Transcript 幻灯片 1 - 信息科学与技术学院 - Sun Yat

10 级计算机科学 2＋2
Circuits and Analog Electronics
电路与模拟电子技术
Prof. Li Chen, School of Information Science and Technology, Sun Yat-sen University
中山大学信息科学与技术学院
陈立副教授
Email: [email protected]
Circuits and Analog Electronics
References：
•
W. H. Hayt, Jr., J. E. Kemmerly and S. M. Durbin, Engineering Circuit
Analysis, McGraw-Hill, 2005, ISBN 978-7-121-01667-7.
•
R. L. Boylestad and L. Nashelsky, Electronic Devices and Circuit
Theory, Pearson Education, 2007, ISBN 978-7-121-04396-3.
•
高玉良, 电路与模拟电子技术, 高教出版社, 2004, ISBN 7-04014536-7.
Circuits and Analog Electronics
Handouts available at:
sist.sysu.edu.cn/~chenli
References：
•
W. H. Hayt, Jr., J. E. Kemmerly and S. M. Durbin, Engineering Circuit
Analysis, McGraw-Hill, 2005, ISBN 978-7-121-01667-7.
•
R. L. Boylestad and L. Nashelsky, Electronic Devices and Circuit Theory,
Pearson Education, 2007, ISBN 978-7-121-04396-3.
•
高玉良, 电路与模拟电子技术, 高教出版社, 2004, ISBN 7-04-014536-7.
Teaching Schedule
Weeks
Chapters
References
1, 2
Basis concepts and laws of electronics
Hayt: Ch 1 2 5
3, 4
Basis analysis methods to circuits
Hayt: Ch 3 4
Basis RL & RC circuits
Hayt: Ch 6
Sinusoidal steady state analysis
Hayt: Ch 7
5
6, 7, 8
9
Midterm
10
Diodes and diodes circuits
Boylestad: Ch 1 2
11, 12, 13 Basic BJT amplifier circuits
Boylestad: Ch 3-6
14, 15, 16 Operational amplifier and Op Amp circuits
Boylestad: Ch 11
17
Review
Circuits and Analog Electronics
Ch1 Basic Concepts and Laws of Electric Circuits
1.1 Basic Concepts and Electric Circuits
1.2 Basic Quantities
1.3 Circuit Elements
1.4 Kirchhoff's Current and Voltage Laws
References: Hayt: Ch1, 2, 5; Gao: Ch1;
1.1 Basic Concepts and Electric Circuits
 Signal processing and transmission
transmitter
Antenna

Amplifiers
Circuits
Speaker
Kinescope
Electrical power conversion and transmission
Power
Supplies
Transmission
Loads
1.1 Basic Concepts and Electric Circuits
 Electrical power conversion and transmission
1.1 Basic Concepts and Electric Circuits
Concept of Abstraction
Question: What is the current
through the bulb?
Solution:
In order to calculate the current, we can replace the bulb with a resistor.
R is the only subject of interest, which serves as an abstraction of the bulb.
1.1 Basic Concepts and Electric Circuits
Resistance: R = V/I, 1 =1V/A, ohm;
Conductance: G = 1/R = 1A/V, siemens (S);
1S = 1A/V, i(t) = G × v(t);
Instantaneous current and voltage at
time t;
• A resistor is a circuit element that transforms the electrical energy (e.g.
electricity  heat);
• Commonly used devices that are modeled as resistors include
incandescent, heaters, wires and etc;
• A circuit consists of sources, resistors, capacitors, inductors and conductors;
• Elements are lumped.
• Conductors are perfect.
Lumped circuit abstraction!
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Transmitter
Receiver
Understanding the AM radio requires knowledge of several concepts
• Communications/signal processing (frequency domain analysis)
• Electromagnetics (antennas, high-frequency circuits)
• Power (batteries, power supplies)
• Solid state (miniaturization, low-power electronics)
1.1 Basic Concepts and Electric Circuits
Example 1: The AM audio system
Example 2: The telephone system
1.1 Basic Concepts and Electric Circuits
The AM Radio System
 A signal is a quantity that may vary with time.
* Voltage or current in a circuit
* Sound (sinusoidal wave traveling through air)
* Light or radio waves (electromagnetic energy traveling through free
space)
 The analysis and design of AM radios (and communication systems in
general) is usually conducted in the frequency domain using Fourier
analysis, which allows us to represent signals as combinations of
sinusoids (sines and cosines).
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Frequency is the rate at which a signal oscillates.
Duration of the signal T, frequency of the signal f = 1/T.
High Frequency
Low Frequency
1.1 Basic Concepts and Electric Circuits
The AM Radio System
 Visible
light is the electromagnetic energy with frequency between
380THz (Terahertz) and 860THz.

