cp2 circuit theory
TRANSCRIPT
CP2 Circuit Theory Prof. Todd Huffman [email protected] http://www-pnp.physics.ox.ac.uk/~huffman/ Aims of this course:
Understand basic circuit components (resistors, capacitors, inductors, voltage and current sources, op-amps)
Analyse and design simple linear circuits
+
– +
+
Circuit Theory: Synopsis
Basics: voltage, current, Ohm’s law…
Kirchoff’s laws: mesh currents, node voltages…
Thevenin and Norton’s theorem: ideal voltage and current sources…
Capacitors:
Inductors:
AC theory: complex notation, phasor diagrams, RC, RL, LCR circuits, resonance, bridges…
Op amps: ideal operational amplifier circuits…
Op-amps are on the exam syllabus
Stored energy, RC and RL transient circuits
Reading List • Electronics: Circuits, Amplifiers and Gates, D V Bugg, Taylor
and Francis Chapters 1-7
• Basic Electronics for Scientists and Engineers, D L Eggleston, CUP Chapters 1,2,6
• Electromagnetism Principles and Applications, Lorrain and Corson, Freeman Chapters 5,16,17,18
• Practical Course Electronics Manual http://www-teaching.physics.ox.ac.uk/practical_course/ElManToc.html Chapters 1-3
• Elementary Linear Circuit Analysis, L S Bobrow, HRW Chapters 1-6
• The Art of Electronics, Horowitz and Hill, CUP
Why study circuit theory?
• Foundations of electronics: analogue circuits, digital circuits, computing, communications…
• Scientific instruments: readout, measurement, data acquisition…
• Physics of electrical circuits, electromagnetism, transmission lines, particle accelerators, thunderstorms…
• Not just electrical systems, also thermal, pneumatic, hydraulic circuits, control theory
Mathematics required
• Differential equations
• Complex numbers
• Linear equations
0ILC
1
dt
dI
L
R
dt
Id2
2
V(t)=V0ejωt
jXRZ Z
VI
V0–I1R1–(I1–I2)R3 = 0
(I1–I2)R3–I2R2+2 = 0
Covered by Complex Nos & ODEs / Vectors & Matrices lectures (but with fewer dimensions)
Charge, voltage, current
Potential difference: V=VA–VB
Energy to move unit charge from A to B in electric field
B
AV dsE VE
B
AQW dsE
Charge: determines strength of electromagnetic force quantised: e=1.62×10-19C [coulombs]
[volts]
Power: rate of change of work
IV
dt
dQV
dt
dVQQV
dt
d
dt
dWP
nAvedt
dQI
No. electrons/unit vol
Cross-section area of conductor
Drift velocity
Charge Q=e
Current: rate of flow of charge
[amps]
[watts]
Ohm’s law Voltage difference current
I
L
A V
A
LR
IRV
R=Resistance Ω[ohms]
ρ=Resistivity Ωm
Resistor symbols:
R
Resistivities
Silver 1.6×10-8 Ωm
Copper 1.7×10-8 Ωm
Manganin 42×10-8 Ωm
Distilled water 5.0×103 Ωm
PTFE (Teflon) ~1019 Ωm
R
1g Conductance
[seimens] conductivity [seimens/m]
1
Power dissipation by resistor: R
VRIIVP
22
Voltage source
V0
Ideal voltage source: supplies V0 independent of current
V0 Rint Real voltage source:
Vload=V0–IRint
Rload
Rload
+ –
battery cell
Constant current source
Ideal current source: supplies I0 amps independent of voltage
Real current source:
Rload
Rload
I0
Rint I0
int
0R
VII load
Symbol: or
AC and DC
DC (Direct Current): Constant voltage or current
Time independent
V=V0
AC (Alternating Current): Time dependent
Periodic
I(t)=I0sin(ωt)
T
2f2
50Hz power, audio, radio…
+ –
+
-
RMS values
AC Power dissipation R
VVIP
2RMS
RMSRMS
2
VV 0
RMS
Why ? 2
T
0
2RMS dttV
T
1V
Square root of mean of V(t)2
Passive Sign Convention Passive devices ONLY - Learn it; Live it; Love it!
