

[@alexs9324]

Well, and what will happen when the circuit hasn't D1 and switch S1 will open... what will be voltage on sides of the inductor :) Will reach voltage 2V and 7V or somthing different value and it will be related to quality factor of inductor and switching rise time?

[@50_fadhil63]

If the circuit has not D1, the circuit could blow up. D1 here is called free wheeling diode, to make sure when S1 is open, there is still path for current to flow.

[@holthuizenoemoet591]

Very pleasant teaching style, thank you. What happens if your load doesn't use the extra current? does the voltage increase? Is a buck converter a current or a voltage source (or neither)?

[@d614gakadoug9]

In a a real circuit feedback to adjust the duty cycle to maintain the output voltage is essential. If the load doesn't require the energy stored in the inductor then the output voltage will rise and can become equal to the input voltage.

[@jasonspitzer1503]

Excellent! Thank you.

[@ashlynnundlall]

Do you have cuk and sepik converters ?

[@我喜歡凤龍]

May I have digestive system and type of tooth video,I hope next will be these two

[@Britishbaptist]

Doesn't the source have to be ac for an inductor to charge? . but great video you are a natural academic

[@KilRBass]

Short answer: Since we switch on and of our input voltage periodically we've got an AC voltage which is rectangular in waveform Long answer: Yes and no: If you connect a constant voltage to an inductor in steady state (after an initial transient phase) not much happens. I think this is what you mean by "A DC voltage doesn't charge an inductor". But like mentioned above there is a transient phase: Let the voltage across the inductor be 0V. You then swith on your DC source and the inductor voltage jumps to V_in (input voltage) but the current that would start flowing through the wire is can't jump in an inductor. It starts increasing exponentionally which builds up a magnetic field. This is (roughly explained) the stored energy in the inductor. This transient phase is (depending on the inductance) very short. But since we switch our DC input voltage on and off quickly this happens often, so the stored energy in the inductor drives a current when the switch is off and the magnetic field in the inductor builds back up if the switch is on.

[@Britishbaptist]

@@KilRBass thanks im just a student nd i have a lot of iterest in this stuff .

[@d614gakadoug9]

If an inductor is fully discharged (no energy is stored as a magnetic field) and you apply a voltage across it, the current through it will start at zero and rise linearly with respect to time. With an ideal inductor with no resistance there is no limit to the current that could be reached if you allowed sufficient time. With practical inductors that doesn't happen. The main limiting factor for inductors used in switch mode power supplies is the "core." The core can only handle a certain amount of magnetizing force before it "saturates." When the inductor is dicharging in the circuit, the current still flows in the same direction but the polarity of the voltage across it reverses. The discharge is also linear change (decline) of current with respect to time. Given enough time the current will drop to zero. di/dt = V/L the differential of current through the inductor with respect to time (or, the rate of change of current in the inductor) is equal to the voltage across the inductor divided by the inductance. i is in amperes, t is in seconds, V is in volts and L is in henries One very important thing that this equation says but maybe isn't really obvious is that you cannot instantaneously change the current through an inductor.

[@d614gakadoug9]

@@KilRBass No. The current in the inductor changes linearly, not exponentially with time when a constant voltage is maintained across the inductor.

[@TheFlxwer]

Could u explain the detail applications of buck converter??

[@puwartech7111]

Vrm section of motherboard

[@zozoconde3854]

Best explanation! Keep it up!

[@imanmomeni2990]

please make more videos about electronic 👍👍👍

[@EnglishforCambodia97]

Thank you

[@satyapalsinghrathore6547]

You mean, duty=1ms/5ms, right?

[@d614gakadoug9]

No. The total period is 5 ms - 1 ms ON and 3 ms OFF.

[@zirkumflexXx]

Why is it VL = Vout - Vin? Should it not be VL = Vin - Vout? Because Vin is equal to VL and Vout,.