The RC integrator is a series connected Resistor-Capacitor network that produces an output signal which corresponds to the mathematical process of integration.
For a passive RC integrator circuit, the input is connected to a resistance while the output voltage is taken from across a capacitor being the exact opposite to the RC Differentiator Circuit. The capacitor charges up when the input is high and discharges when the input is low.
In Electronics, the basic series connected resistor-capacitor (RC) circuit has many uses and applications from basic charging/discharging circuits to high-order filter circuits. This two component passive RC circuit may look simple enough, but depending on the type and frequency of the applied input signal, the behaviour and response of this basic RC circuit can be very different.
A passive RC network is nothing more than a resistor in series with a capacitor, that is a fixed resistance in series with a capacitor that has a frequency dependant reactance which decreases as the frequency across its plates increases. Thus at low frequencies the reactance, Xc of the capacitor is high while at high frequencies its reactance is low due to the standard capacitive reactance formula of Xc = 1/(2πƒC), and we saw this effect in our tutorial about Passive Low Pass Filters.
Then if the input signal is a sine wave, an rc integrator will simply act as a simple low pass filter (LPF) with a cut-off or corner frequency that corresponds to the RC time constant (tau, τ) of the series network and whose output is reduced above this cut-off frequency point. Thus when fed with a pure sine wave an RC integrator acts as a passive low pass filter.
As we have seen previously, the RC time constant reflects the relationship between the resistance and the capacitance with respect to time with the amount of time, given in seconds, being directly proportional to resistance, R and capacitance, C.
Thus the rate of charging or discharging depends on the RC time constant, τ = RC. Consider the circuit below.
For an RC integrator circuit, the input signal is applied to the resistance with the output taken across the capacitor, then VOUT equals VC. As the capacitor is a frequency dependant element, the amount of charge that is established across the plates is equal to the time domain integral of the current. That is it takes a certain amount of time for the capacitor to fully charge as the capacitor can not charge instantaneously only charge exponentially.
Therefore the capacitor current can be written as:
This basic equation above of iC = C(dVc/dt) can also be expressed as the instantaneous rate of change of charge, Q with respect to time giving us the following standard equation of: iC = dQ/dt where the charge Q = C x Vc, that is capacitance times voltage.
The rate at which the capacitor charges (or discharges) is directly proportional to the amount of the resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC integrator circuit is the time interval that equals the product of R and C.
Since capacitance is equal to Q/Vc where electrical charge, Q is the flow of a current (i) over time (t), that is the product of i x t in coulombs, and from Ohms law we know that voltage (V) is equal to i x R, substituting these into the equation for the RC time constant gives:
Then we can see that as both i and R cancel out, only T remains indicating that the time constant of an RC integrator circuit has the dimension of time in seconds, being given the Greek letter tau, τ. Note that this time constant reflects the time (in seconds) required for the capacitor to charge up to 63.2% of the maximum voltage or discharge down to 36.8% of maximum voltage.
We said previously that for the RC integrator, the output is equal to the voltage across the capacitor, that is: VOUT equals VC. This voltage is proportional to the charge, Q being stored on the capacitor given by: Q = VxC.
The result is that the output voltage is the integral of the input voltage with the amount of integration dependent upon the values of R and C and therefore the time constant of the network.
We saw above that the capacitors current can be expressed as the rate of change of charge, Q with respect to time. Therefore, from a basic rule of differential calculus, the derivative of Q with respect to time is dQ/dt and as i = dQ/dt we get the following relationship of:
Q = ∫idt (the charge Q on the capacitor at any instant in time)
Since the input is connected to the resistor, the same current, i must pass through both the resistor and the capacitor (iR = iC) producing a VR voltage drop across the resistor so the current, (i) flowing through this series RC network is given as:
As i = VIN/R, substituting and rearranging to solve for VOUT as a function of time gives:
So in other words, the output from an RC integrator circuit, which is the voltage across the capacitor is equal to the time Integral of the input voltage, VIN weighted by a constant of 1/RC. Where RC represents the time constant, τ.
Then assuming the initial charge on the capacitor is zero, that is VOUT = 0, and the input voltage VIN is constant, the output voltage, VOUT is expressed in the time domain as:
So an RC integrator circuit is one in which the output voltage, VOUT is proportional to the integral of the input voltage, and with this in mind, lets see what happens when we apply a single positive pulse in the form of a step voltage to the RC integrator circuit.
When a single step voltage pulse is applied to the input of an RC integrator, the capacitor charges up via the resistor in response to the pulse. However, the output is not instant as the voltage across the capacitor cannot change instantaneously but increases exponentially as the capacitor charges at a rate determined by the RC time constant, τ = RC.
