Tutorial and Simulations for Micro-Cap IV

ELEC 380 Electronic Circuits II Tutorial and Simulations for Micro-Cap IV By Adam Zielinski (Posted at: http://wwwece.uvic.ca/~adam/) Version: August 22, 2002

ELEC 380 Electronic Circuits II - Tutorial 1-1 TUTORIAL This manual is written for Micro-Cap IV - Electronic Circuit Analysis Program for Macintosh computers. The PC Version of the program is available at: www.spectrum-soft.com. Prior to proceeding please familiarize yourself with the Simulation Tutorial for ELEC 330 posted at: http://www.ece.uvic.ca/~adam/. In this Tutorial we will explore other interesting features of the Micro-Cap IV that are relevant to the material covered in the class. The simulations #1 to #6 are part of preparation to the laboratory sessions and must be completed before the laboratory and obtained presented to the laboratory instructor. 1. AC Analysis The AC analysis allows us to see a frequency response or AC transfer function H(jω) of a linear circuit. You can imagine that a sinusoidal voltage source with amplitude 1 volt is applied to a specified node of a circuit (input) and that voltage and relative phase is measured at a different specified node (output) of the same circuit. The voltage ratio or voltage gain and relative phase shift between these two voltages depend on frequency applied. The gain (often expressed in decibels or dbs) and phase are plotted vs. frequency over the specified frequency range. Frequency often is displayed in logarithmic scale. In such scale distance between two frequencies, one 10 times larger than the other is constant irrespective of absolute frequency and is called a decade. Similarly, distance between two frequencies - one twice the other is constant irrespectively of absolute frequency and is called an octave. Such plots are called frequency responses (amplitude and phase) of a linear circuit. In electronic circuits we frequently encounter nonlinear elements such as transistors. For frequency response analysis (AC analysis) such elements are linearized prior to AC analysis. Any nonlinear circuit can be approximated by a linear circuit if the signal applied is sufficiently small. As an illustration let us consider a simple RC circuit shown in Figure T1. 10k 1 2 E1 0.5u.MODEL E1 SIN (F=32 A=1 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0) Figure T1. RC Circuit The voltage source should be added but will not play a role in AC analysis. The output voltage phasor V(2) at node 2 is equivalent to H(jω), which is a complex quantity. To get the amplitude response, we need to plot magnitude of H(jω) or mag(v(2)) which is most frequently expressed in db. This is reflected in the dialog box shown in Figure T2 that also includes phase response PH(V(@)). The

ELEC 380 Electronic Circuits II - Tutorial 1-2 frequency range set is from 100 khz to 1 Hz (if you think that this is a strange order, I agree) Figure T2. Dialog Box The resulting plot in logarithmic frequency axis is shown in Figure T3 with the cursor. 0.00-8.00-16.00-24.00-32.00-40.00 1 10 100 1K 10K 20*log(mag(v(2))) F 0.00-18.00-36.00-54.00-72.00-90.00 1 10 100 1K 10K PH(V(2)) F Expression Left Right Delta Slope F 0.032K 10.000K 9.968K 1 20*log(mag(v(2))) -3.032-49.943-46.911-4.706m Figure T3. The frequency response; amplitude and phase We can observe that the amplitude frequency response represents a low-pass filter that attenuates signal at higher frequencies. At a certain frequency the response reaches linear asymptote with slope of -20dB/decade. We also can see that a 3dB-point occurs at 32 Hz. This is consistent with so-called 3dB or corner frequency for RC circuit fc = 1/2πRC. This result can be verified in time domain by performing transient analysis for signal frequency at fc=32 Hz with set up as shown in dialog box in Figure T4.

