# Serial Connected RC Networks: Integrator

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In this article of our series, we will discuss Serial Connected RC Networks: Integrator. For a passive RC integrator circuit, the input is connected to a resistor, while the output voltage is taken from a capacitor that is the opposite of the RC Differentiator Circuit. The capacitor charges when the input is high and discharges when the input is low.

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In electronics, the basic series-connectedresistance-capacitor (RC) circuit has many uses and applications, from basic charging/discharge circuits to high-grade filter circuits. This two-component passive RC circuit may seem simple enough, but depending on the type and frequency of the input signal applied, the behavior and response of this basic RC circuit may be very different.

A passive RC network is nothing more than a serially connected resistance with a capacitor. In other words, it is a serially constant resistance with a capacitor with a reassurance due to the decreasing frequency as the frequency increases along the plates. Therefore, at low frequencies, the reactance of the capacitor, Xc is high. At high frequencies, Xc = 1/(2πεC) is low in reassurance due to the standard capacitive reacquest formula and we saw this effect in our tutorial on Passive Low Pass Filters. .

If the input signal is a sinus wave, an rc integrator will act as a simple low pass filter (LPF) above the breakpoint with the cutting or corner frequency corresponding to the RC time constant (tau, τ). Thus, when fed by pure sinus wave, an RC integrator acts as a passive low-pass filter that lowers its output above the cutting frequency point.

As we have seen before, the RC time constant reflects the relationship between resistance and capacitance in proportion to resistance, R and capacitance, C, with the time given in seconds, according to time.

Therefore, the charging or discharge speed depends on the RC time constant, τ = RC. Consider the circuit below.

## RC Integrator

The input signal for an RC integrator circuit is applied to the resistance with the output received through the capacitor. Then VOUT equals VC. Since the capacitor is an element connected to the frequency, the amount of load formed between the plates is equal to the time zone integral of the current. In other words, since the capacitor cannot charge instantly, it only takes a certain amount of time for the capacitor to fully charge, since it is only charged exponencially.

Therefore, the capacitor current can be written as follows:

This basic equation above iC = C(dVc/dt) can also be expressed as the instantaneous rate of change over time of Q, which gives us the following standard equation: iC = dQ/dt where the load is Q = C x Vc, that is, capacitance multiplied voltage.

The loading (or discharge) speed of the capacitor is directly proportional to the amount of resistance and capacitance that gives the circuit's time constant. Thus, the time constant of an RC integrator circuit is the time interval equal to the product of R and C.

Since capacitance is equal to Q/Vc, where the electric charge is the flow of Q over a current (i) time (t). That is, it is the product of ixt in coulomb. We know from the Ohm law that voltage (V) is equal. If we put them in the RC time constant equation in ix R:

## RC Time Constant

Then we can see that while both i and R are taking each other, only T remains, indicating that the time constant of an RC integrator circuit has a time dimension in seconds, given the Letter Tau, τ in Greek. 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 up to 36.8% of the maximum voltage.

## Capacitor Voltage

We have previously said that the output for the RC integrator is equal to the voltage on the capacitor, that is: VOUT equals VC. This voltage is proportional to the load, Q is stored in the capacitor supplied by Q = VxC. The result is that the output voltage is integral to the input voltage with the amount of integration that depends on the R and C values and therefore the time constant of the network.

Above, we found that capacitor current can be expressed as load change rate Q over time. Therefore, from a basic rule of the differential account, the time-based derivative of Q is dQ/dt, and we achieve the following relationship as i = dQ/dt:

Q = ∫idt (Q load on capacitor at any time in time)

Since the input is dependent on resistance, the same current must pass through both the resistance and the capacitor (iR = iC), producing a VR voltage drop along the resistance. Thus, the current (i) flows through this series RC network.

Therefore:

Because i = VIN/R, the following can be given to solve VOUT as a function of time:

In other words, the voltage output on the capacitor from an RC integrator circuit, the VIN proportional to the 1/RC constant, is equal to the time integral of the input voltage. Where the RC represents the time constant, τ.