Our visual system perceives the frequency of the electromagnetic
energy as color:
is 460THz, is 570THz, and
is 630THz.
 An AM
red
green
blue
radio signal has a frequency of between 500kHz and 1.8MHz.
FM radio and TV uses different frequencies.

Mathematical analysis of signals in terms of frequency
Most commonly encountered signals can be represented as a Fourier
series or a Fourier transform. A Fourier series is a weighted sum of
cosines and sines.
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Fourier Series: A Fourier series decomposes a periodic function (or signal) into
the sum of a set of sines and cosines. Given function f(t) with angular frequency
ω and period T, its Fourier series can be written as:
f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) + ···

= A0   Akm sin( kt  k )
k 1


k 1
k 1
 A0   Akm sin kt cos k   Akm cos kt sin  k

 A0   Bkm sin kt  Ckm cos kt 
k 1
1 T
A0   f  t dt
T 0
2 T
Bkm   f  t  sin ktdt
T 0
2 T
Bkm   f  t  cos ktdt
T 0
1.1 Basic Concepts and Electric Circuits
 1,0  t  
f
(
t
)

Example: Given function
during a period:


1
,



t

2



Bkm 
1

2

2
0

3
t
1 
1 2
0 f t dt  2 0 1dt  2   1 dt  0
2
1 
f t  sin ktdt 
 1 sin ktdt    1  sin ktdt
1
A0 
2
For the example :
2
2



0
2
  cos kt 
k

 0 , k is even.
2
1

cos
k



 4
k
  k , k is odd.
0
 0
2
1 2
1 
Ckm   f t  cos ktd t 
1 cos ktd t    1  cos ktd t

0
0



2 
2
  cos ktdt 
sin kt  0  0

 0
k

sin ktdt 

4


1
1
4
f (t )  [sin t  sin 3t  sin 5t  ] 

