AC Analysis. Filters.

Transcription

1 Electrical engineer II - LABORATORY no. 2- Sources: 1. Lucia Dumitriu, Mihai Iordache - "Simularea numerica a circuitelor analogice cu programul Pspice", MatrixROM Andrei Vladimirescu - "Spice", Editura Tehnică Istvan Sztojanov, Sever Pașca - "Ghid practic PSpice", Teora Tutorial PSpice de la University of Texas Arlington, 5. Tutorial PSpice de la Bucknell University, Plot PSpice Data AC Analysis. Filters. What is AC analysis? Spice can be used also for AC analysis (even we speak about one frequency or a frequency domain). To obtain results for DC domain we used the commands ".DC" and ".PRINT DC". For AC we have a similar command ".AC" and ".PRINT AC". Spice computes the frequency answer of linear circuits and of the small signals equivalent circuits. Spice uses for AC analyses the nodal analyses modified in complex domain, such that it operates using phasors. Sources in AC During the last laboratory all sources were DC sources. The syntax for AC domain is similar. In Spice we have the convention that an AC source is a cos(sin) function with known initial phase: The information that should be provided is: x(t)=x max cos(ωt+α). - the name of the source that should stat with letter V or I (for voltage, respectively for current source), 1

2 - the nodes between it is connected; - the maximum value (X max ) and the initial phase (α). Separately, the frequency for all sources is defined using a command of the type ".AC". Examples for defining AC sources: *name nodes_list type value phase (degrees) V1ac 1 0 AC 230V 45 Vba 2 3 AC 240; 0 degrees initial phase I3y 3 4 AC 10.0A -45; -45 degrees initial phase Is1 1 9 AC 35mA; 35 ma at 0 degrees Remark 1. During the lecture/seminar) the used function was: x(t)=x ef 2sin(ωt+α 1 ). Spice, as the American system, uses a cosine function. For equivalence, one can use the known formula: X max =X ef 2 and α=α1 π/2 Anyway, this is not obligatory, since the time variable does not interfere in the AC analysis, and for the value of α one can use the value of variable α1 when we choose the convention corresponding to the sinus shape and not to cosine one. To be similar to the seminar, we ll use from now on the sinus convention. 2. One must specify the type AC when defining the sources because the implicit one is the DC. The phase is measured in degrees. If the phase is not given, it is assumed to be 0. In the previous examples V and A (as unit measures) are optional after the value of the voltage, respectively the current. Circuit elements The resistor is declared the same way as for DC: The linear capacitor in Spice is declared as: The linear coil: R_name N- N+ values_in_ohm C_name N+ N- value_in_farad L_name N- N+ value_in_henri The use of.print AC command In order to use the.print AC command, the command.ac should be activated. 2

3 The.AC command is used to sweep the frequency domain for the given circuit. This corresponds to finding the frequency answer. There are three types to sweep the frequency domain: LIN, DEC and OCT. Examples of.ac commands:.ac LIN 1 50Hz 50Hz.AC LIN ; linear sweep of the domain.ac DEC 20 10Hz 100kHz; decade sweep of the domain The first command performs an analysis for a frequency equal to 50 Hz (we chose the linear domain). Adding the unit measures Hz after the value is optional. The second command sweeps the domain consisting of frequencies 150Hz, 160Hz, 170Hz, 180Hz, 190Hz, 200Hz, 210Hz, 220Hz, 230Hz, 240Hz and 250Hz. The third command sweeps logarithmic the domain using 20 points for a decade for a domain of 4 decades (10Hz-100Hz, 100Hz-1kHz, 1kHz-10kHz and 10kHz-100kHz)..PRINT AC command. To print phasor components (complex numbers) there are 4 expressions corresponding to: - Module (M) - Phase (P) - Real part (R) - Imaginary part (I) Examples:.PRINT AC VM(30,9) VP(30,9); prints the module and the phase of the voltage (V voltage VM the module of the voltage).print AC IR(Rs) II(Rs); real and imaginary part of the current circulating through Rs.PRINT AC VM(13) VP(13) VR(13) VI(13); prints all pieces of information about node 13 To display dome graphs function of frequency, we use the command.probe. To visualize different variations, we use from the menu Trace->Add Trace from which we choose the corresponding quantity or different functions of that quantity (e.g. for phase variations we choose the function P(), for the imaginary part variations we choose the function IMG(), for real part variation we choose R(). ) 3

