Lecture 2 Transistor Modeling

8/7/2019 Lecture 2 Transistor Modeling

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Hardware Modeling S.M.Lambor 1

Hardware Modeling

Modeling of Transistor




What is a Transistor?What is a Transistor?

V G S

≥ V T

R o n

S D

A Switch!

|VGS|

An MOS Transistor




Diode Model of the npn BJT •The diode is controlled by the

voltage at B.

•When the diode is completely on,the switch is closed. This is

the saturation region.•

•When the diode is completely

off, the switch is open. This isthe cutoff region.

•

•When the diode is in between, it

is in the active region.




Switch Model of the npn BJT

Controlstransistor Switch

Circuit that isswitched

Remove the part of the circuit that controls the switch andconsider two possible cases:




Current-Voltage RelationsCurrent-Voltage Relations

A good transistorA good transistor

QuadraticRelationship

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6x 10

-4

VDS

(V)

I D

( A )

VGS= 2.5 V

VGS= 2.0 V

VGS= 1.5 V

VGS= 1.0 V

Resistive Saturation

VDS = VGS - VT




Current-Voltage RelationsCurrent-Voltage Relations

The Deep-Submicron EraThe Deep-Submicron Era

Linear Relationship

-4

VDS

(V)

0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5x 10

I D

( A )

VGS= 2.5 V

VGS= 2.0 V

VGS= 1.5 V

VGS= 1.0 V

Early Saturation


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Schematic Symbols and Sign ConventionsSchematic Symbols and Sign Conventions

C

E

B

I B

I E

I C +

–

+

+

–

–

V BC

V BE

V CE

C

E

B

I B

I E

I C +

–

+

+

–

–

V BC

V BE

V CE

(a) npn (b) pnp



Operations ModesOperations Modes



Forward Active OperationForward Active Operation

x

E B C

W B

Carrier Concentration

Depletion

Regions

0 W

pe0

pc0

n b0

n b(0)



Reverse ActiveReverse Active

x

E B C

W B


0 W

pe0

n b0

n b(0)

pc0

n b(W)



Saturation ModeSaturation Mode

x

E B C

W B

C arrier C oncen tration

0 W

pe0

n b0

n b(0)

pc0QS

QAn b

(W )



CutoffCutoff

x

E B C

W B


0 W

pe0

n b0n b(0) pc0

n b(W)



A Model for Manual AnalysisA Model for Manual Analysis

E

C B

βF I B

I B

+

–

V BE

I B = I S (eV BE /φT – 1)

E

C B

βF I B

I B

+ –

V BE (on )

(a) Forward-active (b) Forward-active (simplified)

E

C B

I B

+ –V BE (sat )

(c) Forward-saturation

+ – V CE (sat )

I C

< βFI B



Capacitive Model for Bipolar TransistorCapacitive Model for Bipolar TransistorC

E

B

QF

QR

C be

C bc

S

C cs

collector-substrate

junction capacitance

base-emitter

base-collector junction capacitances

base charge



Base Charge - Diffusion CapacitanceBase Charge - Diffusion Capacitance



Bipolar Transistors - Secondary EffectsBipolar Transistors - Secondary Effects

• Early Voltage

• Parasitic Resistances

• Beta Variations



Early VoltageEarly Voltage

Forward

Active

Saturation

VA

V CE

I C

V BE3

V BE2

V BE1




Modeling of Amplifier




The BJT is an an excellent amplifier when biasedin the forward-active region.

The FET can be used as an amplifier if operated inthe saturation region.

In these regions, the transistors can provide high

voltage, current and power gains. DC bias is provided to stabilize the operating point

in the desired operation region. The DC Q-point also determines

◦ The small-signal parameters of the transistor◦ The voltage gain, input resistance, and output resistance◦ The maximum input and output signal amplitudes◦ The overall power consumption of the amplifier




A Simple BJT AmplifierA Simple BJT Amplifier

The BJT is biased in the forward active region by dc voltage sources V BE

and V CC = 10 V. The DC Q-point is set at, (V CE , I C ) = (5 V, 1.5 mA) with I B= 15 µ A.

