Transistors Appendix

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Transistors Appendix
2
Transistors are scalable electronic switches,
made from doped silicon. Silicon forms a
crystal, and have no free electrons at low
temperature (around 0K). At higher temperatures
(300K), thermal energy is sufficient to release
some electrons from their covalent bonds. The
availability of conducting electrons is, however,
limited (more quantitative) because the band-gap
energy is high relative to a good conductor
(copper).
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N
P
N- and P-type doped silicon are the components
necessary to build switching devices, like diodes
and transistors. Dopant atoms can integrate with
silicons crystal lattice, and create additional
holes or conducting electrons. Because phosphorus
has five valance electrons, an additional weakly
bound electron is present when it integrates.
Thermal energy frees this electron, producing a
large number of conducing electrons in this
material, known as N-type. Tri-valent Boron
creates a hole upon integration, producing P-type
material.
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Sets up a field
Diffusion
Establishes current equilibrium
N
P
N
P
N
P
Diffusion
Diffusion
Si
Si
Si
Si
Si
Si
Field
P
Si
P
Si
P
Si
Si
B
Si
B-
Si
B-
Si
Si
Si
Si
Si
Si

-
The diode device architecture fuses an N and a
P-junction together. The NP junction with no
voltage applied establishes an equilibrium such
that the force of diffusion, which draws
electrons into the P-type material, is opposed by
the force of the electric field, which draws
electrons back into the N-type material. This
equilibrium electric field represents a
potential difference around 0.76V (proven, p. 18
of Electronic Circuit Design and Analysis).
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N-type
P-type
Relative concentration of majority carrier in
N-type material electrons
Si
Si
P
Si
Si
B
Diffusion force
Si
Si
Relative concentration of minority carrier in
P-type material electrons
The N-region contains a higher concentration
electrons than the P-region. Electrons diffuse
from the N to P-type material.
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Electric field (oriented from positive to
negative), 0.75V
Electric force (on electrons) due to equilibrium
field
N-type
P-type
Electrons
Si
Si
P
Si
Si
B-
Diffusion force
Si
Si
Electrons
Diffusion establishes a charge separation, which
sets up an electric field. The field exerts a
force on the electrons opposite the direction of
diffusion.
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Electric field (oriented from positive to
negative) , 0.75V
Electric force (on electrons)
N-type
P-type
Diffusion force
Electrons equilibrium
Diffusion continues and the charge separation
continues to grow in magnitude, until the force
of the electric field due to the charge
separation equally opposes the force of
diffusion. At this point, there is no net current
flow across the junction, and the electron
concentration at either side of the junction
reaches an equilibrium value.
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Reverse Bias
Applied Field
N
P
Net Field gt Equilibrium
Si
Si
Force
P
Si
Si
B-
Si
Si
Equilibrium Field, 0.75V
-

In reverse bias voltage, the fields are oriented
in the same direction, and the magnitude of the
electric field in the space charge region
increases above the equilibrium value. This holds
back electrons, and no current flows. With
forward bias, the net result is that the eclectic
field at the junction is lower than the
equilibrium value, and electrons diffuse.
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Applied electric field from reverse-bias voltage
polarity
Electric field (oriented from positive to
negative)
Net electric force due to fields
N-type
P-type
Diffusion force
Electrons drawn from P-type material near junction
Electrons
Reverse-bias voltage increases net field at the
junction, opposing diffusion. Charges are drawn
away from the junction by the field.
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Forward Bias
Applied Field
N
P
Net Field lt Equilibrium
Si
Si
Net diffusion
P
Si
Si
B-
Si
Si
Equilibrium Field

