Announcements

22
Announcements • First Assignment posted: – Due in class in one week (Thursday Sept 15 th ) http://www.physics.udel.edu/~jholder/Phys645/index.htm

description

Announcements. First Assignment posted: Due in class in one week (Thursday Sept 15 th ). http://www.physics.udel.edu/~jholder/Phys645/index.htm. Lecture 3 Overview. Loop Analysis with KVL & KCL Mesh Analysis Thevenin/Norton equivalent circuits. - PowerPoint PPT Presentation

Transcript of Announcements

Page 1: Announcements

Announcements• First Assignment posted:

– Due in class in one week (Thursday Sept 15th)

http://www.physics.udel.edu/~jholder/Phys645/index.htm

Page 2: Announcements

Lecture 3 Overview• Loop Analysis with KVL & KCL• Mesh Analysis• Thevenin/Norton equivalent circuits

Page 3: Announcements

Circuit analysis method 2a: KVL and KCL

Kirchoff’s Voltage Law: Loop analysisThe sum of the voltages around a closed loop must be zero

• Draw the current direction in every branch (arbitrary) and label the voltage directions (determined by the defined current direction). Voltage on a voltage source is always from positive to negative end.

• Define either clockwise or counter-clockwise as positive direction for summing voltages. Once the direction is defined, use the same convention in every loop. Voltage across a resistor is +’ve if voltage direction the same as current direction, -’ve otherwise

• Apply KVL

Page 4: Announcements

Kirchoff’s Voltage Law: MultiloopThe sum of the voltages around a closed loop must be zero

• Draw the current direction (arbitrary) and label the voltage directions (determined by the defined current direction).

• Define either clockwise or counter-clockwise as positive voltage direction. Once the direction is defined, use the same convention in every loop.

• Apply KVL

R3

00

32

21

VVVVVr

03322

221

RIRIRIIRIr

Say r=1Ω, R1=3Ω, R2=5Ω, R3=10Ω, ε=3VWhat are the currents?

Page 5: Announcements

Kirchoff’s Current LawThe sum of the current at a node must be zero: Iin=Iout

R3

I=I2+I3 (1)ε=Ir+IR1+I2R2 (2)I3R3-I2R2 =0 (3)

1I- 1I2- 1I3 = 0 (4)4I+5I2+ 0I3= 3 (5)

Say r=1Ω, R1=3Ω, R2=5Ω, R3=10Ω, ε=3V

0I- 5I2+10I3= 0 (6)

Page 6: Announcements

Last note on KCL KVL analysis

• If solutions to currents or voltages are negative, this just means the real direction is opposite to what you originally defined

• To deal with current sources: current is known, but assign a voltage across it which has to be solved

Page 7: Announcements

Another Sample Problem: Multiple Sources

Page 8: Announcements

Method 2b: Mesh AnalysisExample: 2 meshes

(Mesh is a loop that does not contain other loops)

Step 1: Assign mesh currents clockwise

Step 2: Apply KVL to each mesh

• The self-resistance is the effective resistance of the resistors in series within a mesh. The mutual resistance is the resistance that the mesh has in common with the neighbouring mesh

• To write the mesh equation, evaluate the self-resistance, then multiply by the mesh current

• Next, subtract the mutual resistance multiplied by the current in the neighbouring mesh for each neighbour.

• Equate the above result to the driving voltage: taken to be positive if it tends to push current in the same direction as the assigned mesh current

Mesh1: (R1+R2)I1 - R2I2 =ε1-ε2

Mesh2: -R2I1+ (R2+R3)I2 =ε2-ε3

Step 3: solve currents

Page 9: Announcements

Method 2b: Mesh Analysis

Page 10: Announcements

Another mesh analysis exampleFind the currents in each branch

Step 1: Replace any combination of resistors in series or parallel with their equivalent resistance

Step 2: Assign mesh currents clockwise

Step 3: Write the mesh equations for each mesh

Left mesh: 11I1-6I2=9

Right mesh: -6I1+18I2=9

Note:suppressed “k” for each resistor, so answer is in mA

Step 4: Solve the equations

Solution: I1=4/3mA =1.33mA

I2=17/18mA =0.94mA

Page 11: Announcements

Mesh analysis with a current sourceMagnitude of current in branch containing current source is IS , (although if the current flow is opposite to the assigned current direction the value will be negative).