We now know that the rate at which the capacitor either charges or discharges is determined by the RC time constant of the circuit. If an ideal step voltage pulse is applied, that is with the leading edge and trailing edge considered as being instantaneous, the voltage across the capacitor will increase for charging and decrease for discharging, exponentially over time at a rate determined by:
So if we assume a capacitor voltage of one volt (1V), we can plot the percentage of charge or discharge of the capacitor for each individual R time constant as shown in the following table.
|τ||% Charged||% Discharged|
Note that at 5 time constants or above, the capacitor is considered to be 100 percent fully charged or fully discharged.
So now lets assume we have an RC integrator circuit consisting of a 100kΩ resistor and a 1uF capacitor as shown.
The time constant, τ of the RC integrator circuit is therefore given as: RC = 100kΩ x 1uF = 100ms.
So if we apply a step voltage pulse to the input with a duration of say, two time constants (200mS), then from the table above we can see that the capacitor will charge to 86.4% of its fully charged value. If this pulse has an amplitude of 10 volts, then this equates to 8.64 volts before the capacitor discharges again back through the resistor to the source as the input pulse returns to zero.
If we assume that the capacitor is allowed to fully discharge in a time of 5 time constants, or 500mS before the arrival of the next input pulse, then the graph of the charging and discharging curves would look something like this:
Note that the capacitor is discharging from an initial value of 8.64 volts (2 time constants) and not from the 10 volts input.
Then we can see that as the RC time constant is fixed, any variation to the input pulse width will affect the output of the RC integrator circuit. If the pulse width is increased and is equal too or greater than 5RC, then the shape of the output pulse will be similar to that of the input as the output voltage reaches the same value as the input.
If however the pulse width is decreased below 5RC, the capacitor will only partially charge and not reach the maximum input voltage resulting in a smaller output voltage because the capacitor cannot charge as much resulting in an output voltage that is proportional to the integral of the input voltage.
So if we assume an input pulse equal to one time constant, that is 1RC, the capacitor will charge and discharge not between 0 volts and 10 volts but between 63.2% and 38.7% of the voltage across the capacitor at the time of change. Note that these values are determined by the RC time constant.
So for a continuous pulse input, the correct relationship between the periodic time of the input and the RC time constant of the circuit, integration of the input will take place producing a sort of ramp up, and then a ramp down output. But for the circuit to function correctly as an integrator, the value of the RC time constant has to be large compared to the inputs periodic time. That is RC ≫ T, usually 10 times greater.
This means that the magnitude of the output voltage (which was proportional to 1/RC) will be very small between its high and low voltages severely attenuating the output voltage. This is because the capacitor has much less time to charge and discharge between pulses but the average output DC voltage will increase towards one half magnitude of the input and in our pulse example above, this will be 5 volts (10/2).
We have seen above that an RC integrator circuit can perform the operation of integration by applying a pulse input resulting in a ramp-up and ramp-down triangular wave output due to the charging and discharging characteristics of the capacitor. But what would happen if we reversed the process and applied a triangular waveform to the input, would we get a pulse or square wave output?
When the input signal to an RC integrator circuit is a pulse shaped input, the output is a triangular wave. But when we apply a triangular wave, the output becomes a sine wave due to the integration over time of the ramp signal.
There are many ways to produce a sinusoidal waveform, but one simple and cheap way to electronically produce a sine waves type waveform is to use a pair of passive RC integrator circuits connected together in series as shown.
Here the first RC integrator converts the original pulse shaped input into a ramp-up and ramp-down triangular waveform which becomes the input of the second RC integrator. This second RC integrator circuit rounds off the points of the triangular waveform converting it into a sine wave as it is effectively performing a double integration on the original input signal with the RC time constant affecting the degree of integration.
As the integration of a ramp produces a sine function, (basically a round-off triangular waveform) its periodic frequency in Hertz will be equal to the period T of the original pulse. Note also that if we reverse this signal and the input signal is a sine wave, the circuit does not act as an integrator, but as a simple low pass filter (LPF) with the sine wave, being a pure waveform does not change shape, only its amplitude is affected.
We have seen here that the RC integrator is basically a series RC low-pass filter circuit which when a step voltage pulse is applied to its input produces an output that is proportional to the integral of its input. This produces a standard equation of: Vo = ∫Vidt where Vi is the signal fed to the integrator and Vo is the integrated output signal.
The integration of the input step function produces an output that resembles a triangular ramp function with an amplitude smaller than that of the original pulse input with the amount of attenuation being determined by the time constant. Thus the shape of the output waveform depends on the relationship between the time constant of the circuit and the frequency (period) of the input pulse.
An RC integrators time constant is always compared to the period, T of the input, so a long RC time constant will produce a triangular wave shape with a low amplitude compared to the input signal as the capacitor has less time to fully charge or discharge. A short time constant allows the capacitor more time to charge and discharge producing a more typical rounded shape.
By connecting two RC integrator circuits together in parallel has the effect of a double integration on the input pulse. The result of this double integration is that the first integrator circuit converts the step voltage pulse into a triangular waveform and the second integrator circuit converts the triangular waveform shape by rounding off the points of the triangular waveform producing a sine wave output waveform with a greatly reduced amplitude.