3
5


1
sin( 2l  1)t

2
l

1
l 1

1.1 Basic Concepts and Electric Circuits
The AM Radio System
Example-Fourier Series
st
基波＋3
次谐波
1 series + 3rd series
st
基波
1 series (k = 1)
3 次谐波
3rd series (k = 3)
• Signals can be represented in terms of their frequency components.
• The AM transmitter and receiver are analyzed in terms of their effects on the
frequency components signals.
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Transmitter Block Diagram
Signal
Modulator
Source
Power
Amplifier
Antenna
Modulator
The modulator converts the frequency of the input signal from the audio range
(0-5kHz) to the carrier frequency of the station (i.e. 605kHz-615kHz)
5kHz
freq
Frequency domain
representation of input
610kHz
Frequency domain
representation of output
freq
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Modulator: Time Domain
Input Signal
Output Signal
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Power Amplifier
• A typical AM station broadcasts several kW
– Up to 50kW-Class I or Class II stations
– Up to 5kW-Class III station
– Up to 1kW-Class IV station
• Typical modulator circuit can provide at most a few mW
• Power amplifier takes modulator output and increases its
magnitude
Antenna
The antenna converts a current or a voltage signal to an electromagnetic
signal which is radiated through the space.
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Receiver Block Diagram
Antenna
RF
Amplifier
IF
Mixer
Audio
Amplifier
Speaker
IF
Amplifier
Envelope
Detector
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Antenna
• The antenna captures electromagnetic energy and converts it to a small
voltage or current.
• In the frequency domain, the antenna output is
Undesired Signals
0 interferences
Carrier Frequency
of desired station
Desired Signal
interferences
frequency
1.1 Basic Concepts and Electric Circuits
The AM Radio System
RF (Radio Frequency) Amplifier
• RF Amplifier amplifies small signals from the antenna to voltage levels
appropriate for transistor circuits.
• RF Amplifier also performs as a Bandpass filter for the signal
– Bandpass filter attenuates the other components outside the
frequency range that contains the desired station
Undesired
Signals
0
Desired Signal
Carrier Frequency of desired station
frequency
The AM Radio System
IF (Intermediate Frequency) Mixer
• The IF Mixer shifts its input in the frequency domain from the carrier
frequency to an intermediate frequency of 455kHz
Desired Signal
Undesired Signals
0
frequency
455 kHz
IF Amplifier
• The IF amplifier bandpass filters the output of the IF mixer, eliminating
all of the undesired signals.
Desired Signal
0
455 kHz
frequency
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Envelope Detector
• Computes the envelope of its input signal
Input Signal
Output Signal
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Audio Amplifier
• Amplifies signal from envelope detector
• Provides power to drive the speaker
Hierarchical System Models
• Modelling at different levels of abstraction
• Higher levels of the model describe overall function of the system
• Lower levels of the model describe necessary details to implement the
system
• In the AM receiver, the input is the antenna voltage and the output is the
sound energy produced by the speaker.
• In EE, a system is an electrical and/or mechanical device, a process, or a
mathematical model that relates one or more inputs to one or more outputs.
Inputs
System
Outputs
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Top Level Model
Input Signal
AM Receiver
Sound
Second Level Model
Antenna
RF
Amplifier
IF
Mixer
IF
Amplifier
Power Supply
Audio
Amplifier
Speaker
Envelope
Detector
1.1 Basic Concepts and Electric Circuits
The AM Radio System
Low Level Model Envelope Detector.
Half-wave
Rectifier
Low-pass
Filter
Circuit Level Model Envelope Detector
+
+
Vin
R
-
C
-
Vout
1.2 Basic Quantities
Units
• Standard SI Prefixes
– 10-12 pico (p)
– 10-9 nano (n)
– 10-6 micro ()
– 10-3 milli (m)
– 103 kilo (k)
– 106 mega (M)
– 109 giga (G)
– 1012 tera (T)
• Electric charge (q)
– in Coulombs (C)
• Current (I)
– in Amperes (A)
• Voltage (V)
– in Volts (V)
• Energy (W)
– in Joules (J)
• Power (P)
– in Watts (W)

I t  q
V
I
R
IR  V
W  qV  Pt  V  I  t
P  VI
1.2 Basic Quantities
Current
•
A mount of electric charges flowing through the surface per unit time.
q
I
t
• Time rate of change of charge
Constant current
Time varying current
q  I t
t
i(t )  dq(t ) / dt
q (t ) 
 i ( x)dx

Unit
1mA  10 3 A
1A  10 3 mA
(1 A = 1 C/s)
• Notation: Current flow represents the flow of positive charge
• Alternating versus direct current (AC vs DC)
i(t)
i(t)
t
t
AC
Time – varying current
DC
Steady current
1.2 Basic Quantities
Current
Positive versus negative current
2A
or
Positive charge of 2C/s moving
Negative charge of -2C/s moving

-2 A
or
Negative charge of -2C/s moving
Positive charge of 2C/s moving
P1.1, In the wire electrons moving left to right to create a current of 1 mA.
Determine I1 and I2.
Ans: I1 = -1 mA; I2 = +1 mA.
Current is always associated with arrows (directions)
1.2 Basic Quantities
Voltage(Potential)
• Energy per unit charge.
• It is an electrical force drives an electric current.

b 
F  dl

b 
Voltage V  W  a
  E  dl
ab
a
q
q
Units: 1 V = 1 J/C
Vab  Va  Vb
Positive versus negative voltage
+
2V
Two “Do Not (DN)”
–

–
-2 V
+
+/- of current (I) DN tell the actual direction of particle’s movement .
+/- of voltage (V) DN tell the actual polarity of a certain point .
1.2 Basic Quantities
Voltage (Potential)
Example
a
Vab  5V 
b
a
a、b, which point’s potential is higher？


Va  6V  Vb  4V 
b

a
Vab = ?

