RC constant of capacitor in RC circuit as a filter

In summary, 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.
  • #1
waqarrashid33
77
0
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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  • #2
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
 
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  • #3
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 completely..
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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Related to RC constant of capacitor in RC circuit as a filter

What is the RC constant of a capacitor in an RC circuit?

The RC constant of a capacitor in an RC circuit is the product of the resistance (R) and capacitance (C) values in the circuit. It represents the time it takes for the capacitor to charge or discharge to 63.2% of its maximum value.

What is the role of the RC constant in an RC circuit?

The RC constant plays a crucial role in determining the frequency response of an RC circuit. It determines the cutoff frequency, which is the frequency at which the circuit begins to attenuate the input signal.

How do you calculate the RC constant of an RC circuit?

The RC constant can be calculated using the formula RC = R x C, where R is the resistance in ohms and C is the capacitance in farads. Alternatively, it can also be calculated using the time constant formula, which is equal to the product of the resistance and capacitance values in seconds.

What is the significance of the RC constant in a filter circuit?

The RC constant is a crucial parameter in filter circuits as it determines the bandwidth and cutoff frequency of the circuit. It also helps to attenuate unwanted frequencies and shape the frequency response of the circuit.

What happens if the RC constant is increased in an RC circuit?

If the RC constant is increased, the time it takes for the capacitor to charge or discharge will also increase. This will result in a lower cutoff frequency and a wider bandwidth in the filter circuit. Additionally, the circuit will have a slower response time to changes in the input signal.

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