Lecture 06 - 1ELE2110A © 2008

Week 6: Small Signal Analysis for Single Stage BJT amplifiers

ELE 2110A Electronic Circuits

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 , orChap 5.7 of Sedra & Smith

Lecture 06 - 3ELE2110A © 2008

Signal Injection

Transistor is a VCCS:

To cause changes in current, vBE must be changed

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

– Collector is not used as an i/p terminal because even if Early voltage is considered, collector voltage has negligible effect on terminal currents

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=TV

BEvSICi exp

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/pterminal

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

Lecture 06 - 5ELE2110A © 2008

BJT Amplifier FamilyConstraints 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.

Lecture 06 - 6ELE2110A © 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

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, rπ, & re for this DC operating pointEliminate 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

Lecture 06 - 8ELE2110A © 2008

Common Emitter Amplifier

Biased using the four-resistor networkAC input is injected to base through a coupling capacitorAC output is taken from collectorEmitter is ac-grounded

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

Lecture 06 - 10ELE2110A © 2008

AC EquivalentReplace all ∞ capacitors by short circuits, ∞ 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 to analyze AC characteristics of amplifier

Lecture 06 - 11ELE2110A © 2008

Simplified AC Equivalent

kΩ100kΩ3.43

kΩ30kΩ1021

==

==

RCRR

RRBR

Lecture 06 - 12ELE2110A © 2008

Small Signal Performance Parameters

Voltage gainInput resistanceOutput resistance

and sometimesCurrent gainInput signal range

In AC analysis, we are interested to find

Lecture 06 - 13ELE2110A © 2008

Voltage Gain

3RCRorLR =

LRmgbevov

bvcv

vtA −===

Terminal voltage gain between base and collector is:

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

( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+−=∴

===

π

πrBRIR

rBRLRmgvA

ivbev

vtAiv

bev

bevov

ivov

vA

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

ππ

πrRRrBRR

rBR

21xixv

in

)(xixv

===

=

Lecture 06 - 15ELE2110A © 2008

Output Resistance

Output resistance is the total equivalent resistance looking into the output of the amplifier at coupling capacitor C3

To find Rout, input source is set to 0 and test source is applied at output

CRorCRR

mgorC

R

≅==∴

++=

xixv

out

bevxvxvxi But vbe=0.

As ro>> RC.

Lecture 06 - 16ELE2110A © 2008

Example 1

Problem: Find voltage gain, input and output resistances.Given data: β= 65, VA =50 VAssumptions: Active-region operation, VBE=0.7 V, small signal operating conditions.

Analysis: To find the Q-point, dc equivalent circuit is constructed.

µA24566µA24165

µA71.3

==

===∴

BIEIBICI

BI

5)4106.1()1(510 =×+++ BIBEVBI β

V67.30)5()4106.1(4105

=∴

=−−×−−−

CEVEICEVCI

Active region of operation is correct.

Lecture 06 - 17ELE2110A © 2008

Example 1 (Cont.)Next we construct the ac equivalent and simplify it

0.84in

in)3out( −=+

−==⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

RIR

RRRmg

ivov

vA

S31064.9 −×==TVC

Img

kΩ64.6==

CI

TVr

βπ

kΩ223=+

=C

ICE

VA

Vor

kΩ23.6in == πrBRR

kΩ57.9out == orCRR

Lecture 06 - 18ELE2110A © 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

Lecture 06 - 19ELE2110A © 2008

Common-Emitter Amplifier with Emitter R

AC equivalent:

A resistor exists at the emitter of the ac equivalent circuit

Also called emitter degeneratedCE amplifier

RE provides negative feedback

AC ground

OutputInput

Lecture 06 - 20ELE2110A © 2008

Negative Feedback due to RE

Assume iC for any reasoniE (=iC/α) vE

vBE

iC : counter act the original change

More discussion on negative feedback later in the course

Lecture 06 - 21ELE2110A © 2008

Overall Voltage Gain

Thus, the task breaks down into two steps:Find the terminal voltage gainFind the input resistance Rin.

Overall voltage gain from vi to vocan be expressed as:

b

ovt v

vA ≡

)||(||

inBI

inB

i

b

RRRRR

vv

+=

Define terminal voltage gain as:

And it can be observed that:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=≡ivb

v

bvov

ivov

vA

Lecture 06 - 22ELE2110A © 2008

Terminal Voltage GainUse hybrid-π model (neglecting ro):

for gmrπ = β >> 1

( )E

Eb

RriiRirv)1(

)1(++=

++=β

β

π

π

Lo iRv β−=

βπ =mgr

ERr

LR

bvov

vtA)1( ++

−=≡βπ

β

ERmgL

Rmg

ERrmgr

LRrmg

vtA+

−≅++

−=1)1( ππ

π

Using , we have

Effect of RE:For RE=0,– Upper limit of Avt

For gmRE>>1,– Avt becomes less dependent on gm

which varies widely

Increasing RE decreases voltage gain!

LRmgvtA −≅

ERLRvtA /−=

Lecture 06 - 23ELE2110A © 2008

Input ResistanceTo find Rin to find Vb/i:

)1()1(

)1(

ERmgrERrmgrERr

ib

vinR

+≅

++=

++==

πππ

βπ

( )Eb Rriv )1( ++= βπ

for gmrπ = β >> 1

Increasing RE increases input resistance!

