Transcript Chapter 1 and 2

Basic Electronics
CHA PT ER 1
LEC T UR E 1 & 2
E N G R. KA S HI F S HA HZA D
Course Details
 Text Book:
Electronic Principles 7th Edition
by Albert Malvino & David J Bates
 Course Instructor: Engr. Kashif Shahzad
 Email: [email protected]
 Cell: +92 333 5186231
 Course homepage:
 http://groups.yahoo.com/group/prestonindustrialelectronics
Course Breakdown
Assignments:
10%
Quizzes:
10%
Others:
05%
Mid Term:
25%
(2/3)
Terminal:
50%
(8/12)
Course Outline
Chapter #
Chapter Name
Exam Question Number
1
Introduction
1
2
Semiconductors
2
3
Diode Theory
3
4
Diode Circuits
4
5
Special Purpose Diodes
5
6
Bipolar Junction Transistors
6
7
Transistor Fundamentals
7
8
Transistor Biasing
8
9
AC Models
8
Course Outline
Chapter #
Chapter Name
Exam Question Number
10
Voltage Amplifiers
9
12
Power Amplifiers
9
13
JFETs
10
14
MOSFETs
10
15
Thyristors
11
16
Frequency Effects
11
17
Differential Amplifiers
12
18
Operational Amplifiers
12
Introduction
CHAPTER # 1
Three kinds of formulas
The definition:
Invented for a new concept
Q
C=
{defines what capacitance is}
V
{does not require verification}
The law:
Summarizes a relationship that exists in nature
f=K
Q1 Q 2
{verified by experiment}
d2
The derivation:
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
RL
RTH
VTH
Remove
the
load.
Calculate
Remove
or the
measure
source.
RTHterminals.
.
Calculate or measure
VTH
across
the
open
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
RN is the same as RTH.
Short the load to find IN.
RL
IRNN
6 kW
The original
circuit
The Norton
equivalent circuit
72 V
4 mA (IN)
4 kW
3 kW
6 kW (RN)
RL
RL
THEVENIN & NORTON
THEVENIN’S THEOREM:
Consider the following:
A
Network
1
•
B
•
Network
2
Figure 10.1: Coupled networks.
For purposes of discussion, at this point, we consider
that both networks are composed of resistors and
independent voltage and current sources
1
THEVENIN & NORTON
THEVENIN’S THEOREM:
Suppose Network 2 is detached from Network 1 and
we focus temporarily only on Network 1.
Network
1
•A
•B
Figure 10.2: Network 1, open-circuited.
Network 1 can be as complicated in structure as one
can imagine. Maybe 45 meshes, 387 resistors, 91
voltage sources and 39 current sources.
2
THEVENIN & NORTON
THEVENIN’S THEOREM:
Network
1
•A
•B
Now place a voltmeter across terminals A-B and
read the voltage. We call this the open-circuit voltage.
No matter how complicated Network 1 is, we read one
voltage. It is either positive at A, (with respect to B)
or negative at A.
We call this voltage Vos and we also call it VTHEVENIN = VTH
3
THEVENIN & NORTON
THEVENIN’S THEOREM:
• We now deactivate all sources of Network 1.
• To deactivate a voltage source, we remove
the source and replace it with a short circuit.
• To deactivate a current source, we remove
the source.
4
THEVENIN & NORTON
THEVENIN’S THEOREM:
Consider the following circuit.
I2
V3
A
_+
R1
_+
R2
V1
V2
_
+
R3
I1
R4
B
Figure 10.3: A typical circuit with independent sources
How do we deactivate the sources of this circuit?
5
THEVENIN & NORTON
THEVENIN’S THEOREM:
When the sources are deactivated the circuit appears
as in Figure 10.4.
A
R1
R3
R2
R4
B
Figure 10.4: Circuit of Figure 10.3 with sources deactivated
Now place an ohmmeter across A-B and read the resistance.
If R1= R2 = R4= 20 W and R3=10 W then the meter reads 10 W.
6
THEVENIN & NORTON
THEVENIN’S THEOREM:
We call the ohmmeter reading, under these conditions,
RTHEVENIN and shorten this to RTH. Therefore, the
important results are that we can replace Network 1
with the following network.
A

RTH
+
_
VTH
B

Figure 10.5: The Thevenin equivalent structure.
7
THEVENIN & NORTON
THEVENIN’S THEOREM:
We can now tie (reconnect) Network 2 back to
terminals A-B.
A

RTH
+
_
Network
2
VTH

B
Figure 10.6: System of Figure 10.1 with Network 1
replaced by the Thevenin equivalent circuit.
We can now make any calculations we desire within
Network 2 and they will give the same results as if we
still had Network 1 connected.
8
THEVENIN & NORTON
THEVENIN’S THEOREM:
It follows that we could also replace Network 2 with a
Thevenin voltage and Thevenin resistance. The results
would be as shown in Figure 10.7.
A

RTH 1
+
_
RTH 2
VTH 2 _+
VTH 1

B
Figure 10.7: The network system of Figure 10.1
replaced by Thevenin voltages and resistances.
9
THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.1.
Find VX by first finding VTH and RTH to the left of A-B.
4W
12 W
_
30 V +
6W

A
+
2W
VX
_

Figure 10.8: Circuit for Example 10.1.
First remove everything to the right of A-B.
10
B
THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.1. continued
4W
12 W
_
30 V +

