# Lecture 101 Equivalence/Linearity (4.1); Superposition (4.2) Prof. Phillips February 20, 2003.

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lecture 101 Equivalence/Linearity (4.1); Superposition (4.2) Prof. Phillips February 20, 2003

lecture 102 Equivalent Sources An ideal current source has the voltage necessary to provide its rated current. An ideal voltage source supplies the current necessary to provide its rated voltage. A real voltage source cannot supply arbitrarily large amounts of current. A real current source cannot have an arbitrarily large terminal voltage.

lecture 103 A More Realistic Source Model vs(t)vs(t) RsRs The Circuit The Source i(t)i(t) + – v(t)v(t) +–+–

lecture 104 I-V Relationship The I-V relationship for this source model is v(t) = v s (t) - R s i(t) v(t)v(t) i(t)i(t)

lecture 105 Open Circuit Voltage If the current flowing from a source is zero, then the source is connected to an open circuit. The voltage at the source terminals with i(t) equal to zero is called the open circuit voltage: v oc (t)

lecture 106 Short Circuit Current If the voltage across the source terminals is zero, then the source is connected to a short circuit. The current that flows when v(t) equals zero is called the short circuit current: i sc (t)

lecture 107 v oc (t) and i sc (t) v(t)v(t) i(t)i(t) v oc (t) i sc (t)

lecture 108 v oc (t) and i sc (t) Since the open circuit voltage and the short circuit current determine where the I-V line crosses both axes, they completely define the line. Any circuit that has the same I-V characteristics is an equivalent circuit.

lecture 109 Equivalent Current Source is(t)is(t)RsRs The Circuit i(t)i(t) + – v(t)v(t)

lecture 1010 Source Transformation VsVs RsRs IsIs RsRs +–+–

lecture 1011 Source Transformation Equivalent sources can be used to simplify the analysis of some circuits. A voltage source in series with a resistor is transformed into a current source in parallel with a resistor. A current source in parallel with a resistor is transformed into a voltage source in series with a resistor.

lecture 1012 Averaging Circuit How can source transformation make analysis of this circuit easier? + – V out 1k  V1V1 V2V2 +–+– +–+–

lecture 1013 Source Transformations + – V out 1k  V1V1 V2V2 +–+– +–+–

lecture 1014 Source Transformations + – V out 1k  V 1 /1k  1k  V 2 /1k  Which is a single node-pair circuit that we can use current division on!

lecture 1015 Linearity Linearity leads to many useful properties of circuits: –Superposition: the effect of each source can be considered separately. –Equivalent circuits: any linear network can be represented by an equivalent source and resistance (Thevenin’s and Norton’s theorems).

lecture 1016 Linearity More important as a concept than as an analysis methodology, but allows addition and scaling of current/voltage values Use a resistor as for example (V = R I): –If current is KI, then new voltage is R (KI) = KV –If current is I 1 + I 2, then new voltage is R(I 1 + I 2 ) = RI 1 + RI 2 = V 1 + V 2

lecture 1017 Class Example

lecture 1018 Superposition “In any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

lecture 1019 The Summing Circuit + – V out 1k  V1V1 V2V2 +–+– +–+–

lecture 1020 Superposition + – V ’ out 1k  V1V1 + – V ’’ out 1k  V2V2 + +–+– +–+–

lecture 1021 Use of Superposition V’ out = V 1 /3 V’’ out = V 2 /3 V out = V’ out + V’’ out = V 1 /3 + V 2 /3

lecture 1022 How to Apply Superposition To find the contribution due to an individual independent source, zero out the other independent sources in the circuit. –Voltage source  short circuit. –Current source  open circuit. Solve the resulting circuit using your favorite technique(s).

lecture 1023 Problem 2k  1k  2k  12V I0I0 2mA 4mA – +

lecture 1024 2mA Source Contribution 2k  1k  2k  I’ 0 2mA I’ 0 = -4/3 mA

lecture 1025 4mA Source Contribution 2k  1k  2k  I’’ 0 4mA I’’ 0 = 0

lecture 1026 12V Source Contribution 2k  1k  2k  12V I’’’ 0 – + I’’’ 0 = -4 mA

lecture 1027 Final Result I’ 0 = -4/3 mA I’’ 0 = 0 I’’’ 0 = -4 mA I 0 = I’ 0 + I’’ 0 + I’’’ 0 = -16/3 mA

lecture 1028 Superposition Procedure 1.For each independent voltage and current source (repeat the following): a) Replace the other independent voltage sources with a short circuit (i.e., V = 0). b) Replace the other independent current sources with an open circuit (i.e., I = 0). Note: Dependent sources are not changed! c) Calculate the contribution of this particular voltage or current source to the desired output parameter. 2.Algebraically sum the individual contributions (current and/or voltage) from each independent source.

lecture 1029 Class Example

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