Transcript Lecture 15

Lecture #15 Basic amplifiers,
Intro to Bipolar transistors
Reading: transistors (chapter 6 and
14)
10/4/2004
EE 42 fall 2004 lecture 15
1
Topics
Today:
Examples, circuit applications:
• Diode circuits, Zener diode
• Use of dependent sources
• Basic Amplifier Models
10/4/2004
EE 42 fall 2004 lecture 15
2
NODAL ANALYSIS WITH DEPENDENT SOURCES
Finding Thévenin Equivalent Circuits with Dependent Sources Present
Method 1: Use Voc and Isc as usual to find VT and RT
(and IN as well)
Method 2: To find RT by the “ohmmeter method” turn
off only the independent sources; let the dependent
sources just do their thing.
See examples in text (such as Example 4.3) and in
discussion sections.
Pay most attention to voltage-dependent voltage
sources and voltage-dependent current sources. We
will use these only.
10/4/2004
EE 42 fall 2004 lecture 15
3
NODAL ANALYSIS WITH DEPENDENT SOURCES
Example : Find Thévenin equivalent of stuff in red box.
R3
Va
Vc
R2
+

A v Vcs
R6
ISS
With method 2 we first find open circuit voltage (VT) and then we
“measure” input resistance with source ISS turned off.
R 2 (R 6  R 3 )
ISS R 6 (R 2  AR 3 ) R 
Verify the solution:
VTH 
TH
R 2  R 3  R 6 (1 - A)
R 2  R 3  R 6 (1 - A)
10/4/2004
EE 42 fall 2004 lecture 15
4
EXAMPLE: AMPLIFIER ANALYSIS
USING THE AMPLIFIER MODEL WITH Ri = infinity:
A: Find Thévenin equivalent resistance of the input.
B: Find the Ratio of the output voltage to the input voltage (“Voltage
Gain”)
R
AMPLIFIER MODEL
F
VIN
RS
V+
V-
Circuit Model in linear region

A
+
V0
Ri
+

V1
+

AV1
+

V0
Method: We substitute the amplifier model for the amplifier, and
perform standard nodal analysis
You find : RIN and VO/VIN
10/4/2004
EE 42 fall 2004 lecture 15
5
EXAMPLE: AMPLIFIER ANALYSIS
USING THE AMPLIFIER MODEL WITH Ri = infinity:
How to begin: Just redraw carefully!
RF
VIN RS
V+
V-

