Rc integrator-linear waveshaping circuit

Rc integrator-linear waveshaping circuit

When nonsinusoidal waveforms are applied to linear networks, the resultant output waveform may be different from the input waveform.

Hence, the circuit is said to shape the input-voltage waveform or to be a linear wave-shaping circuit. See Fig. 2-1 (a).

Figure 2-1 RC Integrator

Linear waveshaping circuits are classified according to components.

The three basic classes are RC circuits, RL circuits, and RLC circuits.

The RC circuit is the simplest and is classified according to the waveshape of its output voltage.

The two most common of these output-voltage waveshapes are the integrated and the differentiated.

The differentiated output is discussed in section Rc differentiator-linear waveshaping circuit; the integrated output is considered here.

2.1 DEFINITION OF RC INTEGRATOR

An RC integrator circuit is a series circuit which consists of a resistor, a capacitor, and a voltage source and is one in which the output voltage is taken across the capacitor.

The voltage which appears across the capacitor is referred to as the integrator output. When charging, the capacitor combines, or integrates, its original voltage with the new change in voltage. Hence, the term integrator circuit is generally and loosely applied to this circuit configuration.

The output-voltage waveform, shown in Fig. 2-1 (b), is one form of the integrator circuit produced wave. Variations of this waveform are discussed in a subsequent section.

The integrator circuit shown in Fig. 2-1 may also be referred to as a low-pass RC filter. A low-pass filter passes the low frequencies and attenuates the higher frequencies, as its name implies.

This characteristic of this circuit is evidenced by the exponential waveshape of the output voltage.

As pointed out in section The generation and analysis of square waves, the sharp corners of a voltage waveform are caused by the higher-frequency components of the applied step or rectangular waveform.

The study of sinusoidal voltages established the fact that as the frequency increases, the reactance of the capacitor decreases, thus effectively bypassing its higher frequencies to ground.

If the higher frequencies are bypassed to ground, they are not available as part of the output voltage; hence, the typical exponential-shaped output-voltage waveshape is produced.

2.2 WAVEFORM DISPLAY

Refer to Fig. 2-1 (b).

Notice that the input voltage and the output voltage have been plotted vertically and to the same time base.

This placement is extremely important because if the input and the output-voltage waveforms are placed horizontally adjacent, the characteristics of the output voltage as compared with those of the input voltage, at a particular instant in time, are extremely difficult to determine.

The vertical presentation illustrated in Fig. 2-1 (b) simplifies the comparison, and only by waveform comparison does the operation of the circuit become evident.

Therefore, An output voltage and/or current waveform should be drawn above or below the input (or some other reference) waveform and to the same time base.

2.3 DESCRIPTION OF OPERATION

Let us now analyze the operation of the circuit in Fig. 2-1 (a).

At t0 time an input-pulse voltage of +10 V is applied.

From t0 to t1 time, the circuit resembles an RC series circuit to which a 10 V battery has been connected. The instant the voltage is applied, all the voltage appears across the resistor.

This occurs because, in zero time, the capacitor has had no time to charge. A voltage cannot exist across the plates of a capacitor unless there are more electrons on one plate than there are on the other.

Even though electrons move very rapidly through a circuit, they require a finite amount of time to do so.

Therefore, in zero time the electrons have had no time to move ; hence, the capacitor is uncharged and has no voltage drop across it.

Kirchhoff's second law states: the sum of the voltage drops around any closed circuit must equal the sum of the voltage sources.

Hence, the +10 V source must equal the sum of the voltage drop across the capacitor and the drop across the resistor.

See Fig. 2-1 (a).

Because the voltage drop across the capacitor is zero volts at t0 time, the voltage drop across the resistor is forced to equal 10 V at t0 time. See Fig. 2-1 (b).

As time increases from t0 time toward h time, the electrons move through the circuit and charge capacitor C to +10 V if the interval is sufficiently long. See e0 voltage waveform in Fig. 2-1 (b).

At t1 time, the capacitor is charged to +10 V, and the source voltage of + 10 V is removed.

Therefore, the applied voltage is zero volts, and the input-voltage source ideally appears as a short circuit.

Hence, the capacitor starts to discharge through the resistor R and completely discharges to zero volts if time permits.

The charge Eq. 5 describes the capacitor-charge curve.

ec = E(1 - e^-t/RC) Charge Eq. [5]

in which

ec is the instantaneous value of voltage across the capacitor at a specific time t (V).

