The timestamps for the different topics covered in the video. 0:21 What is Voltage Multiplier Circuit? 1:41 Application of Voltage Multiplier Circuits 2:32 Voltage Doubler Circuit 8:40 Voltage Tripler and Quadrupler circuit
Clearest explanation I've seen on how this circuit works. The first stage is easy. It gets a bit harder to see subsequent stages. Your explanation makes it easy.
6:35 brought a smile to my face because I've been searching for an explanation of what happens at the midway point of the half cycles when the diodes begin to discharge. I'm happy you explained it and I now understand perfectly. Thanks!
Excellent tutorial. This is the only correct video on youtube that matches realworld and simulation observations. Other videos such as those by 'Organic Chemistry Tutor' and loads of other guys who have received positive reviews are wrong!! They all claim that diode D1 is forward biased all through a half cycle and capacitor C2 gets doubled in the other half. I was always wondering why diode D2 doesn't conduct in the 1/4th cycle and thanks to this channel for clarifying the correct working principle.
Excellent video. Very detailed and clear instruction on the voltage doubler circuit. It's one of the best out there. Thank you for taking the time to share you knowledge. This is a very important circuit for learning AC circuits. It's used in microwave ovens as well at the output of the step-up transformer to power the magnetron.
@11:41 assuming C2 = C3 during this cycle there is charge sharing therefore C2 does NOT dump ALL of its voltage(2Vm) into C3 in the first cycle, first iteration will have 0.5*(2Vm) volts across both C2 and C3. It takes X amount of iteration for C3 to have a voltage 2Vm depending on input voltage. What happening here is C2 is constantly getting charged from the supply then shares its charge with C3 UNTIL the voltage across C3 = 2Vm. In summary, due to charge sharing between C2 and C3 it takes multiple iterations for C2 and C3 to reach 2Vm. The math is Vc2 = Vc3 = 0.5*(2Vm) -> first cycle Vc2 = Vc3 = 0.5*(2Vm + Vinitial) = 0.5*(2Vm + 0.5*(2Vm)) -> second cycle Vc2 = Vc3 = 0.5*(2Vm + Vinitial) = 0.5(2Vm + 0.5*(2Vm + 0.5*(2Vm))) -> third cycle Vc2 = Vc3 = 0.5*(2Vm + Vinitial) = 0.5*(2Vm + 0.5(2Vm + 0.5*(2Vm + 0.5*(2Vm)))) ->fourth cycle . . . . . Until Vc2 = Vc3 = 2*Vm
Was doubting my basics, and searching a comment that explained what I know, then came across this, and now am satisfied! I don't know why the these small but great details are ignored in videos. These hamper the basics!
Nice video but I think that the analysis that you do for the voltage doubler may be wrong. First of all it seems that you are analyzing the initial transient of the circuit, which means that the initial state of your circuit is characterized by all capacitor discharged. The moment, in which I think you commit the first error is at 4.02. you say that the capacitor C1 charges to Vm when the sinusoidal signal reaches the positive peak and then you say that when the sinusoidal signal becomes negative then the diode 2 turns on and D1 turns off. But what happen between the positive peak of the sinusoidal signal and the negative part of the sinusoidal signal? You forgot to analyze that part. In fact, the diode D2 should turn on immediately after the peak of the positive sinusoidal signal. and D1 should turn off in the same moment. So the two diodes change their states much earlier than the moment in which the sinusoidal signal becomes negative. Missing the analysis of that part, will make your analysis not correct. Furthermore, it seems that in your analysis, C1 remains charged to Vm for all the negative cycle of the sinusoidal signal. what make you think that the voltage of C1 will remain constant during the negative cycle? I did not find any reason why that should happen. Try to simulate the circuit and look at the first period of the initial transient: at the first negative peak of the sinusoidal signal the voltage across C1 is actually 0V and not Vm as you state in your video. Overall the dynamic of the circuit is a bit more complicated than what you claim. I am not convinced by your analysis, which in my opinion is just a method to force the correct result. Anyway, feel free to correct me if I am wrong. :) In general, I like your videos, but I do not think that everything can have a short explanation. Good luck for the next videos
@@alokgupta7239 probably because, you did not setup the LTxpice correctly. To see the initial transient you have to check the command "skip initial operating point solution" when you perform the transient. Try it ;)
11:44 I don't think so capacitor c3 will get charge by that path, because the D1 which you short circuited, has V(k)=2Vm and V(A)=Vm, hence that D1 will act as open circuit not short Circuit.... You have simply put a closed path without analysing... ( Don't mistaken me, I really like the content ❤️)
9:50 Why the capacitors will not charge if they are uncharged. (I think all capacitors were initially uncharged, then with your reason even the capacitor c1 should not charge).
