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transistors for cap charge pump

Hi, I want to build a circuit for charging caps in parallel and then discharging in series. I want to switch up to 24 V @ say about 10 A. Ive never worked with transistors that go in between caps and before the load. Can you please recommend some suitable ones and a biasing method. Ive used opamps and ne555 in basic ways. Ive read people sometimes bias directly from chips. Id probably want to bias a few at a time from one signal. Darlington pair made from P channel fets ??? I think marbe in the past Ive tried npns before a load and the bias resistor form voltage divider at the base and didn't work ?? Is that why they use PNP and P channel??

Thakyou
 

KrisBlueNZ

Sadly passed away in 2015
It would be helpful if you gave us some more information.

How many capacitors?

Is the 24V the voltage across each capacitor, or across all of them in series?

What capacitance values?
 
Thanks for the help so far.
I was going to use two 2200 micro farad 25 V caps charged with 12V source. Charging to about 8 V and discharging at about 16V into another 2200 uF for buffering all up making 3, needing 4 swiches. If this cant output enough amps I would add more caps but keep the same switching arrangement. I am hoping for about 10 amps.
Only ever sucessfully used NPN on negative side of the load and when I put them on the positive side I cant make them work. Will using a P Channel Fet solve this. Can you recommend a suitable one ? Even better yet I am in the market for a darlinton pair so I can use less bias cuurent. Can you recommend 2 Fets that work niely together that way ? Or can BJP signal trans to FET. work alright. Iam hoping for minimal voltage drop after the swicth. If put on the right track can work out the biasing details myself without being to much of a nuisance. Can N channel Fet work on positive side of the load or is it like an NPN that way? Also I have never connected caps with transistors before. Can complications arise from that?
Thanks again for the help.
 
What are you trying to achieve with the circuit?

what pulse voltage,
pulse length
frequency?


Is it purely for education or a real project?

It maybe much simpler to use one large FET and a pulse transformer.
 

KrisBlueNZ

Sadly passed away in 2015
The type of circuit you're describing is called a switched capacitor charge pump. Switching circuits aside, it may not be suitable for what you want to do, and here's why.

Take a 2200 µF, 25V electrolytic capacitor that has been charged up to 12V. Now start drawing a current of 10A from it. The voltage across the capacitor will start to drop, as it discharges.

The formula for this dropping voltage is:

dV/dT = I / C

where dV is the change in voltage across the capacitor, in volts;
dT is the time period over which that voltage change occurs;
I is the current being drawn from the capacitor, in amps, and
C is the capacitance, in farads.

Plugging in the values for I (10) and C (0.0022) gives a rate of voltage change of 4,545 volts per second, or 4.5 volts per millisecond.

In other words, after just one millisecond (1/1000th of a second), the voltage on the capacitor will have dropped by 4.5V.

For the output smoothing capacitor, if you want less than, say, 0.45V peak-to-peak of ripple on the output voltage, you will need to recharge it 10,000 times per second.

Ripple will also occur due to the capacitor's internal resistance or ESR (effective series resistance) but this is less of a contributor, as long as you use low-ESR capacitors.

That might be feasible I suppose, but you also need to consider the ripple current specification of the capacitor as well, which is around 1~2A RMS for a good quality low-ESR 2200 µF electrolytic. Ripple current causes heating and reduces the lifetime of electrolytics, so it's best to be very conservative. Paralleling around ten electrolytics will force them to share the ripple current. You should still use good quality Japanese electrolytics.

Increasing capacitances will also allow you to use a lower switching frequency, which reduces losses due to gate capacitance in the MOSFETs.

The other capacitors in the circuit will also need similarly large capacitances and ripple current ratings.

You should definitely use MOSFETs instead of transistors. There are many differences, most of which make MOSFETs more suitable for this project. The most important is the ON characteristics. A saturated MOSFET behaves like a resistor, and it can have a very low resistance - just milliohms, for the big grunty ones! A transistor, on the other hand, has a minimum Vce(sat) voltage (which is much higher for a Darlington), which guarantees significant voltage loss in the transistor.

An N-channel MOSFET can be compared to an NPN transistor, and a P-channel MOSFET can be compared to a PNP transistor. So you can choose the type of MOSFET according to the circuit position, to make the drive circuitry easier. But driving MOSFETs is quite different from driving bipolar transistors.

