# How to slowdown capacitor discharge

Discussion in 'General Electronics Discussion' started by dante_clericuzzio, Jun 30, 2016.

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1. ### dante_clericuzzio

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Mar 28, 2016
I have tested to charge a capacitor using 3 volt battery and it charges about 1 minute up to 235 volts using step up circuit...so the purpose is to test to power a small motor using a capacitor...but when i connect to the motor it instantly discharge and motor spin a while and there is spark its the charges is gone after that....how can i discharge the capacitor slowly and let the motor run accordingly not instantly. by the way the capacitor rating is 47uF @ 400volt usually i charge up to 250 volt max only

please don't tell me about the danger of 235 volt of the cap discharge because i am aware of this and i am using safety glove to handle and with high safety of awareness. This is not important to me...

2. ### Bluejets

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Oct 5, 2014
Sounds like you expect to get more energy out than you put in, however to answer your question, pwm the output to the motor.

3. ### Alec_t

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Jul 7, 2015
What is the rating (Volts, stall current) of the motor? 47uF doesn't hold enough charge to run a typical mains-rated motor for long, with or without a resistor.

4. ### dante_clericuzzio

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Mar 28, 2016
it run on 2.4 volt DC battery

5. ### BobK

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Jan 5, 2010
The energy stored in a capacitor is 1/2 C V^2.

47/1000000 * 235^2 / 2 = 1.3 Joules.

1 Joule is 1 Watt Second. An AA battery with a capacity of 2000mAH has 3W x 3600 seconds = 10800 Joules.

Do you undersand why you cant run the motor very long off your capacitor now?

Bob

Last edited: Jun 30, 2016
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6. ### dante_clericuzzio

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Mar 28, 2016
Completely understand sir your answer exactly is what i need to know...salute!
So in order to run a motor would be to use higher capacitance cap right base on the calculation?

7. ### BobK

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Jan 5, 2010
You would need a much larger capacitor, a supercapacitor, and it still would not be as size efficient as a battery. Why do want to run a motor off a capacitor? Rechargeable batteries are far more practical.

Bob

8. ### dante_clericuzzio

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Mar 28, 2016
Well actually Bob i want to see how the capacitor behave actually and how long the charges can last...because the cap be be charged so fast..like in 1 minute...so i thought it could be a good source of energy since it is faster..charging and it doesn't i can use my step up circuit to do it so fast sometimes not even a minute to charge 400v

9. ### Alec_t

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Jul 7, 2015
I'm not surprised if you apply 235V or more to a 2.5V motor .
If you added resistance to the circuit to slow the discharge then much of the stored energy in the cap would be wasted as heat in the resistor. Your step-up circuit also wastes energy. It would be more efficient to run the motor directly from whatever powers that circuit, rather than boosting the voltage then dropping it back down again with the added resistance.

Last edited: Jun 30, 2016
10. ### BobK

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Jan 5, 2010
I hate to tell you this, but others have had the same idea. Some even think they can run an electric car off capacitors. They have been working on for about 10 years now, with nothing but missed deadlines and nothing to show (except a list of suckers who invested).

Have a look at this: EEstor

Bob

11. ### dante_clericuzzio

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Mar 28, 2016
Bob ac
Bob i am inspired by this video hand crank the super capacitor and then release the charges to the motor for quite a reasonable amount of time...so i thought if we could hand crank to charge the cap and release it energy slowly we could have light without going through the grid and even in remote areas where there is so much darkness j/k..that is my purest intention !! u know actually i am not the kind of guy who read and believe...i always try no matter what they say until i see it myself then i can confirm and believe it..but you seems very knowledgeable about capacitors and all your reply is the most easiest for me to understand than the rest...and give me some ideas to look for more clues and try it

12. ### Alec_t

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Jul 7, 2015
Google 'dynamo torch'. Those gadgets have been around for years. The dynamo charges a re-chargeable battery (which provides a more stable voltage than supercapacitors do).

13. ### dante_clericuzzio

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Mar 28, 2016
I know that torch like the one Eton makes but it seems to me ....it takes more time and effort as oppose to charging capacitors. Charging capacitors are fast

14. ### Bluejets

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Oct 5, 2014
As you said and as Bob explained, so is the discharge.

