Linear Circuit Thoery

7/27/2019 Linear Circuit Thoery

1/41

Review: Linear Circuit Theory

Linear circuits - Resisitors, capacitors, inductors, linearly dependent

sources and independent sources

Nonlinear circuits - diodes, transistors, etc

- most linear circuits can be modelled as one or more linear circuits

diode MOSFET BJT

independent voltage

sources

independent

current

source

resistor capacitor inductor


2/41

Resistors:

Devices defined by their voltage vs their current (i.e.: voltage across vs currentthrough)

V=IR (Ohms Law)

I = V/R = 10V/10 = 1A

* note: current is positive whenflowing into positive

terminal


3/41

Demonstration of Flashlight

I = V/R = 1V/100= 0.01A = 10mA


4/41

Demonstration of a Flashlight (cont)

-I = 5V/100K = 0.00005A= 50mA

I = -50mA

-I = V/R = 10V/1 = 10A


5/41

Demonstration of a Flashlight (cont)

I = -100V/10 = -10A

I = 10V/10 = 1A


6/41

Graphical Solution

I = V/R ----------- Y = mX + B

m = 1/R Equation of a straight line with

B = 0 slope m and Y-intercept of B


7/41

V = V + V

V = IR + IR V = (R + R )

I =V

(R + R )

I =10V

(10 + 20 )

=10V

30= 0.333A

V = IR = (0.333A)(10 )

= 3.33V

V = IR = (0.333A)(20 )

= 6.67V

S 1 2

S 1 2

S 1 2

S

1 2

R1 1

R2 2

V = V = V I = V(1

10W+

1

20W)

I + I = I I = 10V(3

20W)

V

R+

V

R= I I = 3/ 2A = 1.5A

V

R

+V

R

= I I =V

R

=10V

10W

= 1A

V

R=

10V

10W1A I =

V

R=

10V

20= .5A

V(1

R+

1

R) = I

1 2

1 2

1

1

2

2

1 2

1

1

1

2

2

1 2

=

V2

V1 +

-

V1

V2-

+


8/41

In General:

Linear circuits can be represented by equations of the form

V = AI1 + BI2 +...

or I = CV1 + DV2 + ...

where A, B, C, and D are constants or other linear operators


9/41

Circuit Cellar

Calculate the effective resisitence of the following resistor network:

R =(10

+ 100 )=

100

110= 9.09T

2

)( )

(

100

10


10/41

Kirchhoffs Voltage and Current Laws

1. The sum of voltages encountered around any closed loop in a circuit is

equal to 0 (KVL)

2. The sum of all currents entering any node of a circuit is equal to 0

(KCL)


11/41

Loop A: V - V - V = 0

V = V + V (KVL)

Loop B: V - V - V - V = 0

V = V + V + V (KVL)Node 1: I = I + I

Node 2: I + I = I

From Loop A: V = I R + I R = I (R + R )

I =V

(R + R )= 10V

(10 + 20 )= 10V

30= 0.333A

Similarly from Loop B:

V =

0 1 2

0 1 2

0 3 4 5

0 3 4 5

0 1 2

1 2 0

1 1 1 2 2 1 1 2

1 0

1 2

0

I R + I R + I R + I (R + R + R )

I =

V

(R + R + R ) =

V

(30 + 40 + 50 ) =

10V

120 = 0.0833A

I = I + I = 0.333A + 0.0833A = 0.416A

V = I R = (0.0833A)(40 ) = 3.33V

2 3 2 4 2 5 2 3 4 5

20

3 4 5

0

0 1 2

4 2 4


12/41

Alternately:


13/41

R =(30 )(120 )

(30 + 120 )=

3600

150= 24T

2

I = 10V / 24 = 0.416A0

Alternative Solution (cont)


14/41


15/41

Consider:

Calculate IS


16/41

From N1

I = I + I (N1)

I + I = I (N2)

I = I + I (N3)

I + I = I (N4)

V = V + V = I R + I R (Loop A)

V = V + V = I R + I R (Loop B)

I

S 1 2

1 3 4

2 3 5

4 5 S

S R1 R4 1 1 4 4

S R2 R5 2 2 5 5

4 = I + I

V = V + V + V = I R + I R + I R

I = I - I (from N2)

I = I + I (from N3)

V = I R + I R + I R

V = I R + I R - I R

3 1

S R2 R3 R4 2 2 3 3 4 4

5 2 3

4 1 3

S 1 1 1 4 3 4

S 2 2 2 5 3 5


17/41

V = I + (R + R ) + I R

V = I (R + R ) - I R

V = I R + I R + I (R + R )

I =V - I R

(R + R )= 0.5A - (0.5)(I )

I =V + I R

(R + R )= 0.5A + (0.5)(I )

V = 5 - 5I + 5I + 5 + 5I + 20I

20I = 0

I = 0

I = 0.5A

I = 0.5A

I = 1A

S 1 1 4 3 4

S 2 2 5 3 5

3 1 4 2 2 3 3 4

1S 3 5

1 43

2S 3 5

2 53

S 3 3 3 3

3

3

1

2

S


18/41

Find I3


19/41

I1 = I2 + I3

-5V - 100I1 - 10I3 = 0

-5V - 100I1 - 110I3 = 0

-100I2 - 110I3 = 5V

-210I2 - 100I3 = 5V

I2 =5 + 110I3

-210

-100+ 100I3

-210- 110I3 = 5V

5

2.1+

100

2.1I3 - 110I3 = 5V

-62.4I3 = 5 - 2.38 = 2.62V

I3 = - 0.042A = - 42mA

I2 =5V - 4.2V

-210= - 0.0038A = - 3.8mA

I1 = - 42mA - 3.8mA = - 45.8mA

5


20/41

Alternate Solution

I = V/R = 5V/109.17 = 45.8mA


21/41


22/41

I + I = I

LOOP1: 10V - 10I - 100I = 0

LOOP2: 20V - 10I - 100I = 0

LOOP2: 20 - 110I - 10I = 0

I =-20 + 110I

-10= 2 - 11I

LOOP1: 10 - 10 (2 - 11I ) +100I = 0

10 - 20 + 110I + 100I = 0

201I = 10

I = 10 / 210 = 0.0476A = 47.6mA

I = 2 - 11(0.0476) = 2 - 0.524 = 1.476A

I = I + I = 1.476 + 0.0476 =

1 2 3

2 1

3 1

1 2

21

1

1 1

1 1

1

1

2

3 1 2 1.52A

30V

20

= 1.5A


23/41

If R = 100 ?

