Passive AC charge pump or capacitive voltage divider?

Thread Starter

clone477

Joined Oct 3, 2008
28
Hoping some one can help me understand charge pumps as I have been reading up on them for the last week but need clarification. I want to know how a capacitive charge pump can be used to voltage divide(or reduce the input voltage) of an AC source at the its input with only passive components of capacitors and diodes?

What I want to achieve is a passive charge pump, that can charge in series, but then have each capacitor discharged(at a reduced voltage) in parallel. I wanted to avoid the use of transistors.

I came across these articles, one shoes the exact circuit but with switches, and I assume a dc input, and the other describes how it can be down with simply diodes and capacitors when the input is an AC source.

I'm assuming the charge pump is best fitted, although I'm not 100% certain the difference between this and a capacitance voltage divider.

Thanks guys, appreciate it.Screen Shot 2021-03-09 at 11.06.30 PM.pngScreen Shot 2021-03-09 at 11.45.30 PM.pngScreen Shot 2021-03-09 at 11.06.30 PM.png
 

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shortbus

Joined Sep 30, 2009
8,429
I came across these articles, one shoes the exact circuit but with switches, and I assume a dc input, and the other describes how it can be down with simply diodes and capacitors when the input is an AC source.
You should have shown that AC source. I'm not sure, but would like to know more, that AC can do that. Capacitors don't store AC like they do DC. They can only add a slight delay, ver slight, to an AC signal, but to my knowledge they can't store AC so wouldn't make a charge pump. So please share the resource you found about an AC charge pump.
 

Thread Starter

clone477

Joined Oct 3, 2008
28
Here is another thread I came across talking about the same thing, the first respond in the thread describes it.

https://electronics.stackexchange.c...efficiently-charge-a-capacitor-from-ac-source

This is the link to the screen shot I shared, it is the first couple paragraphs.

https://www.maximintegrated.com/en/design/technical-documents/tutorials/7/725.html

I understood that series capacitor voltage divider can not store charge when input is AC, but can store charge when input is DC. But what I suspected is that the diodes on the output of each individual capacitor in that series voltage divider, would act as a half wave rectifier, outputting pulsed DC, which is what I wanted to achieve. Does this seem correct?
 

shortbus

Joined Sep 30, 2009
8,429
It's time to say what you actually wanting/doing. Getting pulsed DC from mains without a transformer is dangerous, and may be against the forum rules.
 

Thread Starter

clone477

Joined Oct 3, 2008
28
It's time to say what you actually wanting/doing. Getting pulsed DC from mains without a transformer is dangerous, and may be against the forum rules.
Of course, I’m not interested in touching mains. I’m experimenting with air core resonant transformers with the function generator. So source will be HF. I’m trying to understand how a voltage divider would function in this example. In the link from stack exchange, it’s exactly what I want to understand. Charge a capacitance voltage divider, off of one of the capacitors, connect a fast recovery diode, for 1/2 wave rectified dc, and I want to know if it will act as a DC pump of sorts, to charge a large storage capacitor off that? Will this work as I explain, passively, without the need to mosfets or any switching?
 

shortbus

Joined Sep 30, 2009
8,429
So far both of your links, especially the second one are for DC not AC. If you want a DC output why not start with DC? All of the chips shown in your second link are DC.

So source will be HF. I’m trying to understand how a voltage divider would function in this example.
You are not making sense. With both HF and HV you create them at what is needed, they are not easily adjustable or variable. I don't want to seem mean, but you should start to read more on these topics.
 

Thread Starter

clone477

Joined Oct 3, 2008
28
So far both of your links, especially the second one are for DC not AC. If you want a DC output why not start with DC? All of the chips shown in your second link are DC.



You are not making sense. With both HF and HV you create them at what is needed, they are not easily adjustable or variable. I don't want to seem mean, but you should start to read more on these topics.
Thanks for responding, it’s no problem, I’m just trying to solve this problem. So I know the 2nd link was Dc to Dc, the commenter above asked for the source of the one paragraph I posted in the original post, so that was the source. This was the key paragraph from that source.....

“Capacitive voltage conversion is achieved by switching a capacitor periodically. Passive diodes can perform this switching function in the simplest cases, if an alternating voltage is available. Otherwise, DC voltage levels require the use of active switches, which first charge the capacitor by connecting it across a voltage source and then connect it to the output in a way that produces a different voltage level.“

it interested me as it says passive diodes can act as switch in ac capacitive voltage divider.

My eyes are bloodshot and have bags under them from so much reading over the last 3 weeks, lol.

The input is AC unfortunately, this ac voltage divider cit unit seems like it will allow me to charge the bank in series at HF ac, then be able to discharge in parallel in either half wave or full wall rectification through the output diodes. I need series charging of caps, the parallel discharge of the same caps, all passively. I know this can be done somehow. I really want the charging time reduced by series capacitors( charging at a higher voltage but overall series capacitance is great lowered in relation to the quantity of caps) then discharge in parallel at the combined capacity of all caps. I under stand charge storage in capacitor is the same, Q=CV but I need the benefit of quick charge times. Thanks for your help.
 

RPLaJeunesse

Joined Jul 29, 2018
156
Since you are dealing with HF, and I'll assume here that is a sine wave signal, why not use a PI network to passively reduce the voltage such that the final p-p AC matches the DC voltage you desire out?
 

