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MALVINO
Electronic
PRINCIPLES
SIXTH EDITION
Introduction
Chapter 1
Three kinds of formulas
The definition:
Invented for a new concept
Q
C=
V
{defines what capacitance is}
{does not require verification}
The law:
Summarizes a relationship that exists in nature
f=K
The derivation:
Q1Q2
d2
{verified by experiment}
Obtained by manipulating other
formulas using mathematics
Q = CV
An ideal voltage source maintains a constant
output voltage, regardless of the value of RL.
10 V
RL
The ideal model can be called
the first approximation.
VRL= 10 Volts
A real voltage source has
a series resistance.
RS
10 V
RL
VRL< 10 Volts
This model is called the
the second approximation.
When RL is equal to or greater than 100 times RS, a real
voltage source is stiff and the first approximation can be used.
An ideal current source maintains a constant
output current, regardless of the value of RL.
1A
RL
The ideal model can be called
the first approximation.
IRL= 1 Ampere
A real current source has
a shunt resistance.
1A
RS
RL
IRL< 1 Ampere
This model is called the
the second approximation.
When RS is equal to or greater than 100 times RL, a real
current source is stiff and the first approximation can be used.
Thevenin’s theorem can be used to replace any
linear circuit with an equivalent voltage source
called VTH and an equivalent resistance called RTH.
6 kW
72 V
4 kW
3 kW
RR
LV
THTH
Remove
thesource.
load. R open
Remove
the
measure
. terminals.
Calculate or Calculate
measure
Vor
TH across theTH
When working with actual circuits,
please remember this guideline:
The input impedance of
a voltmeter should be at least 100 times
greater than the Thevenin resistance to
avoid loading error.
DMMs are usually not a problem since they
typically have an impedance of 10 MW.
6 kW
The original
circuit
72 V
4 kW
3 kW
RL
6 kW (RTH)
The Thevenin
equivalent circuit
24 V (VTH)
RL
Norton’s theorem can be used to replace any
linear circuit with an equivalent current source
called IN and an equivalent resistance called RN.
6 kW
72 V
4 kW
3 kW
RL IR
NN
RShort
N is the
TH. I .
thesame
load as
to R
find
N
6 kW
The original
circuit
The Norton
equivalent circuit
72 V
4 mA (IN)
4 kW
3 kW
RL
6 kW (RN)
RL
6 kW (RTH)
A Thevenin
equivalent circuit
RN = RTH
The Norton
dual
RL
24 V (VTH)
4 mA (IN)
IN =
VTH
RTH
6 kW (RN)
RL
Troubleshooting
• A solder bridge between two lines
effectively shorts them together.
• A cold solder joint is effectively an open
circuit.
• An intermittent trouble is one that
appears and disappears (could be a cold
solder joint or a loose connection).
An open device
• The current through it is zero.
• The voltage across it is unknown.
• V = zero x infinity {indeterminate}
A shorted device
• The voltage across it is zero.
• The current through it is unknown.
• I = 0/0 {indeterminate}
A troubleshooting example:
Do the two 10 W resistors form
a stiff voltage divider?
10 W
12 V
100 kW
10 W
Why?
100 kW
A troubleshooting example:
What are the expected
voltages in this circuit?
10 W
12 V
100 kW
10 W
100 kW
A troubleshooting example:
What are some causes for
this voltage being too high?
10 W
12 V
100 kW
10 W
100 kW
V
A troubleshooting example:
What are some causes for
this voltage being too low?
10 W
12 V
100 kW
10 W
100 kW
V