The capacitor C1 is charging because diode D1 acts as a short circuit (ideally). If you take the second loop, between diode D1- C3-D3- C2, as all the capacitors are uncharged initially there won't be any transfer of charge. In the first loop, the capacitor C1 is charging because of the input source Vin. I hope it will clear your doubt. If you still have any doubt, do let me know here. Further, I have also shown the simulation result at the end, which might help you in visualizing the things.
I need help designing a Voltage Tripler circuit with an input voltage of 110VAC that will power a 330VDC / 2 watt light bulb. Will the voltage tripler circuit that you have designed at 9 minutes work? And what size caps and diodes should I use? I appreciate any help. Thanks!
Ok so basically c1 capacitor becomes charged like a battery when the current passes through D1 diode. And capacitor c1 acting as a battery plus the actual power source it doubles the voltage almost. So it's like the power source is a 1.5v ( AC) battery, which releases the current through the c1 capacitor and the capacitor becomes charged 1.5v. And after the first cycle, the direction of the current changes in the main power source battery(because it's AC voltage reverses direction of the current), as a result we end up with 1.5v battery attached to 1.5v capacitor, becoming two batteries 1.5v+1.5v=3v attached to each other plus-minu-plus-minus like inside toy battery pack.
Thank you for the video, very useful especially to initial ramp-up section which is not covered usually in other tutorials. I wanted to find out how i can do AC analysis of this doubler? What is the AC equivalent circuit? I am asking since this circuit is usually connected to AC sources and i want to look at the matching problems. Thank you
⭕⭕⭕Check 11:44 here you are telling that the C3 is getting charged through the path of D1. But current cannot flow through D1 from C2 because 11:28 see the circuit, I mean considering C2, D1 is reverse bias condition. Please check and let me know!
thanks for your sharing , a litter confused @7:20 , voltage across c1 is -Vm, Vin is -Vm, according to KVL , Voltage apply for C2 is -2Vm, Why it is -Vm in your presentation ?
Why this circuit we cant use for big loads bro.I mean if capacitor capacity is big then? What about current? Current also added or what? Actually I am a commerce student but also studying little bit electronic plz answer.
It is due to power. (P= V*I). Although the output voltage is increased, but since output power can't be more than input power, the current supplied by the output to the load will be limited.
The capacitor values must be in nf or pf and if you use high voltage pulsed dc as input (the one I have made on my channel go electronics ) 3 TO 6 KV ceramic caps needed and 4kv HF diodes !! ( FOR MORE DETAILS WATCH MY VIDEO WHICH PRODUCES 60000 VOLTS !!)
YES I HAVE CREATED A POWER SOURCE WHICH ACTS AS THE INPUT IN MY VOLTAGE MULTIPLIER VIDEO !! AND IT COULD OUTPUT 60000 VOLTS !! WATCH IT ON GO ELECTRONICS !!
Yes, the capacitors are polarised. For the voltage doubler, I have explained it from 5:39 onwards. Please go through it. And still if you have any doubt then do let me know here.
THE UPPER CAPS PASS AC WHILE THE LOWER ONES STORE DC MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT NEARLY 60000 VOLTS !! WATCH IT ON GO ELECTRONICS !!
So c1 only charges from 0 to positive peak and discharges from positive peak to negative peak? But from negative peak to 0, does c1 remain 0 until it reaches passed 0 to positive peak?
It depends on the maximum power that the input source can deliver. Since we are increasing the voltage, so maximum output current that we can draw from the circuit will reduce proportionally.
@@ALLABOUTELECTRONICS Say you have an ideal input voltage source, your Cs are all the same, you have N stages, The frequency is F, the load resistance is RL. What is the current through RL?
