emitter degenerated

8/20/2019 Emitter Degenerated

1/41

Lecture 06 - 1ELE2110A © 2008

Week 6: Small Signal Analysis forSingle Stage BJT amplifiers

ELE 2110A Electronic Circuits


2/41

Lecture 06 - 2ELE2110A © 2008

Topics to cover …

Family of single-stage BJT amplifiers

Small signal analysis – Common-Emitter Amplifier

Common-Emitter Amplifier with Emitter R

Common-Base Amplifier

Reading Assignment:Chap 14.1 – 14.5 of Jaeger and Blalock , or Chap 5.7 of Sedra & Smith


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Lecture 06 - 3ELE2110A © 2008

Signal Injection

Transistor is a VCCS:

To cause changes in current, vBE must bechanged

– Base or emitter terminals can be used as i/p

terminal

– Collector is not used as an i/p terminal because

even if Early voltage is considered, collector

voltage has negligible effect on terminal

currents

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=T

V BE

v

S I

C i exp


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Lecture 06 - 4ELE2110A © 2008

Signal Extraction

Large changes in iC or iE create large

voltage drops across collector and emitter

resistors

– Collector or emitter can be used as o/p

terminal

– Base terminal is not used as o/p terminal due to

small iB


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Lecture 06 - 5ELE2110A © 2008

BJT Amplifier Family

Constraints for signal injection and extraction yield three

families of amplifiers

Input Output

– Common-Emitter (C-E): Base Collector

– Common-Base (C-B): Emitter Collector

– Common-Collector (C-C): Base Emitter

– All circuit examples in the following slides use the four-resistor

bias circuits to establish Q-point for the various amplifiers.


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Lecture 06 - 6ELE2110A © 2008

Topics to cover …






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Lecture 06 - 7ELE2110A © 2008

Using Small Signal Models in BJT ckt Analysis

Steps for using small-signal models:

Determine the DC operating point of the BJT

– in particular, the collector current

Calculate small-signal model parameters gm, r π, & r e forthis DC operating point

Eliminate DC sources – Replace DC voltage sources with short circuits

– Replace DC current sources with open circuits

Replace BJT with an equivalent small-signal model

– Choose most convenient one depending on surrounding circuitry Analyze


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Lecture 06 - 8ELE2110A © 2008

Common Emitter Amplifier

Biased using the four-resistor network

AC input is injected to base through a coupling capacitor AC output is taken from collector

Emitter is ac-grounded


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Lecture 06 - 9ELE2110A © 2008

DC operating point

All capacitors in original amplifier circuits are replaced by

open circuits, disconnecting vI, RI, and R3 from circuit

Q-point can be found from dc equivalent circuit by using

DC model for BJT


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Lecture 06 - 10ELE2110A © 2008

AC Equivalent

Replace all ∞ capacitors by shortcircuits, ∞ inductors by open circuits

Set all independent DC sources to 0,i.e., replace DC voltage sources by

short circuits and DC current sources

by open circuits

Replace transistor by small-signal

model

Use small-signal AC equivalent toanalyze AC characteristics of amplifier


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Lecture 06 - 11ELE2110A © 2008

Simplified AC Equivalent

k Ω100k Ω3.43

k Ω30k Ω1021

==

==

RC

R R

R R B

R


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Lecture 06 - 12ELE2110A © 2008

Small Signal Performance Parameters

Voltage gain

Input resistance Output resistance

and sometimes

Current gain

Input signal range

In AC analysis, we are interested to find


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Lecture 06 - 13ELE2110A © 2008

Voltage Gain

3 R

C Ror L

R =

L Rm g bevov

bvcv

vt A −===

Terminal voltage gain

between base and collector is:

Overall voltage gain from source vito output voltage across R3 is:

( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+−=∴

===

π π

r B

R I

R

r B

R

L Rm g v A

ivbe

v

vt A

ivbe

v

bev ov

ivov

v A


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Lecture 06 - 14ELE2110A © 2008

Input Resistance

The total resistance looking into the

amplifier at coupling capacitor C1represents total resistance of the

amplifier presented to signal source

π π

π

r R Rr B

R R

r B R

21xixv

in

)(xixv

===

=


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Lecture 06 - 15ELE2110A © 2008

Output Resistance

Output resistance is the total

equivalent resistance looking into the

output of the amplifier at couplingcapacitor C3

To find Rout, input source is set to 0

and test source is applied at output

C Ror C

R R

m g

or C R

≅==∴

++=

xi

xv

out

bevx

vxvxi But v be=0.

As r o>> RC .


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Lecture 06 - 16ELE2110A © 2008

Example 1

Problem: Find voltage gain, inputand output resistances.

Given data: β = 65, V A =50 V

Assumptions: Active-regionoperation, VBE=0.7 V, small signal

operating conditions.

Analysis To find the Q-point, dc

equivalent circuit is constructed.

µA24566

µA24165

µA71.3

==

==

=∴

B I

E I B

I

C

I B

I

5)4106.1()1(510 =×+++ B

I

BE

V

B

I β

V67.3

0)5()4106.1(4105

=∴

=−−×−−−

CE V

E I

CE V

C I

Active region of operation is correct.