E is the value of voltage applied to the circuit (V). e is a constant 2.718 (base of the Napierian logarithm).

t is the time the capacitor is allowed to charge (sec).

R is the value of resistance in the circuit through which the electrons must travel in order to charge the capacitor (12).

C is the value of capacitance in the circuit (F).

The discharge Eq. 6 describes the capacitor-discharge curve.

ec = Ee^-t/RC Discharge Eq. [6]

in which ec is the instantaneous value of voltage across the capacitor at a specific time t (V).

E is the value of voltage to which the capacitor has been charged and is the value of voltage from which it discharges (V).

t is the time the capacitor is allowed to discharge (sec).

e is the constant 2.718 (base of the Napierian logarithm).

R is the value of resistance in the circuit through which the electrons must travel in order to discharge the capacitor (Ω).

C is the value of capacitance in the circuit (F).

The capacitor-charge Eq. 5 indicates mathematically that the capacitor can only charge to the applied voltage in (∞) infinite time. Theoretically, the capacitor may never charge to the source voltage in a finite amount of time.

Practically, the capacitor is said to have charged to the applied voltage when it has charged to 99 percent of that voltage.

The capacitor-discharge Eq. 6 demonstrates the same principle.

Theoretically, the capacitor may never discharge completely. Practically, the capacitor is said to have completely discharged when 99 percent of its charge has been depleted.

2.4 TIME CONSTANT

The product RC appears in both the charge and discharge equations. It is defined as a time constant, symbolized r and expressed in the prime unit of time, seconds. This relationship is expressed as Eq. 7.

r = RC Time Constant Eq. [7]

The following formula manipulation demonstrates why the time constant (r) is expressed in the unit of seconds.

r = RC

Q = quantity of charge (coulomb)

Q = CE

I = current (ampere)

Therefore,

C = Q / E

r = RQ / E

But I = Q/t

Therefore, Q = It

Hence r = RIt / E

Dividing numerator and denominator by I,

r = (RtI/E) / (E/I) = Rt / (E/I)

But R = E/I

Hence r = Rt/R

r = t (sec)

As symbolized in the charge Eq. 5, when t equals t time, the capacitor will have charged to 63.1 percent of the applied voltage. The mathematical proof of this relationship follows:

Eq. 5

since

then

Expressed as a percentage,

ec = 63.1% of E in one time constant

As symbolized in the discharge Eq. 6, when t equals r time, the capacitor will have discharged down to 36.9 percent of its original voltage.

The mathematical proof of this relationship follows:

Eq.6

since

Expressed as a percentage,

ec = 36.9% of E in one time constant

Recall that a practical full charge of a capacitor is considered to be approximately 99 percent of the applied voltage. This constitutes five (5) time constants.

The mathematical proof follows:

Then

or,

expressed as a percentage,

Series RC circuits are classified according to the relationship between pulse duration and the length of the time constant. See Fig. 2-2.

Figure 2-2 RC Integrator

A short time constant circuit is one in which the pulse duration is ten or more times greater than the duration of one time constant.

A medium time constant circuit is one in which the pulse duration may be from one-tenth to ten times that of one time constant.

A long time constant circuit is one in which the pulse duration may be one-tenth or less than one-tenth of the time required for one time constant.

2.5 SAMPLE PROBLEM ANALYSIS

Refer to Fig. 2-3.

Figure 2-3 RC Integrator

This schematic illustrates a medium time constant RC integrator circuit.

From a given square-wave input voltage, predict the amplitude and waveshape of the resultant output voltage, and predict the amplitude and waveshape of the resultant output voltage which would be displayed by an oscilloscope.

Input voltage = 10V peak 500 Hz square wave

The following is the solution for the output voltage when t = 1 msec, as shown in Fig. 2-4.

Figure 2-4

Subproblem (may be solved by slide rule):

At the end of 1 msec, the capacitor will have charged to 3.93V, and at this time the +10V source will have been removed.

Hence, the capacitor will start to discharge from +3.93V toward 0V.

The capacitor is allowed to discharge for 1 msec. The solution for the output voltage, when t = 2 msec, as shown in Fig. 2-4, follows.

At the end of 2 msec, the input-pulse voltage of +10 V is reapplied for a duration of 1 msec.

