This is by far the best explanation ive seen so far. I've been struggling with this for two whole days. Most texts and videos skip all the nuances of current flow like you did and that makes all the difference. Your visuals and explainations are so easy to follow. Thank you!!
i can't express in words how u tought this concept....i saw nearly 10 videos but i did't ge understand from anywhere but u explained it in very easy way.....thank you so much sir
Excellent effort to explain the complexities of this type of circuit. It’s helpful that you explained why the voltage drops across the diodes were left out the calculations for the sake of simplicity, but I think it is also worth explaining that 12V AC power source in this is also being simplified to mean 12V Peak AC, rather than the usual 12V AC being RMS (i.e. an ‘average’ of 12V) where the peak voltage would actually be higher (~16.9V before diode voltage drops). Appreciate you’re trying to simplify things, but its important people understand the actual voltages would be much higher (especially if anyone is thinking of playing with making a real circuit!). Great job 👍
does the voltage across the capacitors ever get higher than double the input voltage? like, if you had 100 stages, would the voltage measured across the capacitor in the 100th stage only be double the input voltage? im trying to understand what capacitors should be used in a circuit like this, that info was absent from the video for some reason. is there a rule of thumb for picking capacitors for votage multiplier circuits, like 4 times the input voltage no matter how many stages you make? also, is there a maximum amount of stages you can make?
With each stage the voltage is multiplied higher so, the capacitor increases with each stage. Over this design's length the cost could be less with the first stages but the capacitors in added stages would be factored because circuit analysis that reveals more information would tell you these values. If starting with High Voltage ⚡ then, its special characteristics should be considered because it's prone to ark.
It would be easy to use two transformers: Input Primary to mains voltage T1-then connect the secondary winding of T1 to secondary winding of T2, and use the primary winding as the AC voltage booster. Transformer 1 120V secondary 12 v to 12 v secondary winding T2 and Primary 230 V to 230 volts AC Use different secondary windings on T2 for desired voltage boost, advantage isolation from mains, disadvantage weight and current limited to secondary T2 rating.
I think the reason would be because the terminal of C2 connected to the cathode of D2 is at a higher voltage than the terminal of C2 connected to the anode of D1. Therefore the diodes would prevent the discharging of C2, since they would be reverse-biased in that situation.
So... if you manage to control the 12Vac from 0 to 12, you're going to have a 0 to 72Vcc controlled output, minus all diodes' voltage drops. Considering a possible high-voltage source to, say, capacitor revival, a sinusoidal generator (60Hz or some more suitable frequency) connected to the input of a "tunned" symmetrical power amplifier with some tens of output voltage and some hundreds of milliamps @ its output and driving a multiplier would then give us a DIY controlled high-voltage, low-current capacitor revival circuit. Also, each diode junction would provide some lower voltage output, that could be connected to a selection switch for max voltage output. Hummm... Sound "delicious"... Of course, capacitors' working voltage MUST be carefully set, or else... KABOOM!!!
current doesn't flow thru a capacitor. if it did it would be called a short. I think what you mean is charge the capacitor. outside of that, thanks. lol. I made one when I was ten. now I know why it didn't work. I used dc. I wonder if that's the same principle behind a fly back circuit. I'm designing an ac inverter for fun and I'm dead set on using the irfp250's that I ordered at 120v. hum? I know lower voltage rectification. but I don't want to. I want to use the 250's at a higher efficiency.
@@JeffreyJamon Hi, you’re welcome, glad to be of help. The voltage multiplies, but the current is halved. There are current doublers which require inductors…I have yet to use the current dabbler in series with a subsequent voltage doubler.
Hi I have a question well a few the first one is do I have to have a transformer on the input the second one is do you need different number diode‘s for each connection because I only have one kind third I’m using a set called snap circuits and the only other numbered diode‘s have lights on them which sucks up a lot of voltage please respond thank you
It was a good lecture though I guess that I'm hazy yet why the current only goes through the nearest diode back to the negative terminal of the AC source and not dividing itself between the diode and the next capacitor along the wire. If you follow my reasoning. For example after the current passes through D1 and charges C1 then why doesn't it charge also C3?
Just a quick answer, because as it charges C1 the voltage is held constant and a silicon diode needs a forward voltage of about 0.7V before it conducts, so D2 isn't conducting. .
@@abdessalemakerma4636 I meant that it was Good that you caught the mistake made in the video. That polarized capacitors can't take a reverse polarity voltage. That's all.
@@Someone-de7wf can't tell if you were being sarcastic but what i meant in actual grades was probably around 11th grade or college introductory type class
It should be easier, and then you use a rectifier to turn it into AC with a total voltage smaller than the peak voltage, but still above like, half of it, specifically Vp/√2
...any AC voltage source is specified as Volts - RMS = Vrms...so a 12Vrms voltage source will charge the first capacitor at 12V*sqrt(2) = 16.97V ~17V (- voltage drop on the diode ~0.7)....~16.3V ...and not to 12V as wrongly explained in the video...
