1 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 ( ), published his famous theorem in 1883.

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

3 Th é venin Equivalent Circuit V Th R Th Thévenin’s Theorem: A resistive circuit can be represented by one voltage source and one resistor: Resistive Circuit

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

5 Simple resistive circuit VoVo R1R1 R2R2 vXvX +_+_ iXiX Illustrate concept with a simple resistive circuit: Any two terminals can be designated as a port. Define port variables v X and i X Our objective: Find the equivalent circuit seen looking into the port i x flows to some load (not shown)

6 VoVo R1R1 R2R2 vXvX +_+_ iXiX Find an equation that relates v x to i x i1i1 i2i2 KVL: i 1 R 1 + i 2 R 2 = V o (Each resistor voltage expressed using Ohm’s Law) Also note: v X = i 2 R 2 KCL: i 1 = i 2 + i X

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

8 R 1 R 2 R Th = ––––––– R 1 + R 2 R 2 R 1 R 2 v X = V o ––––––– – i X ––––––– R 1 + R 2 R 1 + R 2 VoVo R1R1 R2R2 vXvX +_+_ iXiX Examine this last equation: It has the form v X = V Th – i X R Th R 2 V Th = V o ––––––– R 1/ + R 2

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

10 v X = V Th – i X R Th V Th R Th vXvX +_+_ iXiX VoVo R1R1 R2R2 vXvX +_+_ iXiX R 2 R 1 R 2 v X = V o ––––––– – i X ––––––– R 1 + R 2 R 1 + R 2 Choose model parameters V Th and R Th : Actual Circuit:Model: R 2 V Th = V o ––––––– R 1 + R 2 R 1 R 2 R Th = ––––––= R 1 || R 2 R 1 + R 2 and From the point of view of v X and i X, the Thévenin circuit models the actual circuit in every way.

11 VoVo R1R1 R2R2 Actual Circuit: R 1 || R 2 R 2 V o ––––––– R 1 + R 2 PORT vXvX vXvX iXiX iXiX +_+_ +_+_ Thévenin Equivalent:

12 Significance of R Th VoVo R1R1 R2R2 Set all independent sources in the actual circuit to zero. For a voltage source, that means substituting a short circuit. Equivalent resistance Equivalent resistance R Th = R 1 ||R 2 R Th is the equivalent resistance seen looking into the port with all independent sources set to zero.

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

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

15 Setting a Current Source to Zero Voltage between nodes determined by what’s connected Current through branch set to I o IoIo open circuit x x

16 Significance of V Th VoVo R1R1 R2R2 Open Circuit Voltage Connect nothing to the port i X = 0 Open Circuit Voltage V Th represents the open circuit voltage of the actual circuit i X = 0 i X automatically set to zero. Port voltage is called the open circuit voltage. +_+_ V Th R Th +_+_ + 0 V – KVL

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

18 Solution Method: Find the Thévenin Equivalent of the Microphone Network Find Thevenin Equivalent remaining circuit. Reconnect the load. Find v LOAD from simplified circuit. Disconnect the load. R 1 =100 k  R 2 = 30 k  R 3 = 10 k  R 4 =10 k  V mic 10 mV Load Find V Th and R Th

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

20 Step 2: Find the Open Circuit Voltage Analyze the circuit under no-load conditions. R 1 =100 k  R 2 = 30 k  R 3 = 10 k  R 4 =10 k  V mic 10 mV Voltage across port terminals will be V Th From KVL around the inner loop*: v 2 = V mic R 2 /(R 1 + R 2 ) = 2.3 mV Note that no current flows through R 3 and R 4.  Voltage across these resistors is zero. V Th = 2.3 mV +–+– *basically, voltage division

21 Step 3: Reconnect the Load to the Thévenin Equivalent Model R Th = 43 k  V Th 2.3 mV Thévenin equivalent of microphone network From simple voltage division: v LOAD = V Th (R LOAD /(R LOAD + R Th ) = 2.3 mV  (50 k  )/(93 k  ) = 0.54 mV Answer 50 k  + v LOAD – R LOAD

22 More Examples: The Norton Equivalent Circuit

23 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: The load is disconnected A short circuit is connected instead R Th V Th I sc = V Th /R Th

24 Circuit Containing a Current Source Consider the following simple circuit: I1I1 R1R1 What is the Thévenin equivalent circuit seen looking into the port? Port

25 Step 1: Find the open circuit voltage: I1I1 R1R1 + V Th – From Ohm’s Law: V Th = I 1 R 1 (That part is simple…) +–+–  Current is zero Open circuit conditions  All of I 1 flows through R 1 Voltage develops across R 1 with polarity shown.

26 Step 2: Find the equivalent resistance I1I1 R1R1 Trivially, by inspection: R Th = R 1 Set the current source to zero. Find the resistance looking into the port. R Th Set the current source to zero  open circuit

27 The Thévenin Equivalent Circuit: R1R1 I1R1I1R1 Thévenin Equivalent V Th = I 1 R 1 R Th = R 1 Actual Circuit: Done! I1I1 R1R1

28 Norton Equivalent Circuit The Norton and Thévenin equivalents of a circuit are interchangeable. The equivalent resistance is the same: R N = R Th The open circuit voltage is the same: V Th = I N R N Norton Circuit. RNRN INRNINRN ININ RNRN Thévenin Circuit.

29 ININ RNRN I sc = I N + V N = 0 – What about the short-circuit current from a Norton Circuit? Apply a short circuit: 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 I N.

30 Norton Equivalent Circuit The short circuit current is the same in each circuit: I N = V Th /R Th Norton Circuit V Th = I N R N I N = V Th /R Th R Th = R N Thévenin Circuit R N = R Th ININ ININ

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

32 R 2 = 30 k  R 3 = 10 k  R 4 =10 k  R 1 =100 k  V mic 10 mV Step 2: Apply a short circuit to the port and compute the short-circuit current. R 1 =100 k  V mic 10 mV R P = R 2 || (R 3 + R 4 ) = 12 k  I P = V mic /(R 1 + R P ) = 0.9  A From current division: R 2 [R 2 + (R 3 + R 4 )] I SC = I P = 0.9  A 30 k  50 k  = 0.54  A = I SC = 0.54  A I SC

33 Find the Norton Equivalent of the Circuit I SC = 0.54  A R N = 43 k  I N = 0.54  A = 23 mV + v OC – “Open Circuit Voltage” v OC = I N R N = ( 0.54  A)(43 k  ) = 23 mV

34 Construct the Thévenin Equivalent of the Circuit I SC = 0.54  A R Th = 43 k  V Th = I SC R Th = (0.54  A )(43 k  ) = 23 mV V Th = 23 mV R Th = 43 k  This result is the same one obtained in the previous example !

35 A circuit that can be represented by a Thévenin Equivalent can also be represented by its corresponding Norton circuit V Th = I N R N R Th = R N ININ RNRN Norton Equivalent Thévenin Equivalent R Th V Th

36 End of This Module Do the Homework Exercises