CP2 Circuit Theory Dr Todd Huffman [email protected] huffman/ Aims of this course:...
Transcript of CP2 Circuit Theory Dr Todd Huffman [email protected] huffman/ Aims of this course:...
CP2 Circuit TheoryDr 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 now on the exam syllabus
Stored energy, RC and RL transient circuits
Reading List• Electronics: Circuits, Amplifiers and Gates, D V Bugg, Taylor and
FrancisChapters 1-7
• Basic Electronics for Scientists and Engineers, D L Eggleston, CUPChapters 1,2,6
• Electromagnetism Principles and Applications, Lorrain and Corson, FreemanChapters 5,16,17,18
• Practical Course Electronics Manual http://www-teaching.physics.ox.ac.uk/practical_course/ElManToc.htmlChapters 1-3
• Elementary Linear Circuit Analysis, L S Bobrow, HRWChapters 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
0ILC1
dtdI
LR
dtId2
2
V(t)=V0ejωt
jXRZ ZV
I
V0–I1R1–(I1–I2)R3 = 0
(I1–I2)R3–I2R2+2 = 0
Covered by Complex Nos & ODEs / Vectors & Matrices lectures
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
dtdQ
VdtdV
QQVdtd
dtdW
P
nAvedtdQ
I
No. electrons/unit vol
Cross-section area of conductor
Drift velocity
Charge Q=e
Current: rate of flow of charge
[amps]
[watts]
Ohm’s lawVoltage difference current
I
L
AV
AL
R
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 ~1019 Ωm
R1
g Conductance[seimens]
conductivity[seimens/m]
1
Power dissipation by resistor: RV
RIIVP2
2
Voltage source
V0
Ideal voltage source: supplies V0 independent of current
V0Rint
Real voltage source:
Vload=V0–IRint
Rload
Rload
+–
battery cell
Constant current sourceIdeal current source: supplies I0 amps independent of voltage
Real current source:
Rload
Rload
I0
RintI0
int0 R
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)T2
f2
50Hz power, audio, radio…
+–
+
-
RMS values
AC Power dissipation R
VVIP
2RMS
RMSRMS
2
VV 0
RMS
Why ? 2
T
0
2RMS dttV
T1
V
Square root of mean of V(t)2
Passive Sign ConventionPassive 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 whichdirection for I?
Kirchoff’s LawsI 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
+IR1
5V=I(1+3+4)kΩ
mA625.08000
V5I
VR2=0.625mA×3kΩ=1.9V
Series / parallel circuitsR1 R2 R3
Resistors in series: RTotal=R1+R2+R3…
n
nT RR
R1 R2 R3
Resistors in parallel
321
n nT
R1
R1
R1
R1
R1
Two parallel resistors: 21
21T RR
RRR
Mesh currents
9V 2V+ +
R1 R2
R3
I1 I2I3
I1 I2
R1=3kΩR2=2kΩR3=6kΩ
Second job: USE PASSIVE SIGN CONVENTION!!!
First job: Label loop currents in all interior loops
++
+
Third job: Apply KCL to elements that share loop currents
Define: Currents Entering Node are positive
I1–I2–I3 = 0 I3 = I1-I2
Mesh currents
9V 2V+ +
R1 R2
R3
I1 I2I3
I1 I2
R1=3kΩR2=2kΩR3=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 I2I3
I1 I2
R1=3kΩR2=2kΩR3=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
43
2I 35
333
31
++
+Solve simultaneous equations
Node voltages
9V 2V+
+
R1 R2
R3
I1 I2I3
R1=3kΩR2=2kΩR3=6kΩ
VX
0V
-
-
+
-
Step 1: Choose a ground node!
Step 2: Label voltages on all unlabled nodes
Step 3: Apply KCL and ohms law using the tricks
Node voltages
9V 2V+
+
R1 R2
R3
I1 I2I3
R1=3kΩR2=2kΩR3=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 nodesAnd apply DV/R to each current.
V2V mA2k1V
mA2k1
31
61
21
V
mA2mA1mA3RV2
RV9
R1
R1
R1
V
XX
X
21132X
mA31
k6V2
I
2mAk2
V2V2I mA
37
k3V2V9
I
3
21
2V9V+
+
R1 R2
R3
I1 I2I3
VX
0V132
132
9)2(0
0
R
VV
R
V
R
VV
III
XXX
Thevenin’s theorem
VeqReq
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) VL0SS
OSeq I
VR
V0
+R1
RL
I1
Iss
R2
I2 VL
21
20 RR
RVVos
21
21
1
0
21
20
eq RRRR
RV
RRR
VR
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
using Thevenin’s theorem:from Prev. Already know VAB =Veq= 2V
2V+
+
R1=3kΩ R2=2kΩ
R3=6kΩ
A
B
9V
2V+
+
A
B
V9
R2=2kΩma
k
VI 3
3
91
kR 31
kI
VR
maIII
mak
VI
SS
eqeq
SS
1
2
12
2
21
2
+
A
B
2V
1kΩ
Iss
I1 I2
using Norton’s theorem:
2V+
+
R1=3kΩ R2=2kΩ
R3=6kΩ
A
B
9V 1kΩ2mA
A
B
Same procedure: Find ISS and VOC
IEQ = ISS and REQ = VOC/ISS
Superposition
9V 2V+
+
R1 R2
R3
I1 I2I3
9V+
R1R2
R3
IA IBIC
2V+
R1 R2
R3
IE IFIG
=
+
I1=IA+IE I2=IB+IF I3=IC+IG
Important: label Everything the same directions!
9V+
R1R2
R3
IA IBIC
9V+
R1
R2R3
IA
IBIC
k5.1
RRRR
32
32
R1=3kΩR2=2kΩR3=6kΩ
k5.4
RRRR
R32
321
mA2k5.4V9
I A
mA5.1I mA5.0I
V35.45.1
V9RIRI
BC
2B3C
Example:Superposition
2V+
R1 R2
R3
IE IFIG
2V+
R2
R3
IF
IEIG
R1
R1=3kΩR2=2kΩR3=6kΩ
k2RR
RR
31
31
k4
RRRR
R31
312
mA5.0k4V2
IF
mA
31
I mA61
I
V142
V2RIRI
EG
1E3G
9V 2V+
+
R1 R2
R3
I1 I2I3
9V+
R1R2
R3
IA IBIC
2V+
R1 R2
R3
IE IFIG
=
+
I1=IA+IE I2=IB+IF I3=IC+IG
mA37
mA31
2I1
mA2
mA5.05.1I2
mA31
mA61
5.0I3
Matching: maximum power transfer
V0Rin
Rload
L2
Lin
2L2
0L
2L
R1
RR
RV
RV
P
Find RL to give maximum power in load
0R2RR
0RR
RRR2RRV
dRdP
LLin
4Lin
LinL2
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?