nicolaisi 
So i want to supply my input with an amount of 200mA current. And i want to charge my capacitor first then discharge it, so that i wont have a sudden surge from my power supply. I already did the calculation, but i'm not that sure it will work, as for the capacitor, i intend to use super cap or gold cap or ELCD.
Thanks.
I don't really see a question in there. You may want to try restating the problem.
Michael
Torby 
To avoid a high current "inrush" when you turn it on, put a small inductor, or even resistance in series with the capacitor. Or maybe even a current source.
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nicolaisi @Michael: well, i had to admit that i do really expressed my question badly, i wanted to make it brief but at the same time wanting to ask more.
@Torby: wow, you understand what i kind of want to ask. That is the reason why i'm charging and discharging. It is to avoid the high current inrush. I'm planning to put a resistor.
how do i calculate the capacitance of the cap?
Bingo! Problem number one. Your LDO input current will be at least as large as the load current, UNLESS it is a switcher (then, one would not usually call it LDO).
To determine the value of the current limit R: the highest charging current happens when the C is totally discharged. At that point, the voltage drop across the current limiter is 5V. If the current limiter is a resistor, then the value is just 5V/200ma = 25 ohms. For current limiting, the value of the capacitor makes no difference. The capacitor value WILL determine how long it takes to recharge and how long the LDO will be able to power the load.
Jim
Nicolaisi,
In addition to what has been mentioned above, do you control the design of the +5V supply?
If it is a 3-Terminal Linear Regulator, such as the LM7805, it has its own internal "short circuit" current limiter circuitry. If I recall correctly, when the Imax is exceeded it does not shut down the output, it just limits it. Hence current will still flow to charge the 5V output / 3V input capacitor.
The 7805 can provide 1 A, with peaks of 2.1 Amps, so you may not need to limit the current into the cap.
Note, however, that you will want to make sure it has a proper heat sink, based upon the input voltage.
JC
nicolaisi theres two switches on my sketch, first one will be ON to charge, then OFF first swicth and ON 2nd switch to discharge the cap to my LDO.
the objective is to utilise low current to charge the cap and discharge providing higher current. Lets say from 20mA to 250mA.
Ok...
You have not mentioned your overall goal.
I presume you realize that Energy Into the Cap = Energy Out of the Cap.
This means that if you set up a slow charging rate, over a "long" time, to put a certain amount of energy into the cap, and then use up that energy up quickly, (250 mA current draw), there will be time intervals with NO Power provided by the 3 V regulator.
Also, if the goal with the low current input is to save energy, you won't do this expending it in a series resistor.
What you have proposed is doable, but more information as to your overall goal would be helpful.
JC
nicolaisi yeah JC, i was trying to express it, but i'm not good at it. I know what i'm proposing is doable, but it seems like it's odd to do it that way, i sort of google and saw people are doing this kind of circuit with the camera flasher circuit. i read some of the circuit, but was unable to transfer some method. It will be great if you can pin point what to add and how others normally do.
What is the stored energy in the capacitor going to be used for?
A camera flash uses a capacitor because that's basically the only feasible way to deliver a couple of tens to hundreds of amps at a couple of hunderds volts in a instant and still allow the device to use a normal power source.
nicolaisi according to my rough draft, a super cap of 1F will be used here. But after i did the calculation, i dont really need such big value of a capacitance i could settle down with 10mF. i'll include my calculation here next.
I'm not sure a 1F supercap can supply 200mA... these tend to have a fairly high impedance.
nicolaisi JC, i think i didnt state my goal, my goal is to prevent sudden current inrush lets say a jump from 5mA to 200mA for a short time then go back down to 5mA. Due to this problem, i plan to do this charging and discharging, so that instead of a sudden jump, small current will be used to charge the cap and provide my 200mA on the other side.
With a switched mode power supply you could control
the current without loosing energy.
Torby So you want this capacitor to carry the regulator for a while if the power is interrupted?
A 1F capacitor drops 1V per second when you're drawing 1A. Your gadget uses about 200mA? So it will drop 1v in 5 seconds. How long to the dropout of the regulator depends on the difference between the supply voltage and the regulator's dropout voltage.
So your power supply needs to provide the 200mA the circuit is using plus the charge current for the capacitor. With a 1F capacitor, this charge current can be pretty big unless you do something to limit it.
The diode stops inrush current into the capacitor. Figure the resistor to be V/A where V is the supply voltage and A is your desired max charge current. For figuring how long the cap can carry the circuit, remember to subtract the diode drop from V.
If you don't know my whole story, keep your mouth shut.
If you know my whole story, you're an accomplice. Keep your mouth shut.
Torby,
Very Clever!
JC
nicolaisi Thanks for the advice. :D
