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Understanding better RC circuits

Discussion in 'General Electronics Discussion' started by lefam, Dec 1, 2010.

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  1. lefam


    Nov 18, 2010
    To charge a capacitor in a RC arrangement it takes 5xRC seconds, right, where:
    RC = R x C (where R is resistors value in ohms, and C capacitance in faradays)

    So if R = 10k, C=100uF --> RC = 10000 Ohm x 0,0001F = 1s
    and it would take 5x1s = 5s to charge the circuit.

    Again if R=1k, C=5000uF --> RC = 1000 Ohm x 0,001F = 1s
    and it would take 5s to charge the circuit.

    My question is:
    - If I build a classic RC circuit that makes a LED fade out when a switch is opened is there any difference in the behavior of the fade if I use R=10k C=100uF or if I use R=1k C=1000uF?
    According to my understanding in both configurations the fade takes 5s. But would exist a difference in the effect (like one configuration being smoother than the other one)?

    Thanks in advance.
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    It's more complex than that.

    After 1 RC interval, the capacitor is charged up to 63% of the supply voltage.

    After one more RC interval, the voltage increases a further 63% of the 37% that's left (that's another 23%, making a total of 86%)

    After 3 RC intervals, it's about 95% charged.

    After 4 RC intervals, 98%

    After 5 RC intervals, 99.3%.

    But it never stops charging...

    When you discharge, a similar thing happens. 63% gets discharged in the first RC interval, another 23 in the next, 9 in the next, 3 in the next, 1 in the next, and so on. But it is never fully discharged.

    If you want to use a discharging capacitor to power something, the current will drop in proportion to the charge remaining. So at the end of each RC interval, the current is just over a third of what it was at the beginning. After 2 intervals it is a little over 10% of its initial current. For a LED this means 10% of the brightness.

    The resistor value determines the initial current, so a larger capacitor and a smaller resistor will give you a LED that is brighter over the discharge curve, but the fall in brightness (expressed as a ratio) will be exactly the same if the RC constant is the same.

    So will it keep the LED on for 5 RC periods? No, the LED will never actually turn off.

    How long will you be able to see the LED glow? Well depending on the ambient brightness, the efficiency of the LED, the initial current, etc, it may be 1 RC period, 2, 5, or 10 (not likely to be 10 though).
  3. lefam


    Nov 18, 2010
    Hi I have attached the schematics. I designed the schematics using the Yenka application.

    In the circuit, I have a LED protected with a 300 Ohm resistor associated in parallel with a 2000uF capacitor in series with a 300 Ohm resistor.

    I simulated the circuit using Yenka and I found out that using 2000uF capacitor with a 300 Ohm resistor outputs a smoother fade out than using 1000uF capacitor with a 600 ohm capacitor, although the RC constant is the same in both configurations.
    In fact, using a bigger capacitor allows a fade out that starts with a brighter LED. Using a smaller capacitor with a bigger resistor creates a fade out that starts with a not so bright LED.

    I have a question: the capacitor discharge time is equal to the charge time in this case we have in parallel the LED with a 300 ohm resistor? When the RC is discharging the capacitor resistor will be in series with the LED resistor forming a 600 ohm resistor.

    Attached Files:

  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    What do you mean by "smoother fade out"?

    In the circuit you have shown, the LED will be brightest while the capacitor is charging. This is because there is less resistance in series with it. As you release the switch the resistance will suddenly decrease, but it decreases to what I have termed the initial brightness. The brightness during charging is somewhat independent of the RC values as you apply the charging voltage to a mid point in the R of the RC for discharge.

    You would be far better off removing the resistor in the path from the battery to the capacitor, or ignoring the sudden change in brightness as the switch is released.

    Of course all this depends on whether you're just simulating this for understanding, or whether you want to build this for some purpose.
  5. lefam


    Nov 18, 2010
    I call smoother fade out when the brightness doesn't suddenly change when the switch is released.

    I have a couple of questions:
    1. In this configuration the charge time is equal to the discharge time?
    2. What is the trick to instantaneously discharge a capacitor? When I tested this circuit on my breadboard I figured that my 2000uF capacitor takes too long to discharge.

    I am simulating just for understanding.
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