Because the capacitor will have been charged to +2.38 V, the effective change in voltage applied across the capacitor will be ( + 10) — (+2.38) or +7.62 V.

Equation 5 must be modified before it may be used to solve for the voltage across the capacitor at the end of 3 msec.

This is necessary in order to compensate for the initial charge on the capacitor. Modified Eq. 5 becomes Eq. 8.

Charge withinitial charge Eq. [8]

in which

Eo = initial charge of capacitor (V)

The following is the solution for the output voltage when t = 3 msec, as shown in Fig. 2-4.

For the succeeding intervals, the methods for solving for the voltage across the capacitor are the same as those established for the intervals between 1 and 2 msec and between 2 and 3 msec.

The numerical value of the voltage across the capacitor for each millisecond of time is listed in Table I.

These values indicate the exponential settling of the voltage across the capacitor. The output-voltage waveform will remain constant from 14 msec in time to infinity.

If an oscilloscope is connected to the output terminals of the RC integrator circuit shown in Fig. 2-3, it will display and measure the peak-to-peak output voltage (ec = 2.46V) shown in Fig. 2-4.

The lengthy calculation for the solution of this sample problem should prepare one to understand why the simultaneous equations, which are used in an alternate method, are valid.

This alternate method of solution employs a simultaneous solution of the charge and discharge equations.

2.6 TWO-EQUATION METHOD OF ANALYSIS OF SAMPLE PROBLEM

Figure 2-5

The capacitor must discharge from the value of the final charge (EF) to the final initial charge (Ex). Hence the basic discharge Eq. 6 becomes

[6]

But

Hence

The capacitor must start its charge from the final initial charge value (Ei) and must charge to the final charge value (EF). Hence the basic charge Eq. 8 becomes

[8]

The expedience of the two-equation method is demonstrated by the following numerical solution to the sample problem in Fig. 2-3.

This value represents the peak-to-peak settled output voltage shown in Fig. 2-4.

If an oscilloscope were connected to the output of the circuit shown in Fig. 2-3, it would display the settled waveform shown in Fig. 2-4.

In general, the term integrated output refers to a long time constant circuit. See Fig. 2-2.

This circuit produces an approximation of a linear ramp voltage waveform.

The disadvantage of this method of producing a linear ramp sweep of voltage is that the peak-to-peak output voltage is extremely small and generally requires additional amplification for practical use.

The short time constant circuit is used to develop the relationship between the rise time of a square or rectangular voltage waveform and its upper 3 dB frequency.

This is accomplished by expressing rise time in terms of time constants and by using sine wave analysis to establish the upper 3 dB frequency which may then be expressed in terms of time constants.

Time constants, as a common denominator, may then be used to relate rise time to the upper 3 dB frequency of the output-voltage waveform.

2.7 RELATIONSHIP OF RISE TIME TO TIME CONSTANTS

The rise time of the practical square wave of voltage as well as the output of the short time constant integrator circuit may be determined by calculating the difference between the time required for the output voltage to reach 0.9E and the time required for the output voltage to reach 0.1E.

Each may be determined from the basic charge Eq. 5. See Fig. 2-6.

Hence it has been established that the rise time is directly proportional to the time constant of the circuit.

ESTABLISHMENT OF UPPER 3 dB FREQUENCY BY SINE-WAVE ANALYSIS EXPRESSED IN TERMS OF TIME CONSTANTS

There is a direct connection between the rise time of a resultant output-voltage waveform and its upper frequency limit.

This is due to the fact that the high-frequency harmonics of the sine wave determine the squareness of the resultant output-voltage waveform—a fact established in section The generation and analysis of square waves.

As in sine wave analysis, the 3 dB reference is employed to establish the practical limit of the upper frequency.

The upper 3 dB frequency f2 is that at which the output power has dropped to half of the mid-frequency value. Half power may also be expressed as 0.707 of the mid-frequency output voltage.

Refer to Fig. 2-6.

Figure 2-6 Low-Pass RC Circuit

The following derivation establishes the high frequency at which the resultant output voltage would drop to 0.707 of the mid-frequency value if a sine wave of voltage were applied to this circuit.

then

Hence

Since

Substituting the value for

Eq. 10

Solving Eq. 10 for RC,

Substituting this value back into Eq. 9,

Equation 11 may be used to establish

(1) The upper 3 dB frequency of a given amplifier.

