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Transcript

Systematic Circuit Analysis Nodal Analysis Chapter 4 Section 1 3 Fundamental Principles • Ohm’s Law V = I · R, where I enters at the higher voltage side • Kirchoff’s Voltage Law Algebraic sum of voltages around a loop equal zero • Kirchoff’s Current Law Algebraic sum of currents entering and leaving a node equal zero Circuit Simplification • Series Resistors can be combined into a single equivalent resistance • Parallel Resistors can be combined into a single equivalent resistance • Current sources in parallel add algebraically • Voltage sources in series add algebraically • A voltage source is series with a resistor can be replaced by a current source in parallel or vice versa Shortcuts • Voltage Divider – a voltage dividing across a series combination of resistors – Largest voltage is across the largest resistor • Current Divider – a current dividing among a parallel combination of resistors – Largest current is through the smallest resistor Nodal Analysis • Employs the 3 fundamental laws • May not require any circuit simplification • Consists of a straight-forward step-by-step procedure • Involves the solution of a set of linear equations simultaneously • Used by most circuit analysis computer programs like PSpice Example • Find the voltages and currents Step One • Identify all nodes that have at least 3 components connected to them • Choose one of the above nodes as the ground, or reference node, Vo=0 volts • Label the other nodes with V1, V2, etc. Voltage Notation • Remember that voltage is always measured with respect to some other point • So V1 the voltage at node 1 minus the voltage at node 0, or V1-Vo = V10 = V1 • V12 would be the voltage at node 1 minus the voltage at node 2, or V12 = V1-V2 Voltage Notation • V12 could also be obtained by going through any path, say node 3 V12 = V132 where V132 = V13 + V32 But V13 + V32 = (V1-V3) + (V3-V2) And (V1-V3) + (V3-V2) = V1-V2 = V12 Circuit After Step One V1 Vo = 0 volts Step Two • Choose currents for every circuit branch that is connected to any of the unknown voltage nodes – In this circuit, choose currents for the branches connected to V1 – You can also label voltages at other nodes in the circuit (nodes that have only two components), Va and Vb Circuit after Step Two I1 Va + V1 - I3 + + I2 - Vo = 0 volts - Vb Step Three • Find the value of each current or find an expression for each current in terms of the unknown voltages, V1, V2, etc. • The idea is to get all the currents to be expressed in terms of the node voltages, so that a KCL equation can be written at each unknown node, giving us as many equations as unknowns Step Three • If you have a resistor between two nodes, then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohm’s Law) • For this circuit the 10Ω resistor is between nodes 1 and ground (node 0) I2 = (V1 – 0) / 10Ω = V1 / 10Ω Step Three Cont. • If you have a resistor and a series voltage source between two key nodes, determine the voltage at the node between the components and then use Ohm’s law • There is a 6v source and 14Ω resistor in series between node 1 and ground Va = Vo+6v. = 6v. I1 = (Va - V1) / 14Ω = (6v - V1) / 14Ω Step Three Cont. • There is also 5v source and 10Ω resistor in series between node 1 and ground • Determine the voltage at the intermediate node Vb = Vo+5v. = 5v. • Use Ohm’s law to determine the current I3 = (V1-Vb) / 14Ω = (V1-5v) / 10Ω Circuit after Step Three Va=6v I1=(6-V1)/14Ω V1 + - I3=(V1-5)/10Ω + - + I2 = V1/10Ω - Vo = 0 volts Vb=5v Step 4 • Write Kirchoff’s Current Law at each node with an unknown voltage I1 = I2 + I3 • Substitute each value or expression into each KCL equation (6v-V1)/14Ω = V1/10Ω + (V1-5v)/10Ω Step 5 Solve • Solve the set of equations for each unknown voltage (6v-V1)/14Ω = V1/10Ω + (V1-5v)/10Ω • Multiply both sides to clear fractions 70Ω·[(6v-V1)/14Ω = V1/10Ω + (V1-5v)/10Ω] 30v - 5·V1= 7·V1 + 7·V1 - 35v • Group like terms and solve 65v = 19·V1 or V1 = 3.42v Step 5 Solve Cont. • Since V1 = 3.42v • Find each current by substituting the voltage values found into each current equation as appropriate I1 = (6v-V1)/14Ω = (6-3.42)/14 = .184 A I2 = V1/10Ω = 3.42/10 = .342 A I3 = (V1-5v)/10Ω = (3.42-5)/10 = -.158 A Step 6 – Reality Check Va=6v I1=.184A + - I3= -.158A V1=3.42v + + I2= .342A - Vo = 0 volts Vb=5v Example with a Dependent Source • Find the voltages and currents I1 Handling the Dependent Source • The dependent source will be dependent on some voltage or current in the circuit • You will need to express that voltage or current in terms of the unknown node voltages • The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables Step One • Identify all nodes that have at least 3 components connected to them • Choose one of the above nodes as the ground, or reference node at zero volts • Label the other nodes with V1, V2, etc. Circuit after Step 1 I1 V1 Vo = 0 volts Step Two • Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes Circuit after Step 2 I1 V1 + - + I2 - - I3 + Vo = 0 volts Vb Step Three • Find the value of each current or find an expression for each current in terms of the unknown voltages, V1, V2, etc. If you have a resistor between two nodes, then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohm’s Law) I2 = (V1 – 0) / 4Ω = V1 / 4Ω Step Three Cont. • If you have a resistor and a series voltage source between two nodes, determine the voltage at the node between the components and then use Ohm’s law Vb = Vo+3v. = 3v. I1 = (V1-Vb) / 2Ω = (V1-3v) / 2Ω Step Three Cont. • If you have multiple resistors and/or voltage sources between two key nodes, combine them together to get one resistor and one source If there is a dependent part of the source, express it in terms of the unknown node voltages Determine the voltage at the node between the equivalent resistor and source Use Ohm’s law Circuit for Step 3 I1 V1 Va + - 5v + 4Ω·I1 + I2 - - I3 + Vo = 0 volts Vb=3v Vb Step Three Cont. • The combined dependent source is 5v+4Ω·I1 So Va = V1 – (5v+4Ω·I1) But I1 in the dependent source was determined to be (V1-3v)/2Ω So Va = V1–(5v + 4Ω·(V1-3v)/2Ω) = -V1+1 • I3 = (Vo-Va) / 4Ω = (0-(-V1+1)) / 4Ω I3 = (V1 -1v) / 4Ω Circuit for Step 4 Va= 1v-V1 V1 I1=(V1-3)/2Ω + - 5v + 4Ω·I1 + I 2= V1/2Ω + I3=(V1-1)/4Ω Vo = 0 volts Vb=3v Vb Step 4 • Write Kirchoff’s Current Law at each node with an unknown voltage I3 = I2 + I1 • Substitute each value or expression into each KCL equation (V1 -1v) / 4Ω = V1 / 4Ω + (V1-3v) / 2Ω Step 5 Solve • Solve the set of equations for each unknown voltage (V1 -1v) / 4Ω = V1 / 4Ω + (V1-3v) / 2Ω • Multiply