Circuit Analysis Lesson #2joelsd/electronics/classwork/electronics1/BME373... · Circuit Analysis Lesson #2. ... • Used to analyze a circuit with multiple sources. •Steps: 1.

BME 372 Electronics I –J.Schesser

67

Circuit Analysis

Lesson #2


68

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


69

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


70

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.


71

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.


72

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))])


73

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


74

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


75

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


76


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


77


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


78

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.


79

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


80

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


81

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


82

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.


83

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


84

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

-


85

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


86

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


87

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


88

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--


89

Homework

Calculate the current labeled, i.

R1 = 2Ω, R2 = 2Ω, R3 = 2Ω, R4 = 3Ω, Vcc = 2v

i4

R1R2

R3

R4

+VCC

--i


90

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


91

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


92

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

Circuit Analysis Lesson #2joelsd/electronics/classwork/electronics1/BME373... · Circuit Analysis Lesson #2. ... • Used to analyze a circuit with multiple sources. •Steps: 1.

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