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