Rc differentiator-linear waveshaping circuit

Rc differentiator-linear waveshaping circuit

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

The voltage which appears across the resistor is referred to as the differentiator output.

The voltage source is frequently a rectangular waveshape. Recall that, in contrast, an integrator circuit is one in which the output voltage is taken across the capacitor of a long time constant RC series circuit.

The basic RC series circuit is, therefore, common to both. This relationship is illustrated in Fig. 3-1.

Figure 3-1

3.1 DEFINITION OF DIFFERENTIATED OUTPUT

A differentiated output is one in which the output voltage is directly proportional to the slope of the input voltage. The slope is the rate of change of input voltage with reference to time.

Hence, any pulse circuit in which the output voltage is directly proportional to the rate of change of the input voltage is generally and loosely referred to as a differentiator circuit.

For the output voltage to be proportional to the rate of change of the input voltage, however, the time constant of the circuit must be short (r = 0.1tp or less).

Therefore, the short time constant circuit is the one which is technically referred to as the differentiator circuit.

Practically, though, any circuit of this description, independent of time constant, is loosely referred to as a differentiator circuit.

Refer to Fig. 3-2, which illustrates the typical "differentiated" output-voltage waveforms for short, medium, and long time constant circuits.

Figure 3-2

The differentiated circuit shown in Fig. 3-2 may also be referred to as a high-pass RC filter. A high-pass filter passes the high frequencies and attenuates the low frequencies, as the name implies. The characteristic of this circuit is evidenced by the tilt of the long time constant voltage waveform shown in Fig. 3-2, and it is further evidenced by the extreme case of tilt of the short time-constant voltage waveform shown in Fig. 3-2.

3.2 DESCRIPTION OF OPERATION

To understand the operation of this circuit, it is necessary to establish the relationship between time and voltage. Refer to Fig. 3-1.

At t0 time, the input voltage is 0 V, it is +10 V, and it is all values interjacent.

The input voltage immediately preceding t0 time is 0 V and the voltage immediately following t0 time is +10 V.

Therefore, it is necessary to be able to distinguish between the time immediately preceding t0 time (designated t-0 time) and the time immediately following t0 time (designated t+o time).

Note that the circuit in Fig. 3-3 is identical to the circuit in Fig. 2-3.

Figure 3-3

In this section, the emphasis is upon the differentiated output of this circuit, whereas in the sample problem in section Rc integrator-linear waveshaping circuit the emphasis is upon the integrated output.

Depending upon the component, across which the output is taken, this medium time constant circuit is either an integrator or a differentiator.

Hence, the results of the sample problem in section Rc integrator-linear waveshaping circuit are used to explain the differentiated output of this common circuit.

Refer to Fig. 3-3.

Note that at t0 time the voltage across the capacitor is 0 V.

Hence, all the voltage drop is forced to appear across resistor R. This is in accordance with Kirchhoff's voltage law, as explained in section Rc integrator-linear waveshaping circuit.

The input pulse of +10 V is applied for 1 msec. During this interval, the capacitor has had time to charge +3.94 V.

Again, in accordance with Kirchhoff's law, the voltage drop across the resistor must equal the difference between the applied + 10 V and the +3.94 V drop across the capacitor, or at t-1 time the voltage across the resistor is +6.06 V.

By t+1 time the input-voltage pulse has been removed; hence, the input voltage is 0 V, and the source appears as a short circuit.

The capacitor which has charged to 3.94 V appears as a voltage source directly across the resistor.

Note, in the schematic in Fig. 3-3, that the polarity of the charged capacitor is negative on the right-hand plate and positive on the left-hand plate.

Therefore, the voltage drop across the resistor becomes a negative voltage with reference to ground. Hence, the output voltage at t+1 time is —3.94 V.

The capacitor discharges from +3.94 to +2.39 V during the interval from 1 to 2 msec. During this interval, the input voltage is 0 V. Hence, the voltage across the resistor must equal the voltage across the capacitor.

The polarity of the voltage drop across the resistor, however, must be negative with reference to ground. The output voltage at t-2 time, therefore, must be —2.39 V.

At t+2 time, the input-voltage pulse of +10 V is reapplied. Recall that the capacitor is charged to 2.39 V at t+2 time and that a capacitor resembles a voltage source at any instant in time.

The voltage source and the charged capacitor (acting like a source) are in series-opposing.

Hence, the voltage across the resistor equals the difference between the two, and the polarity of the larger, or eR, is equal to +7.61 V at t+2 time.

During the interval between 2 and 3 msec, the capacitor charges from +2.39 V at t+2 time to +5.38 V at t-S time.

By t-3 time, the voltage across the resistor has decayed to an amount equal to the voltage change across the capacitor, or en is equal to +4.62 V at f_3 time.

