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Maintaining DC voltage across capacitor with signal?

GHOSTOWLGRID

Dec 2, 2012
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Hey.

lets say I have 24 volts AC at 25KHz.
I have a transformer and I step it up to 2000 volts.
Then I feed it through a diode and into a capacitor.

The circuit has no load, the capacitor then charges and approaches 2000 volts, then my current draw from my 24 volt AC source decreases, and all I am drawing now is enough to maintain votage across this capacitor.

Assuming I have no load across the capacitor to discharge it while maintaining supply to it, My current draw will be very low, correct?

Question, because I have half wave 25KHz signal as my original AC and only 1 diode making it half wave rectification, can I attach an antenna to the (+) terminal of the capacitor and measure 25KHz half wave with a frequency analyzer in the same room? (Negate external noise and distance from the antenna to the meter).
With antenna attached, my capacitor will still be at 2000 volts?
Will my current draw from the source be very low assuming the capacitor reached full charge and the only thing attached is an antenna and no actual resistance load across the capacitor terminals?

I attached a drawing showing the simple circuit.

I basically want to know if adding the antenna will not create additional load and if I will read 25KHz off that antenna, even thought its on the pulsing DC (+) side.
 

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Laplace

Apr 4, 2010
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With no load or leakage the capacitor will remain fully charged at 2000 volts. The antenna will also be at 2000 volts (with respect to the opposite side of the capacitor); however, due to parasitic capacitance to the environment the antenna itself may behave as though it is grounded (almost). Whatever happens will be due to unseen effects not shown on the schematic diagram. But if the opposite side of the capacitor is grounded, then the antenna will be at 2000 volts above ground. I would not walk too close to the antenna in that case.
 

KrisBlueNZ

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Yeah, what Laplace said. And an antenna connected to 2000V DC won't radiate anything... except possibly a bit of corona, a sizzling sound, and a slightly unpleasant smell, if the humidity is high.
 

GHOSTOWLGRID

Dec 2, 2012
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Okay so the cap will stay at 2000 volts, good.
(I'm assuming the antenna is isolated from atmosphere and the cap doesn't have losses. I will compensate for this later, no need to go over it now.)


But what about the 25 KHz signal from the source? Does is still exist as a half wave signal because there is only the one diode?

Because I use the transformer to get 2000 volts AC, then used one diode to make pulsing DC at 2000 volts 25 KHz. Will I still measure a frequency on the antenna?

EDIT:
This is a home project I am trying, mostly for proof of concept.
 
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Laplace

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Looking at your schematic diagram the answer must be that once the capacitor is fully charged the diode blocks the 25 KHz signal from reaching the antenna. But in reality there are potentially many parasitic leakage paths which could couple a portion of the electrostatic or magnetic high frequency signal to the analyzer pickup. So the question to you is what construction techniques have you employed to suppress these leakage paths?
 

KrisBlueNZ

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Okay so the cap will stay at 2000 volts, good. But what about the 25 KHz signal from the source? Does is still exist as a half wave signal because there is only the one diode?
No, not across the capacitor. And if you connect the bottom end of the capacitor to 0V, then there won't be much, if any, 25 kHZ signal on the antenna.

As you said, the diode allows the capacitor to charge up to 2000V DC. The voltage at the secondary of the transformer (the anode of the diode) will continue to be a 25 kHz AC signal, and the capacitor will continue to have 2000V DC across it. But there won't be any significant AC on the capacitor.
Because I use the transformer to get 2000 volts AC, then used one diode to make pulsing DC at 2000 volts 25 KHz. Will I still measure a frequency on the antenna?
Once you add the capacitor, you no longer have "pulsing DC". You have smoothed, continuous DC. So, no.
 

BobK

Jan 5, 2010
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It would help if you told us what you are trying to do, instead of asking questions about what you think is the solution.

Bob
 

KrisBlueNZ

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It would help if you told us what you are trying to do, instead of asking questions about what you think is the solution.
That's the story of our lives here on Electronics Point :-(
 

(*steve*)

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Oh, and at 25kHz, the optimal length of an antenna is about 3 kilometres (for a 1/4 wave antenna). It would work best if mounted vertically.
 

GHOSTOWLGRID

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I'm not trying to do anything unique. this is just proof of concept that I am retrying. Nothing special.

I was just trying to see if I can maintain a high DC voltage across the capacitor while injecting a high frequency pulse into the capacitor (I want a DC pulse, not an AC pulse).