ELEC 380 Electronic Circuits II - Tutorial 1-3 Figure T4 Dialog Box The results are shown in Figure T5 with cursor activated. We can see that the output waveform - v(2) has reduced amplitude to 0.707 volts, which corresponds to 3 db attenuation as expected. Note also a phase shift between waveforms. 1.00 0.60 0.20-0.20-0.60-1.00 0m 20m 40m 60m 80m 100m v(1) v(2) T Expression Left Right Delta Slope T 42.984m 100.000m 57.016m 1 v(1) 0.705 0.951 0.246 4.318 v(2) 0.705 0.318-0.387-6.783 Figure T5 Time domain responses 2. Spectral Analysis Spectral analysis of a periodic waveform can be performed on time domain data x(t) using Fast Fourier Transform FFT(x) algorithm. You can think of FFT as a Fourier Series of an infinite duration periodic waveform made of infinite repetitions of the time domain waveform of duration T. The fundamental frequency of Fourier Series of such constructed waveform is equal to 1/T. This will determine the frequency resolution of spectral analysis based on FFT, that is F=1/T. In order to obtain valid results using FFT it is important to place complete number of cycles of the waveform within the observation window T.

ELEC 380 Electronic Circuits II - Tutorial 1-4 FFT calculates complex numbers and often only its magnitude is of interest. Function MAG(FFT(x)) calculates the magnitude. Let's illustrate these points using two sinusoidal waveforms f1=1000hz with amplitude 1 and another at f2=2000hz with amplitude 0.5 as shown in Figure T6 V1 10k V2 10k.MODEL V1 SIN (F=1000 A=1 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0).MODEL V2 SIN (F=2000 A=0.5 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0) Figure T6 Two sinusoidal waveforms The dialog box in Figure T7 leads to the results shown in Figure 8 Figure T7 Dialog Box

ELEC 380 Electronic Circuits II - Tutorial 1-5 Figure T8 Spectral representation of two harmonic signals The frequency points are separated by F = 100Hz as expected. Each frequency component is represented by only one point in the spectrum (triangular shape is due to the way the points are joined by lines) and two waveforms are fully resolved. The absolute amplitude of spectral components is related to sampling frequency of the time-domain waveforms the higher the sampling rate, the larger the spectral amplitude. The relative amplitudes and frequency positions of the two spectral components are as expected. 3. Tolerances Value of parameters of any physical electronic component is given within certain limits defined by tolerances. For instance, set of resistors with tolerances 10% (or 10 % lot) means that an actual individual resistor will have a random value between +/- 10% of its nominal value. Simulation allows us to investigate finite tolerances effect on overall performance of circuit built using real components. Several simulations are to be performed and a random value of a component within specified tolerances is assigned at each run. This is so called Monte Carlo method (guess where the name came from?). For Worst Case option the parameter is assigned randomly but only at limits of its tolerances. For N parameters this gives 2^N possible combinations. To establish good confidence level, the number of simulations n > 2^N.

ELEC 380 Electronic Circuits II - Tutorial 1-6 As an illustration let's go back to the simple circuit from Figure T1 but assume that the resistor is from 10% lot. With this modification the circuit becomes as shown in Figure T9. 10k LOT=10% 1 2 E1 0.5u.MODEL E1 SIN (F=32 A=1 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0) Figure T9 RC Circuit with uncertain resistor value We will proceed to investigate its frequency response as in Figure T3. The dialog box for Monte Carlo analysis is shown in Figure T10 and the results are shown in Figure T11. Figure T10 Dialog Box for Random Simulation n=10 Figure T11 Amplitude frequency response for n=10 simulations

ELEC 380 Electronic Circuits II - Tutorial 1-7 4. Temperature effects All real electronic elements change their parameters with temperature changes. This applies to passive elements like resistors as well as to active ones like transistors or Operational Amplifier. Simulation is an ideal and simple method to determine the effect of temperature on a circuit. Consider a simple voltage divider shown in Figure T12. R1 1 2 10 R2.Define R1 100K TC=0.001.Define R2 100K Figure T12 Voltage divider circuit Here we use symbols for resistors that need to be defined. Nominal value for both resistors is 100 kohms but resistor R1 changes its value with temperature as determined by its temperature coefficient TC= 0.001. This coefficient specifies how much the resistance will change from its nominal value at nominal temperature for a one degree Centigrade of the difference between the nominal temperature (27 degrees) and the actual one. We will illustrate this by running a transient analysis with printout. The dialog box is shown in Figure T13. Figure T13 Dialog box for temperature variation

