Voltage Multipliers - Half Wave Voltage Doubler Circuit

[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections

[@AllOfTheAboveChannel]

You're a legend. I study correspondence so not having a in person tutor is sometimes hard, but I just search what Im struggling with and BOOM, you got a video on it, thanks a billion man

[@unjoueurcritique]

I just want to say that I have been trying to understand in detail how a half wave multiplier works, and I finally got the right video !

[@drod1167]

Dude I dont even know what this is or have a class on this, but I'm here for the quality of your videos 🤟🏼

[@winlose3073]

always reliable, simple and professional information are given here very informative channel thank you I always have found the info to be useful on this channel where nowhere else it really is.

[@Mark70609]

Great simple explanation, all the better as it is geared towards visual learners.

[@PunmasterSTP]

Multiply? More like, "now I know why", so many people look up to you as their teacher. Keep up the wonderful work!

[@FatmaMagdy-r1u]

Much love! I really enjoy learning physics with you

[@KAFKUBA]

One thing that might confuse viewers...the second stage serves to smooth the ripple and is not necessarily required, such as with microwave ovens. You double the voltage with one stage (c1 and d1). It's a point of confusion you did not address. Otherwise great video.

[@NHL_B]

it's a AC/DC converter, why would you remove the rectifier stage, it's actually what give us a DC output (R>>).

[@KAFKUBA]

@@NHL_B there's all variations of DC and variations of ripple. Cost and system requirements

[@NHL_B]

@@KAFKUBA you can still try to remove D2 and C2 and keep RL, what's gonna happen according to you ?

[@NHL_B]

actually that 2nd stage is called a peak detector, rectifier is just the diode if we want to be accurate.

[@trone32]

Very good and clear explainatiom

[@pranayjit]

damnnnnnn! That was really really intuitive. but i wanted to understand it through wave form at t1, t2, t3, and so on. but really really thanks

[@tolgadabbagh1877]

very good explanation for begginers like me , thank you .

[@김규진-x6h]

Thanks for your all troubles for teaching!

[@ΓΙΩΡΓΟΣΠΑΤΟΥΛΙΔΗΣ-τ2μ]

Thank you for the explanation. I did not understand it from the book. Be well.

[@xanderk704]

Good explanation! Clear video

[@anjalidas5530]

Thank you man... For helping me clearing the concept... Thanks a lot😊

[@ysvsny7]

The best explanation

[@briantamburelli7573]

Thanks. For the first time I think I understand.😊👍

[@derlyswanson]

Thank you for you videos really help me :)

[@surajmahendraramraj3870]

Thank you so much for such a wonderful video.

[@ysvsny7]

Very good explanation

[@ahmedelsmahypersonal]

thank u my dear friend, its simple, easy and lite

[@Study-ns8ne]

🎉🎉🎉🎉fantastic I was struggling 😢😢😢😢

[@Dickskanal]

130 views, ridiculous, you should have 100 000.

[@CNoteZzz]

People wanna watch how to make a slime

[@biribibka]

Thanks, it really had halped me. Thank you!

[@codex8797]

thanks bro, really helped

[@laloverdad1431]

I studied eng 30 years ago if this video had been at that time things would have been different for me.

[@jackwilliam2226]

same here...

[@brandondunn-moodie1544]

this man out here saving careers

[@johnwrightoverlandingrobotics]

Voltage Doubler Ckt: 2(VAC * 1.414 - diode drop) = VDC Be careful not to confuse VAC with Vp as Vp = VAC * 1.414 Capacitors charge to Vp less and diodes they may have in series in voltage multipliers. So if you have 10VAC, you will see approximately 28VDC on the output. If we assume silicon diodes are used and use .6V as the nominal voltage drop for each: 2(10VAC * 1.414 - .6V) = 27.08VDC

[@viniciusps01]

High-quality content. Thanks :)

[@moussafiradil1700]

YOU JUST ROCK .... THANKS A LOT !!!

[@automateTec]

Rectified voltage becomes the peak value of AC when using capacitors. In other words 240VAC measures at almost 340VDC, or in this case, 12VAC will measure at around 17VDC (less the diode drop.) Thanks for not adding music, btw.

