Transcript Phys 345 Electronics for Scientists

Announcements
• Assignment 2 due now
• Assignment 3 posted, due Thursday Oct 6th
• First mid-term Thursday October 27th
Lecture 9 Overview
• Transistors
Transistors
• Semiconductor device
• First Active circuit element - gain > 1
• Discuss the Bipolar Junction Transistor only
• See Simpson Chapter 5 for more detail.
Bipolar Junction Transistors
Collector
Base
Emitter
• NPN Bipolar Junction transistor shown (PNP also possible)
• 3 terminals: Emitter, Base, Collector
• Contains 2 p-n junctions: emitter-base junction, collector-base junction
• Can be thought of as two back-to-back diodes, but operating
characteristics are very different
• Base region (P-type here) must be thin for transistor action to work
Modes of operation
Collector
Base
“OFF”
“ON”
Emitter
• Use 2 voltage supplies to bias the two junctions
(forward or reverse)
• 3 basic modes: cutoff, active and saturation, correspond
to three different bias conditions.
Active Mode
• Collector-Base junction is reverse biased
• Emitter-Base junction is forward biased
• iC=βiB (β=100 - 500) Active circuit element - gain > 1 !!
• How does collector current flow when collector-base junction is
iC
reverse biased?
+
iB
Base
Collector
Emitter
VBE
iE
-V
CE
Active Mode
-
Direction of current flow
+
• What's happening?
• Emitter-base is forward biased; collector-base reverse biased.
• Forward bias of emitter injects electrons into thin base region
• Majority shoot through the base into the collector region where they encounter the voltage
source on the collector and produce a current.
• Electrons combine with holes in the base region and form negative ions which impede the flow
• Drawing off negative charge via the base lead reduces this effect (“making the base smaller”)
- so the base current controls the flow of electrons into the collector
• Nobel Prize 1956; Shockley, Bardeen & Brattain
Bill Shockley
Born
1910
London, England
Died
1989
Stanford, California
Active Mode
• Can relate iB and iC
iC  iB 
• β is defined mainly by the properties and geometry of the materials
• Ideally constant for any particular type of transistor
• Typically around 500
• "Common-emitter current gain"
• Actually varies with temperature and emitter current
• Collector current is controlled by the base current
Emitter current
• Emitter current is the sum of iB and iC (KCL)
iE  iB  iC
iC  iB 
  1

 iC 
  
  1
iE  iC where     


• α is called the common-base current gain (~1.0)
Circuit Symbols and Conventions
• BJTs are not symmetric devices
• Doping and physical dimensions are different for collector and emitter
• Collector largest, connected to heat sink as most power dissipated there
• Emitter region smaller, and more heavily doped to provide an abundance
of charge carriers
• Base region is very thin (~50nm) to enhance probability that electrons will
cross it
• PNP devices also exist - diode senses are reversed, so bias voltage
polarities must also be reversed
iC
i-v Characteristics
+
iB
• Simplest model for low
frequencies (DC condition)
"Ebers-Moll".
• Relates collector current IC to
base-emitter voltage VBE:
Base
Collector
-V
CE
Emitter
VBE
iE


I C  I S eVBE / VT  1
IC
 I S eVBE / VT
kB T
vT = e
VBE
• IS=Saturation Current
• Similar to Diode Law
• Recall IB=IC/β
• Collector current is controlled by the
base-emitter voltage VBE
BJT Amplifier
"Common emitter" configuration
Note labelling scheme:
iC=IC+ic
• To act as an amplifier, first bias the transistor to get it into active mode
• Then superimpose a small signal vbe on the base
• Under DC conditions: I  I eV V
BE
C
T
S
I E  IC / 
I B  IC / 
  1.01
  100
VCE  VCC  I C RC
Added resistor RC
BJT Amplifier
• DC condition biases the BJT to the
point Q on the plot
• Adding a small voltage signal vbe
translates into a current signal that
we can write as:
iC  I C  ic  I S e (VBE  vbe ) / VT
 ISe
VBE / VT
 IC e
e
vbe / VT
vbe / VT
DC component IC
 vbe 

• If vbe/VT is small: I C  ic  I C 1 
 VT 
So the small signal current is:
ic 
Small signal component ic
IC
vbe  g m vbe
VT
i.e. a voltage-to-current amplifier; small signal collector current determined by base-to
emitter voltage
gm is the "transconductance"; corresponds to the slope at point Q
More on small-signals
• Base current also varies with vbe
iB = I B + ib
IB =DC component of base current iB
iC
1 IC
ib =
vbe
b VT
1
ib = gm vbe
IC
1 IC
iB  

vbe
   VT
b
ib = small signal component of base current iB
So, having small signal voltage at the
base, and small signal current at the
base, consider the small signal
equivalent resistance into the base, rπ:
Alternatively, rearrange to give
small signal resistance between
base and emitter, re:
vbe
vbe
b
rp =
=
=
1
ib
gm
gm vbe
b
vbe
a
1
re =
=
@
ie
gm gm
rp = (b +1)re
Small signal models
Use these relations to represent the small signal model for the
transistor by an equivalent circuit. Hybrid-π model:
e.g. as a voltage controlled
current source: ic  gmvbe
gm 
r 
or as a current controlled
current source ic = b ib
IC
VT

gm
Small signal models
Other small signal models may sometimes be more convenient.
T model:
vbe VT a
1
re =
= =
@
ie I E gm gm
rp = (b +1)re
gm 
IC
VT
Using small signal models
1) Determine the DC operating conditions (in particular, the collector
current, IC)
2) Calculate small signal model parameters: gm, rπ, re
3) Eliminate DC sources: replace voltage sources with shorts and
current sources with open circuits
4) Replace BJT with equivalent small-signal models. Choose most
convenient depending on surrounding circuitry
5) Analyze
Voltage gain with small signal model
ic
+
RC
-
vc
ic
+
vbe
vbe
eliminate
DC sources
and apply Tmodel
re
-
•To convert the voltage
controlled current source into
a circuit providing voltage
gain, connect a resistor to the
collector output and measure
the voltage.
• Find the gain using a small
signal model:
vc  ic RC  
ie

RC  ie RC
vbe  ie re
Voltage gain 
re 

gm

vc  ie RC
R

 C
vbe
ie re
re
1
gm
so volt agegain
vc
  g m RC
vbe
gm =
IC
VT
Summary of useful equations
• Basic DC operating
conditions:
I E  I B  IC
I C  I B
I E  I C
 1


I C  I S eVBE / VT
VC  VCC  I C RC
• Add a small signal:
ic =
IC
vbe = gm vbe
VT
gm =
IC
VT
rp =
vbe
b
=
ib
gm
re =
vbe
a
1
=
@
ie
gm
gm