How Boost Converters Work (DC-DC Step-Up) - Electronics Intermediate 1

[@billysunerson]

It is SO hard to find videos that clearly explain the step by step function of a circuit. Thank you!

[@utsavshah981]

"As it says video gives much more clarity than thousand words". This is the best video on KZbin for boost converter. Thanks.

[@igbandautala6039]

For the first time I actually understand this. Great explanation.

[@cortez31]

Thanks a lot, I'm studying mechanical engineering and whenever I come across electronics I tend to understand very little, but I've fully understood this circuit thanks to your video. Will definitely subscribe and try to learn more about electronics :)

[@freeenergytransformer5842]

Many electronic devices need more power than can be supplied by a battery or power supply. This is where a boost converter comes in. It takes the input voltage and steps it up to a higher voltage, supplying the necessary power to the device.

[@organicfarm5524]

Contrary to that, electrical engineering students understand mechanisms/machinery easily. I think it's because of that fact that mechanical engineering is physically more relatable than electrical engineering.

[@bungle6334]

​@organicfarm5524 Back when I was an engineering technician, I was more talented/interested with electronics. I did a lot of prototyping, including populating PCBs. When I was moving on from the company, the owners told me that I should have more confidence when it comes to the mechanical side of engineering. One of the owners, a mech engineer, said that mechanical engineering is mostly just "common sense." I still think about that. I think the thing about mechanical engineering, is that it's much more applicable and intuitive. You can look at bicycle and easily observe how the pedals turn the rear wheel without needing measurement tools or prior knowledge. You can't really look at a board or circuit and understand it's purpose without prior knowledge or measurement tools.

[@TampaTec]

your channel should have more subs.

[@SimplyElectronicsOfficial]

Thanks :) I am sure it will grow over time.

[@DaniyalX]

i agree!!

[@mrawesome6239]

Tampatec ikr it should have way more than yours. ;)

[@straceycalloway4833]

What power supply is he using to the base of the mosfet?

[@rursus8354]

114K

[@epiendless1128]

Simple way to make one of these - 4093 Schmitt NAND gate. Make the first gate into a Schmitt RC oscillator. Use other 3 NAND gates as inverters connected in parallel, to buffer the oscillator into the MOSFET. Use an NPN transisor to pull one input of the NAND oscillator low, stopping the oscillator when the transistor is on. (Because of the inverters between the oscillator and the MOSFET, the MOSFET will stop OFF - very important). Tap off a small proportion of the output voltage using a resistive divider and use this to turn the NPN on when the output reaches the desired voltage. Use high value resistors to avoid loading the output. I first saw that circuit as a TENS machine. Since used it very successfully as an LCD backlight driver. Make sure the diode is a fast switching diode or schottky rectifier. Ordinary rectifiers line 1N4001 are too slow to turn off, and drain the output capacitor when the MOSFET turns on. Even when just breadboarding, I found a slow rectifier is rubbish.

[@thejohnny2377]

This is the most helpful description of a boost converter I've seen.

[@roycesouthen1505]

I agree

[@freeenergytransformer5842]

A boost converter is a DC-DC step-up converter. It takes in a low voltage and outputs a higher voltage. The boost converter is a popular choice for powering microcontrollers, sensors, and other electronics. There are many different types of boost converters on the market. Some are simple, while others have more features. When choosing a boost converter, you'll need to consider the voltage requirements of your project, as well as the size and weight of the converter. Boost converters can be used to power projects that require 5 volts or more. They're also a good choice for systems that need higher voltages, such as 12 volts or 24 volts.

[@liminal-m3g]

Very clearly explained - which is a real skill when dealing with complex topics such as these - thanks!

[@sk.samiulreza6205]

This channel deserves million subs.

[@bascelik91]

awesome buddy! I am preparing an exam and had a lot of trouble understanding this, now it is easy! well done!

[@coolbionicle]

Holy shit! A boost converter is the same as a ram pump!

[@markspc1]

Hydraulics and electronics have lots in common !

[@kushfromhandsworth6601]

Angel Gonzalez Physics enit I'm guessing

[@evananderson8452]

Just like the great Pyramid

[@erwinconxepts]

that is exactly what i'm thinking too and it got me here

[@justinsampler5506]

Read a guys comment on ram pump video

[@teachertechnology1]

One of the best explications I have ever seen. Period.

