AC Circuit Analysis Lab Experiment - Electrical and Electronic Circuits Laboratory, Lab Reports of Electrical Circuit Analysis

Download AC Circuit Analysis Lab Experiment - Electrical and Electronic Circuits Laboratory and more Lab Reports Electrical Circuit Analysis in PDF only on Docsity! CIRCUITS LABORATORY EXPERIMENT 3 AC Circuit Analysis 3.1 Introduction The steady-state behavior of circuits energized by sinusoidal sources is an important area of study for several reasons. First, the generation, transmission, distribution, and consumption of electric energy occur under essentially sinusoidal steady-state conditions. Second, an understanding of sinusoidal behavior makes possible the prediction of circuit behavior when nonsinusoidal sources are used through the use of techniques such as Fourier analysis and superposition. Finally, by specifying the performance of a circuit in terms of its steady-state sinusoidal behavior, the design of the circuit can often be simplified. Needless to say, the importance of sinusoidal steady-state behavior cannot be overemphasized, and many of the topics in future experiments are based on a thorough understanding of the techniques used to analyze circuits driven by sinusoidal sources. In this experiment, the behavior of several types of circuits will be examined to determine their behavior when excited by sinusoidal sources. First, the behavior of both RC and RLC circuits will be examined when driven by a sinusoidal source at a 3-1 given frequency. Subsequently, the frequency response of both a low-pass filter and a high-pass filter will be considered. 3.2 Objectives At the end of this experiment, the student will be able to: (1) Determine the steady-state behavior of linear circuits driven by sinusoidal sources, (2) Use the oscilloscope to measure the phase difference between two sinusoidal signals, (3) Determine analytically the frequency response of a network, (4) Construct Bode plots relating the magnitude and phase response of the voltage ratio of a linear network as a function of frequency, and (5) Design primitive low- and high-pass filters using one resistor and one capacitor. 3.3 Theory 3.3.1 Sinusoidal Steady-State Analysis As stated previously, the steady-state behavior of circuits that are energized by sin- usoidal sources is an important area of study. A sinusoidal voltage source produces a voltage that varies sinusoidally with time. Using the cosine function, we can write a sinusoidally varying voltage as follows v(t) = Vm cos (ωt + φ ) (3.1a) and the corresponding sinusoidally varying current as i(t) = Im cos (ωt + θ). (3.1b) 3 - 2 φj meVV = φφ sincos mm jVVV += It follows then the problem of finding the steady-state response is reduced to finding the maximum amplitude and phase angle of the response signal since the waveform and frequency of the response are the same as the source signal. In analyzing the steady-state response of a linear network, we use the phasor, which is a complex number that carries the amplitude and phase angle information of a sinusoidal function. Given the sinusoidal voltage defined previously in Equation (3.1a), we define the phasor representation as (3.4a) where V, is the maximum amplitude i.e., the peak voltage and φ is the phase angle of the sinusoidal function. Equation (3.4) is the polar form of a phasor. A phasor can also be expressed in rectangular form. Thus, Equation (3.4a) can be written as (3.4b) In an analogous manner, the phasor representation of the sinusoidal current defined in Equation (3.1b) is defined as I = Im ejθ = Im cos θ + jIm sin θ. (3.5) We will find both the polar form and rectangular form useful in circuit applications of the phasor concept. Also, the frequent occurrence of the exponential function ejθ has led to a shorthand notation called the angle notation such that Vm /φ ≡ Vm e jφ (3.6a) and Im /θ ≡ Im e jθ (3.6b) 3-5 Thus far, we have shown how to move from a sinusoidal function (i.e., the time domain) to its phasor transform (i.e., the complex domain). It should be apparent that we can reverse the process. Thus, given the phasor, we can write the expression for the sinusoidal function. As an example, if we are given the phasor V = 5/36o , the expression for v(t) in the time domain is v(t) = 5 cos(ωt + 36o), since we are using the cosine function for our sinusoidal reference function. Now, the systematic application of the phasor transform in circuit analysis re- quires that we introduce the concept of impedance. In general, we find that the voltage-current relationship of a linear circuit element is given by V = Z I (3.7) where Z is the impedance of the circuit element. The impedance of each of the three linear circuit elements is given as follows: (a) The impedance (ZR) of a resistor is R in rectangular form and R / 0o in angle form. From Equation (3.7), we see that the phasor voltage at the terminals of a resistor equals R times the phasor current. The phasor-domain equivalent circuit for the resistor circuit is shown in Figure 3.2 (a). If it is assumed that the current through the resistor is given by the phasor I = Im / θ , then the voltage across the resistor is V = R I = (R / 0o )(Im /θ ) = R Im /θ (3.8) from which we observe that there is no phase shift between the current and voltage. 3 - 6 Figure 3.2. Phasor Domain Equivalent Circuits (b) The impedance (ZL) of an inductor is jωL in rectangular form and ωL/ 90o in angle form. Again, from Equation (3.7), we see that the phasor voltage at the terminals of an inductor equals jωL times the phasor current. The phasor-domain equivalent circuit for the inductor is shown in Figure 3.2(b). If we assume I = Im /θ , then the voltage is given by V = (jωL) I = (ωL / 90o )(Im /θ ) = ωLIm / θ + 90o (3.9) from which we observe that the voltage and current will be out of phase by exactly 90o. In particular, the voltage will lead the current by 90o or, what is equivalent, the current will lag behind the voltage by 90o. (c) The impedance (ZC) of a capacitor is l/jωC (or -j/ωC) in rectangular form and 1/ωC/ -90o in angle form. Equation (3.7) indicates that the phasor voltage at the terminals of a capacitor equals l/jωC times the phasor current. The phasor-domain equivalent circuit for the capacitor is shown in Figure 3.2(c). 3 - 7 ZC = 1/jωC = -j/ωCZL = jωLZR = R 22 )7.156()100( +=TZ Figure 3.4: Phasor Domain Equivalent Circuit since the resistor, inductor, and capacitor are connected in series. Substituting values for R, ZL, and ZC gives ZT = 100 + j 188.5 - j 31.8 = 100 + j 156.7. (3.15) Converting this to angle form, we have / tan-1(156.7/100) , (3.16) which simplifies to ZT = 185.9/ + 57.5o . (3.17) Thus, the current is equal to I = VS / ZT = (5 / 0o )/(185.9 / 57.5o ) = 0.0269 / -57.5° A. (3.18) and the voltage across the capacitor is VC = ZC I = (31.8 / -90o )(0.0269 / -57.5o ) = 0.855 / -147.5o V. (3.19) 3-10 We can now write the stead-state expressions for I(t) and vC(t) directly: i(t) = 26.9 cos (6283t - 57.5o ) mA (3.20) and vC(t) = 0.855 cos (6283t - 147.5o ) V. (3.21) 3.3.2 Frequency Response of AC Circuits: Bode Diagrams As indicated above, network connections of linear circuit elements used to transmit signals generally result in an output signal with an amplitude and time dependence that differs from that of the input signal. If the input signal is sinusoidal, so too will be the output for a linear network. However, usually both the amplitude ratio and the phase difference between the input and the output signals will depend on the frequency. A particularly convenient graphical representation is obtained if logarithmic scales are used for both the frequency and the amplitude ratio while a linear scale is used for the phase difference. Also, extremely useful approximate graphical representations can be obtained for many networks that result in a series of straight lines on these graphs. Both representations are called Bode Diagrams. In this part of the experiment, the measured response of two networks will be compared to that theoretically predicted and the straight line approximations. In addition, decibel notation will be introduced. The frequency response of the elementary circuit shown in Figure 3.5, where vi(t) corresponds to the input signal and vo(t) to the output signal, will initially be determined. Sinusoidal signals will be assumed. In analyzing this circuit, the 3 - 11 ( ) , 1)( Cj R V Z V I ii ω ω ω + == . 