LOOP2: 20 - 100I -100I = 0

20 - 200I - 100I = 0

I = -20 + 200I-100

= 0.2 - 2I

LOOP1: 10 - 10 (0.2 - 2I ) + 100I = 0

10 - 2 + 120I = 0

120I = -8

I = - 0.0667 = - 66.7mA

3

3 1

1 2

21

1

1 1

1

1

1


24/41

Current Source

VR1 = IR1 = (4A)(10) = 40V

VR2 = (4A)(100) = 400V


25/41

Current Source (cont)

Find VIS

+

VIS-


26/41

Current Source (cont)

I = I + I

I = I + 500mA

LOOP1: 5V - 100I - V = 0

LOOP2: 5V - 100I - 1000I = 0

LOOP2: 5V - 1100I - 50V = 0

I =45V

1100= - 0.041A = 41mA

I = - 41mA + 500mA = 459mA

V = (I )(R ) = (-41mA)(1000 ) = -41V

1 2 3

1 2

1 IS

1 2

2

2

1

IS 2 2


27/41

Characteristics of a VI Port

- linear circuits with straight line graphical solutions


28/41

Characteristics of VI Ports (cont)

ISC = short circuit current = VS/VR

VOC = open circuit voltage = VS

V - I R - V = 0

I = V - V-R

I =-1

RV +

V

R

y = mx + B

S L S L

LL S

S

L

S

LS

S


29/41

Thevenin Equivalent Circuits

Circuits containing multiple resistors, and voltage and current source can

be modelled as a simpler ciruit containing a single voltage source andresistor.

The circuits are identical in every aspect and have same V-I characteristics


30/41

Thevenin Equivalent Circuits (cont)


31/41

I = I + I

Loop 1: V - I R - V = 0

V = V - I R

Loop 2: V - I R = 0

V = I R

I = V / R

From Loop 1V = V - (I + I )R

V = V - (V

R+ I )R

V (1 + R / R ) = V - I R

I = V / R -1

R(1 +

R

RV

I

1 2 L

1 1 1 L

2 1 1 1

L 2 2

L 2 2

2 L 2

L 1 2 L 1

L 1L

2

L 1

L 1 2 1 L 1

L 1 1

1

1

2

L

L

)

= -R + R

R RV + V / R 2 1

1 2

L 1 1



32/41


I = -1

RV +

V

R

1

R=

R + R

R R

R =R R

R + R

V

R=

V

R

V =V

R

R R

R + R2

=V R

(R + R )

L

TH

LTH

TH

TH

1 2

1 2

TH 1 2

1 2

TH

TH

1

2

TH1

1

1 2

1

1 2

1 2


33/41

R || R = 20 || 30 = 12

I =10V

22= 0.4545A

V = (12 0.4545) = 5.45V

I =V

R=

5.45V

30= 0.182A

R =(20 10 )

(20 + 10 )= 6.67

V = (10V)(20 )(20 + 10 )

= 6.67V

2 L

1

L

LL

L

TH

TH

I =V

R + R=

6.67V

(6.67 + 30 )= 0.182A

V = I R = (0.182A)(30 ) = 5.45V

LTH

TH L

L L L


34/41

Thevenin Rules

Thevenin Rules are used to:

- find the Thevenin equivalent of any two-terminal resistive circuit

1. Disconnect and elements not to be included in the circuit

2. Find the open circuit voltage at the port. The Thevenin VTH is equal

to the open circuit voltage.

3. Set the independent sources to zero (voltage source look like short

circuits and current sources look like open circuits).

Use series/parallel resistors combinations to find the net resistance

RTH


35/41

We found V = - 41V

I = - 41mA

R2

R2


36/41

Disconnect R2 and find the open circuit voltage VOC

Set independent sources to 0 and find resistance R0

V = R I = 50V

V = 5V - 50V = - 45V

V = V = - 45V

R1 1 S

OC

TH OC

R = R = R

I =V

R + R

=-45V

1100

= - 0.041A= - 41mA

V = (1000)(-41mA)

= - 41V

0 1 TH

R2TH

TH 2

R2


37/41

Found I1 = 47.6mA


38/41

Using Thevenin

VOC = VTH

I =V + V

R + R=

30V

20= 1.5A

V = IR = (1.5A)(10 ) = 15V

V = V = V - V = 10V - 15V = -5V

1 2

1 2

R1 1

OC TH 1 R1


39/41

R = R || R = 10 || 10 = 5TH 1 3

I =V

R + R

=-5V

105

= - 47.6mATH 2


40/41

Voltage and Current Division

Shortcut used in the solution of resistive circuits

V = V

I =V

R=

V

R + R

V = V = IR =V

R + RR

X R2

0

EQ

0

1 2

X R2 20

1 2

2

What is the voltage at node X?


41/41

Voltage and Current Divsion (cont)

V =R

R + R + R VR2

2

1 2 3

S

Linear Circuit Thoery

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