Circuit Theorems Dr Mustafa Kemal Uygurolu Circuit Theorems

Circuit Theorems Dr. Mustafa Kemal Uyguroğlu

Circuit Theorems Overview l l l Introduction Linearity Superpositions Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer

INTRODUCTION A large complex circuits Simplify circuit analysis Circuit Theorems ‧Thevenin’s theorem ‧Circuit linearity ‧source transformation ‧ Norton theorem ‧ Superposition ‧ max. power transfer

Linearity Property A linear element or circuit satisfies the properties of l Additivity: requires that the response to a sum of inputs is the sum of the responses to each input applied separately. If v 1 = i 1 R and v 2 = i 2 R then applying (i 1 + i 2) v = (i 1 + i 2) R = i 1 R + i 2 R = v 1 + v 2

Linearity Property l Homogeneity: If you multiply the input (i. e. current) by some constant K, then the output response (voltage) is scaled by the same constant. If v 1 = i 1 R then K v 1 =K i 1 R

Linearity Property l A linear circuit is one whose output is linearly related (or directly proportional) to its input. i I 0 Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 5 V will give i = 1 A. V 0 v

Linearity Property - Example i 0 Solve for v 0 and i 0 as a function of Vs

Linearity Property – Example (continued)

Linearity Property - Example Ladder Circuit 3 A 5 A +6 V- 2 A +3 V- 2 A 1 A + + + 14 V 8 V 5 V - - - This shows that assuming I 0 = 1 A gives Is = 5 A; the actual source current of 15 A will give I 0 = 3 A as the actual value.

Superposition l The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

Steps to apply superposition principle 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. l l 2. 3. l Turn off voltages sources = short voltage sources; make it equal to zero voltage Turn off current sources = open current sources; make it equal to zero current Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to the independent sources. Dependent sources are left intact.

Superposition - Problem 2 k. W 4 m. A 12 V – + 2 m. A 1 k. W I 0 2 k. W

2 m. A Source Contribution 2 k. W 2 m. A 1 k. W I’ 0 = -4/3 m. A 2 k. W

4 m. A Source Contribution 2 k. W 4 m. A 1 k. W I’’ 0 = 0 2 k. W

12 V Source Contribution 12 V 2 k. W – + 1 k. W I’’’ 0 = -4 m. A 2 k. W

Final Result I’ 0 = -4/3 m. A I’’ 0 = 0 I’’’ 0 = -4 m. A I 0 = I’ 0+ I’’’ 0 = -16/3 m. A

Example l find v using superposition

one independent source at a time, dependent source remains KCL: i = i 1 + i 2 Ohm's law: i = v 1 / 1 = v 1 KVL: 5 = i (1 + 1) + i 2(2) KVL: 5 = i(1 + 1) + i 1(2) + 2 v 1 10 = i(4) + (i 1+i 2)(2) + 2 v 1 10 = v 1(4) + v 1(2) + 2 v 1 = 10/8 V

Consider the other independent source KCL: i = i 1 + i 2 KVL: i(1 + 1) + i 2(2) + 5 = 0 i 2(2) + 5 = i 1(2) + 2 v 2 Ohm's law: i(1) = v 2(2) + i 2(2) +5 = 0 => i 2 = -(5+2 v 2)/2 i 2(2) + 5 = i 1(2) + 2 v 2 -2 v 2 = (i - i 2)(2) + 2 v 2 -2 v 2 = [v 2 + (5+2 v 2)/2](2) + 2 v 2 -4 v 2 = 2 v 2 + 5 +2 v 2 -8 v 2 = 5 => v 2 = - 5/8 V from superposition: v = -5/8 + 10/8 v = 5/8 V

Source Transformation l A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa

Source Transformation

Source Transformation

Source Transformation l l l 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.

Example 4. 6 l Use source transformation to find vo in the circuit in Fig 4. 17.

Example 4. 6 Fig 4. 18

Example 4. 6 we use current division in Fig. 4. 18(c) to get and

Example 4. 7 l Find vx in Fig. 4. 20 using source transformation

Example 4. 7 l. Applying KVL around the loop in Fig 4. 21(b) gives (4. 7. 1) l. Appling KVL to the loop containing only the 3 V voltage source, the resistor, and vx yields (4. 7. 2)

Example 4. 7 Substituting this into Eq. (4. 7. 1), we obtain Alternatively thus

Thevenin’s Theorem l l Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor. Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis.

Implications l l l We use Thevenin’s theorem to justify the concept of input and output resistance for amplifier circuits. We model transducers as equivalent sources and resistances. We model stereo speakers as an equivalent resistance.

Independent Sources (Thevenin) RTh Voc Circuit with independent sources + – Thevenin equivalent circuit

No Independent Sources RTh Circuit without independent sources Thevenin equivalent circuit

Introduction l l l Any Thevenin equivalent circuit is in turn equivalent to a current source in parallel with a resistor [source transformation]. A current source in parallel with a resistor is called a Norton equivalent circuit. Finding a Norton equivalent circuit requires essentially the same process as finding a Thevenin equivalent circuit.

Computing Thevenin Equivalent l Basic steps to determining Thevenin equivalent are – – Find voc Find RTh

Thevenin/Norton Analysis 1. Pick a good breaking point in the circuit (cannot split a dependent source and its control variable). 2. Thevenin: Compute the open circuit voltage, VOC. Norton: Compute the short circuit current, ISC. For case 3(b) both VOC=0 and ISC=0 [so skip step 2]

Thevenin/Norton Analysis 3. Compute the Thevenin equivalent resistance, RTh (a) If there are only independent sources, then short circuit all the voltage sources and open circuit the current sources (just like superposition). (b) If there are only dependent sources, then must use a test voltage or current source in order to calculate RTh = VTest/Itest (c) If there are both independent and dependent sources, then compute RTh from VOC/ISC.

Thevenin/Norton Analysis 4. Thevenin: Replace circuit with VOC in series with RTh Norton: Replace circuit with ISC in parallel with RTh Note: for 3(b) the equivalent network is merely RTh , that is, no voltage (or current) source. Only steps 2 & 4 differ from Thevenin & Norton!