This process is repeated until a settled output voltage is obtained. An oscilloscope connected to the output of the circuit shown in Fig. 3-3 will display the settled voltage waveform.

The capacitor discharge Eq. 6 is also valid as the equation for the exponential decay of the voltage across the resistor in a series RC circuit and may be modified for this use by substituting eR for ec.

[6]

where E — voltage across R at the beginning of a pulse interval (V).

In the following sample problem, Eq. 6 is used to solve for the differentiated output of the circuit shown in Fig. 3-3.

From t1 to t2 time, the output across R decays from —3.94 V toward 0 V. Hence: er at t-2 time

At t+2 time the input pulse is reapplied.

This process is repeated until a settled output voltage is obtained. The values of voltages are listed in Table II, as well as in Fig. 3-3.

3.3 DIRECT SOLUTION FOR SETTLED WAVEFORM

The application of four simultaneous equations to this problem provides a direct solution for the settled differentiated output. In Fig. 3-4, the settled

Figure 3-4

output is indicated as the general case. If the value of E1 is known, E2 may be expressed in terms of E1 ; similarly, if the value of E3 is known, the value of E4 may be expressed in terms of E3.

Hence, Eq. 6 may express these two relationships.

The total voltage at t1 and at t2 times, indicated in Fig. 3-4, must equal the applied voltage of the circuit E.

Hence:

The numerical value for E1, E2, E3, and E4 may be obtained from the simultaneous solution of Eqs.

(A), (B), (C), and (D).

A mathematical expression for E4 in terms of E2 may be obtained by solving Eq. (C) for E1 and substituting the resultant value for E1 in Eq. (A).

Hence:

A mathematical expression for E4 in terms of E2 may be obtained by solving Eq. (D) for Es and substituting the resultant value for Es in Eq. (B).

Hence:

Because Eqs. (E) and (F) contain the common unknowns E2 and E4 they may be solved simultaneously to establish the numerical values of E2 and E4.

This simultaneous solution method is subsequently used to solve the same sample problem which was previously solved by the incremental method.

Multiplying Eq. (E) by 0.606,

Substituting the value of E4 in Eq. (E),

Substituting the values of E2 and E4 in Eqs. (C) and (D), respectively,

Compare with values shown in Fig. 3-3.

The determination of the output voltage of a differentiator circuit has limited practical value. A more comprehensive knowledge of the differentiator circuit, however, may be gained by a clear understanding of the process involved in predicting the shape and magnitude of the output-voltage waveform.

In general, the term differentiated output refers to a short time constant circuit.

See Fig. 3-2. This typical short time constant output-voltage waveform is utilized in the timing and triggering of multivibrator circuits. These applications of this circuit are discussed in subsequent sections.

3.4 TILT TIME

Before the long time constant "differentiator" circuit may be understood, the term tilt time, tt, must be defined. Refer to Fig. 3-5.

In a long time constant differentiated output-voltage waveform, the exponential decay of the voltage is linear for all practical purposes.

Hence, if this "linear" decay is extrapolated to the zero voltage axis, the time from the application of the pulse to that at which the linear extrapolation reaches 0 V is called tilt time, symbolized tt, as shown in Fig. 3-5.

The fractional tilt, symbolized P, may be expressed in terms of the pulse width tp and tilt time tt, as shown in Eq. 12.

P = tp / tt [12]

where

P = fractional tilt (dimensionless)
tp = pulse width (sec)
tt = tilt time (sec)

Equation 12 is not practical for the determination of the fractional tilt P because the measurement of tilt time tt is infeasible.

The fractional tilt P may also be expressed however, in terms of easily measured voltages (E1 — E2) and V, as shown in Eq. 13.

[13]

Figure 3-5 High-Pass RC Circuit

where

P = fractional tilt (dimensionless)

E1 — maximum voltage excursion (V)

E2 = decayed voltage excursion at tp time (V)

V = maximum voltage change at pulse transition time (V)

Once the value of the fractional tilt P has been determined from Eq, 13, it may be inserted in Eq. 12 and tilt time tt may be established.

3.5 RELATION OF TILT TIME TO TIME CONSTANTS

The lower 3 dB frequency of a square wave may be determined by evolving its relation to tilt time. This is accomplished by expressing tilt time and the lower 3 dB frequency in terms of time constants.

Thus, the common denominator of time constants may be used to express the relation of tilt time to the lower 3 dB frequency.