both sides to clear fractions 4Ω·[(V1 -1v) / 4Ω = V1 / 4Ω + (V1-3v) / 2Ω] V1 - 1v = V1 + 2·(V1-3v) • Group like terms and solve 5v = 2·V1 or V1 = 2.5v Step 5 Solving for Currents • Finding each current: • I3 = (V1 -1v) / 4Ω = (2.5v -1v) / 4Ω = .375 A • I2 = V1 / 4Ω = 2.5v / 4Ω = .625 A • I1 = (V1-3v) / 2Ω = (2.5v-3v) / 2Ω = -.25 A Step 6 - Checking I1= -.25A V1=2.5v + Va= 1v-V1= -1.5v 5v+4Ω·I1=4v + I2= .625A + I3=.375A Vo = 0 volts Vb=3v Vb Class Activity • Find the current equations at node 1 Class Activity • Find I2 in terms of V1 + _ For a resistor between nodes, use Ohm’s law Add polarities if not shown (+ at V1, - at V0) I2 = V1/60Ω Class Activity • Find I3 in terms of its value For a current source between nodes, the current is fixed by the source I3 = 2 A Class Activity • Find I4 in terms of V1 + _ + _ If you have series resistors, combine them and then use Ohm’s law I4 = V1/(30+70)Ω = V1/(100Ω) Class Activity • Find I1 in terms of V1 Combine series resistors, even if not next to each other, then find the voltage between the source and equivalent resistance Finding I1 + - V1 Vo=0v Va=100v + - V1 100+80=180Ω Vo=0v So I1 = (Va-V1)/180Ω = (100-V1)/180Ω Class Activity • Find I5 in terms of V1 + - The voltage on the right side of the 50Ω resistor is known, because of the 150v source I5 = (V1-150v) / 50Ω Class Activity • Write KCL equation at node 1 I1 + I3 = I2 + I4 + I5 Substitute all the current equations: (100-V1)/180+2A=V1/60+V1/100+(V1-150v)/50 Class Activity (100-V1)/180+2A=V1/60+V1/100+(V1-150v)/50 Multiply both sides by 9000Ω: 5000-50V1+18000=150V1+90V1+180V1-27000 Group: 50000 = 470V1 Solve: V1= 106.4 v Class Activity Checking I1=-.036A, I2=1.773A, I3=2A, I4=1.064A, I5=-.872A I1+I3= -.036+2 =1.964 A I2+I4+I5 = 1.773+1.064-.872 = 1.965 A – Yes! Example with 3 Unknown Nodes • Find Voltages and Currents Steps 1 & 2 • Find key nodes, assign one to ground, choose currents in branches I1 + V1 - I2 + Va - + + I4 V3 I3 V2 - I5 + - - Vo=0 volts Step 3 – Current Equations • Currents for resistors between nodes I1=(V1-V3)/4Ω I3=(V2-V3)/7Ω I4=V2/1Ω I5= V3/5Ω • Resistor and source between nodes Va = V1+ 9v I2 = (Va-V2)/3Ω = (V1+9v-V2)/3Ω Step 4 – KCL Equations • KCL Equation at each key node: At node 1: 0 = I1 + I2 + 8A At node 2: I2 = I3 + I4 At node 3: 25A + I1 + I3 = I5 • Substituting for each current At 1: 0 =(V1-V3)/4Ω +(V1+9v-V2)/3Ω +8A At 2: (V1+9v-V2)/3Ω =(V2-V3)/7Ω + V2/1Ω At 3: 25A+(V1-V3)/4Ω +(V2-V3)/7Ω = V3/5Ω At node 1 • 0 =(V1-V3)/4Ω +(V1+9v-V2)/3Ω +8A • Multiply both sides by 12Ω to clear fractions 12Ω·[0 =(V1-V3)/4Ω +(V1+9v-V2)/3Ω +8A] Or: 0 =3·V1-3·V3 +4·V1+36v-4·V2 +96v • Combining terms 7·V1 -4·V2 -3·V3 = -132v At node 2 • (V1+9v-V2)/3Ω =(V2-V3)/7Ω + V2/1Ω • Multiply both sides by 21Ω to clear fractions 21Ω·[(V1+9v-V2)/3Ω =(V2-V3)/7Ω + V2/1Ω ] Or: 7·V1+63v - 7·V2 = 3·V2 - 3·V3 +21·V2 • Combining terms 7·V1 -31·V2 +3·V3 = -63v At node 3 • 25A+(V1-V3)/4Ω +(V2-V3)/7Ω = V3/5Ω • Multiply both sides by 140Ω to clear fractions 140Ω·[25A+(V1-V3)/4Ω +(V2-V3)/7Ω =V3/5Ω] Or: 3500v+35·V1-35·V3+20·V2-20·V3=28·V3 • Combining terms 35·V1 +20·V2 -83·V3 = -3500v Set of 3 Simultaneous Eqs. • • • • 7·V1 -4·V2 -3·V3 = -132v 7·V1 -31·V2 +3·V3 = -63v 35·V1 +20·V2 -83·V3 = -3500v Solve by hand, calculator, or computer V1 = 5.414 v V2 = 7.737 v V3 = 46.316 v