4 Ex. 1.1: R-L series circuit Single frequency examples R1=10Ω, L1=10mH. We ll compute, the current through this circuit for an input signal Vin, of amplitude 1V and the frequency f=100hz. Fig. 1. R-L series circuit The corresponding program is: R-L serie circuit Vin 1 0 AC 1 R L m.AC LIN PRINT AC I(R1) IP(R1) IR(R1) II(R1).END Ex. 1.2: R-C series circuit R1=10Ω ;C1=100µF. Analyze the circuit at the frequency f=318hz. Fig. 2. R-C series circuit R-C series circuit Vin 1 0 AC 1 R C u.AC LIN PRINT AC I(R1) IP(R1) V(R1) V(C1) 4

5 .END Ex.1.3: R-L-C series circuit The R-L-C series circuit is given in Figure 3. Its behaviour in AC state depends on frequency and we can distinguish 3 situations. a). Voltages resonance (resistive behaviour) Fig. 3 R-L-C series This takes place when U L =U C, so X L =XC. This happens for the frequency: ω = ω 0 = 1/ LC At this frequency the circuit is purely resistive, its impedance being minimum: Z = Z min = R ( X X ) R L C = and the current through the circuit has a maximum value equal to b). Capacitive behaviour I = max U Z min U I max = R For ω=ω1<ω0, we have X C > X L and so U C > U L and the circuit will behave as an R-C equivalent group and will determine a phase shift of the current before the voltage. c). Inductive behaviour For ω=ω2>ω0, we have X L > X C and so U L > UC and the circuit will behave as an R-L equivalent group and will determine a phase shift of the current behind the voltage. R=100Ω, L=25.33mH and C=1μF, the resonance frequency has a value of approx. f0=1000hz. Run the following Spice circuit: Circuit R-L-C series Iin 0 1 AC 1 R L m C 3 0 1u 5

6 RC 3 0 1T.AC LIN PRINT AC V(1) VP(1).PRINT AC V VP V(L) VP(L) V VP.END The program line RC 3 0 1T has been introduced because the current source has an infinite resistance and therefore the nodes 1,2 and 3 would not have an access way to the ground (also try the program without RC variant). The resistance RC=1TΩ introduced in parallel to the capacitor ensures this path without modifying the circuit functioning. Run the above analysis also for f1=500hz and f2=2000hz and compare the results. Also as an exercise, change the power source into a sine source with amplitude 1V. Notice in this case that RC resistance in parallel to C is no longer necessary. Why? Ex. 1.4 Maximum power transfer In DC state the maximum power transfer takes place when the load resistance is equal to inner resistance of the equivalent voltage source. For AC state, both the inner impedance of the equivalent source and the load impedance are complex values. It can be proved that, in this situation, the maximum power transfer takes place if the load impedance and the inner impedance are complex conjugated. Figure 4. Circuit with maximum power transfer For the circuit given in Fig. 4, Vin has the amplitude 10V and the frequency 1kHz and Ri=10Ω and Li=25.33mH. Then, for maximum transfer, Rl=10Ω and Cl=1μF. Why? Run the following example: Maximum power transfer Vin 1 0 AC 10 RI LI M RL CL 4 0 1U 6

7 .AC LIN 1 1K 1K.PRINT AC I(RL) IP(RL).END Notice that I(Rl)=0.5A what gives a total absorbed power of 2.5W. Ex. 2.1 RLC series resonance Examples for frequency domains Let s analyze the RLC circuit given in Figure 5. In this case, the voltages resonance takes place at a frequency: 1 f0 = =999.3Hz 2 π LC Figure 5. R-L-C series circuit We will analyze the circuit behaviour for a frequency range 100Hz - 10kHz. Run the program and follow the steps given below: Series resonance Vin 1 0 AC 1 R L m C 3 0 1u.AC LIN Hz 10kHz.PROBE.END 7