Refer to the circuit in Fig. 3-5. If a sine wave of voltage were applied to this circuit, the low frequency at which the resultant output voltage would drop to 0.707 of the mid-frequency value could be obtained by the following derivation:

Dividing numerator and denominator by jωRC,

where f1 is the lower 3 dB frequency.

then

hence

This will occur when (lower 3 dB frequency)

when

but

for square wave

hence

where

P = fractional tilt

f1 = lower 3 dB frequency (Hz)

f = frequency of applied square wave (Hz)

[15]

but

hence

[16]

where

t1 = tilt time (sec)

f1 = lower 3 dB frequency (Hz)

3.6 SAMPLE PROBLEM

If the input voltage to the circuit in Fig. 3-5 were a 5000 Hz square wave and the resultant output-voltage waveform measured by an oscilloscope were V = 18 V and (Ex - E2) = 2 V, what would be the lower 3 dB frequency of the resultant output-voltage waveform?

Solution:

In section Rc integrator-linear waveshaping circuit , the formula used to express the relation of frequency to rise time was derived.

In this section the formula used to express the relation of tilt time to frequency has been derived.

These two formulas are extremely effective tools for the establishment of the upper and/or lower 3 dB frequency limits of a square wave or of an amplifier.

Together, the technician may use them to establish the frequency response of an amplifier. The validity and effectiveness of these tools should become obvious with the careful development and derivation of these waveform and frequency relationships.

LABORATORY EXPERIMENT

RC DIFFERENTIATOR—LINEAR WAVESHAPING CIRCUIT

OBJECT:

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

(a) Short 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 Oscilloscope probe—10X attenuator, 10 Mfi 1 Resistor substitution box (10Ω to 10 MΩ, 1 W)

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

Figure 3-1X

PROCEDURE:

1. Short time constant circuit

(a) Connect the circuit shown in Fig. 3-1X.

R = 100 kΩ C = 0.0001 µF

(b) Set the input voltage to 20 V peak and to a pulse width of 100 µsec.

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

(e) Explain why the amplitude of the output-voltage waveform is small.

2. Medium time constant circuit

(a) Connect the circuit shown in Fig. 3-1X.

R = 100 kΩ C = 0.001 uF

(b) Determine the shape and amplitude of the input- and output-voltage waveform by use of an oscilloscope.

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

(d) Prove the results of your experiment by using the point-by-point method shown in the sample problem in Fig. 3-3.

(e) To check the results of your point-by-point method calculations and to double-check the results of your experiment, directly determine the settled output voltages by the four-equation method.

(f) Explain the reasons for any differences between your experimental circuit results and the results of the point-by-point and/or four-equation method.

3. Long time constant circuit—3 dB low-frequency response of a square wave
(a) Connect the circuit shown in Fig. 3-1X.

R = 100 kΩ C = 0.01 µF

(b) Determine the shape and amplitude of the input- and output-voltage waveform by use of an oscilloscope.

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

(d) Using the four-equation method, determine the settled output voltages, and compare with the experimental results.

(e) Determine the fractional tilt of the resultant output-voltage waveform by use of an oscilloscope.

(f) Determine the lower 3 dB frequency of the resultant output-voltage "square wave."

QUESTIONS AND EXERCISES

1. Define a differentiator output.

2. What is the characteristic voltage waveshape of a differentiator output? Give an example.

3. Explain why the voltage amplitude of a differentiator output may be larger than the voltage amplitude of the input.

4. Refer to Fig. 3-2X. Determine how long it will take the output voltage to reach 20 V after the switch is closed.

Figure 3-2X

Figure 3-3X

5. Refer to Fig. 3-3X. Determine the output voltages at the various times indicated.

(a) What is the value of eo at t+0 time?

(b) What is the value of eo at t-1 time?

(d) What is the value of eo at t-2 time?

(e) What is the value of eo at t+2 time?

6. Define tilt.

7. Why is it necessary to establish the value of the pulse width in order to determine the lower 3 dB frequency of a square wave, even if tp does not appear in Eq. 16?

8. Explain why sine wave analysis is a valid method for the determination of the lower 3 dB frequency of a square wave.

9. If a 1000 Hz square wave is applied to an amplifier with a fractional tilt of 0.4 at its output, what is the lower 3 dB frequency of the amplifier?

10. Refer to Eq. 13. Verify the fact that the lower 3 dB frequency of the amplifier is 500 Hz (as specified by the manufacturer), by establishing the value of (E1 - E2) from the output-voltage waveform. The input is a 15 V peak, 10,000 Hz square wave.

11. Define a long time constant circuit.

12. Explain what is meant by a high-pass filter.

13. Refer to Fig. 3-4. When ein = +15 V, 10 kHz square wave, R = 68 kΩ, and C = 0.002 µF, what will be the amplitudes of e0 (Eh E2, E3, and #4) as seen on the face of an oscilloscope?

14. Explain how the frequency response of an audio amplifier may be determined by the use of square waves.

15. Refer to Fig. 3-4X. How long after the switch S is closed will it take the output voltage eo to reach +20 V?

Figure 3-4X

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