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