Lecture 06 - 24ELE2110A © 2008

Output Resistance

0ixi0i ==⇒=∴ β ∞==∴xixv

outR

To find Rout:Set vi to zeroApply a test source at outputRout = vx/ix

E)iR(βev 1+=BIth RRR //=

KVL at left mesh:

( )

( )

0

0)1(

0

=⇒

=+++

=++

e

ethE

e

eth

v

vrRR

vvrRi

π

π

β

Lecture 06 - 25ELE2110A © 2008

Output ResistanceNow, we also include ro in our analysis:

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++≅≡∴πrthRER

EβRor

xixv

outR 1

KVL along loop 1:

evoβi)rx(ievrvxv +−=+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ += ERπrthRxiev //

Voltage at E:

Current division at Emitter:πrth

RERER

xii++

−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

++++

+=∴E

RπrthRxi

πrthRERE

Rxixiorxv //β

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

for large value of ro

Lecture 06 - 26ELE2110A © 2008

Output Resistance

πβ rmg=Assuming and , withthRERr >>+ )( π ERor >>

orR )1(out += β

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++≅π

βrthRER

ERorR 1out

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=++≅ )//(11out ERrmgorrERERrmg

orR πππ

Effect of RE:For RE = 0, Rout = ro

For RE = ∞, upper limit ofRout is obtained:

Increasing RE increases Rout!

Lecture 06 - 27ELE2110A © 2008

Current Gain

Terminal current gain:

Overall current gain:

β−=≡iLi

itA

ii

inB

Bit

s

L

s

Li RR

RAii

ii

iiA

+==≡

Lecture 06 - 28ELE2110A © 2008

Input Signal Range

321

AC

beT

C

DCC

TbeCVv

CC

vVII

VvIeIi Tbe

+=

+≅= )/1(/

Remember, in deriving the linear small signal model, we assumed vbe << VT :

Typically, vbe is required to be less than 5mV!

Lecture 06 - 29ELE2110A © 2008

Input Signal Range

πβπππ rERERmg

bv

ERrb

vrirbev

/1)1( ++=

++==

V)1(005.01005.0 ERmgERERmg

bv +≅++≤

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

β

1>>ERmgIf , vb can be increased beyond 5 mV limit.

( )Eb Rriv )1( ++= βπ

V025.0

1

=>>∴

>>≅=

TVEREITV

EREI

TVER

CI

ERmgCondition for gmRE >>1 :

Increasing RE increases input signal range!

Lecture 06 - 30ELE2110A © 2008

Summary of Emitter Degenerated Common-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( ERmgr +π

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ + )(1 ERrmgor π

ERmgL

Rmg

+−

1

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++−

in//RB

RI

Rin//R

BR

ERmgL

Rmg

1

β−

Lecture 06 - 31ELE2110A © 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

Lecture 06 - 32ELE2110A © 2008

Common-Base Circuits

AC/Small-signal equivalent:

73 || RRRL =

Output

Input

AC ground

Load

SignalSource

Lecture 06 - 33ELE2110A © 2008

Terminal Voltage Gain

Non-inverting!Magnitude same as the CE amplifier with RE=0.

LRmgevov

ACBvt +=≡

Apply test source to input (E)And use BJT small signal model:

Lecture 06 - 34ELE2110A © 2008

Input Resistance

mgmgr

mgr

r

iev

RCBin

1)1//(1

≅=+

== ππ

π

eme vg

rvi +=π

KCL at emitter:

Rin is small (as gm is usually large)!

(gm=IC/VT)

Lecture 06 - 35ELE2110A © 2008

Overall Voltage Gain

IRmgLRmg

RIR

R

IRRmgL

RmginRR

IR

inRRvtA

ivev

evov

ivov

ACBv

+≅

++=

+===

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

1

4

4)//

4(1

)//4

(

//4

Overall voltage gain is

For ,

This is the upper bound.

For ,

1<<IRmg

LRmgACBv +=

1>>IRmg

IR

LR

ACBv +=

IRR >>4for

For large voltage gain, a very small RI is required!Not a good candidate for voltage amplifier

Lecture 06 - 36ELE2110A © 2008

ExampleProblem: Find overall voltage gain.Given data: β=100, Q-point values: IC=245uA,VCE=3.64V, gm=9.8mS, rπ=10.2kΩ, rο=219kΩ.Assumptions: Small-signal operating conditions.Analysis:

kΩ1873

Ω102/1

==

=≅

RRLRmgRCB

in

176=+= LRmgACBvt

59.84

4)

4(1

+=++

=⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

RIR

R

RIRmgACB

vtACBv

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 IRmgb

v +≤

)1(ebviv IRmg+≅ for R4 >> RI.

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=+

=4

4

44

4

)//(1////

RRR

RRgv

RRRRRvv

IIm

i

inI

inieb

Lecture 06 - 38ELE2110A © 2008

Output Resistance

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++=∴π

βrthR

thRorRCB

out 1

πβ rmg=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +≅ )//(1 thRrmgorRCBout 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.

Lecture 06 - 39ELE2110A © 2008

Current Gain

Terminal current gain:

Current gain from source to load:

1+≅== αeili

ACBit

i

inR for RitARinR

RitA

iei

eili

ili

ACBi >>=≅

+===

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

1

Lecture 06 - 40ELE2110A © 2008

Summary of Common Base Amplifier

Terminal voltage gain: Non-inverting and LargeInput resistance: LowOutput resistance: HighCurrent gain: close to UnityInput range: determined by gmRth

Excellent for use as a current buffer

Lecture 06 - 41ELE2110A © 2008

Current Buffer

RsRL

Io

Ii

Rs Ri RLRoA=1

Current buffer

VoIi

Io

iiLS

So II

RRRI <<+

=

LOSiii

OL

Oi

iS

So

RRRRIAIRR

RAIRR

RI

>><<=≅

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

and for

• Requirement of a current buffer:- Low input resistance- High output resistance- Unity current gain

sL RR >>

Only a small portion of source current is delivered to load!

• With a buffer:

• Without a buffer:

for