Circuits Ci it II EE221 Unit 2 Instructor: Kevin D D. Donohue Review: Impedance Circuit Analysis with nodal, mesh, superposition, source transformation equivalent circuits and transformation, SPICE analyses. Equivalent Circuits Circuits containing different elements are equivalent with respect to a pair of terminals if and only if their voltage and terminals, current draw for any load is identical. More complex circuits are often reduced to Thévenin and Norton equivalent q circuits. Equivalent Circuit (Example) Find and compare the voltages and currents generated in 3 of the following loads across terminals AB: open circuit resistance RL sh t circuit short i it A Is Rth Norton B Vs Rth A Thé i Thévenin B Results - Equivalent Circuit Voltage So Source rce Thévenin Circuit Currentt S C Source Norton Circuit VAB Open I s Rth Sh Short RL IAB 0 Is 0 Is RL Rth RL Rth Is Rth RL Rth What is the Norton equivalent for the Thévenin circuit? VAB Open Sh Short RL Vs IAB Vs 0 0 Vs Rth RL RL Rth Vs 1 RL Rth What is the Thévenin equivalent for the Norton circuit? Finding Thévenin and Norton Equivalent Circuits Identify terminal pair at which to find the equivalent circuit. Find voltage across the terminal pair when no load is present (open-circuit voltage Voc) Short the terminal and find the current in the short (short-circuit current Isc) Compute equivalent resistance as: Rth = Voc / Isc Finding Thévenin and Norton Equivalent Circuits The equivalent circuits can then be expressed in terms of these quantities R th Voc I sc A A Isc Rth B Rth Voc B Source Transformation The following circuit pairs are equivalent wrt to terminals AB. Therefore these source and resistor combinations can be Therefore, swapped in a circuit without affecting the voltages and currents in other parts of the circuit. A Is Rth B Rth A Rth Is Rth Vs B B A Is Vs B A Vs R th A A Is B Rth A Rth B A Vs B B 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. interest This method can work well for simple p circuits with source-resistor combinations as shown on the previous slide. This method cannot be used if dependent sources are present. Source Transformation Example Use source transformation to find the phasor value Vˆc 3 k 50 -j3.5 k 6 k Vˆ c Show Vˆc = 2.8 30 V Nodal Analysis Identify and label all nodes in the system. S l ct one Select n n node d ass a reference f nc n node d (V (V=0). 0) Perform KCL at each node with an unknown voltage, g , expressing p g each branch current in terms of node voltages. (Exception) If branch contains a voltage source One way: y Make reference node the negative g end of the voltage l source and d set node d values l on the h positive i i end d equall to the source values (reduces number of equations and unknowns by one) Another way: (Super node) Create an equation where the diff r nc between difference b t n th the node n d vvoltages lt s on n either ith r end nd of f th the source is equal to the source value, and then use a surface around both nodes for KCL equation. Example Find Fi d th the steady-state t d t t value l of f vo(t) in i the circuit below, if vs(t) = 20cos(4t): 10 vs 0.1 F ix 1H 2 ix Show: v0(t) = 13.91cos(4 Show .9 cos( t - 161.6º) 6 .6 ) + vo - 0.5 H Loop/Mesh Analysis Create loop current labels that include every circuit branch where each loop contains a unique branch (not included by any other loop) and no loops p “crisscross” each other (but ( they y can overlap p in common branches). Perform KVL around each loop expressing all g in terms of loop p currents. voltages If any branch contains a current source, One way: Let only one loop current pass through source so loop current equals the source value (reduces number of equations and unknowns by one) A th way: L Another Lett more th than one loop l pass through th h source and d sett combination of loop currents equal to source value (this provides an extra equation, which