ELEC 380 Electronic Circuits II - Tutorial 1-8 The simulation is run from temperatures 27 degrees to 27 degrees in steps of 27. The numerical results obtained are shown in Figure T14 Micro-Cap IV Transient Analysis Limits of Temperature Date 8/8/02 Time 10:21 PM Temperature= -27 Case= 1 T v(2) (usec) (V) 0.000 5.139 0.200 5.139 0.400 5.139 0.600 5.139 0.800 5.139 1.000 5.139 T v(2) (usec) (V) 0.000 5.068 0.200 5.068 0.400 5.068 0.600 5.068 0.800 5.068 1.000 5.068 T v(2) (usec) (V) 0.000 5.000 0.200 5.000 0.400 5.000 0.600 5.000 0.800 5.000 1.000 5.000 Temperature= 0 Case= 1 Temperature= 27 Case= 1 Figure T 14 The temperature effects We can see that the divider functions properly only for the nominal temperature of 27 degrees but the voltage is higher for other temperatures. This is due to a lower resistance of R2 at lower temperatures.

ELEC 380 Electronic Circuits II Simulation #1 2-1 SIMULATION #1 Small Signal Amplifiers This simulation is part of preparation to the Laboratory Session #1. 1. Design the CE amplifier shown in Figure 1-1 for biasing current IE=1mA and gain of 36 (31.1dB) at frequency 1kHz. Note that components values shown in Figure 1-1 are not unique. Vcc 15 Vs 1uF 56k Vi 910 Vo 2N3904 Ve MV1 10k 1.5k 62uF.MODEL MV1 SIN (F=1K A=5M DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0).MODEL 2N3904 NPN (BF=378.5 BR=2 IS=15.8478P CJC=3.62441P CJE=4.35493P RC=1.00539U VAF=101.811 TF=666.564P TR=173.154N MJC=300M VJC=770.477M MJE=403.042M VJE=1 NF=1.34506 ISE=61.1468P ISC=0.00155473F IKF=14.2815M IKR=35.709 NE=2.02174 RE=1.10494 VTF=10 ITF=9.79838M XTF=499.979M ) Figure 1-1 CE Amplifier 2. Select the proper values for the ac source (10mVp-p, f=1khz) and transistor (beta= BF= 150). 3. Set the proper simulation parameters for transient analysis (see dialog box shown in Figure 1-2) and confirm the dc and ac conditions by simulation. Figure 1-2 Dialog Box

ELEC 380 Electronic Circuits II Simulation #1 2-2 Note that under the Transient Option menu the option of calculating the operating dc-point was selected. This allows us to see the waveforms in steady state. Shown in Figure 1-3 is a result: 5.00m 3.00m 1.00m -1.00m -3.00m -5.00m 0m 1m 2m 3m 4m 5m Vs T 14.25 14.17 14.09 14.01 13.93 13.85 0m 1m 2m 3m 4m 5m Vo T Figure 1-3 Transient Analysis After running transient analysis select the state variables under Transient Analysis Menu. You can read numerical values of dc for all nodes: In this particular case we got: Figure 1-4 State Variables This feature is very convenient to verify the dc-analysis. Alternatively, you can select Node voltages and Node numbers as shown in Figure 1-5

ELEC 380 Electronic Circuits II Simulation #1 2-3 Figure 1-5 Node Voltages and Node Numbers You may observe that a waveform at node 3 (output waveform) do not oscillate exactly around 14V, as we would expect. Can you explain it? 3. To see the gain vs. frequency we should run an ac analysis. Let us select the following parameters shown in Figure 1-6 Figure 1-6 Dialog Box

ELEC 380 Electronic Circuits II Simulation #1 2-4 The result is shown in Figure 1-7 30.00 29.00 28.00 27.00 26.00 25.00 100 1K 10K 100K db(vo/vi) F Figure 1-7 Frequency response You can see from the graph the gain becomes independent of frequency from approximately 1kHz.The lower freq. of operation is frequently defined as frequency when the gain drops by 3dB compare to the flat portion of the frequency response. In this case we have the lower frequency of operation at 150 Hz. 5. Investigate the effects of temperature on the gain by ac analysis and temperature variation from 0 to 100 degrees in 50 degree steps. What parameters in the circuit is temperature dependent? The results are shown in Figure 1-8

ELEC 380 Electronic Circuits II Simulation #1 2-5 30.00 29.00 28.00 27.00 26.00 25.00 100 1K 10K 100K db(vo/vi) F Expression Left Right Delta Slope F 0.100K 100.000K 99.900K 1 db(vo/vi) 25.669 27.355 1.686 16.879u Figure 1-8 Frequency response with a realistic component

