In the previous RC Charging and
RC Discharging tutorials, we saw how a capacitor,
C both charges up and also discharges itself through a series 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 an alternating AC waveform that constantly changes
from a maximum value to a minimum value at a rate determined by its frequency. How would this affect our
RC time constant value and the output RC waveforms?.
We saw previously that the capacitor charges up to 5T when a voltage is applied
and discharges down to 5T when it is removed. This 5T time constant value
always remains true as it is fixed by the resistor-capacitor 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.
Typical RC Waveform
Square Wave Signal
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 frequency matches
that exactly of the 5RC time constant ( 5T ) of the circuit,
then the voltage waveform across the capacitor would look something like this:
A 5RC Input Waveform
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 that of the 5RC
time constant, as ƒ = 1/5RC, allowing the capacitor to fully charge and fully
discharge on every cycle resulting in a perfectly matched RC waveform.
If the time period of the input waveform is made longer (lower frequency, ƒ < 1/RC)
for example a time period equivalent to say "8RC", the capacitor would then stay fully charged longer
and also stay fully discharged longer producing an RC waveform as shown.
An 8RC Input Waveform
If however we reduced the time period of the input waveform (higher frequency, ƒ > 1/5RC),
to "4RC" the capacitor would not have sufficient time to either fully charge or discharge with the resultant
voltage drop across the capacitor, Vc being less than its maximum input voltage would produce an RC waveform
as shown below.
A 4RC Input Waveform
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 circuit that converts 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 the capacitor will become fully charged quickly before the next change in the 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:
Sine Wave Input Signal
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.