# RC constant of capacitor in RC circuit as a filter

waqarrashid33
We know that for RC circuit as filter or in multiplier we use capacitors having larger RC constant than the period of input signal.but I am confused here:

As long as i know the charging and discharging period of a capacitor are equal wether it is fully charged or not...(tell me i am right or wrong).
when in first +ve half cycle the capacitors charges it will discharge completely in next cycle but in RC filter we should not completely discharge the capacitor..how does it happen?

As initially we have a discharged capicator and in first cycle it charges and in next cycle it doen't completely discharges. Please clear me on it...

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If you have an AC signal then a resistor then a capacitor to ground, and take the output as the voltage across the capacitor....

The capacitor will charge up as the input voltage rises, but because of the resistor, it cannot charge up quickly enough to follow the input.

So, it reaches some smaller peak charge after the input voltage starts to drop and eventually gets to be lower than the capacitor voltage.

So, the capacitor can't continue to rise in voltage and it starts to follow the input voltage by discharging through the resistor.

It now acquires a negative voltage but again cannot reach the peak voltage of the input waveform and it reaches some lower value.

Like this:

[PLAIN]http://dl.dropbox.com/u/4222062/RC%20filter.PNG [Broken]

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waqarrashid33
From the book of Electronic devices by thomas L-Floyd i attached a RC filter circuit here..
What i understand there is the point that in charging there is no resistance to the current coming into capacitor so it will charge instantly or take a little time because of the resistance of diode.and in discharging the resistance is with capacitor which increases its RC constant and so it does not dicharge completly..
But I certainly think that this is my wrong concept..can you please explain the first two cycles with details..if you have some book about it then please share it with me.

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