Early Effect & BJT Biasing - Penn Engineeringese319/Lecture_Notes/Lec_4...Early Effect & BJT Biasing...

ESE319 Introduction to Microelectronics

12009 Kenneth R. Laker, update 14Sep11 KRL

Early Effect & BJT Biasing● Early Effect● DC BJT Behavior● DC Biasing the BJT



VBE

VCE

IC

IC

VCE

Saturation region

0

-VA

VBE1

VBE2

VBE3

VBE4

Forward-Active region

Ideal NPN BJT Transfer Characteristic

Early Effect



Early Effect - ContinuedCollector voltage has some effect on collector current – it increases slightly with increases in voltage. This phenomenonis called the “Early Effect” and is modeled as a linear increasein total current with increases in vCE:

iC= I S evBEV T 1 vCEV A

VA is called the Early voltage and ranges from about 50 Vto 100 V.

NMOS transistorn=

1V A



Early Effect - Continued

-VAVCE

VBE = ...

VBE = ...

VBE = ...

VBE = ...IC

VCEVBE

IC

Saturation region

Fwd-Active region

15 V ≤ VA ≤ 150 V

Observed by James Early from BTL

slope=1/ ro




iC= I S evBEV T 1 vCEV A Total (bias+signal) quantities:

iC= I C vBE=V BE vCE=V CE

I C= I S eV BEV T 1V CEV A = I C' 1V CEV A

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

Let's call the idealized collector bias current I'C:

I C' = I S e

V BEV T

iC= I Cic vBE=V BEvbe vCE=V CEvce



We shall define: ro=V AI C

'

I C= I C'

V CEro

I C= I C' 1V CEV A = I C' V CEV A

I C'

Rearranging slightly:


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

MOS transistor

ro=V AI DI C

' = I S eV BEV T

I D=12

k n' W

LV GS−V t

2

=> ro = f(V

BE)



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

iC=iC'

vCEro

We – almost always – will ignore the second term above in hand calculations and use our ideal expression for the bias current:

iC≈iC' =I S e

vBEV T




The Early term adds ro to the large signal model:


IC

IC' = I S e

vBE /V T

VCEVBE VCEro

V CE= I C−I C' ro

I C=I C'

V CEro

VBEro



For typical operating conditions:V A≈50−100 V.

I C' ≈1mA.

ro=V AI C

' ≈100 V10−3 A

=100 k

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 specifically told if you must include ro in your circuit analyses and designs.


100 k



Early Effect Scilab SimulationVA=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 one plot(VCE,ICp); //Current in mA. ro=VA/ICprime;//Add bias Early effect IC=ICp+vCE/ro; plot(VCE,IC)end//Plot load line for 1 KOhm Rc - 12V VccRc=1000;Gc=1/Rc;VCC=12;VCLL = 0:0.01:12;ILoLin=Gc*VCC-Gc*VCLL;plot(VCLL,ILoLin)

I LoLin=I C

V CLL=V CE

I C=1

RCV CC−V CE

c

Dictated by circuit



Simulation Results

I C=1

RCV CC−V CE

ro = ∞ro ≠ ∞

slope = -1/R

C

IC (mA)

VCE

(V)

12

10

8

6

4

2

00 2 4 6 8 10 12

Early Effect

load-line



Active Mode Conditions

Base-emitter diode forward-biased:

Base-collector diode reverse-biased:

V BE≥0.7 V

V BC=V BE−V CE≤0.5V

−V CE≤0.5−V BE⇒V CE≥0.2V

V CE≥0.2V

Forward-Active (ideal cond.)

VBE

> 0V

BC < 0

iE = i

C + i

B

vCE

= vCB

+ vBE



Amplifier Biasing GoalsWe wish to set a stable value of IC so that we can apply asignal voltage or signal current to the emitter-base circuit andobtain an amplified (undistorted) version of the signal betweenthe collector and ground.

The transistor cannot saturate during operation, i.e.

vCE0.2V.

And it cannot cut off during operation, i.e.iC0 mA.



Amplifier DC Bias Problem

iC= I CicvBE=V BEvbevCE=V CEvce

TimevI = v

BE

vO = v

CE

RC

VCC

iC

vO

VCE

VCC

0

Time

vi = v

be

VBE

Timev

o = v

ce

VCEsat

= 0.2 V

vI

0.5 1.51.0

Q

cutoff saturation

fwdactive

slope = Av



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

results in a large ac collector current ( ).

● This yields a large change in the ac collector voltage v

ce.