-
In reverse bias voltage, the fields are oriented
in the same direction, and the magnitude of the
electric field in the space charge region
increases above the equilibrium value. This holds
back electrons, and no current flows. With
forward bias, the net result is that the eclectic
field at the junction is lower than the
equilibrium value, and electrons diffuse.
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Applied electric field from forward-bias voltage
polarity
Electric field (oriented from positive to
negative)
Net electric force due to fields
N-type
P-type
Diffusion force exceeds the net electric force,
electrons diffuse across junction, where electron
concentration is above equilibrium
Electrons diffuse from junction into the bulk
Electrons
Forward-biased voltage reduces net field at the
junction, reducing the width of the depletion
region. The force of diffusion dominates.
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Transfer function
Device schematic
I
Two-state Non-linear response at on threshold
Vin
Input voltage V forward bias V reverse bias
Output current Current No current
Voltage input
Sensitive Only 0.7V on threshold
I Is e(Vin/Vt) -1 Is Reverse bias
current, 510(-14)A Vt Thermal voltage 0.026
The output is highly responsive to increases in
the forward-biased input, above the cut in
voltage threshold 0.7V. At zero or reverse bias
input voltage, the output current is the very
small saturation value, around 510(-14)A.
Because the forward bias voltage is included in
the exponent, the exponential terms increases
with respect to the input voltage. At the cut in
voltage 0.7V, the input voltage driven
exponential term begins to dominate the very
small reverse bias saturation current term,
resulting in the exponential behavior.
Physically, this means that diffusion is
unrestricted once the forward bias voltage
establishes a field that fully counters the
built-in equilibrium field.
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VIR
IVs/R
Energy balance I -(1/R)VdVs/R
R
Diode Vd
Voltage Source
Vd Vs
Vd
KVL (energy balance) VsIRVd I -(1/R)VdVs/R
The diode current and voltage are given by the
intersection between the circuit load line and
the diode performance curve. This intersection,
or Q-point, gives the DC voltage and current for
the forward-biased diode in the circuit this
intersection identifies the feasible diode
operating conditions that also satisfy the energy
balance of the circuit.
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Voltage drop 0.7V
Switch voltage Polarity forward to reverse bias
5V
I
R
4.3V, I4.3/R
0V, I0
R

-
Supply 5V
Time

-
Zero output
A diode circuit can make a switch. If the output
is defined as the voltage drop across the
resistance, then output is zero in the reverse
bias condition (red). This is because the diode
is reverse biased in this condition, resulting in
no circuit current and no resistor voltage drop.
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Supply 5V
R
I
Vi (1)
V (o)
Vi (2)
A diode circuit can make an AND gate. If both
inputs are high (5V), then there is no potential
difference across either diode (no voltage
difference between input and supply). Neither
diode is forward biased, no current flows. Since
there is no current in the circuit, there is no
voltage drop across the resistance and output
voltage is 5V (high). If either input is low,
then the diode is forward biased and the voltage
drop across the diode will be 0.7V. Thus, 4.3V
will drop across the resistance, resulting in a
0.7V (low) output.
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Vi (1)
V (o)
Vi (2)
R
I
A diode circuit can make an OR gate. If either
input is high (5V), the corresponding diode is
forward biased with a 0.7V drop. The output
voltage is 4.3V, and the resistance determines
the output current (4.3V/R).
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N
P
Net diffusion
P
Si
Equilibrium Field
Input Voltage
-

Resistance, R
10V
Supply Voltage, V
Voltage Out Always High
A diode in the circuit will not switch the output
because the applied field from the supply voltage
will over-ride the applied field from the input
voltage. The diode is forward biased with respect
the supply voltage , so current will always flow
through the circuit.
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Current through first diode controlled by input
voltage
Second diode is reverse biased with respect to
supply, so no current
Input Voltage

-
Supply Voltage, V
-

Resistance, R
10V
Voltage Out Always Low
Adding a second diode establishes a
three-terminal device in which the voltage across
the first diode is set by the input and
unaffected by the supply. The second diode is
reverse biased with respect to the supply
voltage, which shields the first diode. However,
no current will flow through the outer circuit
because the second diode is reverse biased.
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N
N
P
An NPN junction produces the effect of two
opposing diodes in the circuit.
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Reverse Bias
Forward Bias
Applied field from supply voltage
Applied field from input voltage
N
P
N
Force
Net diffusion
-
Equilibrium Field
Equilibrium Field

