Module 9 Thvenin and Norton Equivalent Circuits In

Module 9 - Thévenin and Norton Equivalent Circuits In this module, we’ll learn about an important property of resistive circuits called Thévenin Equivalence. M. Leon Thévenin (1857 -1926), published his famous theorem in 1883. 1

Thévenin’s Theorem applies to circuits containing resistors, voltage sources, and/or current sources Thêvenin Equivalent Circuit 2

Thévenin’s Theorem: A resistive circuit can be represented by one voltage source and one resistor: RTh VTh Resistive Circuit Thévenin Equivalent Circuit 3

Definition of a “Port” Port: Set of any two terminals PORT Resistive Circuit PORT 4

Illustrate concept with a simple resistive circuit: • Any two terminals can be designated as a port. • Our objective: Find the equivalent circuit seen looking into the port i. X R 1 Vo + R 2 v. X _ Simple resistive circuit Define port variables v. X and i. X ix flows to some load (not shown) 5

Find an equation that relates vx to ix i 1 R 1 Vo i. X R 2 i 2 KVL: i 1 R 1 + i 2 R 2 = Vo KCL: i 1 = i 2 + i. X Also note: + v. X _ (Each resistor voltage expressed using Ohm’s Law) v. X = i 2 R 2 6

Solve these equations for v. X versus i. X : i 1 R 1 + i 2 R 2 = Vo i 1 = i X + i 2 v. X = i 2 R 2 (i. X + i 2) R 1 + i 2 R 2 = Vo (i. X + v. X/R 2) R 1 + v. X = Vo Rearrange the variables… i. X R 1 + v. X (R 1 /R 2 + 1) = Vo or Vo – i. X R 1 v. X = ––––– 1 + R 1 /R 2 v. X = V o R 2 ––––––– – R 1 + R 2 R 1 R 2 i. X ––––––– R 1 + R 2 7

Examine this last equation: R 1 Vo i. X R 2 v. X = V o R 2 ––––––– – R 1 + R 2 + v. X _ R 1 R 2 i. X ––––––– R 1 + R 2 It has the form v. X = VTh – i. X RTh R 2 VTh = Vo ––––––– R 1/ + R 2 R 1 R 2 RTh = ––––––– R 1 + R 2 8

Constructing the Thévenin Equivalent Circuit RTh +i R – X Th VTh i. X + v. X _ Write down KVL for this circuit: v. X = VTh – i. XRTh “Output voltage = voltage source – voltage drop across RTh” 9

Actual Circuit: R 1 Vo i. X + R 2 v. X = V o R 2 ––––––– – R 1 + R 2 v. X Model: RTh + VTh _ R 1 R 2 i. X ––––––– R 1 + R 2 i. X v. X _ v. X = VTh – i. XRTh Choose model parameters VTh and RTh: R 2 VTh = Vo ––––––– R 1 + R 2 and RTh R 1 R 2 = ––––––= R 1 || R 2 R 1 + R 2 • From the point of view of v. X and i. X, the Thévenin circuit models the actual circuit in every way. 10

Actual Circuit: i. X R 1 + Vo R 2 v. X _ PORT Thévenin Equivalent: R 1 || R 2 Vo ––––––– R 1 + R 2 i. X + v. X _ PORT 11

Significance of RTh R 1 Vo R 2 Equivalent resistance • Set all independent sources in the actual circuit to zero. • For a voltage source, that means substituting a short circuit. • Equivalent resistance RTh= R 1||R 2 üR is the equivalent resistance seen looking into the port with all independent sources set to zero. Th 12

Setting a Voltage Source to Zero Current determined by what’s connected… Voltage between nodes fixed at Vo Vo 13

Setting a Voltage Source to Zero Voltage between nodes fixed at 0 V by short circuit LOAD 14

Setting a Current Source to Zero Current through branch set to Io x Voltage between nodes determined by Ioopen circuit what’s connected x 15

Significance of VTh R 1 + Vo R 2 _ i. X = 0 Open Circuit Voltage • Connect nothing to the port • i. X automatically set to zero. • Port voltage is called the open circuit voltage. RTh i. X = 0 +0 V– VTh KVL + Open Circuit Voltage _ • VTh represents the open circuit voltage of the actual circuit 16

Example: Resistor Network Balanced audio microphone system 50 k = Input resistance of typical audio amplifier. What voltage is developed across a 50 k resistive load? R 1=100 k Vmic 10 m. V R 3 = 10 k R 2 = 30 k + v. LOAD – 50 k R 4 =10 k Microphone network Load 17

Solution Method: Find the Thévenin Equivalent of the Microphone Network • Disconnect the load. • Find Thevenin Equivalent remaining circuit. • Reconnect the load. • Find v. LOAD from simplified circuit. R 1=100 k Vmic R 3 = 10 k R 2 = 30 k 10 m. V R 4 =10 k Find VTh and RTh Load 18

