TRANSIENTS AND RC TIME CONSTANTS The capacitor has a wide range of applications in electronic circuits, some of which are energy storage, dc blocking, filtering, and timing. Thus, it is important for engineering students to understand capacitor operation. This experiment is designed to familiarize the student with the simple transient response of two-element RC circuits, and the various methods for measuring and displaying these responses. Theory RC Transients In normal operation, a capacitor charges part of the time and discharges at other times. These terms are described below in two parts: first, capacitor charging, and then discharging. Consider the circuit shown in Fig. 9.1. For t < 0, switch S2 is closed and S1 is open. At t = 0, switch S1 is closed and S2 is opened. If we assume that the circuit was in a dc steady state before the switching occurred, there was no energy stored on the capacitor and therefore vo(0) = 0.

Fig. 9.1 Series RC circuit

Applying KCL at the upper capacitor node (for t > 0) yields C

∂vo vo − Vs + = 0, ∂t R

or ∂vo v V + o = s , ∂t RC RC The solution is

vo (t ) = Vs (1 − e

−t

RC

)

, t>0

(9.1)

This is the equations describing the capacitor voltage while charging. A graph of the capacitor voltage is shown in Fig. 9.2. The voltage begins at zero, rises rapidly, and approaches Vs as t → ∞. For all practical purposes, vo (t) = Vs when t = 5(RC). In Eq. (9.1), RC is the time constant of the circuit. The quantity symbol for the time constant is τ. Thus, for this RC circuit, τ = RC.

Fig. 9.2 Increasing exponential for the charging case Now consider the discharging process. Suppose that the capacitor voltage has reached Vs ( t >> τ). Then, at t = t1, switch S1 is suddenly opened and S2 closed. KCL applied at the upper capacitor node (for t > t1) yields ∂v o vo + =0 ∂t RC the solution to which is

vo (t ) = Vs e

−

( t − t1 ) RC

, t > t1

(9.2)

Equation (9.2) describes the capacitor voltage while discharging. A graph of Eq. (9.2) for (t - t1) > 0 is shown in Fig. 9.3. The voltage begins at Vs, decays exponentially, and approaches zero asymptotically as t → ∞. Again, vo(t) is considered to be zero after 5 τ, in which τ = RC.

Fig. 9.3 Decaying exponential for the discharging case. Suppose that the capacitor had a nonzero initial value or was not allowed to charge or discharge fully, as shown in Fig. 9.4. In this case, for a single time-constant RC circuit with initial and final values Vi and Vf, the solution can be written as −

t

vo (t ) = V f + (Vi − V f )e τ ,

(9.3)

in which vo(t) is the capacitor voltage and Vi and Vf are the initial and final values, as shown in Figure

9.4.

Fig. 9.4 Typical capacitor voltage waveforms for (a) charging and (b) discharging.

Equation (9.3) applies also to the circuit shown in Fig. 9.5, in which the output voltage vo(t) is taken across the resistor. Vi and Vf are resistor voltages.

Fig. 9.5 Another form of series RC circuit. Prelab Question 1: A 100 µF capacitor is connected in series with a 10 kΩ resistor and a 10-V DC source for 2 seconds. Then it is disconnected quickly and connected to a single resistor of 1 kΩ.

(a)

Write the capacitor voltage, vc(t), for 0 < t < 2 s.

(b)

Write the capacitor voltage, vc(t), for t > 0, where t = t - 2.

(c)

How much voltage is on the capacitor at t = 17 s ?

Time Constants A glance at Eqs. (9.1) and (9.2) reveals that RC, or τ, has the units of seconds. Also, the time constant is the time required for a charging or discharging quantity to come within 1/e of its final value. Equivalently, it is the value of time for which the exponent in Eq. (9.3) has the value -1. This occurs when t = τ. When the exponent equals –2, two time constants have elapsed, and so on. At one time constant, an evaluation of e-1 shows that the charging curve has risen to 63.2% of the maximum amplitude, as shown in Fig. 9.6(a). Also, at one time constant, the capacitor voltage as shown in Fig. 9.6(b), has decreased to 36.8% of its initial amplitude. Figure 9.6 Exponential curves for (a) charging and (b) discharging.