11 1 RCj V Cj R V Cj IZV iiCo ω ω ω + = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == impedance across vi is a function of the operating frequency and can be expressed as Z(ω) = R + 1/jωC . (3.22) Using Ohm's law in the phasor domain, the current in the circuit is (3.23) and the output voltage, which is equal to the voltage across the capacitor, is (3.24) Examination of Equation (3.24) above indicates that for low frequencies (that is, for ω 0 ), the output voltage is approximately equal to the input voltage (i.e. V0 Vi). Also, for high frequencies (that is for ω ∞), the output voltage becomes negligibly small (i.e. V0 0). Since low frequency signals are virtually unaffected, a network of this type is called a low-pass filter. It should be noted that many complex circuits comprised of active elements, such as transistors, may be reduced to an equivalent circuit of this type. The capacitance, C, is often the 3 - 12 R .1 1 1|)(| =≈ωA .707.0 2 1 11 1|)(| 1 ==+ =ωA We can obtain values for |A(ω)| and θ(ω) for the low-pass filter using Equa- tion (3.29) and Equation (3.30). For ω << ωI , we have from Equation (3.29), (3.32) We can determine the phase angle from Equation (3.30), namely, θ ≅ tan -1 (0) = 0o. (3.33) In decibels, we see that AdB = 0 since 20 log10(1) = 0. Now, when ω is an order of magnitude below ω1 (i.e., ω = 0.1ω1), we find that θ = tan -1 (0.1) = -5.710° ≅ - 6 °. (3.34) These relationships between A(ω) and θ(ω) are shown plotted in Figure 3.6. For ω = ω1, we can compute the magnitude of the voltage amplitude ratio from Equation (3.29) as (3.35) Thus, |V0| = 0.707 |Vi| at the critical frequency ω1. In decibels, we can express the magnitude of the voltage ratio as AdB = 20 log10 (1/√2) = -3.0 dB . (3.36) It follows, then, that at ω = ω1 the magnitude of the voltage ratio is reduced by 3 dB. As a result, the critical frequency is often called the -3 dB point. Also, since the power-transfer ratio is proportional to the square of the voltage ratio, i.e., 3-15 , 10 loglog20'' 110 1 10 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =− nn YX ω ω ω ω (1/√2)2, the critical frequency ω1 (or f1 when expressed in Hertz) is also known as the half-power frequency. The phase angle for ω = ω1 is calculated by substituting into Equation (3.30) and is given by θ = - tan -1 (1) = - 45° . (3.37) For ω >> ω1, we determine from Equation (3.29) that |A(ω1)| ≅ ω1 /ω . (3.38) Hence, in this region, |A(ω)| varies inversely with frequency. The rate of change can be determined by considering a decade of frequency. Let X = |A(ωn)| and X dB = 20 log10 X , (3.39) where X is the magnitude of the voltage ratio function at some frequency ωn , and let Y = |A(10ωn)| and Y dB = 20 log10 Y , (3.40) where Y is the magnitude of the voltage ratio function at a frequency 10 times ωn . Then, using Equation (3.38), the difference in voltage ratio (in dB) over a decade of frequency is given by (3.41) which simplifies to X ' - Y ' = 20 log10 (10) = 20 dB . (3.42) 3 - 16 Thus, the gain at 10ωn is 20 dB less than it was at ωn. It follows then, that the rate of fall of the amplitude ratio, where ω >> ω1, is -20 dB/decacde. Finally, from the expression for the phase angle in Equation (3.30), we observe that as ω ∞, θ -90°, but at ω = 10ω1, θ = - tan -1 (10) = -84.289 ≈ - 84° . (3.43) As indicated previously, Figure 3.6 is a Bode plot for a low-pass filter with critical frequency ω1. Notice that the approximate Bode plot makes use of the straight-line segments to approximate the actual response (shown as a dotted line around the critical frequency). Also, note that the maximum amplitude rolloff is -20 dB/decade and the maximum phase shift is - 90°. Equally important to the low-pass filter just considered is a network in which the positions of the capacitor and the resistor are interchanged as shown in Figure 3.7. In this case, since the capacitance is in series with the input voltage source and the output terminals of the network, the voltage v0 becomes very small as the frequency is reduced. For zero frequency (i.e., "dc"), the output voltage is zero since a capacitor blocks dc voltages. At very high frequencies, the capacitor's impedance approaches 3 - 17 the magnitude and phase angle of the voltage ratio is again 3 dB down and +45°, respectively, at the critical frequency. The rate of increase for ω << ω1, is 20 dB/decade and as ω 0, θ + 90°, while at ω = 0.1ω1, θ = + 84°. For ω >> ω1, the magnitude of the voltage ratio is 1 (0 dB) and the phase angle is 0°. Obtaining values for A(ω) and θ(ω) from Equation (3.48) and Equation (3.49), as shown previously for the low-pass filter, is left to the student. 