b

+Q from point b to point a get energy ，Point a is
Positive? or negative ?
1.2 Basic Quantities
Voltage (Potential)
Example
Va  0
Vab  Va  Vb  Vb   IR1 , Vb   IR1
Vc  E1  Vb  E1  IR1
Vc  Vc  Ir1  E1  I ( R1  r1 )
c
c´
d
I
b
Vd  Vc  IR2  E1  I ( R1  r1  R2 )
Vd   Vd  E2
Va  Vd   Ir2  Vd  E2  Ir2  E1  E2  I R1  r1  R2  r2   0
I
E1  E2
R1  r1  R2  r2
d´
a
1.2 Basic Quantities
Voltage (Potential)
Example
Va=?
Va  8.1(V)
Va  1.52(V)
K Open
I
K Close
I
1.2 Basic Quantities
Example
I
I
va  E 
E
 R1
R1  R2
va E1  va E2  va  E3  va



R
R1
R2
R3
va  E1 
 va 
E1  E2
 R1
R1  R2
E1 R2 R3 R  E2 R1R3 R  E3 R1R2 R
R1 R2 R3  R2 R3 R  R1R2 R  R1R3 R
1.2 Basic Quantities
Power
• One joules of energy is expanded per second.
dq
p
(
t
)

dw
(
t
)
/
dt

V
(
t
)
 Vab (t )i (t )
P = W/t
ab
dt
• Rate of change of energy
i(t)
+
v(t)
–
p(t) = v(t) i(t)
v(t) is defined as the voltage with
positive reference at the same
terminal that the current i(t) is entering.
• Used to determine the electrical power is being absorbed or supplied
– if P is positive (+), power is absorbed
– if P is negative (–), power is supplied
1.2 Basic Quantities
Power
Example
+ 2A
P  5  2  10W
-5V
–
+ 2A
P  5  2  10W
5V
Power is supplied. delivered power
to external element.
Power is absorbed. Power delivered
to
–
Note :
+ 2A
–
+5V
-5V
–
+ -2A
Power absorbed .
1.2 Basic Quantities
Power
• Power absorbed by a resistor:
p (t )  v (t )  i (t )
 R  i 2 (t )
 v 2 (t ) / R
 G  v 2 (t )
 i 2 (t ) / G
1.2 Basic Quantities
Power
P1.5 Find the power absorbed by each element in the circuit.
-
+
I1  2A
5
I1
I2
+
-
+
1
+
-
-
3
-
+
2
+
I3
+
4
+
-
Supply energy : element 1、3、4 .
Absorb energy : element 2、5
-
I 3  1A
I 2  1A
V1  4V
V2  8V V3  4V 
V4  7V V5  3V 
P1  I1V1  8W ;
P3  I 2V3  4W ;
P4  I 3V7  7W ;
P2  I1V2  16W ;
P5  I 3V5  3W ;
1.2 Basic Quantities
Open Circuit
R=
R0
I=0, V=E , P=0
E
Short Circuit
R=0
R0
R＝0
E
E
I
R0
V  E  IR0  0
PE  I 2 R0
1.2 Basic Quantities
Loaded Circuit
I
R0
R
I 
E
Ro  R
V  IR  E  IR0
E
VI  EI  I 2 R0
P  PE  P0
1.3 Circuit Elements
Key Words:
Resistors, Capacitors, Inductors, voltage source,
current source
1.3 Circuit Elements
• Passive elements (cannot generate energy)
– e.g., resistors, capacitors, inductors, etc.
• Active elements (capable of generating energy)
– batteries, generators, etc.
• Important active elements
– Independent voltage source
– Independent current source
– Dependent voltage source
• voltage dependent and current dependent
– Dependent current source
• voltage dependent and current dependent
1.3 Circuit Elements
Resistors
Dissipation Elements
l
v=iR
R
S
 v-i relationship
i

P=vi=Ri2=v2/R >0 ，
v


Resistors connected in series:
– Equivalent Resistance is found by Req= R1
+ R2 + R3 + …
Resistors connected in parallel 1/Req=1/R1 +
1/R2 + 1/R3 + …
R1
R1
R2
R2 R3
R3
1.3 Circuit Elements
Capacitors
• Capacitance occurs when two conductors (plates) are separated by a
dielectric (insulator).
• Charge on the two conductors creates an electric field that stores
energy.
• The voltage difference between the two conductors is proportional
to the charge: q = C v
• The proportionality constant C is called capacitance.
• Units of Farads (F) - C/V
1F=106F, 1F=106PF
• 1F= one coulomb of charge of each conductor causes a voltage of
one volt across the device.
1.3 Circuit Elements
Capacitors