Thank you! I spent around 3 hours trying to understand this with different videos, but you helped me realize the key point in just 10 minutes. You're the best! 🎉🫶🏼
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You're literally the only person who didn't skip important steps. Thank you!
By far the best explanation I´ve seen of this. Finally, someone explains in detail every cycle. Thank you so much!
This is by far the best explanation ive seen so far. I've been struggling with this for two whole days. Most texts and videos skip all the nuances of current flow like you did and that makes all the difference. Your visuals and explainations are so easy to follow. Thank you!!
i can't express in words how u tought this concept....i saw nearly 10 videos but i did't ge understand from anywhere but u explained it in very easy way.....thank you so much sir
This is the best explanation I've seen yet. For the first time I understand how this works.
Excellent effort to explain the complexities of this type of circuit. It’s helpful that you explained why the voltage drops across the diodes were left out the calculations for the sake of simplicity, but I think it is also worth explaining that 12V AC power source in this is also being simplified to mean 12V Peak AC, rather than the usual 12V AC being RMS (i.e. an ‘average’ of 12V) where the peak voltage would actually be higher (~16.9V before diode voltage drops). Appreciate you’re trying to simplify things, but its important people understand the actual voltages would be much higher (especially if anyone is thinking of playing with making a real circuit!). Great job 👍
I really appreciate your videos. Your videos are awesome and I can easily understand it. Thank you SO MUCH. You are a good speaker.
✊🏽✊🏽 keep on doing the electronic lessons✨💖💖💖. I really need them. I have downloaded more than 10 videos of Electronic lessos
Thank you sir, Only well explained video on YT.
Sir,I really many more & most thanks because that circuits work on my maintenance department //lot lot of thank you very much
Oh man, the things I would do to this man, for all the help he's given me over my career, insane.
Awesome explanation! I now understand how this circuit works! Thanks!
wow your voice is very clear in this video thanks man keep the good work God bless you million times a day.
Very good tutorial! Helped me a lot!
Thank you so much, i understand easily from your video..
Very clear explanation, thanks!
yo whats that software you use man?
can you make a explanation video on voltage multiplier ckt working a dc source
Why does the current not flow to C2 in the 1st negative cycle?
Good explanation, thanks!! Is it possible to multiply voltage of small signals using this method?
Very good video
does the voltage across the capacitors ever get higher than double the input voltage? like, if you had 100 stages, would the voltage measured across the capacitor in the 100th stage only be double the input voltage? im trying to understand what capacitors should be used in a circuit like this, that info was absent from the video for some reason. is there a rule of thumb for picking capacitors for votage multiplier circuits, like 4 times the input voltage no matter how many stages you make? also, is there a maximum amount of stages you can make?
With each stage the voltage is multiplied higher so, the capacitor increases with each stage. Over this design's length the cost could be less with the first stages but the capacitors in added stages would be factored because circuit analysis that reveals more information would tell you these values.
If starting with High Voltage ⚡ then, its special characteristics should be considered because it's prone to ark.
Good explanation thank you. What type of diode and capacitor could we use? What is the maximum frequency used with this circuit?
Please I want to know if this strategy also work for Ac to AC voltage booster
It would be easy to use two transformers: Input Primary to mains voltage T1-then connect the secondary winding of T1 to secondary winding of T2, and use the primary winding as the AC voltage booster. Transformer 1 120V secondary 12 v to 12 v secondary winding T2 and Primary 230 V to 230 volts AC Use different secondary windings on T2 for desired voltage boost, advantage isolation from mains, disadvantage weight and current limited to secondary T2 rating.
Are AC capacitors needed? Or does it work with DC capacitors?
Non polarized capacitors that can take AC are required.
Can i series the capacitor without the diode ? Will it work ?
Can you explain why D1 and D2 do not discharge the capacitor C2?
I think the reason would be because the terminal of C2 connected to the cathode of D2 is at a higher voltage than the terminal of C2 connected to the anode of D1. Therefore the diodes would prevent the discharging of C2, since they would be reverse-biased in that situation.
Because the second half cycle coming off of source voltage plus charged C1 at 12 volts, makes it a 24volt charge to charge C2.
@@jaysmith3259 Thx
I live in the Netherlands, were we have 220 volt ac.
What kind of capacitors and diodes I need?
Is not really 'as far as you want' because of the losses.
That parallel battery gon blow up
Is this a practical ckt?
Can we run loads on it like LCD?
So... if you manage to control the 12Vac from 0 to 12, you're going to have a 0 to 72Vcc controlled output, minus all diodes' voltage drops. Considering a possible high-voltage source to, say, capacitor revival, a sinusoidal generator (60Hz or some more suitable frequency) connected to the input of a "tunned" symmetrical power amplifier with some tens of output voltage and some hundreds of milliamps @ its output and driving a multiplier would then give us a DIY controlled high-voltage, low-current capacitor revival circuit. Also, each diode junction would provide some lower voltage output, that could be connected to a selection switch for max voltage output. Hummm... Sound "delicious"... Of course, capacitors' working voltage MUST be carefully set, or else... KABOOM!!!