● Base voltage source:● A small ac change in base voltage

results in a large change in the ac collector current (ic = IS exp(vbe/VT)).

● This yields a large change in the ac collector v

ce voltage.

ib

vC

iC

iB

Source vBE

VCC

RC



Voltage Source Input With Collector Load

Solution of the simultaneousequations exists where the twocurves: the exponential (iC,vBE) andthe straight line (iC,vCE) intersect:

iC= I S evBEV T

iC=V CC−vCE

RC

V CC−vCERC

=I S evBEV T

Load Line

BJT

Circuit



Scilab Plot of NPN Characteristic //Calculate and plot npn BJT collector//characteristic using active mode modelVT=0.025;VTinv=1/VsubT;IsubS=1E-14;vCE=0:0.01:10;for vBE=0.58:0.01:0.63 iC=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);



NPN Transistor Load Line

Vce (V.)

Ic (mA.)

0 1 2 3 4 5 6 7 8 9 100.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Plot Outputv BE=0.63V.

v BE=0.62V.

v BE=0.60V.

Load Line

iC=V CC−vCE

RCV CC=10VRC=10k

vBE=0.04V

vCE≈7V

iC (mA)

vCE

(V)



Amplifier ActionNote that as vBE varies from about 0.59 V to 0.63 V, vC varies from about 1 V to 8 V! A 0.04 V peak-to-peak swing of vBE results in an 7 V peak-to-peak swing in vCE - a voltage-gain ratio of 7/0.04, or about 175. The output magnitude is about 175 times the input mag-nitude, for a gain of 175.The input signal has two components: a dc one called the bias voltage, and an ac one called the (small) signal voltage. For proper operation, let:

V BIAS=V BE MAX V BE MIN /2=0.61VvSIGNAL=V BE MAX −V BE MIN /2=0.02V peak



Candidate Bias Configurations

Base voltagesource

Base currentsource

Emitter currentsource



Drive Base With a Base Current Source

Assume: =100

I C= I B=100⋅5⋅10−6

I C=0.5 mA.

For this collector current:

V CE=V CC−RC I C

V CE=10−104⋅0.5⋅10−3=5 V

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

VCE

IC R

c = 10 kΩ

VCC

= 10 VI

B

I = 5 µAQ1



Current Bias Beta DependenceUnfortunately, β is often poorly controlled and may easilyvary from 100 to 200. And beta is also temperature dependent!

I C=100⋅5⋅10−6=0.5 mA.

For β = 100:

The BJT with a VCE = 5 V

For β = 200:I C=200⋅10⋅5

−6=1.0 mA.

The BJT is saturated!

Base current source biasing → BIAS POINT IS UNSTABLE.

V CE=10−104⋅0.5⋅10−3=5V V CE=10−10

4⋅1⋅10−3=0 VV CE=V CC−RC I C



Drive Base with a Base Voltage Source

For an IC of 0.5 mA:

V BE=V T ln I CI S

Given: I S=10−14 A

I C=0.5⋅10−3 Aand:

V BE=0.025 ln 0.5⋅1011

V BE=0.025⋅24.635=0.616 V

OK. Apply 0.616 volts to thebase and we have the desired collector current!

Since VCE

= 5 V the transistor is nearly at the center of the desired operating region!I C= I S e

V BEV T

VCC

= 10 V

Rc = 10 kΩ

VBE

= 0.616 V

Q1



Voltage Bias IS and VCE DependenceUnfortunately, IS is highly temperature-dependent, doublingfor a 5oC increase in temperature. If the base-emitter voltage is chosen to give 0.5 mA collector current at 20oC (68oF), it will double at 25oC and halve at 15oC.IC is also highly sensitive to VBE. Consider two values of collector current, IC and 10IC:

10 I CI C

=I S e

V BE10V T

I S eV BE1V T

V BE10−V BE1=V T ln 10

V BE10−V BE1=0.025⋅2.3025=0.058V.

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



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

I C= I EChanges in I

C for α in the range determined by the extremes

of β are negligible

100200⇒ 100101

200201

There is considerable variation in base current, however, but this is usually of no consequence.

I B=I E

1⇒

I E51

I BI E

201

=

1



ConclusionBiasing a BJT poses potential large bias stability prob-lems, since its characteristics are highly sensitive to temperature and since its electrical properties (princip-ally β) can vary widely from one device to another!

The next lecture sequence will cover some techniques for stabilizing the BJT bias.

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Early Effect & BJT Biasing - Penn Engineeringese319/Lecture_Notes/Lec_4...Early Effect & BJT Biasing...

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