In the previous RC Charging and RC Discharging tutorials, we saw how a capacitor, C has the ability to both charge itself and also discharges itself through a series connected resistor, R at an amount of time equal to 5 time constants or 5T when a constant DC voltage is either applied or removed.

But what would happen if we changed this constant DC supply to a pulsed or square-wave waveform that constantly changes from a maximum value to a minimum value at a rate determined by its time period or frequency. How would this affect the output **RC waveform** for a given RC time constant value?

We saw previously that the capacitor charges up to 5T when a voltage is applied and discharges down to 5T when it is removed. In RC charging and discharging circuits this 5T time constant value always remains true as it is fixed by the resistor-capacitor (RC) combination. Then the actual time required to fully charge or discharge the capacitor can only be changed by changing the value of either the capacitor itself or the resistor in the circuit and this is shown below.

Useful wave shapes can be obtained by using RC circuits with the required time constant. If we apply a continuous *square wave* voltage waveform to the RC circuit whose pulse width matches that exactly of the 5RC time constant ( 5T ) of the circuit, then the voltage waveform across the capacitor would look something like this:

The voltage drop across the capacitor alternates between charging up to Vc and discharging down to zero according to the input voltage. Here in this example, the frequency (and therefore the resulting time period, ƒ = 1/T) of the input square wave voltage waveform exactly matches twice that of the 5RC time constant.

This (10RC) time constant allows the capacitor to fully charge during the “ON” period (0-to-5RC) of the input waveform and then fully discharge during the “OFF” period (5-to-10RC) resulting in a perfectly matched RC waveform.

If the time period of the input waveform is made longer (lower frequency, ƒ < 1/10RC) for example an “ON” half-period pulse width equivalent to say “8RC”, the capacitor would then stay fully charged longer and also stay fully discharged longer producing an RC waveform as shown.

If however we now reduced the total time period of the input waveform (higher frequency, ƒ > 1/10RC), to say “4RC”, the capacitor would not have sufficient time to either fully charge during the “ON” period or fully discharge during the “OFF” period. Therefore the resultant voltage drop across the capacitor, Vc would be less than its maximum input voltage producing an RC waveform as shown below.

Then by varying the RC time constant or the frequency of the input waveform, we can vary the voltage across the capacitor producing a relationship between Vc and time, t. This relationship can be used to change the shape of various waveforms so that the output waveform across the capacitor barely resembles that of the input.

The **Integrator** is a type of **Low Pass Filter** circuit that converts a square wave input signal into a triangular waveform output. As seen above, if the 5RC time constant is long compared to the time period of the input RC waveform the resultant output will be triangular in shape and the higher the input frequency the lower will be the output amplitude compared to that of the input.

From which we derive an ideal voltage output for the integrator as:

The **Differentiator** is a **High Pass Filter** type of circuit that can convert a square wave input signal into high frequency spikes at its output. If the 5RC time constant is short compared to the time period of the input waveform, then the capacitor will become fully charged more quickly before the next change in the input cycle.

When the capacitor is fully charged the output voltage across the resistor is zero. The arrival of the falling edge of the input waveform causes the capacitor to reverse charge giving a negative output spike, then as the square wave input changes during each cycle the output spike changes from a positive value to a negative value.

from which we have an ideal voltage output for the Differentiator as:

If we now change the input RC waveform of these RC circuits to that of a sinusoidal **Sine Wave** voltage signal the resultant output RC waveform will remain unchanged and only its amplitude will be affected. By changing the positions of the Resistor, R or the Capacitor, C a simple first order Low Pass or a High Pass filters can be made with the frequency response of these two circuits dependant upon the input frequency value.

Low-frequency signals are passed from the input to the output with little or no attenuation, while high-frequency signals are attenuated significantly to almost zero. The opposite is also true for a High Pass filter circuit. Normally, the point at which the response has fallen 3dB (cut-off frequency, ƒc) is used to define the filters bandwidth and a loss of 3dB corresponds to a reduction in output voltage to 70.7 percent of the original value.

where RC is the time constant of the circuit previously defined and can be replaced by tau, T. This is another example of how the *Time Domain* and the *Frequency Domain* concepts are related.

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very good webside for student

I used a function generator to input Square wave voltage into a series RC circuit. The function generator was at 40 Hz and the trace on the oscilloscope looked like the typical RC wave form. At first I thought it that the differentiator explained what I saw on my oscilloscope but I didn’t see multiple peaks, I only saw one. Can you explain why the output didn’t look like the output for the 5RC input waveform in the article?

No, I can not see your scope or circuit. For 40Hz, R = 1k0 and C = 5uF or any 5RC combination of the two.

Can you please add some more waveform for RC ckts, because its easy to understand the behaviour with waveform…. 🙂

easy to understand

your ideas

your explanation easy to understand add some more example with numerical explanation…..

Very nice easy to understand articles.

Once again a very nice and informative article

very good website very use full to beginners and its very easy language to understand ……………..

Hi,

I’m a very stupid person who want to answer for a very , uhm, difficult (for me) scenario:

If a band of magnets move over a set of copper wire, that is how a Permanet Magnet Generator(PMG) works, the wire connected to a capacitor (the voltage/amps/ohms etc are not known, at a rate of 1000 rpm, each magnet is spaced at 10mm intervals (1000/10)=100 times / second, how quickly will the capacitor charges and discharges? Will it be possible, in serie format, be able to produce twice or even more current?

Or must the capacitors only be used at the exit where all the current flows came together?

Between magnet flows / before the transformer (if it’s needed), or after the transformer?

A very confused, illiterate question, from the same type of person…

Hello.

In regards to the third example labeled “A Shorter 4RC Input Waveform” … the diagram seems to represent a “A Shorter 2RC Input Waveform”, not a 4RC time interval like the title suggests. (Or am I missing something?)

Hello John, The example shows that if the frequency is increased so that the time constant of the input waveform is equivalent to 4RC, then the half-period of the input signal will be much less than the required 5RC half-period required changing the shape of input waveform.