Circuits II EE221 Unit 2 Instructor: Kevin D. Donohue Review: Impedance Circuit Analysis with nodal, mesh, superposition, source transformation, equivalent circuits and SPICE analyses.

Equivalent Circuits  Circuits containing different elements are equivalent with respect to a pair of terminals, if and only if their voltage and current draw for any load is identical.  More complex circuits are often reduced to Thévenin and Norton equivalent circuits.

Equivalent Circuit (Example) Find and compare the voltages and currents generated in 3 of the following loads across terminals AB:  open circuit  resistance R L  short circuit R th VsVs IsIs A A B B Thévenin Norton

Results - Equivalent Circuit V AB I AB Open0 Short0 RLRL Current Source Norton Circuit V AB I AB Open0 Short0 RLRL Voltage Source Thévenin Circuit What is the Norton equivalent for the Thévenin circuit? 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 V oc )  Short the terminal and find the current in the short (short-circuit current I sc )  Compute equivalent resistance as: R th = V oc / I sc

Finding Thévenin and Norton Equivalent Circuits The equivalent circuits can then be expressed in terms of these quantities R th I sc A B V oc R th A B

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. R th IsIs A B I s R th R th A B VsVs A B A B IsIs A B VsVs A B VsVs A B A B IsIs

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 cannot be used if dependent sources are present.

Source Transformation Example  Use source transformation to find the phasor value  Show = V 3 k  6 k  -j3.5 k 

Nodal Analysis  Identify and label all nodes in the system.  Select one node as a reference node (V=0).  Perform KCL at each node with an unknown voltage, expressing each branch current in terms of node voltages.  (Exception) 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 of the source is equal to the source value, and then use a surface around both nodes for KCL equation.

Example Find the steady-state value of v o (t) in the circuit below, if v s (t) = 20cos(4t): vsvs 10  2 i x 1 H 0.5 H 0.1 F ixix Show: v 0 (t) = 13.91cos(4t º) +vo-+vo-

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 “crisscross” each other (but they can overlap in common branches).  Perform KVL around each loop expressing all voltages in terms of loop currents.  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)  Another way: Let more than one loop pass through source and set 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-state response for v c (t) when v s (t) = 5cos(800  t) V Can be derived with mesh or nodal analysis or source transformation: nF 6 k  3 k  vs(t)vs(t) + v c (t) -

Linearity and Superposition  If a linear circuit has multiple independent 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-state response for v c (t) when v s (t) = 4cos(200t) V and i s (t) = 8cos(500t) A. Can be derived with superposition: 5 mF 6  44 vs(t)vs(t) +vc(t)-+vc(t)- 10 m  is(t)is(t)

SPICE Solution Steady-State Analysis in SPICE is performed using the.AC (frequency sweep) option in the simulation set 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, Phase or radians) Sources in the AC analysis must be set up in “edit simulation model” menu to: 1. Identify source as sinusoidal through the small signal AC and distortion tab. 2. Provide a magnitude and phase and check the USE box

ex16-Small Signal AC-2-Table FREQMAG(V(IVM))PH_DEG(V(IVM)) (Hz)(V)(deg) k SPICE Example Find the phasor for v c (t) for v s (t)= 5cos(2  ft) V in the circuit below for f = 100, 200, 300, 400, 500, … Hz. Note that 400 Hz was the frequency of the original example problem.

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:

Plot of Magnitude Linear Magnitude, Linear Frequency dB Magnitude, Log FrequencyLinear Magnitude, Log Frequency dB Magnitude, Linear Frequency

Plot of Phase Linear Frequency, in Degrees Log Frequency, in Degrees

SPICE Example Find the phasor for v c (t) when v s (t)= 20cos(4t) V in the circuit below (note f = 2/  =0.6366) Extract Line from table of interest FREQMAG(I(VAM))PH_DEG(I(VAM))MAG(V(IVM))PH_DEG(V(IVM)) (Hz) m Voltage of interest