@@aduedc well, in that case, for ideal voltage source, the output current will be N*Vm / RL. Where Vm is the maximum voltage of the input source, N is the number of stages and RL is the load resistor. But practically, you will not get that. Let's say you have 5V source, which can provide maximum 1A current. (Overall 5W) Now, if you increase that voltage by 10 times, then your output voltage is 500V. And if you connect 10 ohm resistor as load then your output current will not be 50A. Because with that current and voltage, if you find the required power then it is 50A*500V = 25000W. But can you get that using 5W source, No. In this case, the maximum current that you can get is 5W / 500V = 10mA. (considering there is no other losses in the N stages) So, accordingly you should select the load resistance. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS This is not the case. Here is an experiment, use 100pf caps, measure the output current say at 10K ohm load. Then use the 10uF cap and measure it again. Also do this experiment with say 1KHz, and do the same experiment at 10KHz. As you know you can build a virtual resistor by charging and discharging a capacitor. So it is not that simple, it has to do with frequency, Capacitor value, diode losses, .... I am searching for paper on that if you find it let me know.
@@aduedc Yes, that's true. But to make such circuits, the RC time constant has to be much larger than the time period of the input signal. So, you need to select R and C accordingly.
The voltage rating of the capacitors should be atleast 3 times more than the desired output voltage. (e.g for 10V, output, use 35V rating capacitors) At the end, around 12:50, I have shown the simulation results and circuit. You can check that for reference.
If you observe, during the negative half cycle, the polarity of the voltage has been reversed. So, basically, it will try to charge the capacitor in the other way. So, the charge across the capacitor C1 will be lost. And it will be transferred to C2. But all this will happen gradually, in a couple of half cycles (more than two cycles) Please check the simulation which I have shown at the end of the video to get more idea.
The same query I have ,during the negative half cycle, since the diode 2 is in reverse direction how will at allow the capacitor 1 to discharge through it??
@11:45 during 2nd positive cycle Vin and Vc1 cancel each other and cathode of D1 is at +2Vp from capacitor C2 so I do not see how D1 is forward biased. Do you understand my question ? Thank you great video as usual.
No, it will reduce. Because the input and output power should remain same. ( Considering no losses). So, as the output voltage increase, the current will reduce.
THIS DEVICE IS A HIGH VOLTAGE LOW CURRENT DEVICE !! I HAVE MADE ONE ON MY CHANNEL WHICH COULD OUTPUT 60000 VOLTS BUT LESS CURRENT MAYBE IN MILLI AMPS !!
No, it won't get reversed. While connecting the capacitors the circuit, one needs to check the proper polarity of the capacitors (For polarised capacitors).
what's the difference between transformer and voltage multiplier? transformers just make the amplitude of the power bigger while voltage multiplier just offsets the power wave?
In the case of the transformer, the output is also an AC signal, while in voltage multiplier it is DC signal. So, convert it into DC you need to rectify it. Also, to generate the same high voltage the size of the transformer will become large. But in voltage multiplier, the output current is usually very small.
When I made connection of multipler . Output of circuit not desirable as when 6 volt AC applied to doubler circuit .It give15 volts output . also in triplet circuit 23 output Why this happened can u help me Thanks.
Please check the couple of things. First, whether the input is 6V peak or 6V RMS. If it is RMS then you need to multiply it with factor 1.4. Second there will be a voltage drop across the diode. So you also need to take into account. And third make sure, RC time constant is much larger than the time period of the input signal. Just increase the RC time constant a little bit. I am sure you will get the desired results.
The timestamps for the different topics covered in the video.
0:21 What is Voltage Multiplier Circuit?
1:41 Application of Voltage Multiplier Circuits
2:32 Voltage Doubler Circuit
8:40 Voltage Tripler and Quadrupler circuit
whats the name of your simulator
I love how you break it up into half-waves and remove diodes that are off. Makes it so much easier to visualize what’s happening
Clearest explanation I've seen on how this circuit works. The first stage is easy. It gets a bit harder to see subsequent stages. Your explanation makes it easy.
becauouse Kirchhoff's voltage law, V_C2=V_C1+V_in; V_C1=V_in=V_m -> V_C2=2*V_m
6:35 brought a smile to my face because I've been searching for an explanation of what happens at the midway point of the half cycles when the diodes begin to discharge.