Total base-emitter voltage is:

Collector-emitter voltage is: This produces a load line.



21

BJT Amplifier (continued)BJT Amplifier (continued)

An 8 mV peak change in v BE gives a 5 µ A change in i B and a

0.5 mA change in iC .

The 0.5 mA change in iC gives a 1.65 V change in vCE .

If changes in operating currentsand voltages are small enough,

then I C and V CE waveforms areundistorted replicas of the inputsignal.

A small voltage change at the base causes a large voltage

change at the collector. Thevoltage gain is given by:

The minus sign indicates a 1800 phase shift between input andoutput signals.




A Simple MOSFET AmplifierA Simple MOSFET Amplifier

The MOSFET is biased in the saturation region by dc voltagesources V GS and V DS = 10 V. The DC Q-point is set at (V DS , I DS ) =

(4.8 V, 1.56 mA) with V GS = 3.5 V.

Total gate-source voltage is:A 1 V p-p change in v GS gives a 1.25 mA p-p change in i DS and a 4 V p-p

Change in v DS . Notice the characteristic non-linear I/O relationship

compared to the BJT.




AC coupling through capacitors is

used to inject an ac input signaland extract the ac output signalwithout disturbing the DC Q-point

Capacitors provide negligibleimpedance at frequencies of

interest and provide opencircuits at dc.

A Practical BJT Amplifier using Coupling andA Practical BJT Amplifier using Coupling and

Bypass CapacitorsBypass Capacitors

In a practical amplifier design, C 1

and C 3are large coupling capacitors

or dc blocking capacitors, theirreactance (XC = |ZC| = 1/wC ) at

signal frequency is negligible. Theyare effective open circuits for thecircuit when DC bias is considered.

C 2is a bypass capacitor. It provides

a low impedance path for ac currentfrom emitter to ground. Iteffectively removes R E (required

for good Q-point stability) from the

circuit when ac signals are



A Practical BJT Amplifier usingA Practical BJT Amplifier using

Coupling and Bypass CapacitorsCoupling and Bypass Capacitors

DC d AC A l i A li i fDC d AC A l i A li ti f




DC analysis:◦ Find the DC equivalent circuit by replacing all

capacitors by open circuits and inductors (if any) byshort circuits.

◦ Find the DC Q-point from the equivalent circuit byusing the appropriate large-signal transistor model.

AC analysis:◦ Find the AC equivalent circuit by replacing all

capacitors by short circuits, inductors (if any) byopen circuits, dc voltage sources by ground

connections and dc current sources by open circuits.◦ Replace the transistor by its small-signal model

◦ Use this equivalent circuit to analyze the ACcharacteristics of the amplifier.

◦ Combine the results of dc and ac analysis

(superposition) to yield the total voltages andcurrents in the circuit.

DC and AC Analysis -- Application ofDC and AC Analysis -- Application of

SuperpositionSuperposition




Summary of useful equationsSummary of useful equations

•Basic DC operatingconditions:

•Add a small signal:




Voltage gain with small signal modelVoltage gain with small signal model

•To convert the voltagecontrolled current source intoa circuit providing voltage

gain, connect a resistor to thecollector output and measurethe voltage.•Find the gain using a smallsignal model:

vbe vbe

+

-

+

-

ic

re

RC

ic

vc

eliminateDC sourcesand applyT-model




Voltage gainVoltage gain

•So, we have a voltage-to-current amplifier (a voltage controlled current source)•To convert the voltage controlled currentsource into a circuit providing voltage gain,connect a resistor to the collector output andmeasure the voltage.

So, small signal voltage gain:




How to build a Common emitterHow to build a Common emitter

amplifieramplifier

•Why bother with 2 voltagesupplies?