3.4 Advanced Preparation The following advanced preparation is required before coming to the laboratory: (1) Thoroughly read and understand the theory and procedures. (2) Solve for the steady-state voltage across C2 in the circuit shown in Figure 3.9 at both frequencies given to you by your instructor. (3) Perform a PSpice Transient simulation to find the steady-state voltage across C2 in the circuit shown in Figure 3.9. Note that the simulation time must be long enough for the transient response to reach steady-state. Make a copy of a segment of the steady-state voltage. Do the simulation for both frequencies. (4) Solve for the steady-state voltage across R4 in the circuit shown in Figure 3.10 at both frequencies given to you by your instructor. (5) Perform a PSpice Transient simulation to find the steady-state voltage across R4 in the circuit shown in Figure 3.10. Note that the simulation time must be long enough for the transient response to reach steady-state. Make a copy of a segment of the steady-state voltage. Do the simulation for both frequencies. (6) Calculate the critical frequency in Hertz (using the values given to you by your instructor) for the low-pass and high-pass filter test circuits shown in Figure 3.11 and Figure 3.12, respectively. 3 - 20 3.5 Experimental Procedure 3.5.1 AC Analysis of a RC Circuit Construct the circuit in Figure 3.9 using the HP33120A Function Generator output as Figure 3.9: RC circuit to be used for performing AC analysis. vT(t). Connect Channel 1 of the scope across vT(t) and Channel 2 across C2 and trigger the scope on Channel 1. Set the output of the HP33120A to produce a sine wave, adjust its peak-to-peak voltage to V1 volts (which is a peak voltage of V1/2 volts) using the scope, and set its frequency to fa Hz. Check using the frequency counter and digital oscilloscope. Now, measure the voltage across C2 by measuring the peak amplitude of the voltage waveform on Channel 2 plus the phase shift of the waveform on Channel 2 with respect to the waveform on Channel 1. Next, set the output of the HP33120A to have a frequency of fb Hz and repeat the above procedure. 3.5.2 AC Analysis of a RLC Circuit In a manner similar to the way the circuit in Figure 3.9 was constructed, assemble the circuit shown in Figure 3.10. Set the output of the HP33120A to produce a sine 3 - 21 vT(t) Figure 3.10: RLC circuit to be used for performing AC analysis. wave, adjust its peak-to-peak voltage to V2 volts (which is a peak voltage of V2/2 volts) using the scope, and set its frequency to fc Hz. Check using the frequency counter and digital scope. Now, measure the voltage across R4 by measuring the peak amplitude of the voltage waveform on Channel 2 plus the phase shift of the waveform on Channel 2 with respect to the waveform on Channel 1. Next, set the output of the HP33120A to have a frequency of fd Hz and repeat the above procedure. 3.5.3 RC Low-Pass Filter In this part, you are to experimentally determine the amplitude and phase angle of the voltage ratio of the RC low-pass filter network shown in Figure 3.11. This should be done over a frequency range from two orders of magnitude below upto two orders of magnitude above the critical frequency f1. Construct the circuit in Figure 3.11 using the HP33120A output vT(t) as vi(t). Connect Channel 1 of the scope to measure vi and Channel 2 to measure v0 and trigger the scope on Channel 1. Next, set the output to be sinusoidal with a peak-to-peak voltage of V3 volts. Use the following procedure to determine the critical frequency and to obtain frequency response data, i.e., values for the amplitude and phase angle of the voltage ratio as a function of frequency. Be sure 3 - 22 3.6 Report 3.6.1 AC Analysis of a RC Circuit 3.6.1.1 Clearly show your analysis to solve for the steady-state voltage across the capacitor C2 in Figure 3.9. In particular, draw the phasor-domain equivalent circuit, show all steps in determining the phasor voltage across C2, and express your answer as a sinusoidal function in the time domain. This should be done for both frequencies used (fa and fb). 