BME 372 Electronics I –J.Schesser
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Circuit Analysis
Lesson #2
BME 372 Electronics I –J.Schesser
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Voltage Division
• The voltage across impedances in series divides in proportion to the impedances.
Z1
Z2
a
b
c
21
2
2
21 Law sOhm' KVL );(
ZZZ
ZZZ
ac
bc
bc
bcabac
VV
IVIVVV
Z1
Zn
a
c
n
i
ac
i
ZZZZ
21VV
BME 372 Electronics I –J.Schesser
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Current Division• The current into impedances in parallel
divides in proportion to the inverse of the impedances.
c
21
2
21
11
11
2121
)/1()/1(1
Law sOhm' KCL );11(
ZZZ
ZZZ
Z
ZZ
ac
ac
II
VI
VIII
)/1()/1()/1()/1(
21 n
i
ac
i
ZZZZ
II
Iac
Z1 Z2
a
I1 I2 Z1 ZnI1 In
…
…
a
c
BME 372 Electronics I –J.Schesser
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Mesh Analysis
1. Define a current in each mesh (loop) of a network. For example, in a 5 mesh network, define 5 current unknowns.
2. Using KVL, write an equation for each mesh using the unknown currents. In our 5 mesh example, you’ll have 5 equations and 5 unknown currents.
3. Solve for the unknown currents and now apply these currents to the network to find the voltages for each impedance in the network.
BME 372 Electronics I –J.Schesser
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Nodal Analysis
1. Define a voltage at each node (junction point) of a network. For example, in a 5 node network, define 5 voltage unknowns.
2. Using KCL, write an equation for each node using the unknown voltages. In our 5 node example, you’ll have 5 equations and 5 unknown voltage.
3. Solve for the unknown voltages and now apply these voltages to the network to find the currents for each impedance in the network.
BME 372 Electronics I –J.Schesser
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Mesh Analysis Example
5Vdc 5Vdc10
5 5
I1I2
51 ;5
110155- );(10550
2Mesh 10155 ;5)(1050
1Mesh
21
12122
21211
II
IIIII
IIIII
5Vdc 5Vdc10
5 5
.2a .2a- 1v ++ 1v -
.4a+
4v
-
Mesh Analysis
I1(15)+I2(-10) = 5
I1(-10)+I2(15) = -5
I1 = 0.2, I2 = -0.2
clear all;R=[15 -10;-10 15];V=[5;-5];I=R\V;grid offaxis offtext(.1,1,'Mesh Analysis')text(.1,.9,['I1(',num2str(R(1,1)),')+I2(',num2str(R(1,2)),') = ',num2str(V(1))]);text(.1,.8,['I1(',num2str(R(2,1)),')+I2(',num2str(R(2,2)),') = ',num2str(V(2))]);text(.1,.7,['I1 = ',num2str(I(1)),', I2 = ',num2str(I(2))])
BME 372 Electronics I –J.Schesser
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Nodal Analysis Example
5Vdc 5Vdc10
5 5
V I2
422 ;0105
55
50
1Node
321
V
VVVVIII
5Vdc 5Vdc10
5 5
.2a .2a- 1v ++ 1v -
.4a+
4v
-
I1
I3
clear all;R=[1/5+1/5+1/10];VS=[10/5];V=R\VS;I1=(5-V(1))/5;I2=(5-V(1))/5;I3=-V(1)/10;grid offaxis offtext(.1,1,'Nodal Analysis')text(.1,.9,['V = ',num2str(V)]);text(.1,.8,['I1 = ',num2str(I1),', I2 = ',num2str(I2),', I3 = ',num2str(I3)]);
Nodal Analysis
V = 4
I1 = 0.2, I2 = 0.2, I3 = -0.4
BME 372 Electronics I –J.Schesser
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Superposition
• Used to analyze a circuit with multiple sources.• Steps:
1. Set all sources except for one to zero (voltage sources are shorted-circuited, current sources are open-circuited)
2. Solve for the currents and voltages for all of the circuit elements
3. Repeat steps 1-2 for the remaining sources.4. Add each of the solutions to obtain the solution for
the entire circuit
BME 372 Electronics I –J.Schesser
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Superposition Analysis Example
5Vdc 5Vdc10
5 5
5Vdc 10
5 5
53
2515
3105
5
33.3310
1550
5105*105||10
11
s
p
I
R
5Vdc 3.33
5
I1s1
5Vdc 105 5
Source 1
I1s1=.6 I2s1=-.4
I3s1=-.2
2.51
53
155
4.52
53
1510
13
12
s
s
I
I
I1I2I3
I1s1I2s1I3s1
BME 372 Electronics I –J.Schesser
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Superposition Analysis Example
5Vdc 5Vdc10
5 5
5Vdc10
55
53
2515
3105
5
33.3310
1550
5105*105||10
22
s
p
I
R
5Vdc3.33
5
I2s2
5Vdc10
55
Source 2
I1s2=-.4 I2s2=.6
I3s2=-.2
2.51
53
155
4.52
53
1510
23
21
s
s
I
I
I1I2I3
I1s2I2s2I3s2
BME 372 Electronics I –J.Schesser
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Superposition Analysis Example
5Vdc 5Vdc10
5 5
.2a .2a- 1v ++ 1v -
.4a+
4v
-
• Summing the results of each solution:
4.2.2.2.6.4.
