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ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 11Sep08 KRL

Early Effect & BJT Biasing

Early Effect

DC BJT Behavior DC Biasing the BJT

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Early Effect

Collector voltage has some effect on collector current itincreases slightly with increases in voltage. This phenomenon

is called the Early Effect and is modeled as a linear increase

in total current with increases in vCE:

iC=I

Se

vBE

VT 1vCEVA

VA

is called the Early voltage and ranges from about 50 V.

to 100 V.

NMOS transistorn=

1

VA

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Early Effect - ContinuediC=I

Se

vBE

VT 1 vCEV

ATotal (bias+signal) quantities:

iC=I

CvBE=V

BE vCE=VCE

iC=IC=ISe

VBE

VT

1

VCE

VA

=IC

'

1

VCE

VA

Consider dc (bias) condition (signal = 0):

Call the idealized collector bias current I'C:

IC

'=I

Se

VBE

VT

iC=ICic vBE=VBEvbe vCE=VCEvce

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We shall define: ro=

VA

IC

'

IC=I

C

'V

CE

ro

IC=I

C

' 1VCEVA

=IC' VCEVA

IC'

Rearranging slightly:

Early Effect - Continued

The dc current due to both emitter and collector voltages is:

MOS transistor

ro=VA

IDIC

'=ISe

VBE

VT

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Although the bias current including the Early effect is bettermodeled as:

IC=I

C

'V

CE

ro

We almost always will ignore the second term above anduse our usual expression for the bias current:

ICI

C

'=I

Se

VBE

VT


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The Early term adds ro

to the large signal model:


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For typical operating conditions:

VA50100V.

IC1mA.

ro=

VA

IC

100

103=100k

We usually can ignore ro

since, in practice, ro

is in parallel

with other resistors, which are much smaller than .For the time being, you will be told if you must include r

oin

your pencil-and-paper circuit designs.


100 k

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Early Effect Scilab Simulation

VsubA=100;vCE=0.01:0.01:12; //array 0.01 to 12 in .01 V. stepsfor ICprime = 2:2:10 //mA.ICp = ICprime*sign(vCE); //ICp value for each vCE oneplot(vCE,ICp); //Current in mA.

ro=VsubA/ICprime;//Add bias Early effectIC=ICp+vCE/ro;plot(vCE,IC)

end//Plot load line for 1 KOhm Rc - 12V VccRc=1;

Gc=1/Rc;Vcc=12;vCLL = 0:0.01:12;ILoLin=Gc*Vcc-Gc*vCLL;plot(vCLL,ILoLin)

ILoLin

VCLL

ILoLin=

1

RC

VCCV

CLL

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Simulation Results

ILoLin=1

RC

VCCVCLL

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Active Mode Conditions

Base-emitter diode forward biased:

Base-collector diode back biased:

vBE0.7V

vBC=v

BEv

CE0.5V

vCE

0.5vBE

vCE

0.2V

vCE0.2V

Forward-Active(ideal cond.)

VBE > 0V

BC< 0

iE

= iC

+ iB

vCE

= vCB

+ vBE

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Amplifier Biasing GoalsWe wish to set a stable value ofI

Cso that we can apply a

signal voltage or current signal to the emitter-base circuit andobtain an amplified (undistorted) version of the signal between

the collector and ground.

The transistor cannot saturate during operation, i.e.

vCE0.2V.

And it cannot cut offduring operation, i.e.iC0mA.

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Amplifier DC Bias Problem

iC=I

Ci

c

vBE=V

BEv

be

vCE=V

CEv

ce

TimevI

= vBE

vO

= vCE

RC

VCC

iC

vO

VCE

VCC

0

Time

vi= v

be

VBE

Time

vo

= vce

VCEsat

= 0.2 V

vI

0.5 1.51.0

Q

cutoff saturation

fwdactive

slope = Av

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Amplifier Action Base current source: A small change in base current

results in a large collector current( ).

This yields a large change in col-

lector voltage.Base voltage source:

A small change in base voltageresults in a large change in collec-tor current (i

C= I

Sexp(v

BE/V

T)).