I'm happy you explained it and I now understand perfectly.
Thanks!
earlier i used to find electronics very difficult ,bt your lucid explanation makes it easy to understand,Thanks a lot sir
I'm from Bangladesh. The way you teach is far far better to me. You clear the topic of a very easy method. Thanks a lot
Excellent tutorial. This is the only correct video on youtube that matches realworld and simulation observations. Other videos such as those by 'Organic Chemistry Tutor' and loads of other guys who have received positive reviews are wrong!! They all claim that diode D1 is forward biased all through a half cycle and capacitor C2 gets doubled in the other half. I was always wondering why diode D2 doesn't conduct in the 1/4th cycle and thanks to this channel for clarifying the correct working principle.
Excellent video. Very detailed and clear instruction on the voltage doubler circuit. It's one of the best out there. Thank you for taking the time to share you knowledge. This is a very important circuit for learning AC circuits. It's used in microwave ovens as well at the output of the step-up transformer to power the magnetron.
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS IN THE FORM OF BLAZING ARCS !!! WATCH IT ON GO ELECTRONICS !!!
@11:41
assuming C2 = C3 during this cycle there is charge sharing therefore C2 does NOT dump ALL of its voltage(2Vm) into C3 in the first cycle, first iteration will have 0.5*(2Vm) volts across both C2 and C3. It takes X amount of iteration for C3 to have a voltage 2Vm depending on input voltage. What happening here is C2 is constantly getting charged from the supply then shares its charge with C3 UNTIL the voltage across C3 = 2Vm. In summary, due to charge sharing between C2 and C3 it takes multiple iterations for C2 and C3 to reach 2Vm.
The math is
Vc2 = Vc3 = 0.5*(2Vm) -> first cycle
Vc2 = Vc3 = 0.5*(2Vm + Vinitial) = 0.5*(2Vm + 0.5*(2Vm)) -> second cycle
Vc2 = Vc3 = 0.5*(2Vm + Vinitial) = 0.5(2Vm + 0.5*(2Vm + 0.5*(2Vm))) -> third cycle
Vc2 = Vc3 = 0.5*(2Vm + Vinitial) = 0.5*(2Vm + 0.5(2Vm + 0.5*(2Vm + 0.5*(2Vm)))) ->fourth cycle
.
.
.
.
.
Until Vc2 = Vc3 = 2*Vm
Was doubting my basics, and searching a comment that explained what I know, then came across this, and now am satisfied!
I don't know why the these small but great details are ignored in videos. These hamper the basics!
Thank you for sharing your demonstration. I am looking forward to learning more about amplifiers.
Thanks for this awesome video. I like how you have analyzed the circuit from fundamentals!
Believe me you explain things better than most university professors. I just understood this nagging topic from you. Many thanks
Damn, there is something about Indian accent English that makes things easy to understand :D, nice vid man
MAKE YOUR OWN MULTIPLIER THAT COULD OUTPUT NEARLY 60000 VOLTS GO WATCH IT ON GO ELECTRONICS !!
i am from ethiopian thanks very much
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS IN THE FORM OF PLASMA ARCS !!! WATCH IT ON GO ELECTRONICS !!
Very well explained, thankyou ❤️
Nicely explained... Thanks!
best video on the topic out there hope friends will find it
Lot of videos of this topi c,,,but u r the best
Superbb way you are explaining...thanku sr...
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS !!! WATCH IT ON GO ELECTRONICS
Thanks for this amazing video 👌👍
GREAT VIDEO. VERY WELL EXPLAINED. 👍👍👍
Great work sir...amazing wonderful.
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS !! WATCH IT ON GO ELECTRONICS
Great Video ! Crystal clear explanation !!!
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS WATCH IT ON GO ELECTRONICS !!!
@@ge5645 watch it and experience it yourself with some very BIG capacitors and touch it go make sure!
Thank you so much for the effort
beautifully explained.