This works only if the current source is not shared by any other meshFor a shared current source, label it with an unknown voltage.

Page 12: Announcements

Mesh analysis with mixed sources• Find Ix • Identify mesh currents and label accordingly• Write the mesh equationsMesh1: I1 =-2Mesh2: -4I1+8I2 -4I4=12Mesh3: 8I3 =-12Mesh4: -2I1-4I2 +6I4=10

I1= -2.0AI2= 1.5AI3= -1.5AI4= 2.0A

Ix=I2-I3

Ix=3.0A

Page 13: Announcements

Method 3: Thevenin and Norton Equivalent Circuits

vTH= open circuit voltage at terminal (a.k.a. port)

RTH= Resistance of the network as seen from port(Vm’s, In’s set to zero)

Any network of sources and resistors will appear to the circuit connected to it as a voltage source and a series resistance

Page 14: Announcements

Norton Equivalent Circuit

Any network of sources and resistors will appear to the circuit connected to it as a current source and a parallel resistance

Page 15: Announcements

Calculation of RT and RN

• RT=RN ; same calculation (voltage and current sources set to zero)• Remove the load.• Set all sources to zero (‘kill’ the sources)

– Short voltage sources (replace with a wire)– Open current sources (replace with a break)

Page 16: Announcements

Calculation of RT and RN continued• Calculate equivalent resistance seen by the load

Page 17: Announcements

Calculation of VT

• Remove the load and calculate the open circuit voltage

SROC VRR

RVV21

22

Page 18: Announcements

Calculation of IN• Short the load and calculate the short circuit current

(R1+R2)i1 - R2iSC = vs

-R2i1 + (R2+R3)iSC = 0

KCL

Page 19: Announcements

Source Transformation

Summary: Thevenin’s Theorem• Any two-terminal linear circuit can be replaced with a voltage source

and a series resistor which will produce the same effects at the terminals

• VTH is the open-circuit voltage VOC between the two terminals of the circuit that the Thevenin generator is replacing

• RTH is the ratio of VOC to the short-circuit current ISC; In linear circuits this is equivalent to “killing” the sources and evaluating the resistance between the terminals. Voltage sources are killed by shorting them, current sources are killed by opening them.

Page 20: Announcements

Summary: Norton’s Theorem• Any two-terminal linear circuit can be replaced with a current source

and a parallel resistor which will produce the same effects at the terminals

• IN is the short-circuit current ISC of the circuit that the Norton generator is replacing

• Again, RN is the ratio of VOC to the short-circuit current ISC; In linear circuits this is equivalent to “killing” the sources and evaluating the resistance between the terminals. Voltage sources are killed by shorting them, current sources are killed by opening them.

• For a given circuit, RN=RTH

Page 21: Announcements

Maximum Power Transfer• Why use Thevenin and Norton equivalents?

– Very easy to calculate load related quantities– E.g. Maximum power transfer to the load

• It is often important to design circuits that transfer power from a source to a load. There are two basic types of power transfer– Efficient power transfer (e.g. power utility)– Maximum power transfer (e.g.

communications circuits)• Want to transfer an electrical signal

(data, information etc.) from the source to a destination with the most power reaching the destination. There is limited power at the source and power is small so efficiency is not a concern.

Page 22: Announcements

Maximum Power Transfer: Impedance matching

LLT

TL R

RRvRip

22

4

2

4

22

4

22

)()(2

)()(

0)(

)(2)(

LT

LTTL

LT

LTT

LT

LTLLTT

L

RRRRvR

RRRRv

RRRRRRRv

dRdp

LT RR L

T

Rvp

4

2

max so maximum power transfer occurs when and

Differentiate using quotient rule:

Set to zero to find maximum:

2vdxdvu

dxduv

vu

dxd

http://circuitscan.homestead.com/files/ancircp/maxpower1.htm