Basis for Thevenin and Norton Equivalent Circuits

Objective of Lecture Describe the differences between ideal and real voltage

and current sources Chapter 8.1 and 8.2 Principles of Electric Circuits

Demonstrate how a real voltage source and real current source are equivalent so one source can be replaced by the other in a circuit. Chapter 6.6 Electric Circuit Fundamentals Chapter 8.3 Principles of Electric Circuits Chapter 4.4 Fundamentals of Electric Circuits Chapter 2.6 Electrical Engineering: Principles and Applications

Voltage Sources Ideal Real An ideal voltage source has

no internal resistance. It can produce as much

current as is needed to provide power to the rest of the circuit.

A real voltage sources is modeled as an ideal voltage source in series with a resistor. There are limits to the current

and output voltage from the source.

Limitations of Real Voltage Source

Real Voltage Source

LLL

SSL

LL

RVI

VRR

RV

/=+

=VL

IL

RL

Voltage Source Limitations (con’t) RL = 0Ω RL = ∞Ω

W0 /

V0

max

==

=

L

SSL

L

PRVI

V

W0PA 0min

==

=

L

L

SL

IVV

Current Sources Ideal Real An ideal current source has

no internal resistance. It can produce as much

voltage as is needed to provide power to the rest of the circuit.

A real current sources is modeled as an ideal current source in parallel with a resistor. Limitations on the maximum

voltage and current.

Limitations of Real Current Source Appear as the resistance of the load on the source

approaches Rs.

Real Current Source

LLL

SSL

SL

RIV

IRR

RI

=+

=VL

IL

RL

Current Source Limitations (con’t) RL = 0Ω RL = ∞Ω

W0 V0min

==

=

L

L

SL

PV

II

W0P

A0

L

max

==

=

SSL

L

RIVI

Electronic Response For a real voltage source, what is the voltage across the

load resistor when Rs = RL? For a real current source, what is the current through

the load resistor when Rs = RL?

Equivalence An equivalent circuit is one in which the i-v

characteristics are identical to that of the original circuit. The magnitude and sign of the voltage and current at a

particular measurement point are the same in the two circuits.

Equivalent Circuits RL in both circuits must be identical.

IL and VL in the left circuit = IL and VL on the left

Real Current Source Real Voltage Source

1

2 VL

IL

RL VL

IL

RL

Example #1 Find an equivalent current source to replace Vs and Rs

in the circuit below.

VL

IL

RL

Example #1 (con’t) Find IL and VL.

mWPmAVVmAVP

PPP

mAkVIRVI

VVkk

kV

VRR

RV

Vs

Vs

RsLVs

L

LLL

L

SSL

LL

36)2)(1218()2(12

26/12/

121836

6

=−+=

+=

=Ω==

=Ω+Ω

Ω=

+=

VL

IL

RL

Example #1 (con’t) There are an infinite number of equivalent circuits that

contain a current source. If, in parallel with the current source, Rs = ∞Ω

Rs is an open circuit, which means that the current source is ideal.

VL

IL

RL

mWPPmWmAVIVP

VkmAVII

IsL

LLL

L

LS

2424)2(12

12)6(2

=====

=Ω==

Example #1 (con’t) If RS = 20kΩ

mWPmAmAVmAVP

IVIVPPPVRIVV

mAmAkkkI

IRRRI

Is

Is

RsRsLLRsLIs

LLIsL

S

LS

SLS

0.32)267.2(12)2(12

12

67.2220

206

=−+=

+=+====

=Ω

Ω+Ω=

+=

Example #1 (con’t) If RS = 6kΩ

mWPmAmAVmAVP

IVIVPPPVRIVV

mAmAkkkI

IRRRI

Is

Is

RsRsLLRsLIs

LLIsL

S

LS

SLS

48)24(12)2(12

12

426

66

=−+=

+=+====

=Ω

Ω+Ω=

+=

Example #1 (con’t) If RS = 3kΩ

mWPmAmAVmAVP

IVIVPPPVRIVV

mAmAkkkI

IRRRI

Is

Is

RsRsLLRsLIs

LLIsL

S

LS

SLS

72)26(12)2(12

12

623

36

=−+=

+=+====

=Ω

Ω+Ω=

+=

Example #1 (con’t) Current and power that the ideal current source needs to

generate in order to supply the same current and voltage to a load increases as RS decreases. Note: Rs can not be equal to 0Ω. The power dissipated by RL is 50% of the power

generated by the ideal current source when RS = RL.

Example #2 Find an equivalent voltage source to replace Is and Rs

in the circuit below.

Example #2 (con’t) Find IL and VL.

mWPmAmAV

mAVPPPP

VmAVRIV

mAI

II

Vs

Vs

RsLVs

L

LLL

L

SL

07.1)714.05(214.0

)714.0(214.0

214.0)300(714.0

714.050300

50

=−+

=+=

=Ω==

=Ω+Ω

Ω=

Example #2 (con’t) There are an infinite number of equivalent circuits

that contain a voltage source. If, in series with the voltage source, Rs = 0Ω

Rs is a short circuit, which means that the voltage source is ideal.

mWPPmWP

mAVIVPmAI

VRVIVVV

VsL

L

LLL

L

LLL

LS

153.0153.0

)714.0(214.0714.0

300/214.0/214.0

===

===

Ω====

Example #2 (con’t) If RS = 50Ω

mWPmAVV

AVPIVIVPPPmARVII

VVV

VRRRV

Vs

Vs

RsRsLLRsLVs

LLVsL

S

LL

SLS

179.0)714.0)(214.025.0(

)714.0(214.0

714.0/

25.0214.0300

50300

=−+

=+=+=

===

=Ω

Ω+Ω=

+=

Example #2 (con’t) If RS = 300Ω

mWPmAVV

AVPIVIVPPPmARVII

VVV

VRRRV

Vs

Vs

RsRsLLRsLVs

LLVsL

S

LL

SLS

306.0)714.0)(214.0418.0(

)714.0(214.0

714.0/

418.0214.0300

300300

=−+

=+=+=

===

=Ω

Ω+Ω=

+=

Example #2 (con’t) If RS = 1kΩ

mWPmAVV

AVPIVIVPPPmARVII

VVkV

VRRRV

Vs

Vs

RsRsLLRsLVs

LLVsL

S

LL

SLS

662.0)714.0)(214.0927.0(

)714.0(214.0

714.0/

927.0214.0300

1300

=−+

=+=+=

===

=Ω

Ω+Ω=

+=

Example #2 (con’t) Voltage and power that the ideal voltage source needs to

supply to the circuit increases as RS increases. Rs can not be equal to ∞Ω. The power dissipated by RL is 50% of the power

generated by the ideal voltage source when RS = RL.

Summary An equivalent circuit is a circuit where the voltage

across and the current flowing through a load RL are identical. As the shunt resistor in a real current source decreases in

magnitude, the current produced by the ideal current source must increase.

As the series resistor in a real voltage source increases in magnitude, the voltage produced by the ideal voltage source must increase. The power dissipated by RL is 50% of the power produced by

the ideal source when RL = RS.