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Lecture 06 - 17ELE2110A © 2008

Example 1 (Cont.)

Next we construct the ac

equivalent and simplify it

0.84

in

in)3out

( −=+

−==

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

R I

R

R R Rm g

ivov

v A

S31064.9 −×==

T V

C I m g

k Ω64.6==

C I

T V

r β

π

k Ω223=+

=

C I

CE V

AV

or

k Ω23.6in

== π r B R R

k Ω57.9out

== or C R R


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Lecture 06 - 18ELE2110A © 2008

Topics to cover …






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Lecture 06 - 19ELE2110A © 2008


AC equivalent:

A resistor exists at the emitter ofthe ac equivalent circuit

Also called e m i t t e r d e g e n e r a t e d CE amplifier

RE provides negative feedback

AC ground

OutputInput


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Lecture 06 - 20ELE2110A © 2008

Negative Feedback due to RE

Assume iC for any reason

iE (=iC/α)

vE vBE

iC : counter act the original

change

More discussion on negativefeedback later in the course


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Lecture 06 - 21ELE2110A © 2008

Overall Voltage Gain

Thus, the task breaks down into two steps:

Find the terminal voltage gain

Find the input resistance Rin.

Overall voltage gain from vi to vocan be expressed as:

b

ovt

vv A ≡

)||(

||

in B I

in B

i

b

R R R

R R

v

v

+=

Define terminal voltage gain as:

And it can be observed that:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎟

⎠

⎞

⎜⎜⎜⎜

⎜

⎝

⎛

=≡i

vb

v

bvov

ivovv A


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Lecture 06 - 22ELE2110A © 2008

Terminal Voltage GainUse hybrid-π model (neglecting r o):

for gmr π = β >> 1

( ) E E b

Rr i

iRir v

)1(

)1(

++=++=

β

β

π

π

Lo

iRv −=

β π =m g r

E Rr

L R

bv

ov

vt A)1( ++

−=≡ β

π

β

E R

m g

L R

m g

E Rr

m g r

L Rr

m g

vt A+

−≅++

−=1)1(

π π

π

Using , we have

Effect of RE:

For RE=0, – Upper limit of Avt

For gmRE>>1,

– Avt becomes less dependent on gmwhich varies widely

Increasing RE decreases voltage gain!

Rm g vt −≅

E R

L Rvt /−=


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Lecture 06 - 23ELE2110A © 2008

Input Resistance

To find Rin to find Vb/i:

)1(

)1(

)1(

E Rm g r

E Rr m g r

E Rr ib

v

in R

+≅

++=++==

π

π π

β π

( ) E b Rr iv )1( ++= β π

for gmr π = β >> 1

Increasing RE increases input resistance!


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Lecture 06 - 24ELE2110A © 2008

Output Resistance

0ixi0i ==⇒=∴ ∞==∴

xixv

out R

To find Rout:

Set vi to zero

Apply a test source at output

Rout = vx/ix

E )iR( β ev 1+=

B I th R R R //=

KVL at left mesh:

( )

( )

0

0

)1(

0

=⇒

=++

+

=++

e

eth

E

e

eth

v

vr R

R

v

vr Ri

π

π

β


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Lecture 06 - 25ELE2110A © 2008

Output ResistanceNow, we also include r o in our analysis:

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++≅≡∴

π r th R

E R

E β R

or

xi

xv

out R 1

KVL along loop 1:

evo β i)r x(ievr v xv +−=+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ += E

Rπ r th R xiev //

Voltage at E:

Current division at Emitter:π r th

R E

R E

R

xii ++−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎟⎟⎟

⎠

⎞

⎜⎜

⎜⎜⎜

⎝

⎛

++

++

+=∴ E

R

π

r

th

R xi

π r th R E R

E R

x

i

x

ior xv // β

Put 2nd and 3rd equations into 1st equation:

for large value of r o


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Lecture 06 - 26ELE2110A © 2008

Output Resistance

π β r m g =Assuming and , withth R

E Rr >>+ )( π E Ror >>

or )1(out +=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++≅ π β

r th

R E

R E R

or R 1out

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ +=

++≅ )//(11

out E Rr m g or r

E R

E Rr m g or R π

π

π

Effect of RE:

For RE = 0, Rout = r o For RE = ∞, upper limit of

Rout is obtained:

Increasing RE increases Rout!


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Lecture 06 - 27ELE2110A © 2008

Current Gain

Terminal current gain:

Overall current gain:

β −=≡ i Li

it A

ii

in B

Bit

s

L

s

Li R R

R Ai

i

i

i

i

i A +==≡


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Lecture 06 - 28ELE2110A © 2008

Input Signal Range

{

321

AC

be

T

C

DC

C

T beC

V v

C C

vV

I I

V v I e I i T be

+=

+≅= )/1(/

Remember, in deriving the linear small signal model,

we assumed vbe


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Lecture 06 - 29ELE2110A © 2008

Input Signal Range

π β π

π π

r E R E Rm g

bv

Rr

bv

r ir be

v

/1)1( ++

=

++

==

V)1(005.01005.0 E

Rm g E

R

E Rm g b

v +≅++≤⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

β

1>> E

RmIf , v b can be increased beyond 5 mV limit.

