Transcript Simple Resistor Circuits

CP2 Circuit Theory
Dr 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:
Stored energy, RC and RL transient
circuits
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
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
2
d I
dt
• Complex numbers

L dt

1
LC
I  0
V(t)=V0ejωt
I 
• Linear equations
2
R dI
V
Z
Z  R  jX
V0–I1R1–(I1–I2)R3 = 0
(I1–I2)R3–I2R2+2 = 0
Covered by Complex Nos & ODEs / Vectors & Matrices lectures
Charge, voltage, current
Charge: determines strength of electromagnetic force
quantised: e=1.62×10-19C
[coulombs]
Potential difference: V=VA–VB
[volts]
Energy to move unit charge from A to B
in electric field
B
E   V
V    E  ds
A
B
W    Q E  ds
A
Charge Q=e
Current: rate of flow of charge
I 
dQ
dt
 nAve
No. electrons/unit vol
Cross-section area
of conductor
Drift
velocity
[amps]
Power: rate of change of work
P 
dW
dt

d
dt
QV  
Q
dV
dt
 V
dQ
dt
 IV
[watts]
Ohm’s law
Voltage difference  current
Resistor symbols:
L
I
R
A
V
V  IR
R 
L
A
R=Resistance Ω[ohms]
ρ=Resistivity Ωm
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
Conductance
[seimens]
g 
1
R
conductivity
[seimens/m]
Power dissipation by resistor: P  IV  I 2R 
 
V
2
R
1

Voltage source
V0
+
–
V0
battery cell
Rload
Ideal voltage source: supplies V0
independent of current
Real voltage source:
Rint
Rload
Vload=V0–IRint
Constant current source
I0
Rload
Ideal current source: supplies I0 amps
independent of voltage
Symbol:
or
Real current source:
I0
Rint
Rload
I load  I 0 
V
R int
AC and DC
DC (Direct Current):
Time independent
+
–
V=V0
AC (Alternating Current):
+
-
Constant voltage or current
Time dependent
Periodic
I(t)=I0sin(ωt)
2
  2 f 
T
50Hz power, audio, radio…
RMS values
2
AC Power dissipation
VRM S 
Why
2 ?
P  IRMS VRMS 
VRMS
R
V0
2
Square root of mean of V(t)2
VRM S 
1
T

T
0
V
2
t dt
Passive Sign Convention
Passive devices ONLY - Learn it; Live it; Love it!
R=Resistance Ω[ohms]
V  IR
Two seemingly Simple questions:
Which way does the current flow, left or right?
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:
I1
I2
I3
Sum of all currents at a
node is zero
I1+I2–I3–I4=0
I
I4
n
 0
(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)
R1
I
1kΩ
V0
5V
3kΩ
R2

Vn  0
4kΩ
R3
Calculate the voltage
across R2
Kirchoff’s voltage law:

Vn  0
-V0+IR1+IR2+IR3=0
+IR1
+
–V0
–
1kW
I
+
–
5V=I(1+3+4)kΩ
V0
+
3kW
–
4kW
+
–
+IR1
I 
5V
8000 W
 0 . 625 mA
+IR2
VR2=0.625mA×3kΩ=1.9V
Series / parallel circuits
R1
R2
R3
RT 
R
n
Resistors in series: RTotal=R1+R2+R3…
Resistors in parallel
R1
R2
R3
1
RT

R
n

Two parallel
resistors:
1
1
R1
n

1
R2

RT 
1
R3

R 1R 2
R1  R2
n
Potential divider
R1
V0
R2
V 0R 2
R1  R2
USE PASSIVE SIGN CONVENTION!!!
Show on blackboard
Mesh currents
I1
+
+
9V
+
+
I1
R2
I3
R3
I2
I2
+
R1
R1=3kΩ
R2=2kΩ
R3=6kΩ
2V
First job: Label loop currents in all interior loops
Second job: USE PASSIVE SIGN CONVENTION!!!
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
Fourth job:
I1
+
+
9V
+
+
I1
R2
I3
R3
I2
I2
+
R1
R1=3kΩ
R2=2kΩ
R3=6kΩ
2V
Apply Kirchoff’s Voltage law around each loop.
Last job: USE Ohm’s law and solve equations.
Mesh currents
R1
-9V+I1R1+I3R3 = 0
–I3R3+I2R2+2V = 0
9V/kΩ = 9I1–6I2
-2V/kΩ = -6I1+8I2
 I2=1 mA
R2
I3
I2
R3
I2
+
I1
+
+
9V
+
+
I1
R1=3kΩ
R2=2kΩ
R3=6kΩ
2V
I3 =I1–I2
Solve simultaneous
equations
I1  5
3
mA
V3  R3 I 3  4 V
I3  2
3
mA
Node voltages
R1
I1
9V
VX
+
+
R1=3kΩ
R2=2kΩ
R3=6kΩ
R2
I3
I2
R3
-
+
2V
-
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
VX
R1
+
I1
9V
+
R2
I3
I2
-
R3
-
R1=3kΩ
R2=2kΩ
R3=6kΩ
+
2V
-
0V
0 
0
I 2  I 3  I1
V X  (  2V )
R2