Then, assuming that the initial load on the capacitor is zero, that is, VOUT = 0, and the input voltage VIN is constant, the output voltage is expressed as follows in the VOUT time zone:

## RC Integrator Formula

In other words, an RC integrator circuit is a circuit in which the output voltage is proportional to the integral of the input voltage of the VOUT. With this in mind, let's see what happens when we apply a single positive pulse in the form of a step voltage. RC integrator circuit.

## Single Pulse RC Integrator

When a single-stage voltage pulse is applied to the input of an RC integrator, the capacitor charges through resistance in response to the pulse. However, since the voltage on the capacitor cannot change instantly, but the output will not be instantaneous as the capacitor increases exponentially while charging at a speed determined by the RC time constant τ = RC.

We now know that the speed at which the capacitor charges or discharges is determined by the circuit's RC time constant. If an ideal step voltage pulse is applied, that is, if the front edge and the rear edge are considered instantaneous, the voltage on the capacitor will increase for charging and decrease for discharge. Exponentially over time, at a rate determined as follows:

### Capacitor Charging

### Capacitor Discharge

Therefore, assuming a volt (1V) capacitor voltage, we can draw the percentage of charging or discharge of the capacitor for each R time constant, as shown in the table below.

Note that on 5 or more time constants, the capacitor is considered 100 percent full or fully discharged. Now let's say we have an RC integrator circuit consisting of 100kΩ resistance and 1uF capacitor, as shown.

## RC Integrator Circuit Example

The time constant of the RC integrator circuit is therefore given as follows: RC = 100kΩ x 1uF = 100ms.

Therefore, if we apply a step voltage pulse to the input, for example, with a duration of two time constants (200mS), we can see from the table above that the capacitor will charge up to 86.4% of its fully charged value. If this pulse has an amplitude of 10 volts, this equates to 8.64 volts before the capacitor ejaculates from resistance to the source again, while the input pulse returns to zero.

Assuming that the capacitor is allowed to fully discharge at a time of 5 time constants or 500mS before the arrival of the next input pulse, the graph of charging and discharge curves will look like this:

## RC Integrator Charging/Discharge Curves

Note that the capacitor discharges not from the 10 volt input, but from a starting value of 8.64 volts (2 time constants). Then, since the RC time constant is constant, we can see that any change in input pulse width will affect the output of the RC integrator circuit. If the pulse width is increased and is equal to or greater than 5RC, the shape of the output pulse will be similar to that of the input, since the output voltage reaches the same value as the input.

However, if the pulse width is reduced below 5RC, the capacitor will only partially charge and cannot reach the maximum input voltage, which will cause a smaller output voltage, since the capacitor cannot charge that much, which leads to an output voltage proportional to the integral of the input. Voltage.

Therefore, assuming an input pulse equal to a time constant, that is, 1RC charges and discharges between 63.2% and 38.7% of the voltage on the capacitor, not between 0 volt and 10 volts at the time of capacitor change. Note that these values are determined by the RC time constant.

## Fixed RC Integrator Time Constant

Therefore, for a continuous impact input, the correct relationship between the periodic time of the input and the RC time constant of the circuit will take place by producing the integration of the input, a kind of ramp output, and then a ramp down exit. But for the circuit to function correctly as an integrator, the value of the RC time constant must be large compared to the periodic time of the input. This is the RC ≫ T, which is usually 10 times larger.

This means that the size of the output voltage (proportional to 1/RC) will be very small between the high and low voltages, which significantly reduces the output voltage. This is because the capacitor has much less time to charge and unload between pulses. However, the average output DC voltage will increase towards half the input, and in our example of impact above this will be 5 volts (10/2).

## Sine Wave RC Integrator

Here, the first RC integrator converts the original pulse-shaped input into an increasing and decreasing triangular waveform, which becomes the input of the second RC integrator. This second RC integrator circuit rounds the points of the triangular waveform into a sinus wave, as it performs an effective pair of integrations on the original input signal with the RC time constant affecting the degree of integration.