This yields a large change in col-lector voltage.

ib

vCE

iC

iB

Source vBE

VCC

RC

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Voltage Source Input With Collector Load

Solution of the simultaneousequations exists where the twocurves: the exponential (i

C

,vBE

) and

the straight line (iC,v

CE) intersect:

iC=

ISe

vBE

VT

iC=

VCCv

CE

RC

VCCv

CE

RC

=ISe

vBE

VT

Load Line

BJT

Circuit

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Scilab Plot of NPN Characteristic//Calculate and plot npn BJT collector//characteristic using active mode modelVsubT=0.025;VTinv=1/VsubT;IsubS=1E-14;

vCE=0:0.01:10;for vBE=0.58:0.01:0.63iC=IsubS*exp(VTinv*vBE);plot(vCE,1000*iC); //Current in mA.

endVcc=10;

Rc=10000;vLoad=0:0.01:10;iLoad=(Vcc-vLoad)/Rc;plot(vLoad,1000*iLoad);

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NPN Transistor Load Line

Vce (V.)

Ic (mA.)

0 1 2 3 4 5 6 7 8 9 10

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0Plot OutputvBE=0.63V.

vBE=0.62V.

vBE=0.60V.

Load Line

iC=

VCCv

CE

RC

VCC=10V

RC=10k

vBE=0.04V

vCE7V

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Amplifier ActionNote that as v

BEvaries from about 0.59 V to O.63 V, v

CE

varies from about 8 V to 1V!A 0.04 V peak-to-peak swing ofv

BEresults in an 7 V peak-

to-peak swing in vCE - A ratio of 7/0.04, or about 175.The output magnitude is about 175 times the input magni-tude, for a gain of 175.

The input signal has two components: a DC one called thebias voltage, and an AC one called the (small) signal volt-

age. For proper operation, let:

VBIAS

=VBEMAXVBEMIN/2=0.61V

vSIGNAL

=VBEMAXVBEMIN/2=0.02V

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Candidate Bias Configurations

Base voltagesource

Base currentsource

Emitter currentsource

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Drive Base With a Current Source

Assume: =100

IC=IB=1005106

IC=0.5mA.

For this collector current:

VCE=V

CCR

CIC

VCE=1010

40.510

3=5V

The transistor is almost right in thecenter of the desired operatingregion!

VCE

IC

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Current Bias Beta DependenceUnfortunately, beta is very poorly controlled and may easilyvary from 50 to 200. And beta is also temperature dependent!

IC=505106=0.25mA.

For beta of 50:

The device with a VCE

= 7.5 V

is close to theIC

cutoff point.

For beta of 200:

IC=2001056=1.0mA.

The device is saturated, VCE

tries to be 0 V!

There are huge problems associated with trying to drive aBJT with a base current source! BIAS POINT IS UNSTABLE.

VCE=101040.25103=7.5V VCE=1010

4

1103=0V

VCE=V

CCR

CIC

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Drive base with a voltage source

For anIC

of 0.5 mA:

VBE=V

Tln

IC

IS

Given: IS=1014

A

IC=0.510

3Aand:

VBE

=0.025ln

0.5

10

11

VBE=0.02524.635=0.6159V

OK. Apply 0.6159 volts to thebase and we have the desiredcollector current!

Again, the transistor is nearly atthe center of the desired operating

region!

IC=ISe

VBE

VT

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Voltage Bias ISand VCEDependenceUnfortunately,I

Sis highly temperature-dependent, doubling

for a 5 degree C increase in temperature. If the base-emittervoltage is chosen to give 0.5 mA collector current at 20 degreesC (68 F), it will double at 25 C and halve at 15 C.

IC

is also highly sensitive to VBE

. Consider two values of

collector current,IC

and 10IC:

10IC

IC

=

ISe

VBE10

VT

ISe

VBE1

VT

VBE10

VVE1=V

Tln 10

VBE10VBE1=0.0252.3025=0.0576V.

Less than a 60 mV change in VBE voltage increases IC by anorder of magnitude (10X). BIAS POINT IS UNSTABLE.

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Emitter Current SourceThis holds collector current close to its desired value since:

IC= I

E

Changes in IC

for in the range determined by the

extremes ofare negligible

5020050

51

200

201

There is considerable variation in base current, however, but

this is usually of no consequence.

IB=

IE

1I

E

51I

B

IE

201

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Conclusion

Biasing a BJT poses large stability problems, sinceits characteristics are highly sensitive to temperature

and since its electrical properties (principally beta)vary widely from one device to another!

The next lecture will cover some techniques for

stabilizing BJT operation.

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