Module 9 - Th é venin and Norton Equivalent Circuits

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

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

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

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

5. R1 iX R2 + _ vX Define portvariables vXand iX Vo Simple resistive circuit 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 ixflows to some load (not shown)

6. i1 R1 iX R2 i2 + _ vX Vo KCL:i1= i2 +iX Find an equation that relates vx to ix (Each resistor voltage expressed using Ohm’s Law) KVL:i1R1 + i2R2 = Vo Also note:vX = i2R2

7. (iX +vX/R2) R1 + vX = Vo R2 R1 R2vX = Vo–––––––– iX –––––––R1 + R2R1 + R2 Solve these equations for vXversus iX : i1R1 + i2R2 = Vo i1= iX +i2 vX = i2R2 (iX +i2) R1 + i2R2 = Vo Rearrange the variables… iX R1 +vX (R1/R2 + 1) = Vo or Vo– iX R1vX = ––––––––– 1 + R1/R2

8. R1 iX R2 + _ vX Vo R2 R1 R2vX = Vo–––––––– iX –––––––R1 + R2R1 + R2 Examine this last equation: It has the form vX = VTh– iX RTh R2 VTh =Vo –––––––R1/ + R2 R1 R2RTh= –––––––R1 + R2

9. + – RTh iXRTh iX VTh + _ vX Constructing the Thévenin Equivalent Circuit Write down KVL for this circuit: vX = VTh– iXRTh “Output voltage = voltage source – voltage drop across RTh”

10. R1 RTh iX iX R2 VTh + _ + _ vX vX Vo R1 R2RTh = ––––––= R1|| R2R1 + R2 and Actual Circuit: Model: R2 R1 R2vX = Vo–––––––– iX –––––––R1 + R2R1 + R2 vX = VTh– iXRTh Choose model parameters VTh and RTh: R2 VTh =Vo –––––––R1 + R2 • From the point of view of vX and iX, the Thévenin circuit models the actual circuit in every way.

11. R1 R1|| R2 R2 Vo Actual Circuit: iX + _ vX PORT Thévenin Equivalent: iX + _ R2 Vo –––––––R1 + R2 vX PORT

12. R1 Equivalent resistance R2 Vo Significance of RTh • Set all independent sources in the actual circuit to zero. • For a voltage source, that means substituting a short circuit. • Equivalent resistance RTh= R1||R2 • RTh is the equivalent resistance seen looking into the port with all independent sources set to zero.

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

14. Voltage between nodes fixed at 0 V byshort circuit LOAD Setting a Voltage Source to Zero

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

16. iX = 0 Open Circuit Voltage R1 RTh Open Circuit Voltage R2 VTh • iX automatically set to zero. Vo + _ KVL Significance of VTh + _ • Connect nothing to the port • Port voltage is called the open circuit voltage. iX = 0 + 0 V– • VTh represents the open circuit voltage of the actual circuit

17. R3 = 10 k R1=100 k + vLOAD – Vmic10 mV R2 = 30 k 50 k R4 =10 k Load Microphone network 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?

18. R3 = 10 k R1=100 k Find VThand RTh R2 = 30 k Vmic 10 mV Load Load R4 =10 k Solution Method: Find the Thévenin Equivalent of the Microphone Network • Disconnect the load. • Find Thevenin Equivalent remaining circuit. • Reconnect the load. • Find vLOAD from simplified circuit.

19. R3 = 10 k R1=100 k Vmic RTh R2 = 30 k RTh 43 k R4 =10 k Note: R1||R2 = (100 k)||(30 k) = 23 k Step 1: Find the Equivalent Resistance • Set the voltage source to zero. (Substitute a short circuit.) • Find the equivalent resistance RTh • RTh = R3 + R1||R2 + R4 = 10 k + 23 k + 10 k = 43 k

20. + – VTh = 2.3 mV 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*: v2 = VmicR2/(R1 + R2) = 2.3 mV *basically, voltage division R3 = 10 k R1=100 k R2 = 30 k Vmic 10 mV R4 =10 k • Note that no current flows through R3 and R4. •  Voltage across these resistors is zero.

21. RLOAD + vLOAD – 50 k Thévenin equivalent ofmicrophone network Answer Step 3: Reconnect the Load to the Thévenin Equivalent Model RTh = 43 k VTh2.3 mV From simple voltage division: vLOAD = VTh (RLOAD/(RLOAD + RTh) = 2.3 mV  (50 k)/(93 k) = 0.54 mV

22. More Examples: The Norton Equivalent Circuit

23. Isc = VTh/RTh 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 RTh VTh

24. Port Circuit Containing a Current Source Consider the following simple circuit: I1 R1 What is the Thévenin equivalent circuit seen looking intothe port?

25. Current is zero  + – Step 1: Find the open circuit voltage: + VTh – I1 R1 • Open circuit conditions  All of I1 flows through R1 • Voltage develops across R1 with polarity shown. • From Ohm’s Law: VTh = I1R1 (That part is simple…)

26. RTh Step 2: Find the equivalent resistance • Set the current source to zero. R1 I1 • Set the current source to zero  open circuit • Find the resistance looking into the port. • Trivially, by inspection: RTh = R1

27. R1 I1R1 I1 R1 Thévenin Equivalent Actual Circuit: RTh = R1 VTh = I1R1 The Thévenin Equivalent Circuit: • Done!

28. RN IN RN INRN Norton Circuit. Thévenin Circuit. Norton Equivalent 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

29. Isc = IN What about the short-circuit current from a Norton Circuit? • Apply a short circuit: + VN = 0 – RN 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.

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

31. RTh orRN Example: Resistive Network Find the Norton Equivalent of the following circuit using the short-circuit currentmethod R3 = 10 k R1=100 k Vmic 10 mV R2 = 30 k R4 =10 k Step 1: Find RTh (same asRN) by setting the source to zero. By inspection,RTh = R3 + R1||R2 + R4 = 10 k + 23 k + 10 k = 43 k

32. R1=100 k ISC = 0.54 A Vmic 10 mV R1=100 k IP = Vmic/(R1 + RP) = 0.9 A Vmic 10 mV RP = R2 || (R3 + R4) = 12 k From current division: 30 k50 k R2 [R2 + (R3 + R4)] = 0.54 A = ISC ISC = IP = 0.9 A Step 2: Apply a short circuit to the port and compute the short-circuit current. R3 = 10 k R2 = 30 k R4 =10 k

33. + vOC – IN=0.54 A RN = 43 k = 23 mV “Open Circuit Voltage” Find the Norton Equivalent of the Circuit ISC=0.54 A RN = 43 k vOC = IN RN= (0.54 A)(43 k) = 23 mV

34. RTh = 43 k VTh = 23 mV Construct the Thévenin Equivalent of the Circuit ISC=0.54 A RTh = 43 k VTh = ISC RTh = (0.54 A )(43 k) = 23 mV This result is the same one obtained in the previous example!

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

36. End of This Module Do the Homework Exercises