[@shahedulislam94]

This intermediate series is really useful, thanks and I hope you continue to make them.

[@freeenergytransformer5842]

Boost converters are a type of DC-DC step-up converter. They are used to increase the voltage of a DC supply. Boost converters are often used to power devices that require a higher voltage than can be obtained from a battery or other DC source. There are many factors to consider when selecting a boost converter. The most important factors are the input and output voltages, and the current requirements of the load. Other factors include efficiency, size, and cost. Boost converters can be divided into two main categories: linear and switching. Linear boost converters are simpler but less efficient than switching boost converters.

[@WhatsUpWithSix]

I am loving the tool you used to show the circuit graphically. What software or service did you use?

[@martinjrgensen6744]

everycircuit

[@Eduardo_Espinoza]

@martinjrgensen6744 Thank you 😊

[@AyushKumarSingh-tm6bp]

Proto, gentleman

[@DrakkarCalethiel]

EveryCircuit? I love that program. Dead simple but actually very powerful for beginners to learn how to draw circuits, design them and simulate them. On a tablet ideal for mobile tinkering.

[@DrDope164]

This is a great channel, you should really have more subs. Will recommend to my fellow ET students.

[@freeenergytransformer5842]

DC-DC step-up or boost converters are a type of electronic converter that are used to increase the voltage of a DC power supply. They are commonly used to power devices that require a higher voltage than what can be supplied by a battery or other power source. Boost converters work by taking in a low voltage input and converting it to a higher voltage output. This makes them a really useful tool for powering devices like Arduino boards, motors, and LED lights.

[@naveedrajab5601]

Most helpfull electronics channel ive seen . You make everything so easy to understand

[@russelibrahim4045]

This is cool. I knew what a boost converter is but I never realized how easy it is to build one

[@thamerali9697]

Great great great channel. I have been looking for you for centuries.

[@Kenshindegozaru]

Thanks for your channel. Im really stupid, but I love electronics. Your videos help me a lot to learn. Thanks.

[@davebutler3905]

Well explained. One thing to help remember inductor behaviour is they attempt to keep the current going in the same direction, when they "collapse".

[@aasthapandey888]

the explaination is so smooth & simple, really worth it watching

[@shivamsonik7127]

OMG what an explanation. Love your work. You are making me inspire to do great work like this.

[@anthonyvolkman2338]

Excellent excellent excellent! I have been building these things for quite a while now, and you are spot on!

[@loverandjoy]

Question. Can this circuit be used to boost power from let us say a 22 volt battery to heat a heating coil resistor to heat 6dl water in a 1 litre volume for a duration of time?

[@benhardi2179]

This is the one of the best video ever that explains about this in very short video.

[@JerryLee..]

Awesome explanation, demonstration, and tools to show the workings.

[@leozendo3500]

Recommend to use a more scientific simulator. Every circuit is great for basic calculation, but it is not gonna provide you with real life acceptable values for your circuit to work properly. In real life, the inductor will ring and cause high voltage spike which could breakdown MOSFET. The diode must be a fast Schottky diode. Input cap must be populated or it won't step up much voltage under load. If there is no feed back and the output cap current peak, the circuit is theoretically unusable. Bad feedback loop will cause unexpected damage.(Popping LED without noticing)

[@bornatona3954]

so which simulator you recommend

[@av6966]

spice

[@bornatona3954]

Monsieur Mohandis spice is engine which is used in almost all sim

[@baladar1353]

Mr. Tech I'm sorry to say but you're wrong. In this particular circuit, there's an inductor, which generates a voltage spike in the exact moment when the FET starts to conduct (shorts the inductor). This spike goes through the diode, then charges the capacitor. This charge is very tiny (depends on the value of the inductor and the frequency of the pulse-generator), but the voltage of the spike is high. Of course you can't overcharge a capacitor, but if it's a - let's say - 200V one, you can be sure it will be charged up to this voltage over time. This is the only reason the inductor has to be in that circuit - the voltage spike it generates when it's shorted. Older engines use the same principle for their ignition, there's a capacitor, an induction coil (ignition coil), a breaker and at the end there's a spark plug. The sparks voltage is high (12.000-30.000V) while the battery voltage is 6 or 12V, the current is low, but it will ignite the fuel-air mixture in the cylinder.

[@radiofun232]

I agree. It is better to make it in real and test it. These computer models are interesting, but far from reality, especially when it comes to these types of circuits.