2.4.6.
23133
22122
21111
ss
ss
ss
IIIIIIIII
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Thevenin and Norton Equivalent Circuits
• Thevenin’s Theorem: Any circuit consisting of passive and active components can be represented by a voltage source in series with an equivalent set of passive components– The value of the voltage source equals the voltage seen
at the output terminal without any load connected to it, i.e., the open-circuit voltage
– The value of the equivalent set of passive components equals the impedance looking back into the terminals with the sources set to zero, i.e., the output impedance.
BME 372 Electronics I –J.Schesser
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Thevenin and Norton Equivalent Circuits
• Norton’s Theorem: Any circuit consisting of passive and active components can be represented by a current source in parallel with an equivalent set of passive components– The value of the current source equals the current seen at the
output terminal shorted and without any load connected to it, i.e., the short-circuit current
– The value of the equivalent set of passive components equals the impedance looking back into the terminals with the sources set to zero, i.e., the output impedance.
• Note that the Thevenin and Norton Equivalents Circuits are equivalent to each other when the value of the Thevenin’s voltage source equals the product of the equivalent impedance times the Norton’s current source
BME 372 Electronics I –J.Schesser
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Thevenin and Norton Examples
5
10Vdc 15
a
b
75.3415
2015515||5
ImpedanceOutput
5.71015515
ab :lsat termina VoltageCircuit Open
o
abOC
R
vV
5
15
a
b
Ro=3.75
7.5Vdc
3.75a
b
BME 372 Electronics I –J.Schesser
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Thevenin and Norton Examples
5
10Vdc 15
a
b
Short Circuit Current at terminals: ab10 25
Output Impedance5 15 155 ||15 3.75
20 4
abSC
o
I a
R
a
b
5
10Vdc 15
a
b
2Adc3.75
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Input and Output Impedance of a Circuit
• Input impedance of a circuit is the impedance looking into the input terminals of the circuit with any load connected to the output and all internal sources are set to zero.
• Output impedance of a circuit is the impedance looking into the output terminals of the circuit without any load connected to the output and all internal and input sources are set to zero.