Nice video but I think that the analysis that you do for the voltage doubler may be wrong. First of all it seems that you are analyzing the initial transient of the circuit, which means that the initial state of your circuit is characterized by all capacitor discharged. The moment, in which I think you commit the first error is at 4.02. you say that the capacitor C1 charges to Vm when the sinusoidal signal reaches the positive peak and then you say that when the sinusoidal signal becomes negative then the diode 2 turns on and D1 turns off. But what happen between the positive peak of the sinusoidal signal and the negative part of the sinusoidal signal? You forgot to analyze that part. In fact, the diode D2 should turn on immediately after the peak of the positive sinusoidal signal. and D1 should turn off in the same moment. So the two diodes change their states much earlier than the moment in which the sinusoidal signal becomes negative. Missing the analysis of that part, will make your analysis not correct. Furthermore, it seems that in your analysis, C1 remains charged to Vm for all the negative cycle of the sinusoidal signal. what make you think that the voltage of C1 will remain constant during the negative cycle? I did not find any reason why that should happen. Try to simulate the circuit and look at the first period of the initial transient: at the first negative peak of the sinusoidal signal the voltage across C1 is actually 0V and not Vm as you state in your video. Overall the dynamic of the circuit is a bit more complicated than what you claim. I am not convinced by your analysis, which in my opinion is just a method to force the correct result. Anyway, feel free to correct me if I am wrong. :) In general, I like your videos, but I do not think that everything can have a short explanation. Good luck for the next videos
Even I was thinking the same but not able to find the decrease in the voltage of capacitor 1 after the one fourth cycle
@@alokgupta7239 probably because, you did not setup the LTxpice correctly. To see the initial transient you have to check the command "skip initial operating point solution" when you perform the transient. Try it ;)
Thanks for the great explanation. Can you include a section on how to calculate the necessary values of the diodes and capacitors?
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS (ARCS) !! WATCH IT ON GO ELECTRONICS
Very good 👍👍
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD SHOOT OUT ARCS OF NEARLY 60000 VOLTS !! WATCH IT ON GO ELECTRONICS !!!
@@ge5645 YES VERY GOOD! TO YOU MAKE SURE YOU TOUCH IT TEST WITH SOME BIG CAPS!
thank u sir
very helpful and nice explanation
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT NEARLY 60000 VOLTS !! WATCH IT ON GO ELECTRONICS !!
Amazing Content 💥🔥🔥🔥👍👍
Thanks
You Slightly Miss An Error At Point When -vm
Yes the circuit used here is slightly incorrect this is not the voltage doubler, its peak to peak detector
very nice explanation
Thanks for making
I am understanding it very well
UNDERSTANDING AND MAKING ARE QUITE DIFFERENT THINGS MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS !! WATCH IT ON GO ELECTRONICS!!
Good explanation, Thank you
MAKE YPUR OWN VOLTAGE MULTIPLIER THAT COULD SHOOT PLASMA ARCS OF AROUND 60000 VOLTS !!! WATCH IT ON GO ELECTRONICS!!!
11:44 I don't think so capacitor c3 will get charge by that path, because the D1 which you short circuited, has V(k)=2Vm and V(A)=Vm, hence that D1 will act as open circuit not short Circuit.... You have simply put a closed path without analysing... ( Don't mistaken me, I really like the content ❤️)
At the end, I have also shown the simulation. Please go through it. It will help you.
Yeah I checked! But please explain me, in 11:44 the path taken to charge C3. How is the diode short circuited when V(cathode)>V(anode)??
Aap Hindi mein bata dete toh achha hota aur aapne kvl bolake explain Kiya toh aapne likh ke diya hota to achha hota . But all concept is clear..
Thank you, you are awesome.
This is like the capacitor and diode increase voltage circuit.
Thank you.
Thank you for explanation. In this exercise, are all capacitor sizes same or different?
All are of same value.
@@ALLABOUTELECTRONICS
Thank you so much good day
awesome sir..,it helps me lot
It's a very good video
good explanation
I HAVE MADE A VOLTAGE MULTIPLIER THAT COULD OUTPUT NEARLY 60000 VOLTS !!! GO WATCH IT ON GO ELECTRONICS !!
9:50
Why the capacitors will not charge if they are uncharged.
(I think all capacitors were initially uncharged, then with your reason even the capacitor c1 should not charge).
The capacitor C1 is charging because diode D1 acts as a short circuit (ideally).
If you take the second loop, between diode D1- C3-D3- C2, as all the capacitors are uncharged initially there won't be any transfer of charge. In the first loop, the capacitor C1 is charging because of the input source Vin.