ELEC 380 Electronic Circuits II Simulation #2 3-1 SIMULATION #2 Large Signal Amplifiers This simulation replaces the Procedures part of the Laboratory Session #3 and should be done prior to the lab. We will introduce here the Fourier Analysis (FFT) in Micro-Cap IV 1. Consider the same circuit as in Simulation #1 and shown in Figure 2-1 C1 R1 Vi RC C3 Vc 2N3904 VCC Vo.DEFINE RL 1K.DEFINE RC 5K.DEFINE RE 1K.DEFINE VCC 9 Ve Vs f=10khz R2 RE C2 RL.MODEL 2N3904 NPN (BF=378.5 BR=2 IS=15.8478P CJC=3.62441P CJE=4.35493P RC=1.00539U VAF=101.811 TF=666.564P TR=173.154N MJC=300M VJC=770.477M MJE=403.042M VJE=1 NF=1.34506 ISE=61.1468P ISC=0.00155473F IKF=14.2815M IKR=35.709 NE=2.02174 RE=1.10494 VTF=10 ITF=9.79838M XTF=499.979M ).MODEL Vs SIN (F=10K A=15M DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0) Figure 2-1 Large Signal Amplifier Design the amplifier shown in Figure 2-1 for the maximum output compliance. Note: the analytic results in this case will not be accurate because of large distortion present for a large signal applied to CE amplifier. Assume: RL=1k, RC=5k, RE=1k and frequency of operation f=10khz. In this simulation the values of some resistors and capacitors are not given and must be found to obtain: Voltage gain: 42.2 or 32.2dB Output compliance: PP=2.2V 2. Simulate the circuit you have designed. Investigate the gain of the amplifier and all dc-voltages in the circuit. First we check the frequency response of the circuit using ac analysis. Result is shown in Figure 2-2.

ELEC 380 Electronic Circuits II Simulation #2 3-2 40.00 36.00 32.00 28.00 24.00 20.00 100 1K 10K 100K db(vo/vi) F Expression Left Right Delta Slope F 0.100K 100.000K 99.900K 1 db(vo/vi) 3.085 28.874 25.790 258.155u Figure 2-2 Frequency Response of the Amplifier As we can see the amplifier has the gain is 29 db, which is less than expected. Investigate and comment of this discrepancy possibly caused by an error in the software. 2. Now perform the transient analysis. 3. The maximum calculated input signal to avoid output clipping is 55 mv p-p but we will drive the input with signal 30mVp-p. In simulation select the dc-point calculation in order to avoid transients due to capacitances in the circuit. The result are shown in Figure 2-3

ELEC 380 Electronic Circuits II Simulation #2 3-3 15.00m 9.00m 3.00m -3.00m -9.00m -15.00m 0u 60u 120u 180u 240u 300u Vs1 T 3.00 2.60 2.20 1.80 1.40 1.00 0u 60u 120u 180u 240u 300u Vc T Figure 2-3 Input and output signals Note that the output waveform is quite distorted. This is due to nonlinear characteristic of the transistor that shows up for large signal operation. The peakto-peak output in this case is 0.8 Vp-p. 4. Spectral (Fourier) Analysis 5. The Fourier analysis performs Fourier series expansion of the analyzed signal using FFT algorithm as discussed in Tutorial. As noted it is important that you select a complete number of cycles to assure smooth boundary between repetitions. If the boundary contains discontinuity, higher order harmonics will be computed which are not present in the actual waveform. Perform the transient analysis and select the following parameters as shown in Figure 2-4. The display will show magnitude of FFT vs. selected range of frequencies. Figure 2-4 Parameters for FFT

ELEC 380 Electronic Circuits II Simulation #2 3-4 The result is shown in Figure 2-5 700.00 560.00 420.00 280.00 140.00 0.00 0K 10K 20K 30K 40K 50K mag((fft(vc))) F Figure 2-5 The Results of FFT We can see a dc-component is present at zero frequency; fundamental frequency component is present at 10kHz and higher harmonics at multiple of 10 khz. We can access the numerical values by selecting the N option in Transient Analysis Limits and the results are shown in Figure 2-6 Figure 2-6 The numerical Results of FFT You can see that the second harmonic distortion in this case is (4.281/52.49) x 100% = 8%. Would you by a stereo with such distortion? What is the acceptable value?

