Capacitor

Discussion in 'General Electronics Chat' started by Sle7in, May 23, 2011.

  1. Sle7in

    Thread Starter New Member

    Jan 13, 2010
    20
    0
    Hello guys,

    Can anyone tell me how this circuit operates??
    Just a simple basic question but i really want to know...I'm learning..

    I'm curios about the capacitor present in the circuit...what exactly the capacitor do??
     
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  2. ke5nnt

    Active Member

    Mar 1, 2009
    384
    15
    Kinda looks to me like the LED on the left side would be lit when power is applied, and the LED on the right would be lit only when the push button is depressed. Most of the time, a capacitor in a circuit like this would be to help smooth out voltage fluctuations from the source, in this case a 12V battery. I have seen some LED circuits similar to this, though slightly different, where current is put into a circuit with a push button switch that charges a capacitor, then when the button is released, the LED remains lit because it is being driven by the discharging capacitor. However, I've only seen that done with a polarized electrolytic capacitor, which in this circuit it definitely is not.

    That's an interesting circuit you've got there, not quite sure what it's supposed to do.
     
  3. Sle7in

    Thread Starter New Member

    Jan 13, 2010
    20
    0
    This is capacitor test circuit...The right circuit to learn the capacitor behavior...but I don't really understand how this circuit operates..
     
  4. radiohead

    Active Member

    May 28, 2009
    474
    31
    It's two LEDs in series, each with its own current limiting resistor. R1 doesn't even need to be there. Assuming the LEDs are typical red LEDs that operate on 2.1 VDC with a current draw of 20mA ( C4SMF-RJS-CT0W0BB2TB-ND ) R2 has to drop the difference between the two LEDs and the supply voltage. 12 VDC - 4.2 VDC = 7.8 VDC. Now divide dropped voltage (7.8 VDC) by LED current (.02A) = 390 Ohms As for the LED wattage, we use Joule's Law. 7.8 VDC x .02 = .156 Watts. Since .156 watts is more than 60% of a 1/4 watt (.250), you should use a 1/2 watt resistor. As Ke5nnt said, the cap is just there to "smooth" out voltage. In my opinion, it should be across the battery terminals, not where it is. However, capacitors across the power rails can be used to "tighten" the rails to ensure stages of a circuit don't interfere with each other.
     
  5. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    I built the circuit to verify my thoughts. When power is first connected, the left LED flashes ON due to the charge current of the cap. After the cap charges, the left LED goes out. As for the right LED and switch, if power is still connected and the switch is pressed, both LEDs light up. However, if after the left LED turns itself off AND power is removed, pressing the button will cause the right hand LED to blink briefly from the energy stored in the cap. If the cap were shorted, the left LED would not blink. If the cap were open, after removing power and pressing the button, the right LED would not blink. Have to agree that it could be a sort of cap testing circuit.
     
  6. Sle7in

    Thread Starter New Member

    Jan 13, 2010
    20
    0
    Sorry guys, can some explain to me how the capacitor helps to smooth out voltage fluctuation??
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    While this might be the function of a capacitor in certain circuits, this is not the purpose of the capacitor in this particular situation. BillB3857 has given a detailed explanation of the circuit operation - he even went to the trouble of building the circuit to see what happened. And not even a thanks from you! Have you read and understood what he wrote?
     
  8. pistnbroke

    Member

    May 9, 2011
    32
    1
    Whilst I agee with TNK re the response and wonder whats wrong with a text book or google .....

    When the supply voltage is on the rise the cap charges from the supply ...when the supply is falling the capacitor discharges and that discharge keeps the supply volts up . Not perfect but reduces the ripple...
     
  9. Sle7in

    Thread Starter New Member

    Jan 13, 2010
    20
    0
    Thanks all pistnbroke, tnk, bill, radiohead, ke5nnt....sorry for my bad...I ask because I don't understand...if I'm understand, I don't be here to ask u all professionals...I try looking in the text books, google it, simulate it but it seem like I don't really get it clearly...I understand a little bit but I need it to be crystal clear so I can proceed to my next topic to learn...I don't care if I'm the stupidest among all of you and I don't care someone called me stupid...as long as I can learn something from you guys and I really appreciate it...I'm just a kid to learn what I like...if I'm not belong here because of my stupidity then I'm sorry...Sorry again if I've shown bad manners in this forum...thanks for all the valuable information...
     
  10. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215
    ..........getting a handle on electronics is not an easy proposition. Bertus says it so well......

    " You don't have to know everything, if you know where to find it.
    When you do ask questions, you may look stupid.
    When you do NOT ask questions, you will STAY stupid."

    Welcome to AAC !!! :D
     
  11. Sle7in

    Thread Starter New Member

    Jan 13, 2010
    20
    0
    haha...I don't mind called stupid but I mind if I ask stupid question and somebody told me to find it myself....T_T...that's why some people choose to stay stupid rather than asking question...
     
  12. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215
    The person, hopefully not on AAC, that told you to go find whatever yourself, would be the exception, not the rule on these forums.
    I have been a little thick at times on here, though the guys and gals here are some of the most helpful on the whole dang internet !!
     
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