Lab 1 - EE 420L
3. Read how to write and submit lab htm report
1. Use LTspice to set up the circuits seen in Figs. 1.21, 1.22, 1.24. And get the similar but complete simulation results.
2. Change the value of resistor or capacitor and hand calculate the circuit's operation.
3. Simulate the changed RC circuit and discuss the differences.
4. Don't forget to backup your report and work directory on your computer or dropbox and upload it to the CMOSedu.com for the future study and discussion.
2.) Second, by doing AC analysis, we can know more about the RC circuit. Specify the amplitude of AC input signal as "1" and get the result of amplitude and phase of the output signal. The following table shows different frequency corresponding to different amplitude and phase for the AC response. And the red row is the amplitude and phase of output signal for the specified Vin at the f=200Hz. The capacitor looks like load capacitor and when the frequency is higher than 100Hz, both amplitude and phase of the output signal are decreasing a lot. So the bandwidth of the RC circuit is 100Hz. Also, that means, the load capacitance or capacitor can pass lower-frequency signal and cut off the higher-frequency signal.
| Frequency | Amplitude(dB) | Phase(degree) |
| 10Hz | 0 | -3.6 |
| 20Hz | 0 | -7.1 |
| 100Hz | -1.5 | -32 |
| 200Hz | -4.1 | -51.6 |
| 300Hz | -6.5 | -61.8 |
| 400Hz | -8.6 | -68.2 |
| 500Hz | -10.3 | -72 |
| 1kHz | -16 | -80 |
| 10kHz | -35 | -89 |
| 100kHz | -55.9 | -89.9 |
| 1MHz | -76 | -89.99 |
3.) Third, for understanding the RC circuit deeply, change the R and C and hand calculate the circuit. Besides, use the simulation result to verify the hand calculations. When R is changing from 1k to 10kohms, through the amplitude equation 1/sqrt(1+(2*pi*f*RC) 2)=1/sqrt(1+(2*pi*200*R*1u)2), we can get that the amplitude will be reduced. For example, the amplitude of output @R=1k is 623mV, but @R=10k, the amplitude becomes 1/sqrt(1+(2*pi*200*10k*1u)2)=80mV. And also, from the phase equation -tan-1 (2*pi*f*R*1u), the R can make the phase shift worse with constant C. For R=10kohms, the phase shift is -85 degree and the corresponding delay is 1.2mslarger than the previous RC @R=1kohms.. The following simulations shows that increasing resistance causes the amplitude reducing and time delay becoming longer.
4.) The increasing capacitance like increasing resistance can also cause the reduction of the amplitude and phase of the output signal. From the calculation equations, we know that enlargering the R or C has the same effect on amplitude and phase. For example, for C=10uF, the phase shift is -85degree and the time delay is 1.2ms. Differently, the larger resistance can further increase the power dissipation of the RC circuit.The simulation result of changing capacitance is shown as below.
Discussions of Fig. 1.22:
1.) First, build up the RC circuit in Fig. 1.22. Add a 2uF capacitor to the sides of R1. Due to the change of transfer function of the RC circuit, the amplitude and phase calculation has been changed. The transfer function can be given by Vout/Vin=(1+jwRC1)/(1+jwR(C1+C2)). So the amplitude of the output signal(@Vin=1V) is sqrt[1+(jwRC1)2]/sqrt[1+(jwR(C1+C2))2]. And the phase response becomes tan-1(2*pi*f*RC1)-tan-1(2*pi*f*R(C1+C2)). Substitute the RC values to the equations, we can get the amplitude of the output signal is 0.6 and the phase shift is -6.82degree and the time delay is -95us. The simulation results is shown as below.
2.) Second, the AC analysis is shown as follow. From the amplitude equation sqrt[1+(jwRC1)2]/sqrt[1+(jwR(C1+C2))2], we notice that the amplitude still goes smaller by increasing frequency. But when the frequency is very large, the |AV| is approximate to C1/(C1+C2)=2/3, equal to 20log 2/3=-3.52dB. So that's why the amplitude looks like the following curve.
As for phase, from the equation tan-1(2*pi*f*RC1)-tan-1(2*pi*f*R(C1+C2)), these two terms has little difference. At first, the second term is a little bigger than the the first one. When the frequency gets higher, these two terms are both closer to pi/2 as shown in the following tan-1 function curve. That's why the phase curve decreases at first and then moves back to the 0 degree.
3.) From the AC simulation result, generate the follow table showing frequency, amplitude and phase. Based on this table, the below plot figure is to illustrate that the amplitude decreases firstly and higher frequency cannot change the amplitude any more. The phase curve is like parabolic curve.
| Frequency | Amplitude(dB) | Phase(degree) |
| 1Hz | 0 | 0 |
| 5Hz | 0 | 0 |
| 10Hz | 0 | -3.5 |
| 20Hz | 0 | -6.5 |
| 50Hz | -1.3 | -11 |
| 60Hz | -1.6 | -11.5 |
| 70Hz | -1.8 | -11.5 |
| 100Hz | -2.5 | -10.5 |
| 200Hz | -3.18 | -6.8 |
| 300Hz | -3.36 | -4.8 |
| 1kHz | -3.5 | -1.5 |
| 10kHz | -3.52 | 0 |
| 1MHz | -3.52 | 0 |
Discissoions of Fig.1.24:
1.) First, build up the circuit in Fig. 1.24. Get the simulation result matching to the Fig. 1.24 in textbook.
2.)Second, hand calucate the RC circuit with the pulse signal input. The output signal can be given by Vout= Vin (1-exp(-t/RC). The time constant is RC value. The delay time is when the Vout=50%|Vin|, td=0.7RC; and the rising time or falling time is the time it takes the output to be from the 10%|Vin| to 90% |Vin|, tr=2.2RC. The following simulations show that the increasing resistance or capacitance can make the time delay and rising or falling time longer.