8 1. Run the program Open the PROBE editor 8

9 Select Axis Settings 9

10 2. Select Log as show in the figure 10

11 3. Select Add Trace 4. Select as in the figure 11

12 5. Notice the maximum of I(R) at 1000Hz We ll add the graph of the current phase 12

13 Choose Add Trace Introduce the current phase 13

14 So, we obtains the graphs as in the figure Choose Add Y Axis 14

15 Add the impedance Z(ω)=V(1)/I(R) To see the cursor see the figure above By activating the cursor, establish the value of the impedance ( ) phase for 3 frequencies: f1=500hz, f0=1000hz and f2=2000hz. V (1) Z ω = and of the I( R) 15

16 (Remark: the cursor A1 can be modified using the left mouse button and the cursor A2 with the right button). For f<f0, the circuit has a capacitive behaviour and for f>f0, it has an inductive behaviour. Passive Filters The passive filters, containing only resistors, coils and capacitors, have many applications. The simplest filters can be obtained by placing in series connection a resistor with a capacitor. Low-Pass (LPF) and High-Pass (HPF) Filters In Fig. 2 we actually have a low pass filter if we consider as output the voltage on the capacitor and a high pass filter if we consider as output the voltage on the resistor. We use the following Spice program: Low pass filter Vin 1 0 AC 1 R k C nF.ac dec 10 1hz 1meghz.probe.end Fig. 6. Determination of the LPF band 16

17 Draw the line of value = and denote the frequency of intersection with the graph. This frequency is called cutting frequency. (Remark: cursor A1 is modified using the left button and the cursor A2 with the right button). Figure 7. Cutting frequency On another graph represent the voltage phase on the capacitor, VP(2). Verify using the other cursor that, at cutting frequency, the phase is -45. As an exercise we ask you to analyze in a similar manner the behaviour of the high pass filter (the output on the resistor). Pass Band Filter of Butterworth type The scheme of Butterworth pass band filter is given in Figure 8. Figure 8. Butterworth pass band filter We use the following program: Pass band filter *de tip Butterworth Vin 1 0 ac 1 l u 17

18 c u c u l u r ac dec k 25k.probe.end After running the program display the curve V(3). The pass band is determined by reading the coordinates of the intersection points with the line It results B=2.252kHz. Figure 9. The band corresponding to the pass band filter The scheme is presented in Figure 10. Stop band filter 18

19 Use the following program: Stop band filter Vin 1 0 AC 1 R R C U L M R ac OCT probe.end Figure 10 Stop band filter After running the program display the curve V(2).The cutting frequencies are determined by reading the coordinates corresponding to the intersection points with the curve 0.707*500mV because the amplitude in the pass bands is 500mV. It results f1=58.4hz and f2=61.7hz. Figure 11. Stop band filter 19

20 Problems Find the currents (module and phase) for the following circuits using Spice (for all circuits the frequency f=50hz): Problem 1.1 Given data: L 1 =100/π mh, L 2 =300/π mh, C=500/π μf, R=10Ω, V 1 =20sin(ωt- π/4) [V], V 2 =60sin(ωt+ π/4) [V] Figure 12. Problem 1.1 Problem 1.2 Given data: L1= 200/π mh, L2=100/ π mh, C= 1000/ π μf, R=10Ω, V1=20*sqrt(2)*sin(ωt π/2)[v], V2=20sin(ωt+3 π/4) [V] Figure 13 Problem 1.2 Problem 1.3 Given data: L= 100/ π mh C=250/ π μf, R1=20Ω, R2=10Ω, R3=20Ω, V1=40sin(ωt+ π/4)[v], V2=80sin(ωt+ π/4)[v] 20

21 Conclusions Figure 14 Problem 1.3 In this laboratory we ve analyzed analogue circuits function of frequency. We took into account both the case when this frequency is unique and the case when the frequency is given by a frequency interval. 21