ECE201 Lect-121 Equivalence/Linearity (5.1); Superposition (5.2, 8.8) Dr. Holbert March 6, 2006

ECE201 Lect-122 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.

ECE201 Lect-123 A More Realistic Source Model vs(t)vs(t) RsRs The Circuit The Source i(t)i(t) + – v(t)v(t) +–+–

ECE201 Lect-124 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)

ECE201 Lect-125 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)

ECE201 Lect-126 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)

ECE201 Lect-127 v oc (t) and i sc (t) v(t)v(t) i(t)i(t) v oc (t) i sc (t)

ECE201 Lect-128 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.

ECE201 Lect-129 Equivalent Current Source is(t)is(t)RsRs The Circuit i(t)i(t) + – v(t)v(t)

ECE201 Lect-1210 Source Transformation VsVs RsRs IsIs RsRs +–+–

ECE201 Lect-1211 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.

ECE201 Lect-1212 Averaging Circuit How can source transformation make analysis of this circuit easier? + – V out 1k  V1V1 V2V2 +–+– +–+–

ECE201 Lect-1213 Source Transformations + – V out 1k  V1V1 V2V2 +–+– +–+–

ECE201 Lect-1214 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!

ECE201 Lect-1215 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).

ECE201 Lect-1216 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

ECE201 Lect-1217 Class Example Learning Extension E5.1

ECE201 Lect-1218 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.”

ECE201 Lect-1219 The Summing Circuit + – V out 1k  V1V1 V2V2 +–+– +–+–

ECE201 Lect-1220 Superposition + – V ’ out 1k  V1V1 + – V ’’ out 1k  V2V2 + +–+– +–+–

ECE201 Lect-1221 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

ECE201 Lect-1222 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).

ECE201 Lect-1223 Problem 2k  1k  2k  12V I0I0 2mA 4mA – +

ECE201 Lect mA Source Contribution 2k  1k  2k  I’ 0 2mA I’ 0 = -4/3 mA

ECE201 Lect mA Source Contribution 2k  1k  2k  I’’ 0 4mA I’’ 0 = 0

ECE201 Lect V Source Contribution 2k  1k  2k  12V I’’’ 0 – + I’’’ 0 = -4 mA

ECE201 Lect-1227 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

ECE201 Lect-1228 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.

ECE201 Lect-1229 Class Example Learning Extension E5.2 Learning Extension E8.15(a) & (b)