was lost because of the unknown voltage drop on current source) Analysis Example Find the steady steady-state state response for vc(t) when vs(t) = 5cos(800t) V + vc(t) 3 k vs(t) 114.86 nF 6 k C be Can b derived d i d with ith mesh h or nodal d l analysis l i or source transformation: t f ti Vˆc 2.5000 - j1.4434 2.8868 30 vc (t ) 2.8868 cos 800t V 6 Linearity and Superposition If a linear circuit has multiple independent sources, sources then a voltage or current anywhere in the circuit is the sum of the quantities produced by the individual sources (i.e. activate one source at a time). This property is called superposition. To deactivate a voltage source, set the voltage equal to zero (equivalent to replacing it with a short circuit). To deactivate a current source, set the current equal to zero (equivalent to replacing it with an open circuit). Analysis Example Find the steady steady-state state response for vc(t) when vs(t) = 4cos(200t) V and is(t) = 8cos(500t) A. 10 m 4 vs(t) is(t) 6 5 mF + vc(t) - C be Can b derived d i d with ith superposition: iti vc (t ) 2.1 cos500t 135 .92cos(200t - 94.4) V SPICE Solution Steady-State Analysis in SPICE is performed using the .AC AC (f (frequency sweep)) option ti iin th the simulation i l ti sett up. It will perform the analysis for a range of frequencies. You must indicate the: 1. Starting frequency 2. Ending frequency 3. Number of stepping increments and scale (log or linear) 4 Scale for the results (linear or Decibel, 4. Decibel Phase or radians) Sources in the AC analysis must be set up in “edit simulation i l ti model” d l” menu to: t 1. Identify source as sinusoidal through the small signal AC and distortion tab. 2 Provide a magnitude and phase and check the USE box 2. SPICE Example Find the phasor for vc(t) for vs(t))= 5cos(2ft) 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. IVm C 114 86n 114.86n R1 3k V R2 6k ex16-Small Signal AC-2-Table FREQ MAG(V(IVM)) (Hz) (V) +100.000 +3.299 +200.000 +3.203 +300 000 +3.059 +300.000 +3 059 +400.000 +2.887 +500.000 +2.703 +600.000 +2.520 +700.000 +2.345 +800.000 +2.182 +900.000 +2.033 +1.000k +1.898 PH_DEG(V(IVM)) (deg) -8.213 -16.102 -23 413 -23.413 -30.000 -35.817 -40.893 -45.295 -49.107 -52.411 -55.285 Plotting Frequency Sweep Results Choices for AC (frequency sweep simulation) For frequency ranges that include several orders of magnitude, g , a logarithmic g or Decade (DEC) ( ) scale is more practical than a linear scale The magnitude Th i d results l can also l be b computed d on a logarithmic scale referred to a decibels or dB defined as: M dB 20 log10 ( M ) Plot of Magnitude Linear Magnitude, Magnitude Log Frequency Circuit1-Small Signal AC-5 +1.000 +10.000 +100.000 dB Magnitude Magnitude, Log Frequency Frequency (Hz) +1.000k Circuit1-Small Signal AC-6 +10.000k +1.000 +10.000 +100.000 Frequency (Hz) +1.000k +10.000k +3.000 +0 000e+000 +0.000e+000 +2.000 +1.000 -20.000 +0.000e+000 MAG(V(IVM)) DB(V(IVM)) Linear Magnitude, Linear Frequency u14ex1.ckt-Small Signal AC-7 +10.000k +20.000k +30.000k dB Magnitude, Linear Frequency u14ex1.ckt-Small Signal AC-8 Frequency (Hz) +40.000k +10.000k +50.000 +3.000 +0.000e+000 +2.000 +1.000 -20.000 20 000 +0.000e+000 MAG(V(IVM)) DB(V(IVM)) +20.000k +30.000k Frequency (Hz) +40.000k +50.000 Plot of Phase Linear Frequency, in Degrees u14ex1.ckt-Small Signal AC-8 +10.000k +20.000k Frequency (Hz) +30.000k +40.000k +50.000 +0.000e+000 -50.000 PH_DEG(V(IVM)) Log Frequency, in Degrees +1.000 +0.000e+000 -50.000 PH_DEG(V(IVM)) u14ex1.ckt-Small Signal AC-9 +10.000 +100.000 Frequency (Hz) +1.000k +10.000k SPICE Example Find the phasor for vc(t) when vs(t)= 20cos(4t) V in the circuit below (note f = 2/ =0.6366) R 10 L 1 vAm V 0 FCCCS L0 .5 Voltage of interest IVm C .1 Extract Line from table of interest FREQ MAG(I(VAM)) PH_DEG(I(VAM)) MAG(V(IVM)) PH_DEG(V(IVM)) +108.440 +13.912 -161.560 (Hz) +636.600m +7.589