Transcript Transistor Detail

Transistors Appendix
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 bandgap energy is high relative to a good conductor (copper).
Dopant :
Phosphorus
Boron
Means :
Add electrons to
conduction band
Remove electrons from
valence shell, holes
created
Creates :
N-type material
P-type material
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 silicon’s 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
Sets up a field
Diffusion
Establishes current equi
Diffusion
Si
Si
Si
Si
Diffusion
Si
Si
P
Si
P+
Si
P+
Si
B
Si
B-
Si
B-
Si
Si
Si
Si
Si
Si
P
N
+
P
N
Field
Si
-
E
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).
N-type
P-type
Relative concentration of majority carrier in N-type material: electrons
Si
Si
P
Si
Si
B
Si
Si
Diffusion “force”
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.
E
Electric field (oriented from positive to negative), ~0.75V
Electric force (on electrons) due to equilibrium field
N-type
Electrons
P-type
Si
Si
P+
Si
Si
B-
Si
Si
Diffusion “force”
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.
E
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.
Reverse Bias
Applied
Field
E
Net Field > Equilibrium
Si
Si
P+
Si
Si
B-
Si
Si
Force
N
P
Equilibrium
Field, ~0.75V
E
-
+
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.
E
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.
Forward Bias
Applied
Field
E
Net Field < Equilibrium
N
Si
Si
Net diffusion
P+
Si
Si
B-
Si
Si
P
Equilibrium
Field
E
+
-
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.
E
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.
Device schematic
Transfer function
I
Input: voltage
V forward bias
V reverse bias
Output: current
Current
No current
Two-state:
Non-linear response
at “on” threshold
Voltage input
Vin
Sensitive:
Only 0.7V “on” threshold
I = Is*[ e^(Vin/Vt) -1 ]
Is = Reverse bias current, 5*10^(-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 5*10^(-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
V=IR
I=Vs/R
R
Voltage
Source
Diode
Vd
Energy balance:
I= -(1/R)*Vd+Vs/R
Vd = Vs Vd
KVL (energy balance):
Vs=IR+Vd
I= -(1/R)*Vd+Vs/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.
Voltage drop
0.7V
Switch voltage
Polarity forward to
reverse bias
5V
I
4.3V,
I=4.3/R
Supply : 5V
R
+
-
0V,
I=0
R
+
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.
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.
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).
E
Net diffusion
P+
E
Input Voltage
Equilibrium
Field
-
Si
P
+
N
10V
Supply Voltage, V
Resistance, R
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.
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
+
+
-
-
10V
Resistance, R
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
N
P
N
An NPN junction produces the effect of two opposing diodes in the circuit.
Forward Bias
Reverse Bias
Applied field from
supply voltage
Applied field
from input voltage
Force
Net diffusion
-
P
Input Voltage
Equilibrium
Field
Supply Voltage, V
-
+
Base current
Equilibrium
Field
N
+
N
10V
Resistance, R
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
Make base thin
P
N
Net diffusion
N
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
Net diffusion
N
Emitter
-
Force
N
Collector
Equilibrium
Field
-
Input Voltage
Supply Voltage, V
+
+
Ib
P
Base
10V
Resistance, R
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
Device
Transfer Function
Energy balance:
Ib= -(1/Rb)*Vbe+Vbase/Rb
Rb
+
Vbase -
I (base) Vbe
Intersection gives
Ib and Vbe
B
Load line
E
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.
iC  I s e
vBE
Energy balance:
Vsource = Vt +IcRc
VT
C
Rc
Rb
+
Vbase -
I (base) Vbe
B
E
Vout
Vt
+
-
Vsource
Ib
 Dp N A W 1 W 2  vBE VT  1  vBE VT iC
iB  I s 

   I se

e

 
 Dn N D LP 2 Dn b 
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
Output
Out (Ic)
Various inputs (Vbe)
C
B
In (Vbe)
E
On
Vt
Off
Vt
Energy Balance:
Vt must be > Vbe, else
–(1/Rc)Vt+Vsource/Rc=Ic
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