[@deankay4434]

Thank you having clear voice and reasonable speeds of explanation. I would think some call this a "Buck to Buck" convertor. I understand automotive electrical systems and a few components plus using a lab scope. As a dealer tech since the mid 70's, I am trying hard to make a simple HVAC actuator control circuit. It is a low current dc motor, plastic gear reduction to drive a door to direct air into or around the heater core. The dash controls are mechanical but I am handy in fabrication. It could be "Heat" but with a linear potentiometer. Since actuators can be simple, this on is as it drives a pot for feed back. Move the dash lever, the motor turns on and moves to new location and stops. Yet, it has to change directions. I was thinking of an OP Amp, the 2 pots as voltage dividers for input. To make it easy, use mosfets in "H" bridge. Just working out on paper, then test and build. Actuator are $10 and I can make a 5 Vdc circuit for mosfet control. Like you, I am into motorcycles but started on lawnmowers to make the $1 it paid in 67 and better than mower with wheel driven blades. Self taught as Yamahas where metric and metric tools fit standards, but not the other way. In the county, word of mouth found Datsuns, Opels and everything else coming to me to fix. It didn't hurt growing 7" to 6'3" at 11 and had tools w/unused machine shed on the farm. I just want to built this as I am updating a 67 C10 PU and auto trans where manual is. Plus updating the engine on the cheap. This way I can have my vehicle to drive to Dr. or store. My wife has a 96 Corolla w/160k so need backup vehicle she can drive and operate. Just simple but reliable. Sad to say she had medical issues and I will have my 4th fusion at L3-L2 and hopes it the last. I don't have a Facebook or cell phone, but get by. If you have an idea for a new video using two op amps to drive H bridge, I will be watching your channel. Thanks again for very good just job on the videos. Subscribed! Omaha.

[@tonyavito208]

Hello, nice explanation! What is the software you are using for simulation?

[@VictorRivarola007]

Thank you for your explanation of the boost converter. Unfortunately, I still have a question on their working. When the mosfet is off, the capacitor is charged by the circuitry. However, when the mosfet is on, there is a direct path from the input power source to the ground. That means that power is still being used, unlike the boost converter. It also means that the boost converter should short circuit. But it doesn't, so there has to be something I cannot figure out... Do you know what is It?

[@andrasszasz4373]

It's actually the inductor, when the mosfet is on the inductor act's as a load and starts converting the electric charge into magnetic field, when the inductor saturates ( reaches the maximum magnetic field it can store, measured in henry) it will be a direct path and the mosfet will get shorted. This is the reason why high frequency converters can use smaller inductor like 10uH because the "on" time is shorter and it will not saturate. Low frequency converters need bigger inductor, if not it will saturate and short out. Most dedicated dc/dc uC controllers have a current feedback as soon the current starts rising because the inductor starts saturating the mosfet will be close to prevent short circuit.

[@nejiniisan1265]

Since you are working with AC, there won't be a short. If the mosfet would be on steadly, then yes, a short would happen. But think as an AC circuit. A 500 uH inductor switched at 100 kHz would act as a 315 ohms impedance. Since you are driving the inductor with a pulse train, the back emf is the responsible for stepping up the voltage. Remembering that V_backemf = -L di/dt Where di/dt is related to the duty cycle and the switching frequency.

[@anandsivathanu8517]

@@andrasszasz4373 If inductor is saturated, mosfet will get shorted. shorted means what? mosfet become faulty?

[@andrasszasz4373]

@@anandsivathanu8517 Shorted means that the power supply's positive and negative rail will be shorted (same thing as if you take the powers supply positive and negative wire and touch them together), this will cause excess current through the mosfet which will result in excess heat and the mosfet will burn out.

[@anandsivathanu8517]

@@andrasszasz4373 Thanks for Your Kindly Reply Mr.Andras Szasz. I understand your reply.

[@rsedsrahi8794]

Thanks.... you have just made it easier to understand the working principle clearly.

[@abegailvino3111]

Hi your video was amazing, may I know what simulator you've used? :)

[@MMarkur]

It’s called Everycircuit.