3.6.1.2 Clearly explain the procedure you used to measure the peak voltage and phase angle of the voltage across the capacitor C2 using the scope. Specifically, you should provide a hardcopy of v0 compared to vs indicating the peak voltages in volts and phase difference in time. Also, show how you converted the phase difference from time to degrees. 3.6.1.3 Discuss the accuracy of your measured results compared to your solution in Problem 1.1 above and indicate the reasons for any errors between the measurements and the theory. 3.6.1.4 Attach a copy of the steady-state voltage across C2 obtained per Section 3.4 (3) and comment on the experimental results versus PSpice simulation results. 3.6.2 AC Analysis of a RLC Circuit 3.6.2.1 Clearly show your analysis to solve for the steady-state voltage across the resistor R4 in Figure 3.10. In particular, draw the phasor-domain equivalent circuit, show all steps in determining the phasor voltage across R4, and express your answer as a sinusoidal function in the time domain. This should be done for both frequencies used (fc and fd). 3 - 25 3.6.2.2 Discuss the accuracy of your measured results compared to your solution in Step 2.1 above and indicate the reasons for any errors between the measurements and the theory. 3.6.2.3 Attach a copy of the steady-state voltage across R4 obtained per Section 3.4 (5) and comment on the experimental results versus PSpice simulation results. 3.6.3 RC Low-Pass Filter 3.6.3.1 Prepare a table presenting your measurements for vi, v0, the amplitude of the voltage ratio |A|, and the phase angle (θ) as a function of frequency for the low-pass filter test circuit in Figure 3.11. Calculate AdB for each of the frequencies and add these values to your table. Generate a Bode plot of AdB and θ as a function of frequency in Hertz on semilog paper. 3.6.3.2 Calculate the theoretical critical frequency, f1, in Hertz for the low-pass filter in Figure 3.11 and draw the straight line approximate Bode plot for the amplitude of the voltage ratio and phase angle on the same graph as the experimental data. 3.6.3.3 Discuss how accurate the straight line approximation is compared to the theory and your experimental data. 3.6.4 RC High-Pass Filter 3.6.4.1 Prepare a table presenting your measurements for vi, v0, the amplitude of the voltage ratio |A|, and the phase angle (θ) as a function of frequency for the high-pass filter test circuit in Figure 3.12. Calculate AdB for each of the frequencies and add these values to your table. Generate a Bode plot of AdB and θ as a function of frequency in Hertz on semilog paper. 3 - 26 3.6.4.2 Calculate the theoretical critical frequency, f1, in Hertz for the high-pass filter in Figure 3.12 and draw the straight line approximate the Bode plot for the amplitude of the voltage ratio and phase angle on the same graph as the experimental data. 3.6.4.3 Discuss how accurate the straight line approximation is compared to the theory and your experimental data. 3.6.5 Design Problem A combination of low-pass and high-pass networks can give a frequency response that is similar to that of a band-pass filter. Consider the network below. Design the network so as to have (1) |Vo/VT| ≈ 0.5 at f = 1 kHz, (2) |Vo/VT| ≤ 0.1 for f ≤ 60 Hz, and (3) |Vo/VT| ≤ 0.1 for f ≥ 16k Hz. The figure above indicates these specified values. Use the load resistance RL specified by the instructor. 3.6.5.1 Identify the values selected for the capacitors (C1 & C2) and resistor (R1). 3.6.5.2 Use PSpice to provide a plot of |Vo / VT| showing the response of your circuit design. Use a log scale for frequency f. 3.6.5.3 Verify that the design meets the requirements by doing a steady–state AC PSpice simulation of the circuit at the three frequencies indicated on the figure above. Present results in tabular form. 3 - 27 ≤ 0.1 VoVT RL + R1 C1 C2 V2 V1 0.5 f(Hz) 1k 16k60 |Vo/VT|