VX
R3

V X  9V
R1
All currents leave all labeled nodes
And apply DV/R to each current.
Only one equation,
Mesh analysis would give two.
USE PASSIVE SIGN CONVENTION!!!
R1
0 
0
I 2  I 3  I1
V X  (  2V )

R2
VX

R3
I1
9V
V X  9V
+
R1
VX
R2
I3
R3
0V
 1
1
1 
9V
2V
 
V X 



 3 mA  1mA  2 mA

R3
R1 
R1
R2
R2
1
1 1
1
VX  
 
 2 mA
6
3  kW
2
VX
1k W
I1 
I3 
 2 mA

9V  2V
7
3k W
2V
6k W


1
3
3
mA
VX  2 V
mA
I2 
2V  2V
2k W
 2mA
I2
+
2V
Thevenin’s theorem
Any linear network
of voltage/current
sources and
resistors

Veq
Req
Equivalent circuit
In Practice, to find Veq, Req…
RL (open circuit)
RL0 (short circuit)
IL0
VL0
Veq=VOS
R eq 
V OS
I SS
Req = resistance between terminals when all voltages
sources shorted – Warning! This is not always obvious!
I1
V0
+
V os  V 0
R1
I2
R2
R2
R1  R 2
V0
Iss
RL
VL
R eq 
I ss 
V0
R1
R2
R1  R2
V0
R1

R 1R 2
R 1 R 2
Norton’s theorem
Any linear network
of voltage/current
sources and
resistors

I0
Equivalent circuit
Req
R1
V0
I eq
Rload
VL=V0–ILR1
 I eq 
Req
I eq  I R  I L 
V0
Rload
VL
R eq
R eq I eq  R eq I L  VL
R eq
R eq  R 1
V0
R

V0
R
R
 IL
A
R1=3kΩ
9V
+
R2=2kΩ
R3=
6kΩ
+
1kΩ
A
2V +
2V
B
B
using Thevenin’s theorem:
from Prev. Already know VAB =Veq= 2V
A
I1
9V
+
R2=2kΩ
R1  3 k W
2V
Iss
+
I2
I1 
9V
3kW
I2 
 3 ma
2V
2 kW
 1ma
I SS  I 1  I 2  2 ma
B
R eq 
V eq
I SS
 1k W
A
R1=3kΩ
9V
R2=2kΩ
+
R3=
6kΩ
+
A
2V
2mA
1kΩ
B
B
using Norton’s theorem:
Same procedure: Find ISS and VOC
IEQ = ISS and REQ = VOC/ISS
Superposition
R1
+
9V
I1
R1
+
9V
IA
I1=IA+IE
R2
I3
I2
R3
+
=
IC
R3
2V
Important: label
Everything the same
directions!
R2
R1
IB
+
I2=IB+IF
IE
R2
IG
R3
I3=IC+IG
IF
+
2V
R2
R1
+
9V
IA
IC
IB
R3
R1=3kΩ
R2=2kΩ
R3=6kΩ
Example:
Superposition
IA
9V
R1
+
IA 
9V
4 . 5k W
R1 
IC
IB
R3
R2
 2 mA
R 2R 3
R2  R3
 4 . 5k W
R 2R 3
R2  R3
 1 . 5k W
I C R 3  IBR 2  9 V
I C  0 . 5 mA
1 .5
4 .5
 3V
I B  1 . 5 mA
R1
IE
IG
IF
R3
2V
+
R1=3kΩ
R2=2kΩ
R3=6kΩ
R2
2V
+
IG
IE
R3
R1
R 1R 3
R1  R3
 2k W
R2 
IF
IF 
4k W
R1  R3
 4k W
R2
2V
R 1R 3
 I G R 3  I ER 1  2 V
 0 . 5 mA
IG  
1
6
mA
2
4
 1V
IE 
1
3
mA
R1
+
9V
I1
R1
+
9V
R2
I3
I2
R3
+
=
IA
IC
R3
2V
R2
R1
IB
+
R2
IE
IG
R3
IF
+
I1=IA+IE
I2=IB+IF
I3=IC+IG
1

I1   2  mA
3

I 2  1 . 5  0 . 5 mA
1

I 3   0 . 5  mA
6


7
3
mA
 2 mA

1
3
mA
2V
Matching: maximum power
transfer
Find R to give maximum
L
power in load
V0
Rin
Rload
2
P 
dP
dR
2
 V0
L
VL
RL
R in
2
2
 V0
R in
RL
1
 RL  RL
2
 R L   2R L R in  R L 
2
R in
 RL 
R in  R L  2 R L  0
4
 0
 R in  R L
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 ElCheap-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