[@rajeshsaha3446]

At DC voltage, inductor gets short(V=L(dI/dt) or ()Reactance offered by L, XL=JWL, W=0, XL=0)....so how does it boost the voltage

[@sadeqsadeq7128]

I think it depends on the voltage across the capacitor

[@mateimarin8190]

As soon as current passes through an inductor, the inductor generates a magnetic field (simply said). So now we have energy stored in the magnetic field of the inductor. And now as soon as the transistor closes, current doesn't flow through the coil anymore and the magnetic field collapses (because it's not being maintained by the flow of electrons) and pushes electrons through the diode and then into the capacitor. The diode makes sure that electrons can't get back to ground (negative) as soon the transistor is closed (so they are therefor forced to stay in the capacitor). This process repeats many many times until the capacitor is fully charged up. (Make sure that the diode can handle that high voltage so it doesn't break) the capacitor should also not be charged over it's rating. Correct me if I'm wrong with anything. Hope I could help

[@amal717]

If the magnetic field is going to accumulate each time the switch closes , then that means the voltage will increase into infinity!!!,actually what limits the voltage to stabilize at 100 volt??

[@julioalvar]

This is usually used with a resistor as load, so this helps the voltage to stabilize somewhere higher than the input voltage. On the other side, in real life, nothing can stop it. The voltage circuit is prone to infinite, but the components can't endure such voltages. Each component has a limit, and when the limit is reached, thing would systematically start to blow up.

[@amal717]

Julio Cesar Alvarez Gomez thanks for reply . but that is so strange that it will stabalize due to load because the load is constant

[@julioalvar]

No problem Mosab. You are right, the load is constant, so the load won't allow the capacitor to just increase its charge. While the Mosfet is ON, the load will get the energy stored in the capacitor, when OFF the load will get the energy from the inductor meanwhile the capacitor also reloads again. From another perspective, when the Mosfet is ON, the charge in the capacitor will decrease depending on the load value, and it will increase as the Mosfet goes Off. With that on mind, the output voltage during a cycle (i.e. Mosfet goes ON , then Off and finally ON again) will oscillate around an average value that remains equal if the ON and OFF Times of the Mosfet also remains equal. That is what I called stable voltage in the output.

[@amal717]

Julio Cesar Alvarez Gomez thank you alot. i got it .

[@gamalmothersil6943]

The capacitor has a limit so it will saturate after the maximum voltage is reached.

[@markbeukers221]

Thanks a lot, this movie is very useful! It describes the operation of the boost converter in a really clear way.

[@137bob3d]

good of you to make this video showing how an L works. esp. the part about the V across the L reversing and the current flowing in the same direction.

[@ducpham2896]

Can you explain me why When the switch if off, the polarity of the inductor changes, and the circuit uses inductor as a source instead of the 9v voltage source ?

[@SimplyElectronicsOfficial]

+Duc Pham because when the switch is off, there is now energy trapped within the inductor in the form of a magnetic field. It wants to get rid of this magnetic field, and the only way it can do that is to get rid of the excess energy in the form of electrical current. Therefore the only path it can go to is through the diode and capacitor. It has to change its own polarity to allow its current to flow to the capacitor.

[@TurbidSugar19]

Very nice explanation. Author should be proud of himself, wish i had doctor like this in my EE study. Ps-how out voltage control implemented practicly in this circuit? As well as clock for MOS (bet it isnt opAmps and crystals)

[@janbosik8953]

This isn't even science or engineering. It's pure magic.

[@abhisekrout8522]

What will make the output voltage settle at 100V?? As far as I know, here the output voltage will keep on increasing till the capacitor bursts. Someone..... Please correct me if wrong. Thank You.

[@kacperfilipek8461]

I'm probably late but, no, it will not. Notice that the voltage source charging the capacitor is a series of consisted of a coil and a battery. The voltage induced by the coil is relatively the same, because the current flowing through the coil when the switch is closed is also the same over time. And the capacitor charges because the sum of voltages on the coil and battery is bigger than the voltage on the capacitor. As the capacitor charges and it's voltage rises, the difference between left part of the circuit and the capacitor will lower, so the current flowing to the capacitor will be lower as well until it is balanced

[@rominhawk3949]

Excellent presentation; thank you. Though, I didn't get what happens to the current? Let's say I am inputting a 12V, 30A from a power supply, and I want to boost the 12V to 24V. How much of the current will be available on the output side? Thanks.

[@kuro2410]

Can someone pls explain to me why when the switch is on the current doesn't flow to the other part of the circuit. Shouldn't the current split between the 2 paths?