•Use a voltage divider R2/R1 to

provide base-emitter voltage to

correctly bias the transistor.




DC condition: the voltage dividerDC condition: the voltage divider

•The voltage divider should provide sufficientvoltage to place thetransistor in active mode(base-emitter forwardbiased):

•Current through resistorsshould be >10 times basecurrent for stability




Amplifier specificationsAmplifier specifications

What other parameters of an amplifier do wecare about?

Voltage gain Dynamic range Frequency response (bandwidth) Input impedance Output impedance




Voltage GainVoltage Gain•Voltage gain•Use small signal model (short Voltage sources and capacitors)

voltage gain

usually r e<<RE

•Voltage gain is defined by resistors RC and RE

ground

ground

αie




Frequency response (Bandwidth)Frequency response (Bandwidth)

•Normally interested in providing a

small, AC signal to the base•Use capacitors to remove ("block")any low frequency (DC) component("capacitively couple the signal to thebase") which could affect the biascondition

•C1 forms a high-pass filter with R1in

parallel with R2 (Assuming the AC

impedance into the base is large).

•Cut off frequency ω0=1/RC, so to

remove frequencies <f min :




Frequency response (Bandwidth)Frequency response (Bandwidth)

•Also worthwhile to place a capacitor

on the output

•C2 forms a high pass filter with RL.

•Cut off frequency ω0=1/RC, so to

remove frequencies <f min :




Dynamic RangeDynamic Range•Maximum voltage output = Vbb

•Minimum = 0

•Beyond this the signal becomes 'clipped' or distorted•To get the maximum possible voltage swing, both positive andnegative, set VC=0.5 VBB

•Maximum 'dynamic range‘- The maximum undistorted voltage swing.

VC



36

Input impedanceInput impedance

r b

r OUT

•Consider the circuit without the voltage

divider resistors. What's the small signal (AC)

input impedance at the base, r b?

•

•

•

•

•

•

•

•

•Including voltage divider resistors in parallel

• Input signal sees a total input

impedance r IN= R1 // R2 // r b

d




Output impedanceOutput impedance

r b

ROUT

•If R L=10kΩ and we want a low frequency cutoff of 20Hz, What is C2?




L q y , 2

•If VBB=15V and IC=2mA what is R C?

DC condition

Frequency response

Impedance

Gain/Dynamic range

I d




ImpedancesImpedances•Why do we care about the input and output impedance?

•Simplest "black box" amplifier model:

RIN

ROUT

VIN AVIN VOUT

•The amplifier measures voltage across RIN, then generates a

voltage which is larger by a factor A

•This voltage generator, in series with the output resistance ROUT , is

connected to the output port.

•A should be a constant (i.e. gain is linear)

dI d




ImpedancesImpedances•Attach an input - a source voltage VS plus source impedance RS

RIN

ROUT

VINAVIN

VOUT

•Note the voltage divider RS + RIN.

•VIN=VS(RIN/(RIN+RS)

•We want VIN = VS regardless of source impedance

•So want RIN to be large.

•The ideal amplifier has an infinite input impedance

VS

RS

dI d




ImpedancesImpedances•Attach a load - an output circuit with a resistance RL

•Note the voltage divider ROUT + RL.

•VOUT =AVIN(RL/(RL+ROUT )

•Want VOUT =AVIN regardless of load

•We want ROUT to be small.

•The ideal amplifier has zero output impedance

RIN

ROUT

VINAVIN VOUT

VS

RS

RL




Capacitor Selection for the CE Amplifier

The key objective in design is to make the capacitive reactancemuch smaller at the operating frequency f than the associated

resistance that must be coupled or bypassed.




Static power dissipation in amplifiers isdetermined from their DC equivalentcircuits.

Amplifier Power DissipationAmplifier Power Dissipation

Total power dissipated in C-Band E-B junctions is:

where

Total power supplied is:

The difference is the power dissipated by the bias resistors.



Questions

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Lecture 2 Transistor Modeling

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