ECE201 Lect-71 Circuits with Resistor Combinations (2.6, 8.7) Dr. Holbert February 8, 2006
ECE201 Lect-72 Solving Circuits with Series and Parallel Combinations The combination of series and parallel impedances can be used to find voltages and currents in circuits. This process can often yield the fastest solutions to networks. This process may not apply to complicated networks.
ECE201 Lect-73 Series and Parallel Impedances Impedances are combined to create a simple circuit (usually one source and one impedance), from which a voltage or current can be found Once the voltage or current is found, KCL and KVL are used to work back through the network to find voltages and currents.
ECE201 Lect-74 1k 2k 1k 2k 1k +–+– 10V + – V1V1 + – V3V3 + – V2V2 Example: Resistor Ladder Find V 1, V 2, and V 3
ECE201 Lect-75 1k 2k 1k 2k 1k 10V + – V1V1 + – V3V3 + – V2V2 Example: Resistor Ladder Find an equivalent resistance for the network with V 1 across it, then find V 1 using a voltage divider. +–+–
ECE201 Lect-76 1k 10V + – V1V1 Example: Resistor Ladder +–+–
ECE201 Lect-77 1k 2k 1k 2k 1k 10V + – 5V + – V3V3 + – V2V2 Example: Resistor Ladder Find an equivalent resistance for the network with V 2 across it, then find V 2. +–+–
ECE201 Lect-78 Example: Resistor Ladder 1k 2k 1k 10V + – 5V + – V2V2 1k +–+–
ECE201 Lect-79 1k 2k 1k 2k 1k 10V + – 5V + – V3V3 + – 2.5V Example: Resistor Ladder +–+–
ECE201 Lect-710 Example: Notch Filter Find V out Use = k 0.1 100 10V 0 + – V out 70.4mH 100 F +–+–
ECE201 Lect-711 V out 1k 0.1 100 10V 0 + – j106 –j6.67 Example: Notch Filter Find the equivalent impedance of the resistor, inductor, and capacitor. +–+–
ECE201 Lect-712 V out 1k 100 10V 0 + – 7.12 –89.99 Example: Notch Filter Combine top resistor and impedance. +–+–
ECE201 Lect-713 Example: Notch Filter V out 1k 10V 0 + – –4.07 +–+–
ECE201 Lect-714 Example #2: Notch Filter Find V out Use = 377 1k 0.1 100 10V 0 + – V out 70.4mH 100 F +–+–
ECE201 Lect-715 Example: Notch Filter V out = 1.23V 0.17
ECE201 Lect-716 Frequency Response
ECE201 Lect-717 Using MATLAB to Solve Circuits MATLAB can perform computations with complex numbers. You can use it as a calculator to compute phasors and impedances for AC steady-state analysis. You can also use it to automate computations of frequency responses.
ECE201 Lect-718 Using MATLAB Entering a complex number: >> 1+2j ans = i Multiplying complex numbers: >> (1+2j)*(3+4j) ans = i
ECE201 Lect-719 Example: Notch Filter Find V out Use = k 0.1 100 10V 0 + – V out 70.4mH 100 F +–+–
ECE201 Lect-720 Compute Impedances >> omega = 377 omega = 377 >> xl = j*omega*70.4e-3 xl = i >> xc = 1/(j*omega*100e-6) xc = i
ECE201 Lect-721 Equivalent Capacitor/Inductor Impedance >> zeq = (0.1+xl)*xc /(0.1+xl+xc) zeq = e e+03i
ECE201 Lect-722 Voltage Divider >> vin = 10 vin = 10 >> vout = vin*1e3 /(100+zeq+1e3) vout = i
ECE201 Lect-723 Magnitude and Angle >> abs(vout) ans = >> angle(vout) ans = >> angle(vout)*180/pi ans =
ECE201 Lect-724 Class Examples Learning Extension E2.14 Learning Extension E2.15 Learning Extension E8.12