Early Voltage

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ESE319 Introduction to Microelectronics

12009 Kenneth R. Laker, update 11Sep12 KRL

Early Effect & BJT Biasing

) Early Effect)

DC BJT Behavior ) DC Biasing the BJT





V BE

V CE

I C

I C

V CE

Saturationregion

0

-V A

V BE1

V BE2

V BE3

V BE4

Forward-Activeregion

Ideal NPN BJT

TransferCharacteristic

Early Effect





Early Effect - Continued

Collector 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 increase

in total current with increases in v CE :

iC " I S e

v BE

V T -1* vCE

V A .

is called the Early voltage and ranges from about 15 V

to 150 V .

NMOS transistor 3n"

1

V A

V A"/V CE

/ I C

I C






-V A

V CE

V BE = ...

V BE

= ...

V BE

= ...

V BE

= ... I

C

V CE

V BE

I C

Saturationregion

Fwd-Activeregion

15 V # V A

# 150 V

Observed by

James Earlyfrom BTL

slope"1! r o





Early Effect - Continued iC " I

S e

v BE

V T -1*vCE

V A .Total (bias+signal) quantities:

iC " I C v BE "V BE vCE "V CE

I C " I S e

V BE

V T

-1*

V CE

V A

." I C

'

-1*

V 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 C *ic v BE "V BE *vbe vCE "V CE *vce





We shall define: r o"V A

I C '

I C " I C ' *V CE

r o

I C " I C ' -1*V CE

V A ." I C ' *V CE

V A

I C '

Rearranging slightly:


The dc current due to both V BE

and V CE

is:

MOS transistor

r o"V A

I D I C

' " I S e

V BE

V T

I D"1

2k n

' W

L-V GS $V t .

2

=> r o = f(V

BE )

r o





Although the bias current is better modeled by including the Earlyeffect

I C " I C

' *

V CE

r o

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

I C % I C ' " I S e

V BE

V T






The Early term adds r o to the large signal model:


I C

I C ' " I S e

v BE !V T

V CE

V BE V

CE r o

V CE "- I C $ I C ' . r o

I C " I C ' *V CE

r o

V BE

r o





For typical operating conditions:

V A%50$100V.

I C ' %1mA.

r o"V A

I C ' %

100V

10$3 A"100 k 0

We usually can ignore r o since, in practice, r

o is in parallel

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

o in your circuit analyses and designs.


100 k 0





Simulation Results

r o = "

r o ! "

slope =-1/R

C

I C (mA)

V CE

(V)

12

10

8

6

4

2

00 2 4 6 8 10 12

Early Effect

load-line

Note: r o is in parallel with R

C .

V CC

I C "

1

RC -V CC $V CE .

RC

load-line

dictated by circuit

V CE

I C





Active Mode Conditions

Base-emitter diode forward-biased:

Base-collector diode reverse-biased:

V BE '0.7V

V BC "V BE $V CE &0.5V

$V CE &0.5

$V BE #V CE '0.2

V

V CE '0.2V

Forward-Active(ideal cond.)

V BE > 0V

BC < 0

i E = i

C + i

B

vCE

= vCB

+ v BE





Amplifier Biasing GoalsWe wish to set a stable value of I

C so that we can apply a

signal voltage or signal current 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.

vCE ,0.2V.

And it cannot cut off during operation, i.e.iC ,0mA.





Amplifier DC Bias Problem

iC " I C *ic

v BE "V BE *vbe

vCE "V CE *vce

Timev I = v

BE

vO = v

CE

RC

V CC

iC

vO

V CE

V CC

0

Time

vi = v

be

V BE

Time

vo = v

ce

V CEsat

= 0.2 V

v I

0.5 1.51.0

Q

cutoff saturation

fwd active

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 vce.

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

results in a large change in the accollector current (i

c = I

S exp(v

be /V

T )).

) This yields a large change in theac collector v

ce voltage.

2 ib

vC

iC

i B

Source v BE

V CC

RC

(ac + dc)

vCE





Voltage Source Input With Collector Load

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

C

,v BE

) and

the straight line (iC ,v

CE ) intersect:

iC " I S e

v BE

V T

iC "V CC $vCE

RC

V CC $vCE

RC

" I S e

v BE

V T

Load Line

BJT

Circuit

(ac + dc)

vCE





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 10

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0Plot Output v BE "0.63V.

v BE "0.62V.

v BE "0.60V.

Load Line

iC "

V CC $vCE

RC

V CC "10V

RC "10k 0

/ v BE

"0.04V

/ vCE %7V

iC

(mA)

vCE

(V) =vC





Amplifier ActionNote that as v

BE varies from about 0.59 V to 0.63 V , v

CE varies

from about 1 V to 8 V !

A 0.04 V peak-to-peak swing of v BE

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 thebias voltage, and an ac one called the (small ) signal

voltage. For proper operation, let:

V BE "V BIAS "-v BE - MAX .*v BE - MIN ..!2"0.61V

vbe"v signal "-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: 2"100

I C "2 I B"100(5(10$6

I C "0.5mA.

For this collector current:

V CE "V CC $ RC I C

V CE

"10$104(0.5(10

$3"5V

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

V CE

I C R

c = 10 k !

V CC

= 10 V I

B

I = 5 µ A

Q1





Current Bias Beta DependenceUnfortunately, # is often poorly controlled and may easily

vary from 100 to 200. And # is also temperature dependent!

I C "100(5(10$6"0.5mA.

For # = 100:

The BJT with a V CE

= 5 V

For # = 200:

I C "200(10(5$6"1.0mA.

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"0V

V CE "V CC $ RC I C





Drive Base with a Base Voltage Source

For an I C 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 "0.025ln -0.5(1011.

V BE "0.025(24.635"0.616V

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

Since V CE

= 5 V the transistor is

nearly at the center of the desiredoperating region!

I C " I S e

V BE

V T

V CC

= 10 V

Rc = 10 k !

V BE

= 0.616 V

Q1





Voltage Bias I S

and V CE

DependenceUnfortunately, I

S is highly temperature-dependent, doubling

for every 5oC increase in temperature.

If the base-emitter voltage is chosen to give I C = 0.5 mA at 20oC (68o F ),

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

I C is also highly sensitive to V

BE . Consider two values I

C and 10 I

C :

10 I C

I C " I S e

V BE10

V 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 V BE

voltage increases I C by an

order of magnitude (10X ). BIAS POINT IS UNSTABLE.





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

I C "1 I E

Changes in I C due to variations in " in the range determined

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

100+2+200#100

101+1+

200

201#0.990+1+0.995

There is considerable variation in base current, however, but

this is usually of no consequence.

I B" I E

2*1#

I E

101+ I B+

I E

201

1" 2

1*2





Conclusion

Biasing a BJT poses potential large bias stability prob-lems, since its characteristics are highly sensitive to

temperature and since its electrical properties (princip-ally #) 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 Voltage

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