I hope it will clear your doubt.
If you still have any doubt, do let me know here.
Further, I have also shown the simulation result at the end, which might help you in visualizing the things.
@@ALLABOUTELECTRONICS Hi, is there a way to calculate how fast the capacitors will really charge?
Excellent
Please make videos on Transducers and sensors
Shukria Bai
In RF energy harvesting system, voltage multiplier is used extensively..
I need help designing a Voltage Tripler circuit with an input voltage of 110VAC that will power a 330VDC / 2 watt light bulb. Will the voltage tripler circuit that you have designed at 9 minutes work? And what size caps and diodes should I use? I appreciate any help. Thanks!
Plz make video on low voltage low power adders and multipliers for mtech students
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS IN FORM OF ARCS !! WATCH IT ON GO ELECTRONICS !!!
pls make videos on transistors
Ok so basically c1 capacitor becomes charged like a battery when the current passes through D1 diode. And capacitor c1 acting as a battery plus the actual power source it doubles the voltage almost. So it's like the power source is a 1.5v ( AC) battery, which releases the current through the c1 capacitor and the capacitor becomes charged 1.5v. And after the first cycle, the direction of the current changes in the main power source battery(because it's AC voltage reverses direction of the current), as a result we end up with 1.5v battery attached to 1.5v capacitor, becoming two batteries 1.5v+1.5v=3v attached to each other plus-minu-plus-minus like inside toy battery pack.
Excellent video
Plz, can this curcuit operate motor 2 amp 100 volt ? Help
If we apply a voltage directly to a capacitor only, does it conducts any current? If not then why?
it conducts
MAKE YOUR OUR VOLTAGE MULTIPLIER THAT COULD OUTPUT NEARLY 60000 VOLTS!! WATCH IT ON GO ELECTRONICS !!
@@ge5645 YES GE, AND YOU SHOULD USE BIG CAPS FOR THE VOLTAGE. WHEN YOU BUILD IT, TOUCH THE ENDS TO VERIFY ITS GOOD
What software did you use for your circuit simulation?
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT 60000 VOLTS !!! WATCH IT ON GO ELECTRONICS !!
Can we store the voltage that is multiplied by voltage multiplier?
YES !! LEARN TO MAKE A HIGH FREQUENCY HIGH VOLTAGE (60000 VOLTS) VOLTAGE MULTIPLIER !!
Thank you for the video, very useful especially to initial ramp-up section which is not covered usually in other tutorials. I wanted to find out how i can do AC analysis of this doubler? What is the AC equivalent circuit? I am asking since this circuit is usually connected to AC sources and i want to look at the matching problems. Thank you
pls make a video explaining the bjt operations ... thank you
⭕⭕⭕Check 11:44 here you are telling that the C3 is getting charged through the path of D1. But current cannot flow through D1 from C2 because 11:28 see the circuit, I mean considering C2, D1 is reverse bias condition.
Please check and let me know!
thanks for your sharing , a litter confused @7:20 , voltage across c1 is -Vm, Vin is -Vm, according to KVL , Voltage apply for C2 is -2Vm, Why it is -Vm in your presentation ?
Why this circuit we cant use for big loads bro.I mean if capacitor capacity is big then? What about current? Current also added or what? Actually I am a commerce student but also studying little bit electronic plz answer.
It is due to power. (P= V*I). Although the output voltage is increased, but since output power can't be more than input power, the current supplied by the output to the load will be limited.
THIS DEVICE IS A HIGH VOLTAGE LOW CURRENT ONE I HAVE MADE ONE ON MY CHANNEL GO ELECTRONICS !!! WHICH COULD OUTPUT 60000 VOLTS
for the simulations, how do we calculate the values of te capacitors and diodes?
The RC time constant should be much larger than input signal time period.
@@ALLABOUTELECTRONICS INPUT SIGNAL CAN BE AC OR ANY HIGH FREQUENCY ONE !!!
The capacitor values must be in nf or pf and if you use high voltage pulsed dc as input (the one I have made on my channel go electronics ) 3 TO 6 KV ceramic caps needed and 4kv HF diodes !! ( FOR MORE DETAILS WATCH MY VIDEO WHICH PRODUCES 60000 VOLTS !!)