ELEC 380 Electronic Circuits II Simulation #3 4-1 SIMULATION #3 Frequency Response This simulation is part of preparation to the laboratory Session #4. 1. Consider the circuit similar to that used in Simulation #2 and shown in Figure 3-1: 50 Vs1 C1 R1 Vi RC Vc 2N3904 VCC C3 Ve Vo Vs f=10khz R2 1K C2 1K.MODEL 2N3904 NPN (BF=378.5 BR=2 IS=15.8478P CJC=3.62441P CJE=4.35493P RC=1.00539U VAF=101.811 TF=666.564P TR=173.154N MJC=300M VJC=770.477M MJE=403.042M VJE=1 NF=1.34506 ISE=61.1468P ISC=0.00155473F IKF=14.2815M IKR=35.709 NE=2.02174 RE=1.10494 VTF=10 ITF=9.79838M XTF=499.979M ).MODEL Vs SIN (F=10K A=15MV DC=0 PH=0 RS=50 RP=0 TAU=0 FS=0) Figure 3-1 Amplifier The input is Vs1 and the output is Vo. The resistor Rs=50 represents the internal resistance of the driving source. 2. Design the amplifier for 3dB lower frequency fl=10 khz and midband gain Vo/Vs of 20 (or 26dB). Assume and set the following parameters for the transistor. 3. Confirm you design by simulation. Simulations. Run the ac analysis. Figure 3-2 shows what you might obtain

ELEC 380 Electronic Circuits II Simulation #3 4-2 30.00 26.00 22.00 18.00 14.00 10.00 1K 10K 100K 1M db(vo/vi) F Expression Left Right Delta Slope F 0.001M 1.000M 0.999M 1 db(vo/vi) -2.735 23.182 25.917 25.942u Figure 3-2 Frequency Response of the amplifier 4. Assume the transistor parameters as given in the model and predict the upper frequency of operation of your amplifier. Compare it with the result obtained by simulation and shown in Figure 3-3: 30.00 26.00 22.00 18.00 14.00 10.00 1K 10K 100K 1M 10M 100M 1G db(vo/vi) F Expression Left Right Delta Slope F 0.199M 1000.000M 999.801M 1 db(vo/vi) 23.178 8.186-14.991-14.994n Figure 3-3 Lower and upper frequency of operation

ELEC 380 Electronic Circuits II - Simulation #4 5-1 SIMULATION #4 Differential Amplifiers This simulation is part of preparation to the laboratory Session #5. Consider the differential amplifier that will be used in the laboratory and shown in Figure 4-1: Vi 1.5K 2N3904 2N3904 Vo 15 10MV 47 47 2.2K -15.MODEL 10MV SIN (F=1K A=10MV DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0).MODEL 2N3904 NPN (BF=378.5 BR=2 IS=15.8478P CJC=3.62441P CJE=4.35493P RC=1.00539U VAF=101.811 TF=666.564P TR=173.154N MJC=300M VJC=770.477M MJE=403.042M VJE=1 NF=1.34506 ISE=61.1468P ISC=0.00155473F IKF=14.2815M IKR=35.709 NE=2.02174 RE=1.10494 VTF=10 ITF=9.79838M XTF=499.979M ) Figure 4-1 Differential amplifier 1. Perform the transient and frequency analysis using Vi as the input Vo as the output. The results are shown in Figure 4-2 and Figure 4-3 10.00m 6.00m 2.00m -2.00m -6.00m -10.00m 0m 0.60m 1.20m 1.80m 2.40m 3m Vi T 10.36 10.31 10.26 10.21 10.17 10.12 0m 0.60m 1.20m 1.80m 2.40m 3m Vo T Figure 4-2 Transient analysis