[@rohithnechikkunnan8374]

When switch is in on position ( closed) circuit gets completed through the inductor to switch(transistor) back to battery. At this stage negative voltage from battery gets passed through the switch to inductor , this voltage is negative, and as u see diode is placed in such a way that it can conduct only positive voltage to the other end and hence charge the capacitor. Diodes banded side is negative that goes to +ve terminal of capacitor , a diode can pass the voltage in only one direction so when the capacitor gets charged it cant discharge back to diode since +ve is connected to -ve of diode.

[@kuro2410]

@@rohithnechikkunnan8374 I researched a little bit and i think it's actually because the capacitor already has 9 volts after the initial startup and current flows from higher voltage to lower voltage. Since they have the same voltage the current won't actually go to the capacitor

[@xiaohe8594]

i havent seen a such clear video ever ! thank you so much

[@bassfear_9952]

people will always wonder why this wasnt teached it school cause this is a good qestion for every beginer of electronics.

[@gstube1]

First videos that I really understand electronics!

[@TheFlacker99]

I hope to see more of these. I just finished all of them by the looks of it.

[@fredparkinson1289]

At 5:15 you say that when the switch is opened polarity of the inductor switches so that negative is on the left and positive is on the right. Then you go on to say that current flows from left to right. But current flows from a higher potential to a lower potential, so why isn't it flowing from right to left?

[@JackJackie-xk7pl]

You are always just showing us how they work we also need to know how to design them

[@michaelc.4321]

I learned something today, it’s not often that I can say that for sure.

[@flanksinatra3785]

Great video. Very straightforward and understandable explanation.

[@simonhollad5511]

when does the voltage stop raising and why?

[@zdrux]

There's a point where the capacitor is fully saturated and can no longer store any more charge.

[@russelibrahim4045]

If the system is conservative, Vout = Vin * (1/(1-D)) where D is the duty cycle of the switching.

[@ginebro1930]

I never was explained why but heres my thoght, the inductor can saturate (when every magnetic domain is polarized in the same direction) after this the main principle wich generates the voltage won't increase. Other reason could be that the capacitor exceeds it's maximun admisible voltage, then a internal discharge occurs, for this reason you must take care to use a well insulated capacitor or don't use a extreme conmutation frecuency. The first of the 2 reason that happens first will stop the voltage grow, in the second case it will also destroy the capacitor.

[@Brian-uq6jm]

Say if you're using a standard 9V battery as power source on this boost converter, even though the voltage you generate is a lot higher, there's just not enough power overall (which means a lot lower current) to hurt you right?

[@rikka0_059]

If the output capacitor stores enough energy at sufficiently high voltage, even the boost converter and battery is not powerful, it can still kill a person in an instant. So always discharge before touch.

[@theinvisibleboy2973]

so, now the charges stored in that capacitor needs to be discharged. Or else what happens ??

[@d614gakadoug9]

In a real circuit there will be feedback so that the voltage is regulated. One way or another you must adjust the duty cycle of the switch so that the amount of charge put into the capacitor is just equal to that being taken out by the load. This might be done by directly controlling the duty cycle with a fixed frequency or by using a variable frequency with fixed ON or OFF time, or by "hysteritic" control where you completely disable switching until the voltage has fallen to some threshold, the re-enable it until the voltage has risen to some threshold. Most converters use fixed frequency and variable duty cycle. Some will also reduce the frequency with very light loads to improve the efficiency. If you don't use feedback, the voltage on the capacitor could rise to the point where something would "break." The switching transistor or diode might fail from over-voltage or the capacitor might rupture. You might be be able to get a thousand volts from a 9 volt battery. You would need to have a well-designed circuit so that the switching and other losses wouldn't be limiting. Trying to go much over 1 kV limits your choices in components, but it could be done.

[@boog567]

The maximum unloaded charge voltage that the capacitor reaches will be determined by the size of the inductor and the point at which its core saturates.

[@d614gakadoug9]

@@boog567 No. Once the inductor is charged to any extent when the switch is ON, that charge is going to go somewhere when the switch is turned off. If the switch, diode and capacitor can take it the charge will go into the capacitor. As the capacitor voltage increases the voltage across the inductor increases during discharge time, shortening the time it takes for it to discharge. Discontinuous inductor current is easily possible meaning core saturation isn't happening. di/dt - V/L (i in amperes, t in seconds, V in volts, L in henries) If you had 10 V in and a 50% duty cycle, the inductor would be able to fully discharge each cycle once the capacitor voltage got above 20 volts since the voltage across the inductor during discharge is the cap voltage minus the supply voltage. As long as the discharge volt-second product is greater than the charge volt-second product inductor current will be discontinuous so saturation won't happen. The cap voltage would continue to climb until something couldn't withstand the voltage.