As previously stated, the curves have practically reached their final values after 5τ, the charging curve has risen to 99.3% of its final value, and the discharging curve has decreased to 0.67% of its initial value. If a curve of some unknown function of time is known to be exponential, a graphical technique may be employed to determine the time constant, thus enabling one to write the mathematical description of the curve. For the discharging case, consider the curve shown in Fig.9.7. A tangent line is drawn at any arbitrary point y(to) on the curve. The intercept of the tangent line with the time axis yields a second point, tb. If the time axis units are known, the time constant can be determined from τ = tb - ta.

Figure 9.7 Graphical determination of the time constant τ. For a charging, or increasing, exponential curve, the tangent line intercepts the asymptote to the curve. Projecting this intercept to the time axis establishes the time interval (ta - tb) and hence τ.

RC Circuit Response to a Periodic Step-Voltage Excitation With its inertia-less electron beam, the oscilloscope is particularly adapted for the display of voltage waveforms that are repetitive. The oscilloscope can continuously display some portion of a periodic input waveform. A transient waveform, however, occurs only once, and is therefore not repetitive. It can be displayed conveniently only on an oscilloscope with memory. For the oscilloscopes in the EEL 3303L laboratory, it is necessary to apply a repetitive “step” voltage to the input of the RC circuit to display the transient response of the circuit. A good approximation of the transient response may be obtained using a square-wave excitation since it is periodic and may be regarded as a series of positive and negative step voltages. For a periodic square wave with a reasonably long half-period (T/2 > 5 τ), the exponential growth and decay during a single half-period of the square wave will be practically complete. Thus, the oscilloscope display of a periodic step voltage will appear very similar to that of a single step input to the RC circuit, as is shown in Fig. 9.8.

Figure 9.8 Series RC circuit response to a “zero-centered” periodic step voltage input.

Simple RC Series Integration and Differentiation Circuits Consider the circuit shown in Fig. 9.9(a). By KCL,

C

∂vo vo − vin + = 0, ∂t R

If the time constant τ = RC is large compared to T/2 so that vo is much less than vin during the charging time of T/2, then ∂vo v ≅ in , RC ∂t or vo ≅

1 vin ∂t , RC ∫

Thus, the circuit is approximately an integrating circuit. Since the integral of a constant is a ramp (straight line), the integration effect is that of the straight lines shown in Fig. 9.9(b).

Fig. 9.9 (a) RC integrating circuit with (b) input and output waveforms.

Within limits, for the circuit shown in Fig. 9.10(a), the output voltage vo is approximately equal to the derivative of the input voltage vin. For this operation the time constant τ = RC must be small compared to half-period T/2. Then the circuit transforms a rectangular wave into a series of short pulses, as shown in Fig. 9.10(b). Mathematically, vo ≅ RC

∂vin , ∂t

Fig. 9.10 (a) RC differentiating circuit with (b) input and output waveforms.

Fig. 9.11 Circuit for transient response.

Transient Response Prelab Question 2: For the circuit shown in Fig. 9.11, R = 1 MΩ, C = 1 nF, and vin is a periodic square-wave voltage of 10 V, peak-to-peak. Calculate τ (Don’t forget the 1-MΩ oscilloscope input resistance), and the frequency necessary for a reasonable display of the charging and discharging transients. (Hint: Use Thevenin’s equivalent circuit to calculate τ.)

Procedure: Connect the circuit as shown in Fig. 9.11 and adjust the input frequency to the calculated value. Now, adjust the sweep of the oscilloscope to display 1) the growing exponential, then 2) the decaying exponential. Graphically determine the time constant. Compare it with the calculated value. For the function generator, select the Large Amplitude Waveforms tab and use the output on DAC0.

Differentiating Circuit 1. Connect the circuit shown in Figure 9.10 (a) for R = 1kΩ and C = 1 nF. Apply a square wave having a peak voltage of 5 V and a frequency of 6 kHz. 2. Examine the output voltage relative to the source voltage. Repeat for C = 10 nF, C = 0.1 µF, and C = 1 µF.

Integrating Circuit 1. Connect the circuit shown in Figure 9.11 and with C = 1 nF, examine the output relative to the source for R = 1 kΩ, R = 120 kΩ, and R = 1 MΩ. 2. For the same circuit with R = 1 kΩ and C = 0.1 µF, examine the output trace relative to the source for 10 Hz, f = 100 Hz, f = 1 kHz, f = 6 kHz.