I realize once the capacitor is fully charged, the pulsing will decrease, so I will incorporate a pause every so many AC cycles to allow capacitor voltage to dissipate enough that I can pulse again and have my signal during the recharge (recharge meaning losing about 10% voltage then going back up to about 100%)

In the drawing I forgot to ad 2 inductors, one on each side of the capacitor. my bad, I used an online drawing thing and was to concentrated on figuring it out. I should have just used paint.
 

KrisBlueNZ

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I was just trying to see if I can maintain a high DC voltage across the capacitor while injecting a high frequency pulse into the capacitor (I want a DC pulse, not an AC pulse).
Can you explain to us what you think the difference is between a "DC pulse" and an "AC pulse"?

You still haven't told us what you hope to achieve by doing this.
 

GHOSTOWLGRID

Dec 2, 2012
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AC pulse as in AC signal, I dont want alternating voltage across the capacitor.

I'm not trying to achieve anything other than just having a DC voltage and keeping the top half of an AC sin wave (by using the diode).

Like I said, its just proof of concept, I'm simply doing this to learn how things operate and react. Learning how the capacitor will charge and discharge in real conditions as opposed to math on a piece of paper.
That's it. I don't understand why you keep thinking I have some master plan to conquer the world. lol :p

some people play with toy cars, I play with electricity. Yah its dangerous, but I find it fun.


EDIT:
I want to charge it with DC pulsing, and see how the DC pulse disapears as it charges.
I want to try and measure it for fun, voltage and the pulsing.
I want to see if the DC pulse creates a measurable signal.
I want to see how long I can keep the DC pulse going before the capacitor approaches saturation or whatever.

stuff like that.

I'm obviously going to try it when I buy some parts, I'm just curious what people have to say about it first.
 

KrisBlueNZ

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I don't think you have a plan to conquer the world. It seems that you don't really have any plan at all. You can't prove a concept if you can't even define that concept meaningfully, and you need to ask meaningful questions if you want meaningful answers. Have fun mucking around :)
 

GHOSTOWLGRID

Dec 2, 2012
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Trying to learn about how the capacitor will charge and how the charge will look and how the half rectified DC pulsing will look, and if I will be able to pick it up on a sniffer.

Or do you have a better way to demonstrate how to maintain a DC voltage across a capacitor while inputting a half wave signal while maintaining the DC voltage?

EDIT:
Well I'm sorry. I thought this was a place to get help and learn.
I guess I was wrong.
 
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Supercap2F

Mar 22, 2014
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I would advise you to get a electronics simulation program. You could draw up this circuit in one of them and you can see all the different waveforms in the circuit, how the capacitor charges up etc. Look at LTSpice and Tina TI, there both free.

Regards

Dan
 
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(*steve*)

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You need to have some sort of grasp of fundamentals. The fact that you don't is pretty obvious from the fact that you haven't realized we've answered your question.

If I asked you for a tricycle, but said I wanted a two wheeled tricycle not a three wheeled one, and I definitely didn't want a bicycle, how would you react?

Sure we help a lot of people but it's hard to help people who have wrong ideas without convincing them their understanding is faulty. Unfortunately we have no other choice than to become more and more blunt. At this point you either dig in and complain about how we make fun of your ideas or the penny drops and you open your mind to us helping you.

The choice is yours. We won't hold it against you if you take the second path.
 

BobK

Jan 5, 2010
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AC pulse as in AC signal, I dont want alternating voltage across the capacitor.

I'm not trying to achieve anything other than just having a DC voltage and keeping the top half of an AC sin wave (by using the diode).

Like I said, its just proof of concept, I'm simply doing this to learn how things operate and react. Learning how the capacitor will charge and discharge in real conditions as opposed to math on a piece of paper.
That's it. I don't understand why you keep thinking I have some master plan to conquer the world. lol :p

some people play with toy cars, I play with electricity. Yah its dangerous, but I find it fun.


EDIT:
I want to charge it with DC pulsing, and see how the DC pulse disapears as it charges.
I want to try and measure it for fun, voltage and the pulsing.
I want to see if the DC pulse creates a measurable signal.
I want to see how long I can keep the DC pulse going before the capacitor approaches saturation or whatever.

stuff like that.

I'm obviously going to try it when I buy some parts, I'm just curious what people have to say about it first.
If that is all you want to do, why does it need to be done with lethal voltages? Nothing you have described here required 2000V.

Bob
 
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