PHYS 536 RC Circuits Introduction

PHYS 536 RC Circuits

Introduction

The response of an electronic circuit depends on the frequency of the input signal. Although there are many variations, two circuits are primarily responsible for amplitude change as a function of signal frequency, one for high frequencies and another for low frequencies. The ﬁlter circuit that attenuates low frequencies is called a High Pass Filter and the circuit that attenuates high frequencies is called a Los Pass Filter. The frequency response (gain) and phase as a function of frequency, or ω, can be viewed graphically in a Bode plot. A typical Bode plot plots gain on the ordinate and frequency on the abscissa. Both axes are generally logarithmic. Likewise the phase angle can be plotted on the y-axis (linear) and the frequency on the x-axis.

High Pass RC Filters

The purpose of a high pass ﬁlter is to pass high frequency signals. Recall that the impedance of a capacitor is given by

1 ZC = (1) ωC At low frequencies, the impedance of a capacitor is large, and as frequency increases, the impedance will become increasingly small. If a capacitor is placed in series with a resistor, at low frequencies most of the voltage will be dropped across the capacitor, while the voltage across the resistor will be small. At high frequencies, the voltage drop across the resistor will be large. So, if the voltage across the resistor is used as an output voltage, as shown in Fig. 1, the RC Vo circuit will function as a high pass ﬁlter. The gain, Vi , is given in Fig. 2. As you can see, the gain is low at low frequencies, and high at high frequencies. The gain for this circuit can be written

vo 1 g = = vi 2 fb 1+ f where the break frequency, fb, is given by Eq. 5. At the extremes of frequency, the gain is given by

1 ⎧ f ⎪ ,f fb ⎨ fb 1 g(f)= √ ,f= fb (2) ⎩⎪ 2 1,f fb

The phase angle of the output voltage is given by fb φ =tan−1 (3) f

◦ ◦ ◦ This shift is 90 when f fb,45 when f = fb,and0 when f fb.

Figure 1: RC Circuit functioning as a high pass circuit

Figure 2: High pass ﬁlter gain

2 Break Frequency

When considering RC ﬁlters, it is useful to be able to characterize their frequency response. One typical way to do this is to specify the break frequency, also called the corner frequency. Looking at Fig. 2, it can be seen that there are two regimes for the circuit, that when f fb and when f fb, which produce two asymptotes. When f fb, the gain is essentially constant, while when f fb, the gain increases linearly with frequency. The frequency at which these asymptotes meet is called the break frequency. For example, for the high pass circuit shown in Fig. 1, the gain is given by

1 g = (4) 2 1 1+ 2πfRC

At high frequencies,

g ≈ 1

At low frequencies,

g ≈ 2πfRC

Equating the two asymptotes,

1=2πfbRC (5)

The break frequency for the high pass is thus

1 fb = (6) 2πRC

LowPassRCFilter

To obtain a low pass RC ﬁlter, the roles of the capacitor and resistor are reversed from Fig. 1, so that the output voltage is now obtained from the capacitor, as shown in Fig. 3. Since the voltage drop across the capacitor is large at low frequencies and small at high frequencies, this circuit is a low pass ﬁlter. A graph of the gain as a function of frequency can be seen in Fig. 4 This circuit’s gain is given by

1 g = (7) 2 f 1+ fb

3 At the extremes of frequency, the gain is given by ⎧ ⎨⎪1,f fb 1 √ ,f= fb g(f)=⎪ 2 (8) ⎩ fb f ,f fb

And the phase shift is given by f φ = − tan−1 (9) fb

Figure 3: RC Circuit functioning as a low pass circuit

Figure 4: Low pass ﬁlter gain

4 Opposite Break Circuit

Figure 5: Opposite break circuit

Figure 6: Gain for the opposite break circuit

An opposite break circuit is a circuit with two break frequencies. It attenuates at high frequency but not at low frequency as shown in Fig. 6. It is used to prevent oscillation in negative feedback circuits. The break frequencies are given by

1 f1 = (10) 2πR2 (C1 + C2) and

5 1 f2 = (11) 2πR2C2

The signal is not aﬀected when its frequency is below the ﬁrst break, f1.The gain above, f2 is given by

C2 g = (12) C1 + C2

In general, the gain is given by 2 f 1+ f2 g = (13) 2 f 1+ f1

Using Eq. 13, it can be shown that

vi vo(f = f1)=√ (14) 2 and

√ C2 vo(f = f2)=vi 2 (15) C1 + C2

This circuit is used to reduce the amplitude of a signal at high frequency without causing high phase shift. The maximum phase shift occurs at f = f1f2 (16) and is given by 1 f1 1 f2 tan φmax = − (17) 2 f2 2 f1

Clipping

The pulse response of many common electronic circuits can be understood using two principles:

−t 1. The time dependent term in all RC changing problems has the form e RC

6 2. The voltage across a capacitor cannot change quickly when there is a resistor that restricts the ﬂow of charge to the capacitor. These two principles are helpful in understanding the circuit shown in Figs. 7 and 8.