-
Supply Voltage, V
Input Voltage
Base current
-

Resistance, R
10V
Voltage Out Always Low
No current flows in the outer circuit because all
electrons entering the P-type region exit through
P-region lead, and because no current can flow
across the reverse biased PN junction Electrons
diffuse from N to P-type material, recombine with
holes, and exit through the conductor in the
P-type material.
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N
N
P
Make base thin
Net diffusion
Force
The electron concentration across the P region
varies from high at the forward biased junction
through which electrons are passing to low at the
reverse biased junction, at which no current is
flowing. Transistors are designed to allow
diffusion of electrons across this concentration
gradient, from the emitter across the base and
into the collector, thus completing the outer
circuit. This is done by making the base thin.
The thin base leads to a sharper concentration
gradient, and reduces the likelihood of
recombination.
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N Collector
N Emitter
P Base
Net diffusion
Force
-
Equilibrium Field

-
Supply Voltage, V
Input Voltage
Ib
-

Resistance, R
10V
Voltage Out Contollable
In a bi-polar junction NPN transistor, electrons
are injected into and diffuse through the base to
the collector, completing the outer circuit. Many
of the electrons will not recombine with holes in
the base (P-region) for two reasons. First, the
emitter is heavily doped and the base is lightly
doped. Second, the P-region is thin. Because
there is a concentration gradient across the
region, electrons diffuse towards the
reverse-biased base-collector junction. The
electrons will be captured by the strong electric
field at this junction, and will flow into the
collector.
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Device
Transfer Function
Energy balance Ib -(1/Rb)VbeVbase/Rb
Intersection gives Ib and Vbe
Rb
B
Load line

E
Vbe
I (base)
Vbase
-
Ib
The transistor can be de-coupled into two parts,
first one being the base-emitter, which functions
like a diode. Diode performance can be determined
(along with the base current) by the load line
intersection with the diode performance curve.
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Energy balance Vsource Vt IcRc
C
Rc
Vout
Rb
Vt
B


E
Vbe
I (base)
-
-
Vbase
Vsource
Ib
The current exiting the collector, Ic, is
determined by the voltage across base-emitter
junction (the input) only. This is because
electron injection to the base from the emitter
is the limiting factor on the current through the
circuit, and base voltage control the degree of
electron injection. As a result, Ic is
independent of the reverse-bias voltage polarity
across the BC junction if, for example, Rc
decreases, the voltage drop across the
transistor, Vt, will increase (energy balance).
The voltage drop across the BE junction is fixed
at 0.7V, for a forward biased diode. Thus, the
reverse bias voltage across the BC junction will
have to increase. However, this has no affect on
the output current. The collector current is
related to the base current by a factor B, or
gain, which is between 50 200 for transistors.
The above design is a common emitter (emitter is
the common connection) bi-polar junction
transistor.
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Output
Out (Ic)
Various inputs (Vbe)
C
B
On
E
In (Vbe)
Vt
Vt
Off
Energy Balance (1/Rc)VtVsource/RcIc
Vt must be gt Vbe, else there is no reverse
bias across the base-collector
Two elements are necessary for the shared
emitter, bi-polar junction NPN transistor. First,
there must be reverse bias voltage polarity
across the base-emitter junction to capture
diffusing electrons in the base. This is
represented on the x-axis. The output is weakly
dependent upon the degree of reverse bias,
reflected by weak slope of each line. But the
output will drop off rapidly as if the voltage
across the transistor is too low to maintain
reverse bias across BE junction. Second, there
must be forward-biased input voltage. As the
input voltage increases (represented by each
curve), the output current increases. The lower
line represents zero input voltage, and output is
zero, as expected.