store energy in an electric field

v-i relationship
t
1
v(t )   i ( x)dx
C 
dv
dq
C
i(t ) =
dt
dt
The
rest
of
the
circuit
i(t)
+
v(t)
-
vC(t+) = vC(t-)



dv
p  iv  cv
dt
1 2
Energy stored w   pdt   cvdv  cv
2
Capacitors connected in series:
– Equivalent capacitance is found
by 1/Ceq=1/C1 + 1/C2 + 1/C3 + …
series
Capacitors connected in parallel
Ceq= C1 + C2 + C3 + …
parallel
1.3 Circuit Elements
Capacitors
P1.7
For (1) :
circuit
i(t)
1 t
v  t    i  t  dt   v  t0 
C t0
t0  0 , v  0   0
+
0.2F
v(t)
-
i(t)
1A
2s
t
1A
1 1
1 dt  0  5  1  0   5

0
0.2
1 2
v  2 
1 dt  5  5   2   1   5  0

1
0.2
0  t  1s
v 1 
1s
1 1
v t  
1 dt   0  5t , v 1  5
0.2 0
1s  t  2s
v(t)
5V
1s
(1)
2s
t
1 1
v t  
1 dt   5  10  5t , v  2   0

0
0.2
1.3 Circuit Elements
Capacitors
P1.7
i(t)
circuit
+
0.2F
v(t)
For (2) :
w  t    Pdt  C  v
i(t)
1A
2s
t
1A
t
t
t0
t0
dv
dt
dt
1
 C  vdv  C v 2  t   v 2  t0  
t0
2
t
For (1)、(2) :
1s
1
If v  t0   0 , w  t   C v 2  t 
2
Now : v  0   0 , v 1  5 , v  2   0 .
w (t)
2.5J
w 1  0.1   2.5  2.5
1s
2s
(2)
t
w  2   0.1   0  0
1.3 Circuit Elements
Inductors
store energy in a magnetic field that is
created by electric passing through it.

i(t)
v-i relationship

v(t)
di
dt
Energy stored: wL (t )  1 Li 2 (t )
2
t
t di
i t 
L 2
w(t )   Pdt  L  i dt  L  idi  i t   i 2 t0 
t
t
v t 
dt
2
Inductors connected in series: Leq= L1 + L2 + L3 + …
Inductors connected in parallel: 1/Leq=1/L1 + 1/L2 + 1/L3 + …
P  iv  Li
0

L
-
iL(t+) = iL(t-)

+
t
1
i (t ) 
v( x ) dx

L 
di (t )
v(t )  L
dt
circuit

0
0
1.3 Circuit Elements
Independent voltage source
RS＝0
Ideal
v
VS
VS
+
i
practical
VS  IRs
V  VS  IRs
1.3 Circuit Elements
Independent current source
Ideal
v
RS＝ ∞
I
IS
practical
i
I S  V / RS
I  I S  V / RS
1.3 Circuit Elements
Voltage source connected in series:
n
VS  VSk
k 1
Voltage source connected in parallel:
n
I S   I Sk
k 1
RS  RS 1 // RS 2 //  // RSn
1
1
1
1



