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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 band-
gap 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- 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
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.
NN PP
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).
NN PP
Si
P
Si
Si
Si
Si
B
Si
NN PP
Si
P+
Si
Si
Si
Si
B-
Si
DiffusionDiffusion
FieldField
Diffusion
NN PP
Si
P+
Si
Si
Si
Si
B-
Si
+ -
Sets up a field Establishes current equilibrium
DiffusionDiffusion
E
The N-region contains a higher concentration electrons than the P-region. Electrons diffuse from the N to P-type material.
Relative concentration of minority carrier in P-type material: electrons
Diffusion “force”
Relative concentration of majority carrier in N-type material: electrons
Si
P
Si
Si
Si
Si
B
Si
P-typeN-type
Diffusion establishes a charge separation, which sets up an electric field. The field exerts a force on the electrons opposite the direction of diffusion.
P-typeN-type
Diffusion “force”
Si
P+
Si
Si
Si
Si
B-
Si
Electric force (on electrons) due to equilibrium field
Electrons
Electrons
Electric field (oriented from positive to negative), ~0.75VE
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.
Diffusion “force”
P-typeN-type
Electrons : equilibrium
Electric force (on electrons)
Electric field (oriented from positive to negative) , ~0.75VE
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.
NN PP
Si
P+
Si
Si
Si
Si
B-
Si
-+
AppliedField
Reverse Bias
EquilibriumField, ~0.75V
ForceForce
Net Field > Equilibrium
E
E
Reverse-bias voltage increases net field at the junction, opposing diffusion. Charges are drawn away from the junction by the field.
P-type
Net electric force due to fields
N-type
Diffusion “force”
Electrons
Electric field (oriented from positive to negative)
Applied electric field from reverse-bias voltage polarity
Electrons drawn from P-type material near junction
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.
NN PP
Si
P+
Si
Si
Si
Si
B-
Si
EquilibriumField
+-
AppliedField
Net diffusionNet diffusion
Forward Bias
Net Field < Equilibrium
E
E
Forward-biased voltage reduces net field at the junction, reducing the width of the depletion region. The force of diffusion dominates.
P-type
Net electric force due to fields
N-typeDiffusion “force” exceeds the net electric force, electrons diffuse across junction, where electron concentration is above equilibrium
Electrons diffuse from junction into the bulk
Electrons
Electric field (oriented from positive to negative)
Applied electric field from forward-bias voltage polarityE
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 counters the built-in equilibrium field.
Two-state:Non-linear response at “on” threshold
Sensitive:Only 0.7V “on” threshold
Voltage input
Transfer function
Input: voltageV forward biasV reverse bias
Device schematic
Output: currentCurrent
No current
I = Is*[ e^(Vin/Vt) -1 ]Is = Reverse bias current, 5*10^(-14)AVt = Thermal voltage = 0.026
Vin
I
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.
RR
Vd
KVL (energy balance):Vs=IR+VdI= -(1/R)*Vd+Vs/R
VoltageSource
DiodeVd
V=IREnergy balance:I= -(1/R)*Vd+Vs/R
Vd = Vs
I=Vs/R
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.
RR
Supply : 5V
RR
+
-
-+
Voltage drop0.7V
4.3V,I=4.3/R
0V,I=0
5V
I
Time
Switch voltagePolarity forward to
reverse bias
Zero output
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)
Vi (2)
RR
V (o)
Supply : 5V
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).
Vi (1)
Vi (2)
RR
V (o)
I
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.
NN PP
Input Voltage +-
Supply Voltage, V10V
Resistance, RResistance, R
VoltageOut = Always High
P+
SiEquilibriumField
Net diffusionNet diffusion
E
E
+-Supply Voltage, V
10V
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.
+-
Input Voltage
Resistance, RResistance, R
VoltageOut = Always Low
Current through first diode controlled by input voltage
Second diode is reverse biased with respect to supply, so no current
An NPN junction produces the effect of two opposing diodes in the circuit.
NNNN PP
+-Supply Voltage, V
10V
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.
+-
NNNN PP
-
Input VoltageResistance, RResistance, R
VoltageOut = Always Low
Base current
Applied field from input voltage
EquilibriumField
ForceForceNet diffusionNet diffusion
EquilibriumField
Applied field from supply voltage
Reverse BiasForward Bias
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.
NNNN PP
Make base thin
ForceForceNet diffusionNet diffusion
+-Supply Voltage, V
10V
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.
+-
NCollector
NCollector
NEmitter
NEmitter
PBase
PBase
-
Input VoltageResistance, RResistance, R
VoltageOut = Contollable
Ib
EquilibriumField
ForceForceNet diffusionNet diffusion
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.
RbRb BB
EEI (base)Vbase
Vbe Load line
Ib
Energy balance:Ib= -(1/Rb)*Vbe+Vbase/Rb
Intersection gives Ib and Vbe
+-
Device Transfer Function
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.
RbRb+-
BB
EE
CC
-+
RcRc
I (base)
Vt
Energy balance:Vsource = Vt +IcRc
Vbase Vsource
T
BEV
v
sC eIi
Vbe
Ib
Vout
21 1
2
BE BE
T T
v vp V V CA
B s sn D P n b
D iN W Wi I e I e
D N L D
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.
Vt
Various inputs (Vbe)
Energy Balance:–(1/Rc)Vt+Vsource/Rc=Ic
Vt must be > Vbe, else there is no reverse bias
across the base-collector
BB
EE
CC
In (Vbe) Vt
Out (Ic) Output
Off
On