12:44 which software did sir use for stimulation ?
Multisim Live
@@ALLABOUTELECTRONICS arigato sensai🥰
What if our input of voltage multipler is continuously fluxing 'varying' ...?what type of multipler is should use?
YES I HAVE CREATED A POWER SOURCE WHICH ACTS AS THE INPUT IN MY VOLTAGE MULTIPLIER VIDEO !! AND IT COULD OUTPUT 60000 VOLTS !! WATCH IT ON GO ELECTRONICS !!
@@ge5645 watch it and experience it yourself with some very BIG capacitors and touch it go make sure!
why did you not consider capacitor voltages when finding the diode states? You have just considered cycle states.
Are polarized capacitors used for these circuits? Doesn’t ac pass through capacitors? How does the charge remain during different cycles?
Yes, the capacitors are polarised. For the voltage doubler, I have explained it from 5:39 onwards. Please go through it. And still if you have any doubt then do let me know here.
THE UPPER CAPS PASS AC WHILE THE LOWER ONES STORE DC MAKE YOUR OWN VOLTAGE MULTIPLIER THAT COULD OUTPUT NEARLY 60000 VOLTS !! WATCH IT ON GO ELECTRONICS !!
How to calculate the value of capacitor in voltage doubler circuit?
So c1 only charges from 0 to positive peak and discharges from positive peak to negative peak? But from negative peak to 0, does c1 remain 0 until it reaches passed 0 to positive peak?
I have provided the simulation at the end of the video. Please go through it, your doubts will get clear.
Tks
Do you know, how to calculate output current for a load of say R, or maximum current output? Thanks.
It depends on the maximum power that the input source can deliver. Since we are increasing the voltage, so maximum output current that we can draw from the circuit will reduce proportionally.
@@ALLABOUTELECTRONICS Say you have an ideal input voltage source, your Cs are all the same, you have N stages, The frequency is F, the load resistance is RL. What is the current through RL?
@@aduedc well, in that case, for ideal voltage source, the output current will be N*Vm / RL. Where Vm is the maximum voltage of the input source, N is the number of stages and RL is the load resistor. But practically, you will not get that.
Let's say you have 5V source, which can provide maximum 1A current. (Overall 5W)
Now, if you increase that voltage by 10 times, then your output voltage is 500V. And if you connect 10 ohm resistor as load then your output current will not be 50A. Because with that current and voltage, if you find the required power then it is 50A*500V = 25000W.
But can you get that using 5W source, No.
In this case, the maximum current that you can get is 5W / 500V = 10mA. (considering there is no other losses in the N stages)
So, accordingly you should select the load resistance.
I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS This is not the case. Here is an experiment, use 100pf caps, measure the output current say at 10K ohm load. Then use the 10uF cap and measure it again. Also do this experiment with say 1KHz, and do the same experiment at 10KHz. As you know you can build a virtual resistor by charging and discharging a capacitor.
So it is not that simple, it has to do with frequency, Capacitor value, diode losses, .... I am searching for paper on that if you find it let me know.
@@aduedc Yes, that's true. But to make such circuits, the RC time constant has to be much larger than the time period of the input signal. So, you need to select R and C accordingly.
sir, how the process stop after doubleing voltage ?
I can’t understand why D2 is off when vin is increased again and reached at Vin again. C2 is charged and voltage difference between C2 is 2Vm.
Please mention the values of the capacitors and their voltages. Thanks.
The voltage rating of the capacitors should be atleast 3 times more than the desired output voltage. (e.g for 10V, output, use 35V rating capacitors) At the end, around 12:50, I have shown the simulation results and circuit.
You can check that for reference.
How capacitor C1 discharged through voltage source and charge C2 to Vm , please explain 6:42
If you observe, during the negative half cycle, the polarity of the voltage has been reversed. So, basically, it will try to charge the capacitor in the other way. So, the charge across the capacitor C1 will be lost. And it will be transferred to C2. But all this will happen gradually, in a couple of half cycles (more than two cycles)
Please check the simulation which I have shown at the end of the video to get more idea.
The same query I have ,during the negative half cycle, since the diode 2 is in reverse direction how will at allow the capacitor 1 to discharge through it??
