Understanding RC circuits better

Thread Starter


Joined Nov 30, 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.
While the time constants are the same for both R and C combinations, the addition of an LED to the circuit will drastically change the charge and discharge rates.

Do you have a schematic of how you are going to connect the circuit? When you consider charge and discharge time, don't forget to consider the current that must go to the LED!! It can slow down the charge and speed up the discharge...

In general, larger capacitors hold more charge which can translate into longer durations of powering the LED before the capacitor discharges to a level where it no longer lights the LED.

Thread Starter


Joined Nov 30, 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.



Joined Feb 4, 2008
The discharge time is greater than the charge time because the LED protective resistor is in series with the capacitor resistor.

The LED is brighter when you are using the 2000uF capacitor because the discharge current is twice the discharge current when using the 1000uF capacitor. However, the voltage variation across the capacitor is exactly the same in both cases.

In your first post, you said that the capacitance is measured in Faradays. Although, this unit is named after the name of Michael Faraday, it is more common to say Farad and not Faraday.