ELEC 380 Electronic Circuits II - Simulation #4 5-2 30.00 26.00 22.00 18.00 14.00 10.00 1K 10K 100K 1M 10M 100M db(vo/vi) F Figure 4-3 AC Analysis 2. Modify the circuit as shown in Figure 4-4 and repeat the measurements: 1.5K 2N3904 2N3904 Vo 15 Vi 47 47 10MV 2.2K -15.MODEL 10MV SIN (F=1K A=10MV DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0).MODEL 2N3904 NPN (BF=378.5 BR=2 IS=15.8478P CJC=3.62441P CJE=4.35493P RC=1.00539U VAF=101.811 TF=666.564P TR=173.154N MJC=300M VJC=770.477M MJE=403.042M VJE=1 NF=1.34506 ISE=61.1468P ISC=0.00155473F IKF=14.2815M IKR=35.709 NE=2.02174 RE=1.10494 VTF=10 ITF=9.79838M XTF=499.979M ) Figure 4-4 Modified Circuit

ELEC 380 Electronic Circuits II - Simulation #4 5-3 10.00m 6.00m 2.00m -2.00m -6.00m -10.00m 0m 0.60m 1.20m 1.80m 2.40m 3m Vi T 10.36 10.31 10.26 10.21 10.16 10.11 0m 0.60m 1.20m 1.80m 2.40m 3m Vo T Figure 4-5 Transient analysis 30.00 26.00 22.00 18.00 14.00 10.00 1K 10K 100K 1M 10M 100M db(vo/vi) F Expression Left Right Delta Slope F 0.001M 100.000M 99.999M 1 db(vo/vi) 21.834 16.438-5.395-53.952n Figure 4-6 AC analysis 3. Modify the circuit as shown in Figure 4-7 and perform the time and the frequency analysis (note that we have to increase amplitude of the input signal).

ELEC 380 Electronic Circuits II - Simulation #4 5-4 1.5K 2N3904 2N3904 Vo 15 Vi 47 47 2.2K -15 1000MV.MODEL 1000MV SIN (F=1K A=1000MV DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0).MODEL 2N3904 NPN (BF=378.5 BR=2 IS=15.8478P CJC=3.62441P CJE=4.35493P RC=1.00539U VAF=101.811 TF=666.564P TR=173.154N MJC=300M VJC=770.477M MJE=403.042M VJE=1 NF=1.34506 ISE=61.1468P ISC=0.00155473F IKF=14.2815M IKR=35.709 NE=2.02174 RE=1.10494 VTF=10 ITF=9.79838M XTF=499.979M ) Figure 4-7 Modified Circuit 1.00 0.60 0.20-0.20-0.60-1.00 0m 0.60m 1.20m 1.80m 2.40m 3m Vi T 10.57 10.44 10.30 10.17 10.04 9.91 0m 0.60m 1.20m 1.80m 2.40m 3m Vo T Figure 4-8 Transient analysis

ELEC 380 Electronic Circuits II - Simulation #4 5-5 0.00-4.00-8.00-12.00-16.00-20.00 1K 10K 100K 1M db(vo/vi) F Expression Left Right Delta Slope F 0.001M 1.000M 0.999M 1 db(vo/vi) -9.612-9.602 0.010 10.186n Figure 4-9 AC analysis 4. Interpret all the results obtained and compare them with calculations

ELEC 380 Electronic Circuits II - Simulation #5 6-1 SIMULATION #5 Instrumentation Amplifier using Op. Amp This simulation is part of preparation to the Laboratory Session #7. We will investigate the effects of finite tolerances on the circuit performance. The basic data for a general purpose Op. Amp like LM 741 and for comparison for a better performance LM 107 are given in Figure 5-1 Figure 5-1 LM741 and LM 107 Data Sheets Task: An instrumentation amplifier with differential gain of 10 is required to operate in the frequency band from 1kHz to 10kHz. Design such an amplifier using 1% resistors and 741 Op. Amp with finite tolerances of its parameters. Confirm is operation and specify the tolerance of the differential gain and the minimum CMRR your amplifier can provide within the specified band.