15. ### BobK

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Jan 5, 2010
Charging capacitors is fast if you can supply a lot of power.

In that video, there are 10 supercapacitors. If they are 1F each, then it is over 200000 times as much capacitance as your 47uF.

I suspect that the capacitors were already charged when that video was taken, the few seconds of cranking would not have supplied the energy to charge them.

There was another video posted here recently of a world class bicycle racer running a pedal powered generator He pedaled for maybe 10 minutes then collapsed utterly exhausted. And produced about enough energy to charge a cellphone once.

Bob

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16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
One important consideration if using high voltage capacitors is to step the voltage down before powering the motor.

The most efficient way is to use a DC-DC converter. However, finding one with an output voltage as low as 2.4V and which also allows an input voltage of 250V will be an interesting challenge. You might be able to modify a small mains powered power supply, but note that they already probably have more than 47uF of storage on the high voltage DC side and they don't last long once you remove the power.

For very low power situations you have the problem that the current required to power the electronics (from the high voltage side) may be higher than is required (again, from the high voltage side) to provide the amount of power to your load.

Using a high voltage is not such a bad idea, since the power stored in a capacitor is proportional to the square of the voltage. Because the thickness of the dielectric is proportional to the voltage, and the capacitance is inversely proportional to distance, things balance out such that the total energy storage is pretty much linearly related to volume for capacitors using a given dielectric. There are other factors which vary, but this is a disappointing rule of thumb.

Super capacitors have a dielectric which is far more "efficient" in terms of space. However these are only available in very low voltages necessitating multiple capacitors in series to withstand a larger voltage.

Before you think that 2 of these in series can withstand twice the voltage (true) and thus store 4 times the energy (false), remember that placing n equal capacitance capacitors in series reduces the total capacitance to 1/n. Once you combine the fact that with n capacitors in series you can withstand nv volts with a combined total capacitance of c/n you'll discover that the total stored energy is proportional to the number of capacitors.

Having a single high voltage capacitor has one significant advantage, you do not need to worry about balancing the voltage across multiple capacitors. As the capacitance of the parts of a series string of capacitors rises, the amount of energy which must be transferred to maintain the balance also increases. If one does not wish to simply dissipate this energy, an active solution is required which can be quite complex. (zener diodes are a simple, cheap, and relatively efficient method for moderate power levels).

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17. ### dante_clericuzzio

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Mar 28, 2016
That means if capacitors connect in series it will reduce the capacitance...this is quite interesting to know...would you mind the reason why the capacitance decrease when it is arrange in series

18. ### BobK

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Jan 5, 2010
One way to look at is is energy conservation.

Say you have two capacitors of 1F each. Each is charged to 10V. The amount of energy in these two capacitors is then 1/2 * 1 * 10^2 or 50 Joules each, so 100 Joules total.

If you put them in series, there would be 20V across the two capacitors. If the capacitance stayed the same we would have 1/2 * 1 * 20^2 or 200 Joules. This can't be right, so the capacitance must go to 1/2 or we are creating free energy.

Another way of looking that capacitance is defined as Q / V. A larger capacitor will store more charge as the same voltage. If we charge two capacitors to V then there will be Q / C charge on each plate of each. Now put them in series. There will be the same charge Q on the outer two plates, but V is now doubled. Q / V is cut in half.

Bob

19. ### dante_clericuzzio

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Mar 28, 2016
Bob the guys video is quite credible and i have made one circuit based on his video and it works perfectly like he explained...so somewhat i quite believe what he does in most of his video. So i was thinking of getting the cap and crank it to proof and confirm his idea as i always do

20. ### BobK

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Jan 5, 2010
I suppose it is possible that the short cranking provided all the energy to run that motor for that long. But that is a very small motor. On the other hand, there are so many fake videos out there, I do not trust any of them.

If you want to no how much effort it takes to produce useful power, just go the gym and hop on a treadmill or elliptical. They will show you how much energy you are expending in Watts. (At least the ones around here do.)

Bob