@@sc-yardhuka9438 d2 in forward bias in negative half cycle
In same circuit, two different explanations of circuit.
@11:45 during 2nd positive cycle Vin and Vc1 cancel each other and cathode of D1 is at +2Vp from capacitor C2 so I do not see how D1 is forward biased. Do you understand my question ? Thank you great video as usual.
I simulated in LTspice and works just as you described. I will figure out my doubt using the simulator. Thank you.
I also think the explanations around 11:45 are wrong: d1 would not conduct
It will be charged by 2vm + v source - Vm which will peak at 2 vm
At 9.56, why is there no current on C2 and C3?
Sir Why will the Vc2 capacitor not charge upto 3Vm in the next cycle?
Hi, would you please mention the timestamp, where you are referring to in the video?
Is the Ampere at the output will remain same as input
No, it will reduce. Because the input and output power should remain same. ( Considering no losses).
So, as the output voltage increase, the current will reduce.
THIS DEVICE IS A HIGH VOLTAGE LOW CURRENT DEVICE !! I HAVE MADE ONE ON MY CHANNEL WHICH COULD OUTPUT 60000 VOLTS BUT LESS CURRENT MAYBE IN MILLI AMPS !!
During the negative half cycle equivalent circuit outer capacitor polarities are reversed why
Does it possible in practical circuits
No, it won't get reversed. While connecting the capacitors the circuit, one needs to check the proper polarity of the capacitors (For polarised capacitors).
THE UPPER CAPS PASS AC WHILE LOWER CAPS STORE DC AND CAN OUTPUT NEARLY 60000 VOLTS !! WATCH IT GO ELECTRONICS !!
Hello sir will you please know me what software did you use for simulation??
Multisim live
MAKE YOUR OWN VOLTAGE MULTIPLIER THAT OUTPUTS 60000 VOLTS IN FORM OF ARCS!! WATCH IT ON GO ELECTRONICS
I don't understand when C2 chargues C3 , its current is going against the diode polarity it shouldnt conduce that way my brain is a mess xD
Please check the simulation of the circuit which I have shown at the end of the video.
I don't understand how you can put a polarized capacitor in an ac circuit. C1 gets the ac signal and should explode anyway?
🤣
what's the difference between transformer and voltage multiplier? transformers just make the amplitude of the power bigger while voltage multiplier just offsets the power wave?
In the case of the transformer, the output is also an AC signal, while in voltage multiplier it is DC signal. So, convert it into DC you need to rectify it. Also, to generate the same high voltage the size of the transformer will become large. But in voltage multiplier, the output current is usually very small.
@@ALLABOUTELECTRONICS Oh! I see now! Thank you very much, sir. I didn't know voltage multipliers are already rectified.
in which software did you do your simulation?
Multisim Live
In v multiplier current also will multiply or it will reduce...
It will reduce.
Its better if you show the actual
Why did the direction of capacitor C2 change
At the 4 min mark your schematic shows c2 with the cathode facing towards the positive rail ?. Ami wrong LOL.
what is the software that you are using?
which software did you used for simulation
Multisim live
Good video.. But a not good idea of providing subtitles as it is hidding content which your writting 😐
On desktop, you can drag the subtitles anywhere on the screen and if required, you can also turn it off manually.
@@ALLABOUTELECTRONICS no fucks given🤣
The theory is not enough, should do practical also...
When I made connection of multipler . Output of circuit not desirable as when 6 volt AC applied to doubler circuit .It give15 volts output . also in triplet circuit 23 output
Why this happened can u help me
Thanks.
Please check the couple of things. First, whether the input is 6V peak or 6V RMS. If it is RMS then you need to multiply it with factor 1.4.
Second there will be a voltage drop across the diode.
So you also need to take into account.
And third make sure, RC time constant is much larger than the time period of the input signal.
Just increase the RC time constant a little bit.
I am sure you will get the desired results.
Thanks u so much
Will it work with AC and DC both?
Here its Ac voltage source
I had hard time understanding your accent you speak too fast. But was good explanation.
Uncle koi smjh nhi aiii
Arigatho
What software do you use to simulate the circuit? Is it paid?
Multisim. The web version ( online simulator) is free, and has some basic functionalities.