RPLaJeunesse

Joined Jul 29, 2018
156
He was after a charge pump, aren't PI pads usually for impedance matching? And they also aren't "adjustable" another value he wanted.
I view the charge pump as a solution to a problem, but not the problem the OP has. For an AC signal he wanted a low-loss solution for changing the voltage, and a pi-network is very low loss and can handle significant power (i.e. a ham's kilowatt linear amp output). As for adjustability, a pi-net can be very adjustable. Depends on frequency and the components (fixed or variable) used. The OP needs to add more constraints and goals for any of us to provide a "great" solution.
 

Thread Starter

clone477

Joined Oct 3, 2008
28
I view the charge pump as a solution to a problem, but not the problem the OP has. For an AC signal he wanted a low-loss solution for changing the voltage, and a pi-network is very low loss and can handle significant power (i.e. a ham's kilowatt linear amp output). As for adjustability, a pi-net can be very adjustable. Depends on frequency and the components (fixed or variable) used. The OP needs to add more constraints and goals for any of us to provide a "great" solution.
Hoping some one can help me understand charge pumps as I have been reading up on them for the last week but need clarification. I want to know how a capacitive charge pump can be used to voltage divide(or reduce the input voltage) of an AC source at the its input with only passive components of capacitors and diodes?

What I want to achieve is a passive charge pump, that can charge in series, but then have each capacitor discharged(at a reduced voltage) in parallel. I wanted to avoid the use of transistors.

I came across these articles, one shoes the exact circuit but with switches, and I assume a dc input, and the other describes how it can be down with simply diodes and capacitors when the input is an AC source.

I'm assuming the charge pump is best fitted, although I'm not 100% certain the difference between this and a capacitance voltage divider.

Thanks guys, appreciate it.View attachment 232372View attachment 232373View attachment 232372
I should try to clarify as best I can do you guys have a good idea of my goals. I assumed a charge pump or capacitor voltage divider was the best way to achieve it but I could be mistaken. My goals.....

1. To charge a set of series capacitors with higher voltage, then have the topology change to discharge each individual capacitor at a reduced voltage in parallel.

2. I want to sum up all the capacitor discharging circuit in parallel to achieve a lower discharge voltage (compared to input voltage) but a combined higher discharge current.

3. One of the main benefits of this will achieve is very quick capacitor charge times as the total series capacitance will be minimized the more capacitors there are in series. It will also allow for all the summer paralleled capacitance to be additive, and much larger. Even though I understand overal capacitors charge is the same at series compared to parallel discharge.

3. Input voltage ideally will be AC at HV(7kv- 10kv), so that is why I want to avoid transistors that will be exposed to this voltage on the input, as highest rated transistors are 4.7kv, much to low. High voltage diodes easily are available or assembled. I visualized ac constantly pumping the series circuit, while the overflow, so to speak is drawn from the circuit it parallel off each individual capacitor, then parallel all the capacitor ends to mass up the current at the same voltages. If this is not possible, I do have the option of a full wave rectifier to bring the input voltage to DC first.
I hope this helps give a clearer view of my thought. I appreciate all the brainstorming.
 

RPLaJeunesse

Joined Jul 29, 2018
156
If the goal is recharging a "lower" (whatever that is) voltage capacitor quickly, consider that c=q/v or restated cv=q. To charge a given cap C to a given voltage V requires a given charge Q. It doesn't matter how many steps you go through, the charge to bring a cap to a given voltage is going to be the same. Running it through an excess of stages at some ungodly voltage makes no sense, it just costs. Build a proper charging supply, with adequate current out and a safe voltage limit, and be done with it.
 

Thread Starter

clone477

Joined Oct 3, 2008
28
If the goal is recharging a "lower" (whatever that is) voltage capacitor quickly, consider that c=q/v or restated cv=q. To charge a given cap C to a given voltage V requires a given charge Q. It doesn't matter how many steps you go through, the charge to bring a cap to a given voltage is going to be the same. Running it through an excess of stages at some ungodly voltage makes no sense, it just costs. Build a proper charging supply, with adequate current out and a safe voltage limit, and be done with it.
You misquoted me, so I assume that’s why you said “recharging a lower voltage capacitor”.
Let me be clear. I need quick very quick charge times, capacitors charged in series have much lower total capacitance then the same caps in parallel. If I have 100uf 100v capacitor, and I charge 100 of these in series, the total capacitance will be 1uf and I will need 10kv input voltage to fully charge these. If I discharge all these, in parallel, all summed up, I should theoretically get a discharge voltage of 100v at a capacitance of 10,000uf.
Nothing confusing here to understand, so you misunderstood what I was saying.
It is much much quicker to charge a cap bank at 1uf then at 10,000uf. As I mention in my post before I understand Charge is the same. If you have a constructive solution, that would be great. Thank you.
 

shortbus

Joined Sep 30, 2009
8,429
(i.e. a ham's kilowatt linear amp output).
I'm not familiar with how linear's work so this may be a dumb question. Where or what are used in the Pi networks for the resistors at those voltages the TS is after(7 - 10 KV)? Not trying to troll but to increase my knowledge.
 
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