If every one of the capasitors are only rated at 12 volts will this system still work? Or do I need higher rated capasitors all the way down the line?
You would need an appropriately rated capacitor for the voltage potential it will be storing.
How would you hook up the input and output
Input could be a battery and output could be an arc
Could u make an explanation vid on integrated circuit?
current doesn't flow thru a capacitor. if it did it would be called a short. I think what you mean is charge the capacitor. outside of that, thanks. lol. I made one when I was ten. now I know why it didn't work. I used dc. I wonder if that's the same principle behind a fly back circuit. I'm designing an ac inverter for fun and I'm dead set on using the irfp250's that I ordered at 120v. hum? I know lower voltage rectification. but I don't want to. I want to use the 250's at a higher efficiency.
Current always flows during capacitor charging until finished with the voltage you're charging it with.
Is he using conventional current flow?
What you illustrate in this video is known as a Cockcroft-Walton Multiplier...
Thank you! This is what I was searching for in the comments!
@@JeffreyJamon Hi, you’re welcome, glad to be of help. The voltage multiplies, but the current is halved. There are current doublers which require inductors…I have yet to use the current dabbler in series with a subsequent voltage doubler.
But In the mosquito bat this formula used with different style???
If you discharge the circuit, will only the last capacitator be discharged or all at once? I would bet only the last given how a capacitator works.
Hi I have a question well a few the first one is do I have to have a transformer on the input the second one is do you need different number diode‘s for each connection because I only have one kind third I’m using a set called snap circuits and the only other numbered diode‘s have lights on them which sucks up a lot of voltage please respond thank you
It was a good lecture though I guess that I'm hazy yet why the current only goes through the nearest diode back to the negative terminal of the AC source and not dividing itself between the diode and the next capacitor along the wire. If you follow my reasoning. For example after the current passes through D1 and charges C1 then why doesn't it charge also C3?
Just a quick answer, because as it charges C1 the voltage is held constant and a silicon diode needs a forward voltage of about 0.7V before it conducts, so D2 isn't conducting. .
@@karhukivi thanks!
why dont we take the positive half cycle first
the polarized capacitor cant run by Ac !!
ha good catch
@@inventorbrothers7053 what do u mean
@@abdessalemakerma4636 I meant that it was Good that you caught the mistake made in the video. That polarized capacitors can't take a reverse polarity voltage. That's all.
ah okey thanks
Actually, the polarities on the capacitors are not changing
Where are the positive and negative ends of the output voltage?
Positive/point D and negative/point A.
Best !!!
What grade is this?
Probably college electromagnetism or AP chem
Koop Bar Oh wow, thanks so much
@@keepmehomeplease probably more of electronic engineering.
@@Someone-de7wf can't tell if you were being sarcastic but what i meant in actual grades was probably around 11th grade or college introductory type class
@@keepmehomeplease oh, I'm in 8th grade lol. I enjoy learning about electricity and circuits.
Why do capacitors charge up one at a time ? Wouldn't they all charge at the same time ?
Only possible if capacitors are already charged to the DC voltages already before the input voltage is applied.
why in the first negative half cycle current isn't flowing through c2-d3-c3-c1 etc.
Because current flows the path to Source with the least resistance with the voltage you're starting out with.
wait parallel batteries dont subtract thier voltages. You either add the volts, or add the Amps......
No increase in energy but no waste in skin effect of magnetic induction.
Is there any simple formula for this circuit?
Why only ac? Why doesn't anyone talk about a DC version?
It should be easier, and then you use a rectifier to turn it into AC with a total voltage smaller than the peak voltage, but still above like, half of it, specifically Vp/√2
Joule thief uses batteries as a power source or DC input.
A square wave would work for the OP. A square wave input is a DC input.
...any AC voltage source is specified as Volts - RMS = Vrms...so a 12Vrms voltage source will charge the first capacitor at 12V*sqrt(2) = 16.97V ~17V (- voltage drop on the diode ~0.7)....~16.3V ...and not to 12V as wrongly explained in the video...
Look into peak detectors
He is explaining the first half cycle, not the second half cycle.
¿And the ground is connected 1/2 wave to positive, and 1/2 wave to negative???? don't produce short circuit with the rest of circuit??
Current it flows only from negative to positive dude...
It flows from positive to negative based on conventional flow. I think you're basing it on electron flow?
Charging of C3 is explained very badly, he should have simply said we got +36V and -12V charging it to 24V.
I watched it.
I wish know about frequence response of this kind of circuit, if someone did, plz awsner this coment.
Thank you! I spent around 3 hours trying to understand this with different videos, but you helped me realize the key point in just 10 minutes. You're the best! 🎉🫶🏼
Please I want to know if this strategy also work for Ac to AC voltage booster