Lecture 05 - 1ELE2110A © 2008

Week 5: BJT Biasing and Small Signal Model

ELE 2110A Electronic Circuits

Lecture 05 - 2ELE2110A © 2008

Topics to cover …BJT Amplifier Biasing Circuits

Small Signal Operation and Equivalent Circuits

Reading Assignment: Chap 13.1 – 13.6 of Jaeger and Blalock , orChap 5.5 - 5.7 of Sedra & Smith

Lecture 05 - 3ELE2110A © 2008

BJT as Amplifier: Example 1

V. and off cut is Q V,. For 570 =≤ OI vv

• Problem: Determine the dc voltage transfer characteristic of the circuit for 0 < vI <5 V• Analysis:

.saturation in is transistor the V,. for Now, V.. be to found is V . for voltage input The

919120

>

==

I

IO

vvv

..)sat( and V . for valid is equation This

kk

.)( or

is voltage output The

k.

that so mode, active the in is and on turns Q V,. For

Vvvv

vv

RiVRiVv

vR

Vvi

v

CEOI

IO

CBCCO

I

B

BEIB

I

2070

4100

701005

10070

70

=≥≥

Ω⎥⎦

⎤⎢⎣

⎡Ω

−−=

−=−=

Ω−

=−

=

>

++ β

Lecture 05 - 4ELE2110A © 2008

Conceptual Bias Circuit for BJTKeep the transistor in the active mode;Establish a Q-point near the center of the active region;Couple the time-varying signal to the base.

Lecture 05 - 5ELE2110A © 2008

Two Obvious Bias Circuits

Fix VBE Fix IB

Wide variations in IC and hence in VCE

“Bad” biasing schemes

IC=βIB , but β is a poorly controlled parameter

IC is an exponential function of VBE and thus VCC

Lecture 05 - 6ELE2110A © 2008

Biasing Circuit in ERG2810 Experiment A3

Adjusting the variable resistor to give a desired amount of IB such that Vo is at about 8V (half the supply voltage).The transistor operates at the middle of active region.

BiasingIB

Vo

Lecture 05 - 7ELE2110A © 2008

Classical Four-Resistor Bias Circuit

VB established by the voltage divider formed by R1 and R2.RE added reduce sensitivity to supply voltage, process, and temperature variations

to be discussed laterCoupling capacitor CC:– open circuit to DC, isolating the signal

source from the dc biasing current. – short circuit to AC signal (if the signal

frequency is large enough and CC is large enough).

The Q-point is usually specified by (IC, VCE) for npn transistor or (IC, VEC) for pnp transistor.

Lecture 05 - 8ELE2110A © 2008

Example 2

21

1RR

RCCVEQV

+=

21

21RR

RREQR

+=

EIERBEVBIEQREQV ++=

µA68.261023.1V7.0-V4

)1(000,167.0000,124

=Ω×

=∴

+++=

BIBIBI β

KVL at loop 1:

Thevenin equivalent circuit at the base:

Assume BJT in active mode

Problem: find Q-point.

µA201== BICI β µA204)1( =+= BIEI β Q-point is (201 µA, 4.32 V)

V32.4=+−=

−−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

CIF

FRC

RCCV

EIERCICRCCVCEV

α

Given β = 200.

Lecture 05 - 9ELE2110A © 2008

Example 2 (Cont’)All calculated currents > 0,

VBC = VBE - VCE = 0.7 - 4.32 = - 3.62 VHence, base-collector junction is reverse-biased, assumption of active region operation is correct.

CICIERC

RCCVCEV 200,3812 −=+−=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

α

IB = 2.7 µA

intersection point Q-point.

Load-line for the circuit is:

Lecture 05 - 10ELE2110A © 2008

Example 2 (Cont’)

21

1RR

RCCVEQV

+=

21

21RR

RREQR

+=

ERBI

EQRBEV

EQV

EI−−

=

VCC Temp. Process (β)

EIERBEVBIEQREQV ++=

Lecture 05 - 11ELE2110A © 2008

Design Objectives: Q-point Insensitive to PVT Variations

1) IE linearly (not exponentially) related to VCC2) For IE to be less sensitive to IB (thus β):

Need small ReqLarge currents through R1 and R2 (I2 >> IB)

3) For IE to be less sensitive to VBE (due to temperature change):

VEQ >> VBE

)( BEVEQ

VBIEQR −<<

ER

BI

EQR

BEV

EQV

EI−−

=

Lecture 05 - 12ELE2110A © 2008

Design Objectives: Low Power and Large Signal Swing

But …For low power consumption:

needs small I2contradict with constraint (2)set I2 = 10IB typically

For large output signal swing:VEQ cannot be too high(because VC∈ [VEQ, VCC])

VEQ =1/3 VCC typically

I2

Lecture 05 - 13ELE2110A © 2008

Design GuidelinesChoose Thevenin equivalent base voltage

Select R1 to set I1 = 9IB

Select R2 to set I2 = 10IB

RE is determined by VEQ and desired IC

RC is determined by desired VCE

24CCV

EQVCCV≤≤

BIEQ

VR

91 =

BIEQ

VCC

VR

102

−=

CI

BEVEQ

V

ER−

≈

ERC

ICE

VCC

VCR −

−≈

Lecture 05 - 14ELE2110A © 2008

Example 3• Problem: Design a 4-resistor bias circuit with given parameters.• Given data: IC =750 µA, VCE = 5 V, β = 100, VCC = 15 V, VBE = 0.7 V• Unknowns: VB, voltages across RE and RC; values for R1, R2, RC and RE.• Analysis: A common approach is to divide (VCC - VCE) equally between RE andRC. Thus, VE =5 V and VC =10 V.

