Early Effect & BJT Biasing - Penn Engineeringese319/Lecture_Notes/Lec_4_BJTBias… · DC Biasing...

25
ESE319 Introduction to Microelectronics 1 2009 Kenneth R. Laker, update 11Sep12 KRL Early Effect & BJT Biasing Early Effect DC BJT Behavior DC Biasing the BJT

Transcript of Early Effect & BJT Biasing - Penn Engineeringese319/Lecture_Notes/Lec_4_BJTBias… · DC Biasing...

Page 1: Early Effect & BJT Biasing - Penn Engineeringese319/Lecture_Notes/Lec_4_BJTBias… · DC Biasing the BJT. ESE319 Introduction to Microelectronics 2009 Kenneth R. Laker, update 11Sep12

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

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VBE

VCE

IC

IC

VCE

Saturation region

0

-VA

VBE1

VBE2

VBE3

VBE4

Forward-Active region

Ideal NPN BJT Transfer Characteristic

Early Effect

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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 evBE

V T 1 vCE

V A is called the Early voltage and ranges from about 15 V to 150 V.

NMOS transistorn=

1V A

V A=V CE

I CI C

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Early Effect - Continued

-VA

VCE

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

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Early Effect - Continued

iC= I S evBE

V T 1 vCE

V A Total (bias+signal) quantities:

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

I C= I S eV BE

V T 1V CE

V A = I C' 1V CE

V A

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

Let's call the idealized collector bias current (no Early Effect) I'C, i.e.

I C' = I S e

V BE

V T

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

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We shall define: ro=V A

I C'

I C= I C'

V CE

ro

I C= I C' 1V CE

V A = I C'

V CE

V A

I C'

Rearranging slightly:

Early Effect - Continued

The dc current due to both VBE

and VCE

is:

MOS transistor

ro=V A

I DI C' = I S e

V BE

V T

I D=12

k n' W

LV GS−V t

2

=> ro = f(V

BE)

ro

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Although the bias current is better modeled by including the Early effect

I C= I C'

V CE

ro

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

I C≈ I C' = I S e

V BE

V T

Early Effect - Continued

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The Early term adds ro to the large signal model:

Early Effect - Continued

IC

IC' = I S evBE /V T

VCEVBE VCEro

V CE= I C−I C' ro

I C=I C'

V CE

ro

VBEro

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

I C' ≈1mA.

ro=V A

I 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.

Early Effect - Continued

100 k

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Simulation Results

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

Note: ro is in parallel with R

C.

V CC

I C=1

RCV CC−V CE

RC

load-line

dictated by circuit

V CE

I C

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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

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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.

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Amplifier DC Bias Problem

iC= I Cic

vBE=V BEvbe

vCE=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

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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

(ac + dc)

vCE

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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 evBE

V T

iC=V CC−vCE

RC

V CC−vCE

RC=I S e

vBE

V T

Load Line

BJT

Circuit

(ac + dc)

vCE

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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);

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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

RC

V CC=10VRC=10k

vBE=0.04V

vCE≈7V

iC (mA)

vCE

(V) =vC

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Amplifier ActionNote that as vBE varies from about 0.59 V to 0.63 V, vCE 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 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 BE=V BIAS=vBE MAX vBE MIN /2=0.61Vvbe=vsignal=vBE MAX −vBE MIN /2=0.02 V peak

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Candidate Bias Configurations

Base voltagesource

Base currentsource

Emitter currentsource

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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

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Current Bias Beta DependenceUnfortunately, β is often poorly controlled and may easilyvary from 100 to 200. And β 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=10−104⋅1⋅10−3=0 VV CE=V CC−RC I C

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Drive Base with a Base Voltage Source

For an IC of 0.5 mA:

V BE=V T ln I C

I 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 BE

V T

VCC

= 10 V

Rc = 10 kΩ

VBE

= 0.616 V

Q1

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Voltage Bias IS and VCE DependenceUnfortunately, IS is highly temperature-dependent, doublingfor every 5oC increase in temperature.

If the base-emitter voltage is chosen to give IC = 0.5 mA at 20oC (68oF),

it will be 2x at 25oC and 0.5x at 15oC.

IC is also highly sensitive to VBE. Consider two values IC and 10IC:

10 I C

I C=

I S eV BE10

V T

I S eV BE1

V 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.

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

I C= I E

Changes in IC due to variations in α in the range determined

by the extremes of β are negligible, i.e.

100200⇒ 100101

200201

⇒0.9900.995

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

I B=I E

1⇒

I E

101 I B

I E

201

=

1

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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.