IRV
R=Resistance Ω[ohms]
Which way does the current flow, left or right?
Two seemingly Simple questions:
Voltage has a ‘+’ side and a ‘-’ side (you can see it on a battery) on which side should we put the ‘+’? On the left or the right?
Given V=IR, does it matter which sides for V or which direction for I?
Kirchoff’s Laws I Kirchoff’s current law: Sum of all currents at a node is zero
I1
I3
I2
I4
I1+I2–I3–I4=0
0In
(conservation of charge)
Here is a cute trick: It does not matter whether you pick “entering” or “leaving” currents as positive. BUT keep the same convention for all currents on one node!
II Kirchoff’s voltage law: Around a closed loop the net change of potential is zero (Conservation of Energy)
V0
I
R1
R2
R3
0Vn
5V
3kΩ
4kΩ
Calculate the voltage across R2
1kΩ
Kirchoff’s voltage law:
V0
I 1kW
3kW
4kW
-V0+IR1+IR2+IR3=0
0Vn
+
+
+
+
–
–
–
–
–V0
+IR1
+IR2
+IR3
5V=I(1+3+4)kΩ
mA625.08000
V5I
W
VR2=0.625mA×3kΩ=1.9V
Series / parallel circuits
R1 R2 R3
Resistors in series: RTotal=R1+R2+R3…
n
nT RR
R1 R2 R3
Resistors in parallel
321
n nT
R
1
R
1
R
1
R
1
R
1
Two parallel resistors: 21
21T
RR
RRR
Potential divider
R1
R2
V0
21
20
RR
RV
USE PASSIVE SIGN CONVENTION!!!
Show on blackboard
Mesh currents
9V 2V + +
R1 R2
R3
I1 I2 I3
I1 I2
R1=3kΩ R2=2kΩ R3=6kΩ
Second job: USE PASSIVE SIGN CONVENTION!!!
First job: Label loop currents in all interior loops
+ +
+
Third job: Apply KCL to nodes sharing loop currents
Define: Currents Entering Node are positive
I1–I2–I3 = 0 I3 = I1-I2
Mesh currents
9V 2V + +
R1 R2
R3
I1 I2 I3
I1 I2
R1=3kΩ R2=2kΩ R3=6kΩ
Last job: USE Ohm’s law and solve equations.
Fourth job: Apply Kirchoff’s Voltage law around each loop.
+ +
+
Mesh currents
9V 2V + +
R1 R2
R3
I1 I2 I3
I1 I2
R1=3kΩ R2=2kΩ R3=6kΩ
-9V+I1R1+I3R3 = 0 I3 =I1–I2
–I3R3+I2R2+2V = 0
9V/kΩ = 9I1–6I2
-2V/kΩ = -6I1+8I2
I2=1 mA VIRV
mAmAI
4
32I
35
333
31
+ +
+ Solve simultaneous equations
Node voltages
9V 2V +
+
R1 R2
R3
I1 I2 I3
R1=3kΩ R2=2kΩ R3=6kΩ VX
0V
-
-
+
-
Step 1: Choose a ground node!
Step 2: Label V’s and I’s on all nodes
Step 4: Apply KCL and ohms law using the tricks
Step 3: Apply KVL to find DV across resistors
Node voltages
9V 2V +
+
R1 R2
R3
I1 I2 I3
R1=3kΩ R2=2kΩ R3=6kΩ VX
0V
132
132
9)2(0
0
R
VV
R
V
R
VV
III
XXX
USE PASSIVE SIGN CONVENTION!!!
-
-
+
-
Only one equation, Mesh analysis would give two.
All currents leave all labeled nodes And apply DV/R to each current.
