How would it be possible to accumilate DC energy over a few seconds and release it...?

Question

For my application I need to accumilate DC energy over a few seconds and release it in a very short amount of time... to pulse the power. What sort of device would allow me to do this using DC electricity? I need to accumilate and pulse the DC energy to get a high power in a short amount of time relative to the low input power over a long amount of time. Please Help!! Thanks!

Answer 1

You, sir, need a capacitor. I'm puzzled how you managed to learn enough about electronics to need high power levels without hearing about them.

Answer 2

Hey Maria.. Get an electrolytic capacitor use this personal sheet to calculate eh time..I made this up several years ago read about the calculation to find the discharge time of a fixed capacitor on the net...DC caps will charge with observed polarity being used...

pay attention to this you can work these all around to find RC or initial voltage or final voltage or in this case time

V= final voltage

Vo=start voltage

e is the numerical log x or Ln

t= time value generally start is t=0

R=resistance kilo ohms

C= capacitance microF 10^-6



8888888888*********don't forget to convert to farads and the resistance**********



the example.....



An RC circuit has R=8 k? and C=3. mF. The capacitor is at voltage 100 V at t=0 when the switch is closed. How long does it take the capacitor to discharge to 20 V?



RC Circuits



V=Voe^(-t/RC)



20=100e^(-t/(8kOHM*3mF)

TO GET TO NEXT STEP DIVIDE BOTH SIDES BY 100,, GETS RID OF 100

0.2=e^(-t/(24s)

TO GET RID OF e log x we need to value in ln of .2

ln(0.2)=-t/(24s)



t=-24ln(0.2)

t=38 seconds

lots a blurb have a good one.From the E...

Answer 3

capacitors work with high voltages but low current. by building a Multi stage voltage multiplier you can generate high voltages at low currents. its THe trade off. but you need a type of oscillator as the Pre-stage for the multiplier. you need an ac for the input of the multiplier but the output will be high voltage DC.



capacitors will allow AC to pass through them. but DC the capacitors will take time to charge up to near full value.

capacitors will hold a charge and large capacitors will hold a charge for a long time. ( the capacitor in the high voltage of a CRT can keep a charge for several years if not discharged properly. seen a guy that got knocked on his rear end by shorting out that capacitor. he had to go to the hospital but was OK)

so a capacitor can act like a voltage battery. but the current levels are low. the time it takes for a capacitor to charge is the RC time constant. it takes about 5 time constants for a capacitor to be what is called charged but will not be or will be 100% of input voltage. it charges to 63.2% of total voltage in the first TC (time constant) then for the 2ND TC and additional 63.2% of total.



say 100 volts is the total voltage

1st TC @ 63.2% = 63.2% = 63.2v

2ND TC @ 63.2% = 86.5% = 86.5v

3rd TC @63.2% = 95.0% = 95.0v

4TH TC @63.2% = 98.2% = 98.2v

5TH TC @ 63.2% = 99.3% = 99.3v



the discharge time is the reverse.

1st TC @ 63.2% = 36.8% of max = 36.8v

2ND TC @ 63.2% = 13.5% = 13.5v

3rd TC @63.2% = 5.0% = 5.0v

4th TC @63.2% = 1.8% = 1.8v

5th TC @63.2% = 0.7% = 0.7v



the time constant is determined by TC = R*C

the time constant = resistance multiplied by the capacitance.

so as long as your time for The discharge is greater than the 5th TC then you will have 99.3% of your total voltage from the capacitors.



the voltage muliplyer i am talking about uses a DC source converts it to AC and then is put through the multiplyer stage which converts it back to DC. but again it is low output current.



you still can not get more out then what you put in. the power level will be less at the output then the input due to natural losses.

Answer 4

Capacitors work that way ONLY with DC.

(They do have AC applications, but that's not one of them.)

I don't know what you consider "high power", but if you're working with over 12V.

stop and learn more before you hurt yourself.