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Berkeley ELENG 42 - Lecture 6 methods of solving circuits

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Slide 1Nodal analysisBranches and NodesNode VoltagesNodal Analysis MethodExampleSlide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 189/13/2004 EE 42 fall 2004 lecture 6Lecture #6 methods of solving circuits• Nodal analysis• Graphical analysis•In the next lecture we will look at circuits which include capacitors•In the following lecture, we will start exploring semiconductor materials (chapter 2).Reading: Malvino chapter 2 (semiconductors)9/13/2004 EE 42 fall 2004 lecture 6Nodal analysis•So far, we have been applying KVL and KCL “as needed” to find voltages and currents in a circuit.•Good for developing intuition, finding things quickly…•…but what if the circuit is complicated? What if you get stuck?•Systematic way to find all voltages in a circuit by repeatedly applying KCL: node voltage method9/13/2004 EE 42 fall 2004 lecture 6Branches and NodesBranch: elements connected end-to-end,nothing coming off in between (in series)Node: place where elements are joined—includes entire wire9/13/2004 EE 42 fall 2004 lecture 6Node VoltagesThe voltage drop from node X to a reference node (ground) is called the node voltage Vx. Example:a bgroundVaVb+_+__+9/13/2004 EE 42 fall 2004 lecture 6Nodal Analysis Method1. Choose a reference node (aka ground, node 0) (look for the one with the most connections, or at the bottom of the circuit diagram)2. Define unknown node voltages (those not connected to ground by voltage sources).3. Write KCL equation at each unknown node. –How? Each current involved in the KCL equation will either come from a current source (giving you the current value) or through a device like a resistor. –If the current comes through a device, relate the current to the node voltages using I-V relationship (like Ohm’s law).4. Solve the set of equations (N linear KCL equations for N unknown node voltages).* with “floating voltages” we will use a modified Step 39/13/2004 EE 42 fall 2004 lecture 6Example•Choose a reference node.•Define the node voltages (except reference node and the one set by the voltage source).•Apply KCL at the nodes with unknown voltage.•Solve for Va and Vb in terms of circuit parameters.0RVVRVRVV3ba2a11anode voltage setVaVb reference nodeR4 V1R2+ -IS R3R1What if we used different ref node?S4b3abIRVRVV9/13/2004 EE 42 fall 2004 lecture 6ExampleV2V1R2R1R4R5R3I1Va152a4a11aIRVVRVRVV9/13/2004 EE 42 fall 2004 lecture 6Load line method• I-V graph of circuits or sub-circuits• I-V graph of non-linear elements• Using I-V graphs to solve for circuit voltage, current (The load-line method)• Reminder about power calculations with nonlinear elements.9/13/2004 EE 42 fall 2004 lecture 6Example of I-V GraphsResistors in Series1K4KIf two resistors are in series the current is the same; clearly the total voltage will be the sum of the two IR values i.e. I(R1+R2). Thus the equivalent resistance is R1+R2 and the I-V graph of the series pair is the same as that of the equivalent resistance. Of course we can also find the I-V graph of the combination by adding the voltages directly on the I-V axes. Lets do an example for 1K + 4K resistorsI+-V1V2V+-+-1K4KCombined 1K + 4KI24(ma)V (Volt)59/13/2004 EE 42 fall 2004 lecture 6VSIExample of I-V GraphsSimple Circuit, e.g. voltage source + resistor.2V2KIf two circuit elements are in series the current is the same; clearly the total voltage will be the sum of the voltages i.e. VS + IR. We can graph this on the I-V plane. We find the I-V graph of the combination by adding the voltages VS and I R at each current I. Lets do an example for =2V, R=2KCombined 2K + 2V+-2K2V I+-VV (Volt)I24(ma)59/13/2004 EE 42 fall 2004 lecture 6Example of I-V GraphsNonlinear element, e.g. lightbulbIf we think of the light bulb as a sort of resistor, then the resistance changes with current because the filament heats up. The current is reduced (sort of like the resistance increasing). I say “sort of” because a resistor has, by definition a linear I-V graph and R is always the same. But for a light bulb the graph kind of “rolls over”, becoming almost flat. Consider a 100 Watt bulb, which means at at the nominal line voltage of 117 V the current is 0.85A. (117 V times 0.85 A = 100W).V (Volt)I0.40.8(A)100I=0.85A at 117VI+-V9/13/2004 EE 42 fall 2004 lecture 6I-V Graphs as a method to solve circuitsWe can find the currents and voltages in a simple circuit graphically. For example if we apply a voltage of 2.5V to the two resistors of our earlier example: We draw the I-V of the voltage and the I-V graph of the two resistors on the same axes. Can you guess where the solution is?At the point where the voltages of the two graphs AND the currents are equal. (Because, after all, the currents are equal, as are the voltages.)1K4K2.5V+-Combined 1K + 4KI24(ma)V (Volt)52.5VSolution: I = 0.5mA, V = 2.5VI+-VThis is called the LOAD LINE method; it works for harder (non-linear) problems.9/13/2004 EE 42 fall 2004 lecture 6 Another Example of the Load-Line MethodLets hook our 2K resistor + 2V source circuit up to an LED (light-emitting diode), which is a very nonlinear element with the IV graph shown below. Again we draw the I-V graph of the 2V/2K circuit on the same axes as the graph of the LED. Note that we have to get the sign of the voltage and current correct!! (The sing of the current is reversed from page 3)I24(ma)V (Volt)5Solution: I = 0.7mA, V = 1.4VI+-V+-2V2KLEDLEDAt the point where the two graphs intersect, the voltages and the currents are equal, in other words we have the solution.9/13/2004 EE 42 fall 2004 lecture 6Power of Load-Line MethodWe have a circuit containing a two-terminal non-linear element “NLE”, and some linear components.The entire linear part of the circuit can be replaced by an equivalent, for example the Thevenin equivalent. This equivalent circuit consists of a voltage source in series with a resistor. (Just like the example we just worked!).So if we replace the entire linear part of the circuit by its Thevenin equivalent our new circuit consists of (1) a non-linear element, and (2) a simple resistor and voltage source in series. If we are happy with the accuracy of graphical solutions, then we just graph the I vs V of the NLE and the I vs V of the resistor plus voltage source on the same axes. The intersection of the two graphs is the solution. (Just like the problem


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Berkeley ELENG 42 - Lecture 6 methods of solving circuits

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Lecture 3

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