[@Aaron-xn7dg]

I'm curious as to whether all of this is self taught or you went to university and did an undergrad in Physics/Applied Mathematics/ or some type of Engineering? Im genuinely curious as to how you're so knowledgeable on all these topics

[@ygoldberg1287]

Basic electronics 101 Associate level

[@Aaron-xn7dg]

@@ygoldberg1287 I'm talking about the thousands of videos he has on various subjects

[@hanniebertquinalagan3049]

His a genius, coz he knows everthing... his channel saves me during my calculus and physics and other applied mathematics subjects examination.. I’m an electrical Enrg. student and I make this channel as my reference in studying. thanks to this man behind this voice... God bless him.

[@maryama5828]

Thank you so much.

[@tracelandreth8704]

during the positive part of the cycle, why doesn't capacitor C1 discharge as current flows through it, through D2, and then charge C2? In the video you say that after the negative part of the cycle, C1 remains at 11.7 volts across it.

[@tylercoombs1]

Wait until everyone realizes that electrons don't "flow" at all, in fact, they barely move. lol

[@alexhammerer4076]

Very good Video thank you!!!

[@rahafmoh9402]

thank you!

[@MaximumUtmost]

The current will inherently be higher while the applied voltage is higher, but the voltage will drop very quickly, and the resultant system energy will indeed remain the same.

[@jaxpher]

you the best

[@CNoteZzz]

Thank you very much sir

[@shreeshkulkarni6953]

Sir I beg you please make a video on how to write decomposition reactions including polyatomic ions like how to write for calcium carbonate. But not binary compounds like silver chloride

[@mr.helpful3844]

great video

[@jonelberdejo9446]

I'm confused what will happen if the ac starts with positive half cycle

[@louco2]

In the first half cycle nothing special will happen C2=AC, in the second half cycle C1=AC, in the 3rd half cycle the magic happens, AC+C1 in séries will charge C2 with double the voltage. (This is all ignoring the voltage drop across the diodes )

[@supoyu6893]

thanks a lot!!!

[@successthruknowledge]

does the capacitor voltage polarity switch for C1 when the AC source is in the positive cycle?

[@manipulativer]

Your tempo and the insertion of additional information is on point! Is there a way to use a jouel thief and instead of lighting a led light, it would charge a capacitor and depending on the inductive level you should - in my understanding - get around 100 volts out of an AA batery? I have almost zero experience into what i should be focused on, but perhaps you can take the challenge. At least an understanding of what is the most efficient potential riser... either with condensators with ac pumping, or dc to dc inductive pumpking. Perhaps its better to start with ac and multiply the voltage that way?

[@shubhamchourasiya4243]

Why did we start the circuit analysis from negative half cycle?

[@Zapgod]

How do you determine the neg and pos sides of the capacitors in a ckt? Is it determined by the diode?

[@surendersingal2192]

Thank you sir for thru n thru explanation. It looks like C2 n Rl will make its own circuit @2V n RL. Where such ckt us used?😊

[@rottennewtonsapple]

I thought that lower load resistance causes the capacitor to discharge faster, leading to smaller voltage drops and lower peak-to-peak ripple voltage?

[@InfoDav]

Oh, maybe the load resistor will prevent C1 from charging enough in reverse polarity to be a problem?

[@arthurharrison1345]

Nice!

[@fernandocruz3457]

could i plug this in to a moter that spins another moter that creates a larger voltage and start a chain reaction and have a self running genorater?

[@SuperPutzPutz]

How can the current drop by one half when the voltage doubles? Isn't that trade off for working with 1:2 ratio stepup transformers that involves electromagnetism?

[@sydwelglobal1439]

That has to do with the low of conservation of energy! Voltage and current are directly proportional to each other!

[@ninjafishing7120]

Is there any other way i can double then voltage not using AC power Source but from battery . Thank You 🥰

[@sanghyeonban9989]

thanks!

[@krisLight594]

ok can you explain to me how much this have influence on watts? If i have for example transformer 400w 12volts 30amps can this circuit deliver 24volts on same amps 30amps its not the same watts anymore?

[@InfoDav]

If the signal starts with the positive half-cycle, won't the capacitor C1 be charged in reverse polarity and therefore be damaged?

[@juls1145]

Is it possible for a voltage regulator and voltage multiplier exist on the same power supply circuit?

[@mohammadibrahim6253]

I am an electrical engineering student and you have me as fuck ... fantastic videos and amazing way of explaining (better than my teachers hahaha)

[@---_UwU_---]

Why - of cap (excessive electrons) and + of source (lack of electrons) not annihilate?