( ) E b Rr iv )1( ++= β π

V025.0

1

=>>∴

>>≅=

T V

E R

E I

T V

E E I

T V

E C I

E Rm g

Condition for gmR E >>1 :

Increasing RE

increases input signal range!


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Lecture 06 - 30ELE2110A © 2008

Summary of Emitter DegeneratedCommon-Emitter Amplifier

Terminal voltage gain: Inverting and Large

Overall voltage gain: Inverting and Large

Input resistance: Large

Output resistance: Large

Terminal Current gain: Large

Effects of the resistor at Emitter: – Voltage gain decreased, but more stabilized (less sensitive to gm)

– Input signal range increased

– Input and output resistance increased

)1( E

Rm g r +π

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ + )(1 E

Rr m g or π

Rm

g

L Rm g

+−

1

⎥⎥

⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢

⎣

⎡

++

−

in //R B R I R

in //R

B R

E Rm g

L R

m g

1

−


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Lecture 06 - 32ELE2110A © 2008

Common-Base Circuits

AC/Small-signal equivalent:

73 || R R R L =

Output

Input

AC ground

Load

SignalSource


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Lecture 06 - 33ELE2110A © 2008

Terminal Voltage Gain

Non-inverting!

Magnitude same as the CE

amplifier with RE=0.

L Rm g

ev

ov

ACBvt +=≡

Apply test source to input (E)

And use BJT small signal model:


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Lecture 06 - 34ELE2110A © 2008

Input Resistance

m g m g

r

m g r

r

i

ev

RCBin

1)

1//(

1

≅=

+

== π π

π

eme v g r

vi +=

π

KCL at emitter:

Rin is small (as gm is usually large)!

(gm=IC /VT)


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Lecture 06 - 35ELE2110A © 2008

Overall Voltage Gain

I Rm g L

Rm g R I R

R

I R Rm g

L Rm g

in R R

I R

in R R

vt A

ivev

evov

ivov ACBv

+≅

++

=

+===

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎟⎟

⎟

⎠

⎞

⎜⎜⎜⎜

⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

1

4

4

)//4(1

)//4

(

//4

Overall voltage gain is

For ,

This is the upper bound.

For ,

1> I

Rm g

I R

L

R

ACBv +=

I R R >>

4for

For large voltage gain, a

very small RI is required!

Not a good candidate for

voltage amplifier


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Lecture 06 - 36ELE2110A © 2008

Example

Problem: Find overall voltage gain.

Given data: β=100, Q-point values: IC=245uA,VCE=3.64V, gm=9.8mS, r π=10.2k Ω, r ο=219k Ω.

Assumptions: Small-signal operating conditions.

Analysis:

k Ω1873

Ω102/1

==

=≅

R R L R

m g RCBin

176=+= L

Rm g ACBvt

59.8

4

4

)4

(1+=

++=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

R I

R

R

R I

Rm g

ACBvt ACBv


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Lecture 06 - 37ELE2110A © 2008

Input Signal Range

For small-signal operation,

Relative size of gm and RI determine signal-handling limit.

V)1(005.0

I

Rm g

b

v +≤

)1(eb

vi

v I

Rm g +≅ for R4 >> R I .

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

++=

+=

4

4

44

4

)//(1//

//

R R

R

R R g

v

R R R

R Rvv

I I m

i

in I

inieb


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Lecture 06 - 38ELE2110A © 2008

Output Resistance

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++=∴

π

β

r th

Rth

R

or RCBout 1

π β r m=

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ +≅ )//(1th

Rr m

g o

r R

CB

out π

Using

Rout here is equivalent to the Rout of

CE amplifier with RE=Rth and resistance

at base equal to zero.

Rout is large.


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Lecture 06 - 39ELE2110A © 2008

Current Gain

Terminal current gain:

Current gain from source to load:

1+≅== α

ei

l i

ACBit

i

in R for Rit A Rin

R

Rit Aie

i

eil

i

il

i

ACBi >>=≅+=== ⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎟

⎠

⎞

⎜⎜⎜

⎜

⎝

⎛

1


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Lecture 06 - 40ELE2110A © 2008

Summary of Common Base Amplifier

Terminal voltage gain: Non-inverting and Large

Input resistance: Low

Output resistance: High

Current gain: close to Unity

Input range: determined by gmRth

Excellent for use as a current buffer

C t B ff


41/41

Lecture 06 - 41ELE2110A © 2008

Current Buffer

RsRL

Io

Ii

Rs Ri RLRo A=1

Current buffer

VoIi

Io

ii

LS

S o I I

R R

R I >

Only a small portion of sourcecurrent is delivered to load!

• With a buffer:

• Without a buffer:

for

emitter degenerated

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