Figure 7: Clipping circuit

Figure 8: Integrating circuit

A step pulse is shortened by a clipping circuit because

− t vo = vie τ (18) where

τ = RC (19)

Integrating Circuit

The circuit shown in Fig. 8 can be used to delay the peak of a square wave signal. The time required for a pulse to reach its maximum value is indicated by the time constant. This simple circuit is employed in a variety of diﬀerent applications. The output voltage is given by −t vo = vi 1 − e τ (20)

7 1 High and Low Break Circuit

Many practical circuits, such as the one in Fig. 9 have a high-break and a low- break. The general pattern can be illustrated with the preceding circuit. The two time constants have been selected so that the breaks are far apart. Also, R1 R2, so the gain between the breaks is minor. The two break frequencies are given by

1 fl = (21) 2π (R1 + R2) C2

1 fh = (22) 2π (R1 R2) C1

And the expected gain by

1 g = (23) 2 f f 1+ − l fh f

Figure 9: High and low break circuit

1. Calculate the two break frequencies using Eqs. 21 and 22 and the compo- nent values shown in Fig. 9. 2. Measure the two break frequencies by ﬁnding the points where g = 0.7. Although any amplitude input signal can be used a good recommended signal is 10 V (p-p). 3. The measured break points may be diﬀerent than those calculated. Why? f o i 4. Measure the gain between v and v at fb = 0.1, 0.5, 1,2, and 10 for both the high and low break frequencies. Record the values and then enter the data in a spread sheet program. Do not measure the actual phase shift at each frequency, but make note of the general behavior of the phase shift as frequency changes.

8 5. Plot the gain as a function of frequency using logarithmic axes, i.e. make the Bode plot. Before you leave this section you could plot some values to verify that the measurements are correct. Add the expected gain. Describe how the phase changed as a function of frequency.

2 Common Variations

In this section, you will explore common variations of RC ﬁlters. For these circuits,

1 fb = (24) 2πτ where τ is the RC time constant for the circuit. For Fig. 10,

R2 1 g = R1 + R2 2 fb 1+ f

τ =(R1 + R2) C

For Fig. 11,

R2 1 g = 2 R1 + R2 f 1+ fb

τ =(R1 R2) C

For Fig. 12

C1 1 g = C1 + C2 2 fb 1+ f

τ =(C1 + C2) R 1. For the circuit in Fig. 10, do the following: (a) Calculate the break frequency using Eq. 24.

9 (b) Experimentally determine the coeﬃcient in front of the square root term in the gain equation by measuring the gain in the region where the square root term is approximately 1. For example, for the circuit in Fig. 10, this would be the region where f fb. (c) Calculate the expected gain at the break frequency by multiply the √1 coeﬃcient determined in the previous step by 2 =0.707, the value of the square root term at f = fb. Record the expected gain and include it in your lab report. (d) Measure the break frequency by determining the point at which the gain matches the expected value calculated in the previous step.

(e) Measure the gain at f =0.1fb and f =10fb.

(f) Calculate the expected gain at f =0.1fb,fb, and 10fb using the theoretical equation for gain. (g) Describe the eﬀect of the circuit on the input signal. 2. Repeat the measurements and calculations completed in step 1 for the circuit in Fig. 11. 3. Repeat the measurements and calculations completed in step 1 for the circuit in Fig. 12.

Figure 10: RC ﬁlter variation 1

3 Opposite Break Circuit

This circuit is used to attenuate a signal at high frequency without causing large phase shift. 1. Calculate the two break frequencies for the circuit shown in Fig. 13 using Eqs. 10 and 11.

2. Measure the gain of this circuit for f = 0.1f1, f1, f2 and 10f2.

10 Figure 11: RC ﬁlter variation 2

Figure 12: RC ﬁlter variation 3

3. Calculate the maximum phase shift and the frequency at which it occurs using Eqs. 17 and 16. 4. Measure maximum phase shift and the frequency at which it occurs.

4 Basic Pulse Circuits

In this section the change in a square wave produced by clipping and integration will be investigated.

1. Observe vo using the oscilloscope for each circuit shown in Fig. 14 and measure the time constant. Sketch the observed waveform and include this in your lab report.

Required Components

• Resistors: (1) 100 Ω, (1) 1 kΩ, (2) 2 kΩ, (1) 24 kΩ, (1) 100 kΩ • Capacitors: (1) 100 pF, (2) 1 nF, (1) 10 nF

11 Figure 13: Opposite break circuit

Figure 14: Integrating and clipping circuits