[@harikan701]

Then what happens to the first positive half cycle?since diode D1 conducts on negative half cycle

[@annguyendang8388]

Can you show the reverse voltage when the diode is OFF?

[@AlexFettweis]

wouldn't an increased frequency on the supply side also reduce the ripple effect on the output?

[@annguyendang8388]

Dear Sir, It seems that the DC output voltage = 2*Vp is not limited by the input power level, but in practical, the DC ouput value gets saturated as the input level is high enough. Can you or someone explain about this?

[@clintcallender9502]

Still a bit confusing with regards to the second cycle (+ve cycle)

[@hsen0099]

may I asked in which reference I can find this type of question ❓ please I need to know immediately

[@hasibuddin2437]

Is that capacitor r polarised?

[@NHL_B]

I have the impression that if we start with the positive half cycle it can't work as it is explained. So What happens to the first positive half cycle ? since diode D1 conducts on negative half cycle ??? In all videos I watched, we always start with AC's negative half cycle so C1 get charged on the right terminal and with the positive half we get Vin + Vc (C1 charged) so 2.Vin. But actually only with C1 and D1, if we start with the positive cycle, D1 is reverse biased so no current flows to C1 If I am right because it's like D1 acts like an open circuit. So it's like the behavior we are looking for only starts when we have the first negative cycle so C1 get charged etc...

[@jackwilliam2226]

throw in electrolytic caps as they use in real life and it gets even more confusing...

[@allthings1337]

Please tell me about the reason of changing polarity of AC ??

[@joffito2495]

it's about electron abundace. think of the ac source as pump. one half wave it pumps electrons into the circuit --> makes one terminal more negative than gnd. the other halfwave it sucks electrons out -> makes it more positive with reference to ground

[@matheokoning4456]

I don't understand how you can put a polarized capacitor in an ac circuit. C1 gets the ac signal and should explode anyway?

[@jackwilliam2226]

that's exactly my point too, but I have seen videos of electrolytic caps in voltage doublers, but how is this possible?

[@jackwilliam2226]

Are the capacitors electrolytic? I didn't think AC and electrolytic caps went well together.

[@emmetray9703]

but in the negative half cycle , current can go trough D1, then trough D3 and carge C2 ? why not? Please someone answer me.

[@bixenbaer]

Current always chooses the path of least resistance back to the source. Why bother going an extra way through D2 when home sweet home is near behind C1 ;)

[@emmetray9703]

@@bixenbaer Thanks

[@enzofitzhume7320]

Something is not right with explanation about C2. As long as the capacitor is rated at the input voltage (AC). Then it doesn't need to be rated at the voltage after D2. Search for Cockcroft-Walton and also Cockcroft-Walton.

[@clueless_andy]

how can this be a half wave doubler when we have two diodes?

[@crazylemons6197]

Why dose input have to be ac ?

[@withIn40]

excuse me sir, i want to ask. how many the capacitance of the capacitor??

[@joffito2495]

you can use 10 - 100 uF caps with appropriate voltage rating

[@Bulent-Keskin]

Thank you for your great videos but why are you repeating the fact that electrons move in the reverse direction of the conventional current in almost all of your videos? :) Anyway, I just hit the right key to skip that 5-seconds part. :)

[@Kmathaddict]

Courage to all watching this video because they have an exam tomorrow 🌝🌝🌝🌝

[@qzorn4440]

Electron flow is less confusing.

[@viramchandhu3200]

Bro how to find unknown resistance of the circuit

[@michaelcostello6991]

We get DC across RL . I just find it confusing as the AC supply is connected directly to RL so you would think RL would have an alternating voltage +- on its lower leg. Can anyone explain. Thanks.

[@joffito2495]

the diodes of course rectify the ac to dc (walf wave recitification here)

[@Dickskanal]

I meant, Likes

[@sasikadayanande2369]

🙏

[@ygoldberg1287]

This is not organic chemistry but electronic basics 101. Associate level.

[@DasAntiNaziBroetchen]

According to EEVblog, you don't need a cap with a voltage rating as high as your output voltage.

[@hamzakhalid4613]

Are u an electrical engineer?

[@engineergaming1669]

What in the world is this, how did I get here