V2V mA2k1
V
mA2k
1
3
1
6
1
2
1V
mA2mA1mA3R
V2
R
V9
R
1
R
1
R
1V
XX
X
21132
X
W
W
mA 3
1
6
2
2mA 2
22mA
3
7
3
92
3
21
W
W
W
k
VI
k
VVI
k
VVI
2V 9V +
+
R1 R2
R3
I1 I2 I3
VX
0V 132
132
9)2(0
0
R
VV
R
V
R
VV
III
XXX
Thevenin’s theorem
Veq Req
Any linear network of voltage/current sources and resistors
Equivalent circuit
In Practice, to find Veq, Req…
RL (open circuit) IL0 Veq=VOS
RL0 (short circuit) VL0 SS
OSeq
I
VR
V0 +
R1
RL
I1
Iss
R2
I2 VL
21
20
RR
RVVos
21
21
1
0
21
20
eqRR
RR
R
V
RR
RV
R
R2
I2
Req = resistance between terminals when all voltages sources shorted – Warning! This is not always obvious!
1
0
R
VI ss
Norton’s theorem
Req
Any linear network of voltage/current sources and resistors
Equivalent circuit
I0
V0 R1
Req eqIRload Rload
VL=V0–ILR1
LLeqeqeq
L
eq
LLReq
VIRIR
IR
VIII
1eq
eq
0eq
RR
R
VI
V0 R
R RV0
IR IL
using Thevenin’s theorem: VOC =Veq= 2.0V
2V
+
+
R1=3kΩ R2=2kΩ
R3=6kΩ
A
B
9V
2V +
+
A
B
V9
R2=2kΩ
mak
VI 3
3
91
W
W kR 31
W
W
kI
VR
maIII
mak
VI
SS
OCeq
SS
0.1
2
12
2
21
2
+
A
B
2.0V
1.0kΩ
Iss
I1 I2
using Norton’s theorem:
2V
+
+
R1=3kΩ R2=2kΩ
R3=6kΩ
A
B
9V 1kΩ 2mA
A
B
Same procedure: Find ISS and VOC IEQ = ISS and REQ = VOC/ISS
Superposition
9V 2V
+
+
R1 R2
R3
I1 I2 I3
9V +
R1 R2
R3
IA IB IC
2V +
R1 R2
R3
IE IF IG
=
+
I1=IA+IE I2=IB+IF I3=IC+IG
Important: label Everything the same directions!
9V +
R1 R2
R3
IA IB IC
9V +
R1
R2 R3
IA
IB IC
W
k5.1RR
RR
32
32
R1=3kΩ R2=2kΩ R3=6kΩ
W
k5.4
RR
RRR
32
321
mA2k5.4
V9IA
W
mA5.1I mA5.0I
V35.4
5.1V9RIRI
BC
2B3C
Example: Superposition
2V +
R1 R2
R3
IE IF IG
2V +
R2
R3
IF
IE IG
R1
R1=3kΩ R2=2kΩ R3=6kΩ
W
k2RR
RR
31
31
W
k4
RR
RRR
31
312
mA5.0k4
V2IF
W
mA3
1I mA
6
1I
V14
2V2RIRI
EG
1E3G
9V 2V
+
+
R1 R2
R3
I1 I2 I3
9V +
R1 R2
R3
IA IB IC
2V +
R1 R2
R3
IE IF IG
=
+
I1=IA+IE I2=IB+IF I3=IC+IG
mA3
7
mA3
12I1
mA2
mA5.05.1I2
mA3
1
mA6
15.0I3
Matching: maximum power transfer
V0 Rin
Rload
L2
Lin
2L2
0
L
2L
R
1
RR
RV
R
VP
Find RL to give maximum power in load
0R2RR
0RR
RRR2RRV
dR
dP
LLin
4
Lin
LinL
2
Lin20
L
Lin RR
Maximum power transfer when RL=Rin
Note – power dissipated half in RL and half in Rin
Circuits have Consequences
• Problem:
– My old speakers are 60W
speakers.
– Special 2-4-1 deal at El-
Cheap-0 Acoustics on 120W
speakers!!
• (“Offer not seen on TV!”)
– Do I buy them?
• Depends! 4W, 8W, or 16W
speakers?
• Why does this matter?
And Now for Something Completely Different