[@boog567]

@@d614gakadoug9 The inductor current never reverses nor reaches zero during the entirety of a duty cycle.

[@d614gakadoug9]

@@boog567 I agree that the direction of current flow never changes, ignoring rining that can occur at the end of discharge, but the ringing really isn't of any importance in energy delivery. The current in the inductor absolutely can go to zero. That is how discontinuous inductor current mode (DCM) is defined. Once the energy stored as a magnetic field has been "delivered" the current drops to zero. One of the fundamental problems with boost converters is that if they are operated in continuous inductor current mode a "right half plane zero" is introduced into the transfer function. Unlike a normal zero in the left half plane, you cannot compensate for a RHP zero with a pole in the error amplifier. Your only option is to accept low bandwidth and the consequent poor dynamic performance. The same issue exists with flyback converters in CCM. (in a flyback converter you can easily tell exactly what is going on with current flow by looking at the _voltage_ on the input winding of the inductor) I haven't designed a lot of boost converters because they are not really of very much interest, but I have designed active power correction input stages for industrial switchmode power supplies. Active PFC uses a sophisticated boost converter in which the the instantaneous input current is made proportional to the instantaneous input voltage (for good power factor) but the average input current is inversely proportional to the average input voltage because of the negative input resistance characteristic of a switchmode converter.

[@briant1550]

I have such an appreciation for this video. Thank you

[@PreludeSon]

What software are you using to build the circuit in the video demo?

[@CondensedComments]

InfinityOrchid Someone said in another comment it's called "Every Circuit."

[@wb5rue]

Great video with an excellent simplified explanation.

[@engineer1075]

sir you are doing really good job using every circuit....please do more videos on power electronic converters thank you

[@schloumf]

Hello, could you tell me please, what timer 100µs you put at the bottom of the circuit--- and the way you have connected it ?Thank's !

[@alhafed11]

Good explanation. Where’s the mosfet gate connected to pls? Thanks

[@DD1316]

Brilliant! Do you plan on providing actual components/values for something common like Input 5-9V 6A, Output 6-30V? Thanks for this vid!

[@pochipooch]

This is very helpful, thank you for your work.

[@mrsemifixit]

When the transistor is off and polarity across the inductor becomes reversed, does that interfere with the input polarity somehow?

[@mmdnaderi7183]

this channel is AMAZING !!!!!!!!

[@extrememod_ahu]

Great video. The only thing that's missing is the loss of amperage mechanics. As voltage goes up, the amps go down dramatically.

[@ZSilver55]

by having the meter on the end of the circuit wouldnt it discharge the capacitor?????, nice vid

[@francou61]

That's a very good point. Any load you put in parallel to the capacitor will draw current from it and discharge it until an equilibrium between the charges coming in (from the inductor) and out is reached. The higher the current going through your load, the lower the voltage will be. Voltmeters have very high internal resistance so the current going through them is minimal. The measured voltage will therefore be very close to the theoretical maximum.

[@westinritter233]

That's a voltemeter no current flows through voltemeters

[@MsLia32]

If the transistor for some reason doesn't swith fast enough, it seems we get a short circuit on the input voltage?

[@anthonyvolkman2338]

MsLia32 Very much so.

[@cjdelphi]

MsLia32 yes, because the inductor is just a wire, any wire will short inc an inductor

[@robertocardenosa493]

Can you make a video about buck and another about buck-boost converters? That would be very helpful :D this channel has been amazing so far

[@alirezaebrahimi9449]

hello, thanks for your amazing video. Is that application, on which you're running the circuit simulation, a paid app? what is its name?

[@K-Luxuriant]

which software is this?

[@enesguler1797]

what is the name of the program you use? thank you.

[@foa2373]

What software you are using?

[@hollywoodbollywoodtrailers4119]

foa . software name is every circuit

[@vamsi1997]

Every circuit

[@SR-kp8zu]

thanksss this was great! quick question, why does the voltage stop stepping up at 100v?

[@Freepalestineforever23]

really helpful! if you ever do continue the series can you post a video on buck converters?

