EE221 Circuits II Unit 2: Impedance Circuit Analysis and Equivalent Circuits, Study notes of Electrical and Electronics Engineering

Download EE221 Circuits II Unit 2: Impedance Circuit Analysis and Equivalent Circuits and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! Circuits II EE221 Unit 2 Instructor: Kevin D. Donohue Review: Impedance Circuit Analysis with nodal, mesh, superposition, source transformation, equivalent circuits and SPICE analyses. Nodal Analysis  Identify and label all nodes in the system.  Select one node as a reference node (V=0)  Perform KCL at each non-reference node, expressing each branch current in terms of node voltages  If branch contains a voltage source  One way: Make reference node the negative end of the voltage source and set node values on the positive end equal to the source values (reduces number of equations and unknowns by one)  Another way: (Super node) Create an equation where the difference between the node voltages on either end to source is equal to the source value and then use a surface around both nodes for KCL. Equivalent Circuit (Example) Find and compare the voltages and currents generated in 3 of the following loads across terminals AB:  open circuit  resistor RL  short circuit Rth Vs Is Rth A A B B Thévenin Norton Results - Equivalent Circuit RL 0Short 0Open IABVAB thsRI sI thL thL s RR RR I + thL th s RR R I + Current Source Norton Circuit RL 0Short 0Open IABVAB sV th s R V thL L s RR RV + thL s RR V + 1 Voltage Source Thévenin Circuit What would the voltage source in the Thévenin have to be for equivalence to the Norton circuit? What would the current source in the Norton have to be for equivalence to the Thévenin circuit? Equivalent Circuits  Circuits containing different elements are equivalent, if their response with respect to a pair of terminals is the same.  For the two previous circuits to be equivalent, what would have to be true about their source and resistance values? Source Transformation The following circuit pairs are equivalent wrt to terminals AB. Therefore, these source and resistor combinations can be swapped in a circuit without affecting the voltages and currents in other parts of the circuit. RthIs A B Is Rth Rth A B Vs Rth A B Rth A B th s R V Is Rth A B Vs A B Vs A B A B Is Source Transformation  Some equivalent circuits can be determined by transforming source and resistor combinations and combining parallel and serial elements around a terminal of interest.  This method can work well for simple circuits with source-resistor combinations as shown on the previous slide.  This method is limited, if dependent sources are present. Analysis Example The steady-state response for vc(t) when vs(t) = 5cos(800πt) V Can be derived with mesh or nodal analysis or source transformation: V 6 800cos8868.2)( 302.8868 j1.4434 - 2.5000 ˆ      −=⇔−∠== ππ ttvV cc  114.86 nF 6 kΩ 3 kΩ vs(t) + vc(t) - ex16-Small Signal AC-2-Table FREQ MAG(V(IVM)) PH_DEG(V(IVM)) (Hz) (V) (deg) +100.000 +3.299 -8.213 +200.000 +3.203 -16.102 +300.000 +3.059 -23.413 +400.000 +2.887 -30.000 +500.000 +2.703 -35.817 +600.000 +2.520 -40.893 +700.000 +2.345 -45.295 +800.000 +2.182 -49.107 +900.000 +2.033 -52.411 +1.000k +1.898 -55.285 SPICE Example Find the phasor for vc(t) for vs(t)= 5cos(2πft) V in the circuit below for f = 100, 200, 300, 400, 500, …..1000 Hz. Note that 400 Hz was the frequency of the original example problem. V R2 6k R1 3k IVm C 114.86n Plotting Frequency Sweep Results Choices for AC (frequency sweep simulation)  For frequency ranges that include several orders of magnitude, a logarithmic or Decade (DEC) scale is more practical than a linear scale  The magnitude results can also be computed on a logarithmic scale referred to a decibels or dB defined as: )(log20 10 MM dB = Plot of Magnitude Linear Magnitude, Linear Frequency dB Magnitude, Log FrequencyLinear Magnitude, Log Frequency dB Magnitude, Linear Frequency MAG(V(IVM)) Frequency (Hz)Circuit1-Small Signal AC-5 +0.000e+000 +1.000 +2.000 +3.000 +1.000 +10.000 +100.000 +1.000k +10.000k DB(V(IVM)) Frequency (Hz)Circuit1-Small Signal AC-6 -20.000 +0.000e+000 +1.000 +10.000 +100.000 +1.000k +10.000k MAG(V(IVM)) Frequency (Hz)u14ex1.ckt-Small Signal AC-7 +0.000e+000 +1.000 +2.000 +3.000 +10.000k +20.000k +30.000k +40.000k +50.000k DB(V(IVM)) Frequency (Hz)u14ex1.ckt-Small Signal AC-8 -20.000 +0.000e+000 +10.000k +20.000k +30.000k +40.000k +50.000k