This is a standard method for the establishment of the high-frequency response of a hi-fi amplifier. A square wave of voltage is applied to the input of an amplifier and the rise time of the resultant output-voltage waveform is measured.

The upper 3 dB frequency of the amplifier may be determined by using the value of the measured rise time in Eq. 11.

(2) The upper 3 dB frequency limit necessary for a proposed amplifier to reproduce a given rectangular voltage waveform.

To design an amplifier which will faithfully amplify a given rectangular voltage waveform, it is necessary to know its upper frequency limit.

The measured value of the rise time of the voltage waveform to be amplified is used in Eq. 11 to establish this frequency.

LABORATORY EXPERIMENT

RC INTEGRATOR—LINEAR WAVESHAPING CIRCUIT

OBJECT:

1. To analyze the RC integrator circuit and to prove that the practical circuitry verifies the theory

(a) Long time constant circuit

(b) Medium time constant circuit

MATERIALS:

1 Square-wave generator (20 Hz to 200 kHz)

1 Oscilloscope, dc time base; frequency response, dc to 450 kHz; vertical sensitivity, 100/mV/cm

1 Resistor substitution box (10 Ω to 10 MΩ, 1 W)

1 Capacitor substitution box (0.0001 to 0.22 µF, 400 V)

Figure 2-1X

PROCEDURE:

1. Long time constant circuit

(a) Connect the circuit shown in Fig. 2-1X.

R = 100 kΩ C = 0.01 µF

(b) Set the input square-wave voltage to 10 V peak and to a pulse duration of 100 µsec.

(d) Draw the input- and output-voltage waveforms on two respective graphs, on the same sheet of paper, with the same time base, and label completely.

(e) Calculate the output voltage, using the direct two-equation method to verify the measured output voltage.

2. Medium time constant circuit

(a) Repeat Steps (a) through (e) of Part 1.

Change the circuit values to

R = 100 kΩ

C = 0.001 µF

(b) Calculate the output voltage, using the charge and discharge equations, to verify the measured output voltage. Show all work. Include a graph which depicts the calculated points. (See sample Fig. 2-4.)

3. Short time constant circuit

(a) Connect the circuit shown in Fig. 2-1X.

R = 100 KΩ C = 0.0001 µF

(b) Set the input square-wave voltage to 10 V peak and to a pulse duration of 100 µsec.

(d) On graph paper, draw and label the input- and output-voltage waveforms.

(e) Measure the rise time tr and record it on the graph drawn in Step (d).

(f) Calculate the upper 3 dB frequency of the square wave.

QUESTIONS AND EXERCISES

1. Define an integrator circuit.

2. What is the characteristic waveshape of the output voltage of an integrator circuit? Make a sketch of this waveshape.

3. Determine the value of R for the circuit illustrated in Fig. 2-2X, in which the voltage across the capacitor is 30 V, 3 msec after the switch is closed.

Figure 2-2X

4. Refer to Fig. 2-3X.

Figure 2-3X

Determine how long it will take the output voltage to reach +5 V after the switch is closed if the capacitor has an initial charge as indicated.

5. In Fig. 2-4X, what will be the output voltage at the end of 12 msec if the input is a square wave with a 5 msec pulse width and an amplitude of 35 V?

Figure 2-4X

6. Explain the relationship between rise time and the upper 3 dB frequency of a rectangular voltage waveform.

7. Explain why a capacitor may never charge completely.

8. Explain the significance of the relationship between the time constant t and the pulse duration tp.

9. In Fig. 2-6, if tr = 2 µsec, what would be the upper 3 dB frequency limit of an amplifier which could reproduce this voltage waveform?

10. What would be the minimum rise time of a 1000 Hz square wave applied to a hi-fi amplifier if the upper 3 dB frequency of the amplifier were 20,000 Hz? (Express answer in microseconds.)

11. Define a short time constant circuit.

12. Using the circuit illustrated in Fig. 2-1X, if R = 27 kΩ and C = 0.003 µF, when a positive 5000 Hz, 15 V peak square wave is applied, what will be the amplitude of the settled output-voltage waveform?

13. Explain what is meant by a low-pass filter.

14. Explain why the rise time is directly proportional to the time constant of the circuit.

15. If the rise time of a rectangular pulse were 0.5 µsec, what would be the upper 3 dB frequency limit of an amplifier which could reproduce this voltage waveform?

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