Get some application notes on MOSFETs from companies that make them - go to http://www.digikey.com/product-search/en/discrete-semiconductor-products/fets-single and look through the manufacturers, go to their web sites, and look for design support and application information on MOSFETs in general, and applications similar to yours.

A charge pump voltage booster that generates nominally twice the input voltage needs only one switched capacitor. On one half of the cycle, the capacitor is charged from the input voltage, and on the other half, it is added to the input voltage and discharged into the output smoothing capacitor. You can connect a large capacitance across the input voltage as well, but that capacitor doesn't need to be switched.

Sorry for the disorganised nature of this reply :-(

Please keep us informed and post your design for us to check.
 
I was hoping to charge the caps through the load in parallel then discharge them in series back through the same load as an attempt to save my batteries when I go camping. Only for loads that are a bit forgiving of some voltage fluctuations. Lighting maybe DC fans
.
Using a low voltage dropout regulator I can get over 11 volts output from the ne555 as pulse voltage.

Dedicating it to a particular load I was hoping to get a volt or three over battery voltage by ajusting the frequency.

The outcome will raise my stanard of living if it works but either way it will be a learning experience. I saw the guidlines to this circuit a while back but it didnt come with transistor types only symbols of switches.
 

KrisBlueNZ

Sadly passed away in 2015
I was hoping to charge the caps through the load in parallel then discharge them in series back through the same load as an attempt to save my batteries when I go camping.
Oh. That's different. The only way to save your batteries is to minimise inefficiencies. You cannot create energy out of nothing. If you charge capacitors in series with the load, the energy that goes into those capacitors is not available to the load. There's no such thing as a free lunch.
Only for loads that are a bit forgiving of some voltage fluctuations. Lighting maybe DC fans
Your best option here would be a buck switching regulator with adjustable output voltage. These can be 80% efficient or more. Adjust the output voltage for the minimum brightness, or fan speed, that you can tolerate. If that reduces the load's power consumption more than the loss of energy in the regulator due to inefficiency, you will save some power.
Using a low voltage dropout regulator I can get over 11 volts output from the ne555 as pulse voltage. Dedicating it to a particular load I was hoping to get a volt or three over battery voltage by ajusting the frequency.
You can increase voltages, but you cannot create energy. You can feed 12V into a boost converter and get 24V out, but voltage is not energy. You need to calculate using power, not voltage. Power is the product of voltage and CURRENT. You need to take current into account.

Sorry, I'm not explaining things very clearly tonight!
The outcome will raise my stanard of living if it works but either way it will be a learning experience. I saw the guidlines to this circuit a while back but it didnt come with transistor types only symbols of switches.
You haven't been very specific about what you hope to do, but I'm pretty sure it's based on misunderstandings and it won't work because it can't work. If you can give a more detailed explanation, I can try to explain more clearly.
 
To KrissBlue NZ
Thankyou for the reply. I appreciate the time and effort taken. There is alot fo good info in there and 1 will copy that into my notes.
The post above, I made at the same time you were making yours but you sneaked yours in first.
Now that my true intentions have been revealed Ill understand if you give up on me.
These types of circuits sometimes annoy people. Im not that savy with the math, but I only chose that cap value because they are the biiggest I can get my hands on over the counter.(JayCar) I suspected I would need more and bigger but thankyou for the formulae, Its very handy to Know what to expect in advance.
Ill see what they have in the way of P channel fets and have a play around and see what kind voltage drops I do get. Being a beginner things happen quite slowly for me but if I have any joy I.ll certianly let you know. Infact Ill try to get something down on paper replacing the fets and how they are biased with switch symbols. Thanks again the info on the caps was great.

To Richardb
I was inteding to do it in two phases. While one lot was charging the other would be discharging. Feeding the discharge back into the load at a higher voltage was so that it would force the current through bfore the battery could
.
I think you guys are probsbly right otherwise everyone would be doing it
 
read up conservation of energy


energy stored in a capacitor is
0.5*c*v*v


two capacitors in parallel is the same energy as 2 capacitors in series if each capacitor has the same voltage on it

and if you still don't believe us maybe try playing with ltspice, link below

http://www.linear.com/designtools/software/

it will save a lot of time and money compared to building your project.
 
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