Step 1: Find the Equivalent Resistance • Set the voltage source to zero. (Substitute a short circuit. ) • Find the equivalent resistance RTh • RTh = R 3 + R 1||R 2 + R 4 = 10 k + 23 k + 10 k = 43 k R 1=100 k Vmic R 3 = 10 k R 2 = 30 k RTh 43 k R 4 =10 k Note: R 1||R 2 = (100 k )||(30 k ) = 23 k 19

Step 2: Find the Open Circuit Voltage • Analyze the circuit under no-load conditions. • Voltage across port terminals will be VTh • From KVL around the inner loop*: v 2 = Vmic. R 2/(R 1 + R 2) = 2. 3 m. V *basically, voltage division R 1=100 k R 3 = 10 k + Vmic R 2 = 30 k 10 m. V VTh = 2. 3 m. V – R 4 =10 k • Note that no current flows through R 3 and R 4. Voltage across these resistors is zero. 20

Step 3: Reconnect the Load to the Thévenin Equivalent Model RTh = 43 k + VTh RLOAD 2. 3 m. V v. LOAD – 50 k Thévenin equivalent of microphone network From simple voltage division: v. LOAD = VTh (RLOAD/(RLOAD + RTh) = 2. 3 m. V (50 k )/(93 k ) = 0. 54 m. V Answer 21

More Examples: The Norton Equivalent Circuit 22

Short Circuit Current Another important parameter of a circuit is its short circuit current The short circuit current of a port is defined as the current that will flow if: ü The load is disconnected ü A short circuit is connected instead RTh VTh Isc = VTh /RTh 23

Circuit Containing a Current Source Consider the following simple circuit: I 1 R 1 Port What is the Thévenin equivalent circuit seen looking into the port? 24

Step 1: Find the open circuit voltage: Current is zero + I 1 – R 1 + VTh – • Open circuit conditions All of I 1 flows through R 1 • Voltage develops across R 1 with polarity shown. • From Ohm’s Law: VTh = I 1 R 1 (That part is simple…) 25

Step 2: Find the equivalent resistance • Set the current source to zero. I 1 RTh • Set the current source to zero open circuit • Find the resistance looking into the port. • Trivially, by inspection: RTh = R 1 26

The Thévenin Equivalent Circuit: R 1 I 1 R 1 Thévenin Equivalent Actual Circuit: RTh = R 1 VTh = I 1 R 1 üDone! 27

Norton Equivalent Circuit RN IN RN Norton Circuit. INRN Thévenin Circuit. • The Norton and Thévenin equivalents of a circuit are interchangeable. • The equivalent resistance is the same: RN = RTh • The open circuit voltage is the same: VTh = INRN 28

What about the short-circuit current from a Norton Circuit? • Apply a short circuit: + IN RN V N = 0 – Isc = IN • The voltage across the Norton resistance becomes zero. • No current flows through the Norton resistance (I = V/R). • All the current flows through the short circuit. • The short circuit current is the source current IN. 29

Norton Equivalent Circuit IN IN = VTh/RTh RN = RTh Norton Circuit VTh = INRN IN RTh = RN Thévenin Circuit • The short circuit current is the same in each circuit: IN = VTh/RTh 30

Example: Resistive Network Find the Norton Equivalent of the following circuit using the short-circuit current method R 1=100 k Vmic 10 m. V R 3 = 10 k R 2 = 30 k RTh or RN R 4 =10 k Step 1: Find RTh (same as RN) by setting the source to zero. By inspection, RTh = R 3 + R 1||R 2 + R 4 = 10 k + 23 k + 10 k = 43 k 31

Step 2: Apply a short circuit to the port and compute the short-circuit current. R 1=100 k Vmic R 3 = 10 k R 2 = 30 k 10 m. V ISC = 0. 54 A R 4 =10 k R 1=100 k Vmic 10 m. V IP = Vmic/(R 1 + RP) = 0. 9 A RP = R 2 || (R 3 + R 4) = 12 k From current division: ISC = IP R 2 30 k = 0. 54 A = ISC = 0. 9 A 50 k [R 2 + (R 3 + R 4)] 32

Find the Norton Equivalent of the Circuit ISC = 0. 54 A RN = 43 k + IN = 0. 54 A RN = 43 k v. OC = 23 m. V – “Open Circuit Voltage” v. OC = IN RN = (0. 54 A)(43 k ) = 23 m. V 33

Construct the Thévenin Equivalent of the Circuit ISC = 0. 54 A RTh = 43 k VTh = ISC RTh = (0. 54 A )(43 k ) = 23 m. V RTh = 43 k VTh = 23 m. V This result is the same one obtained in the previous example! 34

A circuit that can be represented by a Thévenin Equivalent can also be represented by its corresponding Norton circuit IN RTh RN VTh Norton Equivalent Thévenin Equivalent VTh= INRN RTh = RN 35

End of This Module Do the Homework Exercises 36