ELEC 380 Electronic Circuits II - Simulation #5 6-2 1. We will start by designing a simple instrumentation amplifier and check its differential gain using the circuit below. The first stage serves only as an inverter to generate inverted signal needed to drive the amplifier with the differential signal only. You might check that it does not introduce any error in the frequency band of interest. 10K 1K 18 10K V Vs 10K 18 LM301A 1K LM741 Vo 18 10K 18.MODEL LM741 OPA (LEVEL=2 ROUTAC=50 ROUTDC=75 IOFF=20N IBIAS=80N VEE=-18 VCC=18 VPS=16 VNS=-16 CMRR=31.6228K).MODEL LM301A OPA (LEVEL=2 TYPE=3 A=160K ROUTAC=50 ROUTDC=75 VOFF=2M IOFF=3N IBIAS=70N VEE=-18 VCC=18 VPS=14 VNS=-14).MODEL V SIN (F=1MEG A=1 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0) Figure 5-2 Instrumentation Amplifier Its frequency response is shown in Figure 5-3.

ELEC 380 Electronic Circuits II - Simulation #5 6-3 30.00 26.00 22.00 18.00 14.00 10.00 100 1K 10K 100K db(vo/vs) F Figure 5-3 Frequency Response 2. Proceed by allowing finite tolerance in the components used to built the amplifier. This is done by specifying the value of a component (resistors in our case) from a 5% LOT Proceed with the simulation. If only one run is selected, the nominal values for components are assumed. For M runs tolerance limits are randomly selected. For N parameters this gives 2^N possible combinations. To establish a good confidence level M>>2^N. The result obtained for 30 runs is shown in Figure 5-4.

ELEC 380 Electronic Circuits II - Simulation #5 6-4 30.00 26.00 22.00 18.00 14.00 10.00 100 1K 10K 100K DB(Vo/Vs) F Figure 5-4 Frequency response with random parameters 3. Modify the circuit as shown in Figure 5-5 to drive it with common mode signal only and perform the ac analysis:

ELEC 380 Electronic Circuits II - Simulation #5 6-5 1K 10K 10K 18 V Vs 10K 18 LM301A 1K LM741 Vo 18 10K 18.MODEL LM741 OPA (LEVEL=2 ROUTAC=50 ROUTDC=75 IOFF=20N IBIAS=80N VEE=-18 VCC=18 VPS=16 VNS=-16 CMRR=31.6228K).MODEL LM301A OPA (LEVEL=2 TYPE=3 A=160K ROUTAC=50 ROUTDC=75 VOFF=2M IOFF=3N IBIAS=70N VEE=-18 VCC=18 VPS=14 VNS=-14).MODEL V SIN (F=1MEG A=1 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0) Figure 5-5 Common Mode Signal The Common Mode AC response is shown in Figure 5-6. -35.00-50.00-65.00-80.00-95.00-110.00 100 1K 10K 100K db(vo/vs) F Figure 5-6 Common Mode response 4. From your plots deduce all the required parameters of the inst. amp and comment on the results obtained.

ELEC 380 Electronic Circuits II Simulation #6 7-1 SIMULATION #6 Design of a low pass filter This simulation replaces Laboratory Session #10. Tasks: 1. Design the 3 rd order LP Butterworth filter with 3dB bandwidth of 10kHz and gain of 10 2. Check your design by simulation with the exact component values. 3. Select components with 5% tolerances and check the envelope of the frequency response for 50 runs. 4. Apply square waveform of 8kHz to your filter and observe the output. Shown in Figure 6-1 is a sample circuit: 10K 40K 10K 10K 18 18 LM741 10K LM741 Vo Vs 10K 10K 18 V 18 1.59NF 1.59NF 1.59NF.MODEL V SIN (F=1MEG A=1 DC=0 PH=0 RS=1M RP=0 TAU=0 FS=0).MODEL LM741 OPA (LEVEL=2 ROUTAC=50 ROUTDC=75 IOFF=20N IBIAS=80N VEE=-18 VCC=18 VPS=16 VNS=-16 CMRR=31.6228K) Figure 6-1 Low-pass filter

ELEC 380 Electronic Circuits II Simulation #6 7-2 The frequency response for the exact components values is presented in Figure 6-2. 22.00 18.00 14.00 10.00 6.00 2.00 1K 10K 100K db(vo/vs) F Figure 6-2 Frequency response of the filter Filter response with the 5% components and 50 runs with randomly varying parameters is shown in Figure 6-3 22.00 18.00 14.00 10.00 6.00 2.00 1K 10K 100K db(vo/vs) F Figure 6-3 Frequency response with finite tolerances