[@ElectroRestore]

At 100 v out, what would the current capability be? Could you get 10 amps at 100 v from a DC low votage?

[@JimRyan-AKA-Pigeon_Kicker]

Question.... I have two 5 pin adjustable voltage converters. One has the LM2577S, and is a purchased board. The other is the IRU 1150, a recovered component from another converter that met its fate early :) Yes, it works fine.... My question is, using a 5 volt supply to both of the Converters, create a Positive 12 supply from one, and get the other to make negative 12 volts ? Ending up with ground, +12, and -12. Possible ? EDIT: I need the switching Op Amp ability of 12 volts for audio pre amps.

[@nejckitak410]

But does this decreases the current... Or does it only increases voltage.

[@aghasbsb]

Amazing description ! Thanks a bunch

[@VFMIAB]

Thanks for the best Boost Converter explanation on the internet ! One question though: If I have an alternating load on the end of this circuit (let's say a color changing LED-Stripe), wouldn' the alternating current draw of this load require an adaption of the switching frequency on the MOSFET's gate to keep the voltage steady? Thanks in advance ! :)

[@nicolasraimbaud3947]

Amazing, really useful for students man. Thanks ;)

[@barrykery1175]

What a great video. That was very interesting. It's the same principal as an ignition coil works in a point type ignition system in a car. Thanks for posting. I get a simple yet great education. Barry, KU3X

[@biologicalagent]

Could you do another add-on/follow-up video that explains how you might use the same techniques to create a split-rail DC supply?

[@alexeykokh8240]

Hello. Very good video. I realized that I am looking for a similar boost converter, maybe you could recommend me a chip?

[@jeffreylielferrer3862]

​ @Simply Electronics does a boos converter charge batteries on a loop cycle?

[@fobmongrel9925]

Hello, may I ask what "power supply" we are talking about, or better yet, what is the part/function/manipulation inside a power supply that would allow the transistor to switch on and off quickly? New to electronics here...

[@jan-zumwalt]

Very good clear audio.

[@saberline152]

if you stopped opening and closing would it stay on that voltage or would it just stop? also does the speed of switching on and of influence the output voltage? I've heard on another video that the output is almost entirely a function of the duty cycle for a given circuit.

[@krishna2133]

Very good explanation and the animation is on point. Thank you very much!!

[@leeorshimhoni8949]

simply brilliant. easy to understand. explicit .

[@jayasuriyars1865]

Which software are you use for the simulation purpose sir???

[@srinubaburuppa5719]

Awesome!! What simulator are you using?

[@CNoteZzz]

I love you man cuz you explain the circuit very good

[@PrismWarp]

Stupid question but does that mean the 10uf capacitor here need to be rated for 100v or rated for the 9v supply he is using? I would be leaning towards it being 100v since the vom will eventually achieve 100v that he mentions so I assume the capacitor would also be at the potential.

[@atulpandey3938]

Can you tell me which simulation tool you are using ? is it available online?

[@000Enough]

Notice that you sacrifice some water to rise up the water (in the video in the below link), the same for electricity, you need to lose some current to rise up the voltage, the wasted current goes back to the battery through the switch.

[@NateEngle]

Boost converters are more efficient now than they used to be but it's worth bearing in mind that a considerable amount of current gets ditched to ground when you're charging the coil and that can be an issue if you're looking to source much current from your boosted voltage., Buck (stepdown) converters are a LOT more efficient. Back in the day when phones actually had real bells in them I built a simulator box that needed to supply the base tip-ring on-hook voltage of -48VDC which wasn't that hard, but the ringer voltage to make a princess desktop phone's bell ring was more like 80-100VAC at about 20Hz and that part required some serious attention to the breakdown voltages on everything in that part of the circuit. It was fun and I learned a lot but I never got shocked as much by any other construction project as I did by that thing. I think I still have it in the basement somewhere.

[@edselnarvaez9667]

I loved it, thank you !!!! Very insightful

[@Cara.314]

inductors seem to store inertia and it's facinating!

[@DrHarryT]

Use two of these circuits paralleling the inputs and outputs. Have the mosfet hi speed switches alternate so that one o0f the inductor coils is always collapsing generating current, thus a more stable and higher current output potential.

[@Poshsans]

DrHarryT what current will u get?

[@mystwalker479]

Definitely subscribing to this channel! I'm a visual learner so i'm expecting more on this channel