Full Wave Voltage Doubler Circuit
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- เผยแพร่เมื่อ 21 ก.ย. 2024
- This electronics video tutorial explains the workings of the full wave voltage doubler circuit. This circuit converts an AC power source into a DC voltage with an output that is nearly twice the voltage of the input.
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Full wave? More like "hip hip hooray!", for all of this free learning. Thanks again!
The *#1 Math Teacher* on TH-cam with some *Electrical Science* video lately...🔌⚡
I still love it! 😍
All engineering in this channel
This guy single handedly made Millions of Engineers
What category is this channel?
*yes*
you make things way much easier than they actually look like.
God bless you.
sir may you add some videos on how to use the steam table ( thermodynamics)
You've modelled the scenario very nicely but haven't made it clear that you've introduced AC peak voltage, in this case 12V ( 8.4V RMS ), this may lead to confusion.
Thank you so much! This particular circuit is used in the Kenwood Trio KW33-L tube receiver power supply and i could not figure out how to solve the residual ripple on the output. Now i have a hunch on why it still has so much ripple on the circuit.
Thank you for posting. I was trying to grasp the mosfet bootstrap circuit and the charge pump circuit.
Nice. Simple. Maybe explain how the amperage of the circuit changes throughout the doubling process?
Thxs for your excellent and clear explanation. Kind regards from Mexico amigo!
Would you use electrolytic capacitors? (Polarized to prevent flow through C1 to C3 and R1 during the negative cycle)
12V must be RMS right?, not the Peak value ??
Nice! Just a remark the threshold diode in example is 0.3V. About charging C2, the voltage coming from AC source is 12V. The diode is after C2, should not be C2 charging until 12 V in this case?
So, if D2 is AFTER C2 then how does it reduce the voltage on C2? Is that due to C2's internal resistance? Anyway still a great video though I'm curious how the circuits were devised. Is this one way of many? It must be.
That circuit will save current if you connect two of the same resistors at load wired in series. 11.7 volts will drop over each resistor and the 23.4 volts will push the same current needed for one resistor thru both running two devices for the price on one if the higher voltage doesn't damage the 12 volt device.
i'd say the voltage would be at 2Vsup-Vd, since in every half cycle, the supply voltage is connected in parallel to one of the caps at Vsup-Vd, and in series to the other cap at the same Vsup-Vd, thus adding the voltage up to Vsup+Vsup-Vd = 2Vsup-Vd. voltage should however be more stable at 2Vsup-2Vd, depending on the Power of Vsup, since Vsup drops under load
Surely you mean that the 12VAC will charge C1 and C2 to 16.6V each? Unless you specify Vp-p when you say VAC, most of us enginerds will assume Vrms, no?
Yes, I agree!
Yes, that’s what I was thinking - though I’m assuming maybe the explanation of “12v AC” was being ‘simplified’ as it’s peak voltage for the purpose of this explanation, as towards the end it refers to 2 x “peak voltage” less the diode drops.
For anyone reading this and not following - Normal convention is that AC voltages are expressed as RMS (Root Mean Squared) values - so the ‘average’ voltage of 12V AC is at 12V but the peak value (as per the formula) works out at about 1.414x higher (square root of 2), therefore 16.9V peak. So C1 and C2 each charge to about 16.6V each, giving a total combined output voltage of 33.2v (now 23.4v).
This guy is great and awesome we would like to see you
How do you know what size capacitors to use based on load current? Thanks.
@OrganicChemistryTutor , How are functions found ? Say if a scientist was making a mathematical model of a radio signal. How do they figure out the long complicated trig functions?
What happens when you ground the other end of the AC side? How could I get +/- DC double output voltage using a singe AC winding as the source?
C3 is not really needed. Also, Caps charge to the Vp of the AC signal input (VAC).
Vp = VAC * 1.414 - .3V (.3V is used to represent the Ge Diode nominal voltage drop).
Quote: _"C3 is not really needed."_
C3 is a bypass cap. A large enough current draw will create ripple.
I have done this experiment and used 8v... But I got out put 65-66v.. I didn't use c3.
Can you post some videos about physical chemistry? gas law and more. Btw is your name Jake. G ?
thanks
The BIL on my transformer can handle the voltage from the voltage doubler. But can you please let me know what about the extra current from the charged capacitors? Will a higher amperage flowing back into my secondary damage the transformer? I have a 10 kva 14.4 Kv 120/240 volt transformer 60 hz. If I used 500 uf capacitors that hold a lot of joules. That’s a lot of extra current flow about to be discharged right into my secondary coil. If the ampacity of my secondary wires are thick enough to handle the extra current will my system be able to run it as long as I would like? Online it doesn’t explain how much current I can run back into my secondary in a DC system very well. Please help friend.
I’m actually wanting to do a full wave voltage tripler. Same exact issue though. On falstad it looks like more voltage and more current would be flowing back into my system.
What are the capacitance of C1, C2 and C3?
I think his explanation was focused more on the concepts than on specific values. If I'm not mistaken, C3 just has to be "relatively large" compared to C1 and C2.
You. Are. AWESOME!
What is the use of C3? If C3 is omitted, the voltage across RL is 23.4V, isn't it?
It s used as a filter cap to really provide the energy for the load and smooth out the ripple
I watched it.
Nice!
How are the capacitors getting charged if they are reversed??
C1 and c2 get charged depending on the half cycle. They don’t lose their charge during the opposite half wave. C3 will kind of add up the voltages of C1 and 2
Face reveal pls, if not though, just continue what you do. You save a lot of people through your videos. Thank you very much.
Please do a Face Reveal at 1.5 subs 🙏🙏
Yo
Do a face reveal