Lecture 36 1

Thevenin Equivalent Circuits

Lecture 36 2

Review: Independent Source(s)

Circuit with one or more independent

sources

RTh

Voc

+

-

Thevenin equivalent circuit

Lecture 36 3

Review: No Independent Sources

Circuit without independent sources

RTh

Thevenin equivalent circuit

Lecture 36 4

Finding the Thevenin Equivalent

• I-V characteristics are straight lines (a consequence of the linearity of resistor models).

• Circuits with independent sources:

– Intercepts are voc(t) and isc(t)

• Circuits w/o independent sources:

– Slope is RTh

Lecture 36 5

No Independent Sources

v(t)

i(t)

Slope is RTh

Lecture 36 6

Finding RTh

• With no independent sources, Voc and Isc are both zero.

• To find the slope, we apply a test source with a given voltage (or current) and measure the current (voltage) produced by the source.

• This gives the slope of the I-V line.

Lecture 36 7

Example: CE Amplifier

1kVin

+

-2k

+10V

+

-Vo

Lecture 36 8

Small Signal Equivalent

1kVin 100Ib

+

-

Vo

50

Ib

2k+

-

Lecture 36 9

Thevenin Equivalent @ Input

100Ib

50

Ib

2k

RTh

Lecture 36 10

Apply a Test Current Source

100Ib

50

Ib

2k1mA

+

-

Vx

Lecture 36 11

Solve for Vx

Use Nodal Analysis

100Ib

50

Ib

2k1mA

Vx V1

Lecture 36 12

Solve for Vx

mA150

1 VVx

0k2

100-50

11

VI

VVb

x

50

1VVI x

b

Lecture 36 13

Compute RTh

V05.202xV

k202mA1

xTh

VR

Lecture 36 14

Apply a Test Voltage Source

100Ib

50

Ib

2k1V

+

-

V1

Lecture 36 15

Solve for Ib

0k2

100-50

V1 11

VI

Vb

50

V1 1VIb

V999753.01 V

Lecture 36 16

Compute RTh

A949.4 bI

k202V1

bTh I

R

Lecture 36 17

AC Steady State

• Thevenin’s theorem also applies to AC steady state analysis.

• An arbitrary linear circuit can be replaced by an equivalent source and impedance.

• The determination of source and impedance values is essentially the same as for resistor circuits.

Lecture 36 18

Independent Source(s)

Circuit with one or more independent

sources

Voc

+

-

Thevenin equivalent circuit

sc

ocTh I

VZ

Lecture 36 19

No Independent Sources

Circuit without independent sources

ZTh

Thevenin equivalent circuit

Lecture 36 20

Example: IC Interconnects

0.07pF

500Vs

+

-

15 15 15

0.07pF 0.07pF

Find the Thevenin equivalent circuit:

Lecture 36 21

Find Impedances, then Voc and Isc

= 1010

500Vs

+

-

15 15 15

-j1.43k-j1.43k -j1.43k

Lecture 36 22

Source Transformations

515 15 15

-j1.43k

515sV

-j1.43k -j1.43k

Lecture 36 23

Source Transformations Cont.

15 15 sj V319.885.

-j1.43k -j1.43k

+

-

455-j164

Lecture 36 24

Source Transformations Cont.

15

sj

jV

164470

319.885.

-j1.43k

-j1.43k

470-j164

Lecture 36 25

Source Transformations Cont.

15

sj V479.652.

-j1.43k

348-j250+

-

+

-

Voc

Lecture 36 26

Source Transformations Cont.

sj

jV

250363

479.652.

-j1.43k

363-j250 +

-

Voc

Lecture 36 27

Source Transformations Cont.

sj V503.446. 251-j267+

-

Voc

+

-

Lecture 36 28

Voc

Voc = (0.446-j0.503) Vs

Lecture 36 29

Isc

Isc = (0.00184-j0.00005) Vs

Lecture 36 30

ZTh

267251 jsc

ocTh I

VZ