6W

Figure 10.9: Circuit for finding VTH for Example 10.1.
(30)(6)
VAB 
 10V
6  12
Notice that there is no current flowing in the 4 W resistor
(A-B) is open. Thus there can be no voltage across the
resistor.
11
A
B
THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.1. continued
We now deactivate the sources to the left of A-B and find
the resistance seen looking in these terminals.
4W
12 W

A
RTH
6W

Figure 10.10: Circuit for find RTH for Example 10.10.
We see,
RTH = 12||6 + 4 = 8 W
12
B
THEVENIN & NORTON
THEVENIN’S THEOREM: Example 10.1. continued
After having found the Thevenin circuit, we connect this
to the load in order to find VX.
RTH
8W
VTH
+
_
10 V
A

+
2W
VX
_
B

Figure 10.11: Circuit of Ex 10.1 after connecting Thevenin
circuit.
13
(10)( 2)
VX 
 2V
28
6 kW (RTH)
A Thevenin
equivalent circuit
RL
24 V (VTH)
VTH
RN = RTH
The Norton
dual
IN =
4 mA (IN)
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}
Semiconductors
CHAPTER # 2
Core diagrams for copper and silicon:
One valence electron
Copper
+1
Four valence electrons
Silicon
+4
The nucleus plus the inner electron orbits
Silicon atoms in a crystal share electrons.
Valence saturation: n = 8
Because the valence electrons are bound, a silicon
crystal at room temperature is almost a perfect insulator.
Inside a silicon crystal
Some free electrons and holes are created by thermal energy.
Other free electrons and holes are recombining.
Recombination varies from a few nanoseconds to several microseconds.
Some free electrons and holes exist temporarily, awaiting recombination.
Silicon crystals are doped to
provide permanent carriers.
Free electron
(n type)
Pentavalent dopant
Hole
(p type)
Trivalent dopant
This crystal has been doped with a pentavalent impurity.
The free electrons in n type silicon support the flow of current.
This crystal has been doped with a trivalent impurity.
The holes in p type silicon support the flow of current.
Note that hole current is opposite in direction to electron current.
Semiconductors
The most popular material is silicon.
Pure crystals are intrinsic semiconductors.
Doped crystals are extrinsic semiconductors.
Crystals are doped to be n type or p type.
An n type semiconductor will have a few minority
carriers (holes).
A p type semiconductor will have a few minority
carriers (electrons).
Doping a crystal with both types
of impurities forms a pn junction diode.
Junction
P
Negative
ion
N
Positive
ion
Some electrons will cross the junction and fill holes.
A pair of ions is created each time this happens.
As this ion charge builds up, it prevents further
charge migration across the junction.
The pn barrier potential
Electron diffusion creates ion pairs called dipoles.
Each dipole has an associated electric field.
The junction goes into equilibrium when the barrier potential prevents
further diffusion.
At 25 degrees C, the barrier potential for a silicon pn junction is about
0.7 volts.
Each electron that migrates across the
junction and fills a hole effectively
eliminates both as current carriers.
P
N
Depletion layer
This results in a region at the junction that is
depleted of carriers and acts as an insulator.
Forward bias
The carriers move toward the junction
and collapse the depletion layer.
If the applied voltage is greater than
the barrier potential, the diode conducts.
Reverse bias
The carriers move away from the junction.
The depletion layer is reestablished
and the diode is off.
Diode bias
Silicon diodes turn on with a forward bias of approximately 0.7 volts.
With reverse bias, the depletion layer grows wider and the diode is off.
A small minority carrier current exists with reverse bias.
The reverse flow due to thermal carriers is called the saturation current.
Diode breakdown
Diodes cannot withstand extreme values of reverse bias.
At high reverse bias, a carrier avalanche will result due to rapid motion of the minority
carriers.
Typical breakdown ratings range from 50 volts to 1 kV.
Energy levels
Extra energy is needed to lift an electron into a higher orbit.
Electrons farther from the nucleus have higher potential energy.
When an electron falls to a lower orbit, it loses energy in the form of
heat, light, and other radiation.
LED’s are an example where some of the potential energy is converted to
light.
The p side of a pn junction has trivalent
atoms with a core charge of +3. This core
attracts electrons less than a +5 core.
Abrupt junction
Energy
Conduction band
Valence band
P-side
N-side
In an abrupt junction, the p side bands
are at a slightly higher energy level.
Real diodes have a gradual change from one material
to the other. The abrupt junction is conceptual.
Energy bands after the depletion layer has formed
Energy
Conduction band
Energy hill
Valence band
P-side
N-side
To an electron trying to diffuse across the junction,
the path it must travel looks like an energy hill. It
must receive the extra energy from an outside source.
Junction temperature
The junction temperature is the temperature inside the diode, right at
the pn junction.
When a diode is conducting, its junction temperature is higher than the
ambient.
There is less barrier potential at elevated junction temperatures.
The barrier potential decreases by 2 mV for each degree Celsius rise.
Reverse diode currents
IS, the saturation current, doubles for each 10 degree Celsius rise in
temperature. It is not proportional to reverse voltage.
The surface of a crystal does not have complete covalent bonds. The
holes that result produce a surface-leakage current that is directly
proportional to reverse voltage.