A
+
RF
VIN
RS
V-
-
V0
V+
+
V1 +

V0
AV1
Method: We substitute the amplifier model for the amplifier, and
perform standard nodal analysis
- AR F
R F  (1  A)R S
Verify the solution: RIN =
VO/VIN =
R F  (1  A)R S
1 A
10/4/2004
EE 42 fall 2004 lecture 15
6
Bipolar transistors
• Bipolar transistors are made from two PN junctions that
are very close together.
• The name bipolar comes from the fact that both carrier
types play roles in its function
• The connection to the middle slice, called the base, can
control the current without having to supply much itself
N
10/4/2004
P
N
EE 42 fall 2004 lecture 15
7
Symbol
• The symbol for an NPN transistor is:
C
B
E
10/4/2004
Most of the current in an bipolar
transistor flows between the
collector and emitter, but the
amount of current is controlled by
the base-emitter junction.
In an NPN transistor, the current is
conducted by electrons moving
from the emitter (emitter of
electrons) to the collector
(collector of electrons)
EE 42 fall 2004 lecture 15
8
Reversed biased Base-Collector
junction
• To understand how the transistor
works, lets just look at the basecollector junction under a reverse
bias.
• The collector is doped more lightly
than the base, so to balance the
charge, the depletion extends
further into the collector than it does
into the base.
• The electric field is as shown,
holding back the electrons and holes
from the junction
collector
electrons
holes
base
10/4/2004
EE 42 fall 2004 lecture 15
9
Electrons in the base
& depletion region
• What would happen if we were
to “beam in” some electrons
into the base?
– They would be swept into the
collector by the electric field.
• If we were to inject these
electrons by absorbing light,
this device would be a light
detector. (Photodiode)
• If we inject these electrons by
another junction, it is a NPN
bipolar transistor
10/4/2004
EE 42 fall 2004 lecture 15
collector
electrons
-
-
holes
base
10
Minority carriers diffuse
• Outside the depletion region, there is no
electric field, so the carriers move by
diffusion only.
• Diffusion is just the name for the effect that
happens when randomly moving particles
spread from areas where they are
concentrated to areas where there are
fewer of them.
10/4/2004
EE 42 fall 2004 lecture 15
11
Injection of minority carriers
• To see how we would inject minority
carriers into a region, consider a forward
biased junction
• If we forward bias a PN junction, then
the internal field is reduced, holes are
injected into the N side, and electrons
are injected into the P side.
• Since we only desire to have electrons
injected into the base, we heavily dope
the emitter so that most of the current
comes from electrons going from the
emitter into the base, rather than from
holes going into the emitter
10/4/2004
EE 42 fall 2004 lecture 15
base
Emitter
12
How a bipolar transistor works
• So a one sentence description of how a
bipolar transistor works is:
• A forward biased junction injects minority
carriers which can then go through a
nearby reverse biased junction.
10/4/2004
EE 42 fall 2004 lecture 15
13
Gain
• How does a bipolar transistor act as an
amplifier?
• The Emitter-Base junction current is just the
current of a PN diode, i.e. the voltage from the
base to the emitter will give a large current as
soon as the voltage exceeds .7 volts.
• But since most of the carriers (99%) go to the
collector, the base only needs to deliver 1% of
the total current in order to control the voltage
(and therefore the current)
10/4/2004
EE 42 fall 2004 lecture 15
14
Electron flow
• So the forward bias on the
emitter-base junction
induces the electrons to
flow, but most of them
make it across to the
collector instead of stopping
in the base and flowing to
the base terminal
10/4/2004
EE 42 fall 2004 lecture 15
Collector
Base
Emitter
15
Device model
• As long as the Base-collector junction is
reverse biased, and the Emitter-base
junction is forward biased, a good model
of the NPN transistor is:
Collector
Base
I c  I b
Emitter
10/4/2004
EE 42 fall 2004 lecture 15
16
Modes of operation
Cut-off:
• If the Emitter-base junction (the one controlling the
current) is not forward biased, then the transistor is said
to be in cut-off.
• A small amount of current will still flow, usually negligible
Saturation:
• If the Base-collector junction sees so much current flow
that it is no longer forward biased, then the device will no
longer behave as described.
Breakdown:
• If a high enough voltage is applied, the transistor
junctions will break down, and a high current can flow.
10/4/2004
EE 42 fall 2004 lecture 15
17
IV curve
• Since the transistor is a three terminal device is a three
terminal device, you might think that 6 variables would
be important:
• Vbc – the voltage between the base and the collector
• Vbe – the voltage between the base and the emitter.
• Vce- The voltage between the collector and the emitter.
• Ib- the current into the base.
• Ie- the current into the emitter.
• Ic- the current into the collector.
• But the transistor has no net charge, so Ib+Ie+Ic=0
• And of course if you know any two of the voltages you
can calculate the third.
10/4/2004
EE 42 fall 2004 lecture 15
18
Transistor circuit configurations
• Typically we will want to use the transistor as a
device which has an input and an output. Since
one of the terminals must be shared, we call that
a common terminal
• The voltages with respect to the common
terminal are then used to describe the operation
of the transistors
• There are three types of connections:
– Common emitter,
– Common collector,
– Common base
10/4/2004
EE 42 fall 2004 lecture 15
19
IV curve for common emitter
• To show the IV curve for
Ic
a NPN transistor in a
common emitter
configuration, we plot the
voltage from the collector
to the emitter Vce vs the
current from the emitter Ic
• The base current is
shown by setting several
values and then plotting a
curve for each of them
(called steps)
Saturation
10/4/2004
EE 42 fall 2004 lecture 15
Breakdown
Forward
Active
• Cutoff
Vce
20