The main objective of this design project is to help students to design and implement solutions for electrical engineering problems by making them familiarize with different types of electronic circuits consisting of several components such as ICs, MOSFETs, resistors and capacitors in order to know how to design, construct and test these circuits using the following two different methods. First of all, we use the Pspice programming to analyze and design several electronic circuits where in this project we use the Pspice programming in order to design and analyze the circuit in figure 1 as shown in pages 5 & 6. Second, we implement our electronic circuit for this project in the Electronics II Lab, and we verify our results as shown in the images taken from the oscilloscope. Ultimately, we apply what we have learned so far from the lectures in implementing and designing our circuit design for this project.
Figure 1: Circuit Design using Pspice Software
Figure 2: Waveforms Simulation using Pspice Software
* First, we implement the circuit given in figure 1 using R1=1kΩ and C2=0.1uF. By measuring the cutoff frequency for the circuit which is equal to 400kHz.
* Second, we measure the cutoff frequency values for two cases:
Case 1: using the clock frequency (10 kHz), the obtained cutoff frequency for case 1 is 172Hz
Case 2: using the clock frequency (100 kHz), the obtained cutoff frequency for case 1 is 421Hz
Replacing the first resistor (R1) by the switch capacitor in the above figure 2 will change the cutoff frequency values where increasing the clock frequency value such as using 100kHz will increase the value of cutoff frequency from 172Hz to 421Hz in the ratio of about two times. Also, changing the clock frequency will change the cutoff frequency values as a result because the clock signals are out of the phase where the clock frequency = 1/Tc such as if the clock frequency equals 10kHz, then the period will equal to 1/10kHz=0.0001 and by increasing the clock frequency to be equal to 100kHz, the period will equal to 1/100kHz=0.00001. Ultimately, increasing the period Tc , the equivalent resistance will increase according to Req=Tc/C1, and as a result the C1 will be decreasing. Therefore, the value of cutoff frequency will be increasing as the capacitor decreasing because the cutoff frequency is inversely proportional to the capacitor as follows:
f=(1/(2*pi*T_p) where T_p=Req*C.
In this part of this project, we design and implement a second order band pass filter as given in Figure.3 below with f0 =10 kHz, Q = 20 and unity center frequency gain. Moreover, we verify the frequency response of circuit given by Figure.3 using component values for our design. Furthermore, we replace resistor Rg by switched capacitor circuit in order to get the same pass band for the filter, and we verify the pass band and Q values by using clock frequency Φ1 =200kHz.
The following are the design calculations:
Let C1 = 10 nF = C2. Q = 20 -> BW = Fo/Q = 10K / 20 = 500 Rg = 1/(2*pi*R*BW) = 31.8 kΩ. Av (max) = Rg/R and Av = 1 (unity gain) -> Rg = R = 31.8 kΩ. Fo = 1/(2*pi*C*sqrt(R2*R1)), R2 = 10k Ω-> R1 = 253 Ω. Let R3 = R4 = 1.2kΩ.
Circuit Design Results
With the switched capacitor our Fo = 9.9 kHz. BW = 1.3 K -> Q = 7.6 This is because the impedance of the switched capacitor was much lower than the Rg.