RS RS 1 RS 2
RSn
n
RS   RSk
k 1
1.3 Circuit Elements
Voltage controlled (dependent) voltage source (VCVS)
+
vS
+
_
v  vS
_
Current controlled (dependent) voltage source (CCVS)
iS
+
_
Q: What are the units for  and r?
v  riS
1.3 Circuit Elements
Voltage controlled (dependent) current source (VCCS)
+
vS
i  gvS
_
Current controlled (dependent) current source (CCCS)
iS
Q: What are the units for  and g?
i  iS
1.3 Circuit Elements
Independent source
Can provide power to the circuit;
Excitation to circuit ;
Output is not controlled by external.
dependent source
Can provide power to the circuit;
No excitation to circuit;
Output is controlled by external.
1.3 Circuit Elements
Review
• So far, we have talked about two kinds of circuit elements:
– Sources (independent and dependent)
• active, can provide power to the circuit.
– Resistors
• passive, can only dissipate power.
The energy supplied by the active
elements is equivalent to the energy
absorbed by the passive elements!
1.4 Kirchhoff's Current and Voltage Laws
Key Words:
Nodes, Branches, Loops, KCL, KVL
1.4 Kirchhoff's Current and Voltage Laws
Nodes, Branches, Loops, mesh
Node: point where two or more elements are joined (e.g., big node 1)
Branch: Component connected between two nodes (e.g., component R4)
Loop: A closed path that never goes twice over a node (e.g., the blue line)
The red path is NOT a loop
Mesh: A loop that does not contain any other loops in it.
1.4 Kirchhoff's Current and Voltage Laws
Nodes, Branches, Loops, mesh
P1.8
•
•
A circuit containing three nodes and five branches.
Node 1 is redrawn to look like two nodes; it is still one nodes.
1.4 Kirchhoff's Current and Voltage Laws
KCL
• sum of all currents entering a node is zero
• sum of currents entering node is equal to sum of currents leaving node
KCL Mathematically
i1(t)
i5(t)
i2(t)
i4(t)
i3(t)
n
 i (t )  0
j 1
j
n
I
j 1
j
0
1.4 Kirchhoff's Current and Voltage Laws
KCL
• sum of all currents entering a node is zero
• sum of currents entering node is equal to sum of currents
leaving node
P1.9
iA  iB  iC  iD
In   I
Out   I
iA  iB  iC  iO  0
1.4 Kirchhoff's Current and Voltage Laws
KCL
P1.10
KCL-Christmas Lights
Is
120V
+
-
50* 1W Bulbs
• Find currents through each light bulb:
IB = 1W/120V = 8.3mA
• Apply KCL to the top node:
IS - 50IB = 0
• Solve for IS: IS = 50 IB = 417mA
1.4 Kirchhoff's Current and Voltage Laws
KCL
P1.11
We can make supernodes by aggregting node.
Leaving 2 : i1  i6  i4  0
Leaving 3 :  i2  i4  i5  i7  0
Adding 2 & 3 : i1  i2  i5  i6  i7  0
1.4 Kirchhoff's Current and Voltage Laws
KCL
In case of parallel :
1 1
1
I
I
 
, G  G1  G2 , V= 
R R1 R2
R G
Current divider

+
I
I1
N
I1  VG1 
I2
I
G1
 G1 
I
G
G1  G2
V
G1
G2
-
Ik 
Gk
n
I 2  VG2 
G2
I
G1  G2
I
 Gk
k 1
I1 
V
1
RR
1
 I R  I  1 2 
R1
R1
R1  R2 R1
I2 
R1
I
R1  R2
1.4 Kirchhoff's Current and Voltage Laws
KVL

sum of voltages around any loop in a circuit is zero.
KVL Mathematically
n
 v (t )  0
j 1
j
n
V
j 1
j
0
• A voltage encountered + to - is positive.
• A voltage encountered - to + is negative.
1.4 Kirchhoff's Current and Voltage Laws
KVL

KVL is a conservation of energy principle
A positive charge gains electrical energy as it moves to a point with
higher voltage and releases electrical energy if it moves to a point
with lower voltage
b
 Vcd 

c

LOSES W  qVab
AB
a

VA
 Vab 

q
B VB
C

VB
q
V B
q
B
V
W  q(VB  VA )
d
GAINS W  qVcd
q

VA  VCA 
VC
q(VAB  VBC  VCA )  0
If the charge comes back to the same Initial point the net energy gain Must be zero.
1.4 Kirchhoff's Current and Voltage Laws
KVL
P1.13 Determine the voltages Vae and Vec.
Vae  10  24  0
4 + 6 + Vec = 0
16  12  4  6  Vae  0
1.4 Kirchhoff's Current and Voltage Laws
KVL
Voltage divider
+
+
R1
N
+
V
R2
-
V1
V2
-
R1
R1  R2
R2
V2  IR2  V
R1  R2
V1  IR1  V
Vk 
Rk
n
R
k 1
V
k
Important voltage Divider
equations
1.4 Kirchhoff's Current and Voltage Laws
KVL
Voltage divider
P1.14
Example: Vs = 9V, R1 = 90kΩ, R2 = 30kΩ
R1  15k 
Volume control?