kΩ67.6=−

=C

IC

VCC

VCR

V7.5

kΩ60.6

=+=

==

BEVEVBVEIEV

ER A5.6791

A75102µµ

====

BIIBII

A5.7 µβ == CIBI

kΩ124102

kΩ4.8491

=−

=

==

BIBV

CCV

R

BIBV

R

Now choose I2 = 10IB:

Lecture 05 - 15ELE2110A © 2008

Example 4: Two-Resistor Bias Network • Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with

given parameters.• Given data: βF = 50, VCC = 9 V• Assumptions: Forward-active operation region, VEB = 0.7 V• Analysis:

V18.2V88.2)(10009

mA01.650

µA120000,69

V7.0V9

)51(1000000,189)(1000000,189

=

=+−=

==

=Ω

−=∴

++=∴

+++=

BCVBICIECV

BICIBI

BIBIEBVBICIBIEBV

Q-point is : (6.01 mA, 2.88 V)

Lecture 05 - 16ELE2110A © 2008

Current Source Biasing

IE = I independent of the value of β and temperature

Lecture 05 - 17ELE2110A © 2008

Current Source Implementation: Current Mirror

Use collector current of a transistor in active modeNeglect Early effect, IC2 is independent of VCE2 as long as VCE2 > VCEsat (i.e. BJT in active mode)For matched Q1 and Q2 , i.e., having identical IS, β, VA), we have

RBE

VBB

VREFI

−=

REFC II =2

Lecture 05 - 18ELE2110A © 2008

Topics to cover …BJT Amplifier Biasing Circuits

Small Signal Operation and Equivalent Circuits

Lecture 05 - 19ELE2110A © 2008

BJT as an Amplifier

If an ac+dc input signal the total vBE becomes

beBEBE vVv +=The collector current becomes

TbeTbeTBE

TbeBE

VvC

VvVVS

VvVSC

eIeeI

eIi///

/)(

==

= +

Superposition of DC with AC signal

ConceptualAmplifier circuit:

Lecture 05 - 20ELE2110A © 2008

Small-signal TransconductanceFor small ac signal, i.e., vbe << VT :

The ac (or signal) component of the collector current is:

beT

Cc v

VIi =

gm is called the small signal transconductance. It represents the slope of iC-vBE curve at the Q-point.

321

AC

beT

C

DCC

TbeCVv

CC

vVII

VvIeIi Tbe

+=

+≅= )/1(/

T

C

be

cm V

Ivig =≡ We define:

CCbe IiBE

C

vbe

cm v

ivi

g=→

∂∂

=≡ 0

Lecture 05 - 21ELE2110A © 2008

Signal Component of Base Current

Total base current: 43421

AC

beT

C

DC

CCB v

VIIii

βββ1

+==

bem

beT

Cb vgv

VIi

ββ==

1

rπ is the small-signal input resistance between base and emitter, looking into the base.

B

T

mb

be

IVr

givr ==≡ ππ

β or

Signal component of base current:

Define:

Lecture 05 - 22ELE2110A © 2008

Signal Component of Emitter Current

The total emitter current iE: 43421

AC

beT

C

DC

CCE v

VIIii

ααα1

+==

beT

Ebe

T

Ce v

VIv

VIi ==

α1

re is the small-signal input resistance between base and emitter, looking into the emitter.

E

Te

me

bee I

Vrgi

vr ==≡ or α

It is easy to find out that eebe rriir )1()/( +== βπ

Define:

Signal component of emitter current:

Lecture 05 - 23ELE2110A © 2008

Small Signal I-V Expressions

E

Te

e

be

IVr

iv

==B

T

b

be

IVr

iv

== πT

Cm

be

c

VIg

vi

==

(Hybrid-π small signal model of BJT)

Can be modeled by equivalent circuits:

Lecture 05 - 24ELE2110A © 2008

Another model: T-model

T-model

E

Te

e

be

IVr

iv

==B

T

b

be

IVr

iv

== πT

Cm

be

c

VIg

vi

==

Lecture 05 - 25ELE2110A © 2008

Early Effect: CI

CEV

Hybrid-π model including Early effect

C

A

C

CEA

vCE

Co

IV

IVV

vir

BE

≈+

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂∂

=

−1

fixed

Define

Include Early Effect in the model:

)1(/

A

CEVvSC V

veIi TBE +=

Lecture 05 - 26ELE2110A © 2008

Graphic Analysis

Can be useful to understand the operation of BJT circuitsFirst, establish DC conditions by finding IB (or VBE)– Input load line:

Second, figure out the DC operating point for IC– Output load line:

BBBBBE iRVv −=

CECC

CCCCCCCCE v

RRViRiVv 1 −=⇒−=

Lecture 05 - 27ELE2110A © 2008

Graphic Analysis (Cont.)

Apply a small signal input voltage and see ibSee how ib translates into VCECan get a feel for whether the BJT will stay in active region ofoperation– What happens if RC is larger or smaller?