BME 372 Electronics I –J.Schesser
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Examples
I2 I1 1 2
1
1
1 1
2 4 24
1 2 21 4 42 8
kBRANCH
kSOURCEBRANCH
CURRENTSOURCE kBRANCH kSOURCEBRANCH
I I a
V a VV a VV V V V
- 2V + + 2V - + 2V
-
-4V
+4Adc
1
1
11
1
+ 2V
-
-8V
+
Find the currents and voltages in these circuits
1
1
1
11
4Adc
BME 372 Electronics I –J.Schesser
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Examples
5Vdc R4=1
4Adc
R1=1
R3=1 R2=1 I1I2
5Vdc1
1
1 1
1
4Adc
1
1 1 I1s1I2s1I1s2I2s2
0
67.135
23
2122
s
ss
I
AII
AAkkI
AAkkI
AI
s
s
s
33.1431
67.2432
4
11
12
13
I3
I3s2 I3s1
5Vdc 1
4Adc
1
1 1
I1=3I2=-1
I3=4
VVVVVVV
VVVVV
AIAI
AI
RRRRCECURRENTSOU
R
R
RR
10334414
111313
440167.167.2
367.133.1
124
4
3
21
3
2
1
+ 3V-
+ 3V -- 1V + -4V+
+ 10V
-
BME 372 Electronics I –J.Schesser
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ExamplesR2
1k
R1
1k
V=1Vac C1
1n
C2
1n
a
b
c
ZR1=R1
ZC1=1/jωC1
a
b
ZR2=R2
ZC2=1/jωC2
+
-V
c
VZZZZZZZ
ZZZVZ
ZZZZZZ
ZZZZZZ
V
ZZZZZZZZZ
ZZZZ
VZZ
ZV
VZZ
ZV
RccRRcc
Rcc
RRcc
Rcc
Rcc
Rcc
cb
Rcc
RccRcc
Rcccb
Rcb
cbcb
cbcR
cab
)()()(
)(
)(
)()(||
ofn combinatio series with theparallelin
division voltageusing (2) ;
division voltageusing (1) ;
2211221
221
1221
221
221
221
221
221221
221
1
22
2
Find the voltage Vab in terms of V
BME 372 Electronics I –J.Schesser
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ExamplesR2R1
1k1k
V=1Vac C1
1n
C2
1n
a
b
c
)(111
11
)()()(
(1) into (2) ngSubstituti
22211121212
21
2
1
1
1
2
2
21
21
2121112111
21
2211211
221
22
2
CRCRCRjRRCCVV
RRCj
RCj
RCj
RCjCj
CjCjV
VZZZZZZZZZZ
ZZ
VZZZZZZZ
ZZZZZ
ZV
ab
RRcRcRRccc
cc
RccRRcc
Rcc
cR
cab
Find the voltage Vab in terms of V
ZR1=R1
ZC1=1/jωC1
a
b
ZR2=R2
ZC2=1/jωC2
+
-V
c
BME 372 Electronics I –J.Schesser
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Homework
• Voltage and Current division– How does the voltage divide across two capacitors in series? Show
your results.– How does the current divide among two capacitors in parallel?
Show your results.
• Calculate the Currents and Voltages for the following circuits:
1
10Adc
2
1
25Vdc
2Vdc
1
1 10Adc2
1
10Adc
2
1
2Vdc
BME 372 Electronics I –J.Schesser
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HomeworkCalculate the current labeled i and the voltage labeled v in the following circuit R1 = 1Ω, R2 = 2Ω, R3 = 1Ω, R4 = 1Ω, R5 = 2Ω, R6 = 2Ω, R7 = 2Ω, Vcc = 4v
R1
R7
R2
R3
R4
R6
R5+VCC
--i
+v--
BME 372 Electronics I –J.Schesser
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Homework
Calculate the current labeled, i.
R1 = 2Ω, R2 = 2Ω, R3 = 2Ω, R4 = 3Ω, Vcc = 2v
i4
R1R2
R3
R4
+VCC
--i
BME 372 Electronics I –J.Schesser
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Homework
An electrode is connected to an oscilloscope which has a purely capacitance input impedance, CIN. Find and plot the output voltage Vab(jω) as function of ω. Use Matlab to perform the plot.
RS 500RD 29.5k
CD 53n
a
b
1Vac CIN 5.3n
BME 372 Electronics I –J.Schesser
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Homework
• Repeat the analysis of this circuit using Mesh and Nodal Analysis. That is find and plot Vab as a function of frequency. Use Matlab to perform the plot.
a
b
R2
5k
R1
1k
1VacC1
1n
C2
5n
BME 372 Electronics I –J.Schesser
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Homework• Repeat the analysis of this circuit. That is find and
plot Vout / Vin as a function of frequency. Assume Va = 0; C=10nF,R1=2.2MΩ, R2 =330kΩ, R3=2.7MΩ. Perform the plot using Matlab
VinVoutR1
R2 R3
C
C
Va