lecture 20 bipolar junction transistors (bjt): part 4 small signal

Lecture 20Lecture 20

Bipolar Junction Transistors (BJT): Part 4

Small Signal BJT Model

Reading:

Jaeger 13.5-13.6, Notes

ECE 3040 - Dr. Alan DoolittleGeorgia Tech

Further Model Simplifications(useful for circuit analysis)( y )

Ebers-MollForward Active

Mode

Neglect Small Terms

TEB

TEB

TCB

TEB

VV

SCRV

V

FFCV

V

RV

V

FFC eIIIeIIeIeII

0000 111


Modeling the “Early Effect” (non-zero slopes in IV curves)

iB1< iB2< iB3IC •Base width changes due

to changes in the base-

iB3 (theory)

collector depletion width with changes in VCB.

•This changes T, which

iB1 (theory)iB2 (theory)

VAVCE

g T,changes IC, DC and BF

A

Major BJT Circuit Relationships

TEB

TEB

TEB

VV

FO

S

F

CB

A

CEFOF

A

CEVV

SCV

V

SC eIi

iVv

Vv

eIieIi

11


Small Signal Model of a BJT

•Just as we did with a p-n diode, we can break the BJT up into a large signal analysis and a small signal analysis and “linearize” the non-linear behavior of the Ebers-Moll model.

S ll i l M d l l f l f F d i d•Small signal Models are only useful for Forward active mode and thus, are derived under this condition. (Saturation and cutoff are used for switches which involve very large voltage/current swings from the on to off states.)

•Small signal models are used to determine amplifier characteristics (Example: “Gain” = Increase in the magnitude of a signal at the output of a circuit relative to it’s magnitude atof a signal at the output of a circuit relative to it s magnitude at the input of the circuit).

•Warning: Just like when a diode voltage exceeds a certainWarning: Just like when a diode voltage exceeds a certain value, the non-linear behavior of the diode leads to distortion of the current/voltage curves (see previous lecture), if the i t / t t d t i li it th f ll Eb M ll d l


inputs/outputs exceed certain limits, the full Ebers-Moll model must be used.

Consider the BJT as a two-port Network

i1 iTwo Port Network

+V1

-

i1 i2+V2

-

i1=y11v1 + y12v2 ib=y11vbe + y12vce

General “y-parameter” Network BJT “y-parameter” Network

i2=y21v1 + y22v2 ic=y21vbe + y22vce



iib=y11vbe + y12vce

ic=y21vbe + y22vce



o is most


often taken as a constant, F

Alternative Representations

CT

Cm

V

IVI

yg 21

1

40

Transconductance

CEA

m

o

C

To

VVgI

Vy

r

11

1

1 Input Resistance

C

CEAo I

VVy

r

22

1Output Resistance

Y-parameter Model Hybrid-pi Model

v1


Alternative Representations

bobmbem iirgvg

Voltage Controlled Current source version of Hybrid pi

Current Controlled Current source version of Hybrid pisource version of Hybrid-pi

Modelsource version of Hybrid-pi

Model


Single Transistor Amplifier Analysis: Summary of Procedure Steps to Analyze a Transistor Amplifier

Important!

1.) Determine DC operating point and calculate small signal parameters (see next page)

Step

Step 1

2.) Convert to the AC only model.•DC Voltage sources are shorts to ground•DC Current sources are open circuits•Large capacitors are short circuits

Step 2

•Large capacitors are short circuits•Large inductors are open circuits

3.) Use a Thevenin circuit (sometimes a Norton) where necessary. Ideally the

Step 3

Norton) where necessary. Ideally the base should be a single resistor + a single source. Do not confuse this with the DC Thevenin you did in step 1.

Step 4

4.) Replace transistor with small signal model5.) Simplify the circuit as much as necessary

Step 5


necessary. 56.) Calculate the small signal parameters (r, gm, ro etc…) and then gains etc…

Single Transistor Amplifier AnalysisDC Bias Point

Important!Step 1 detail

Thevenin

Vbe Ie

Ib

RTH

3V=IERE+Vbe+IBRTH

3V=IC((o+1)/o) Re+0.7V+IBRTH3V IC((o 1)/o) Re 0.7V IBRTH

3V=IB o((o+1)/o) Re+0.7V+IBRTH

3V= IB(100+1)1300+0.7+ IB7500


B B

IB=16.6 uA, IC= IB o=1.66 mA, IE=(o+1) c/ o=1.67 mA

Single Transistor Amplifier AnalysisCalculate small signal parameters

I C

Step 6 detail Important!

gIV

yr

SIVI

yg

o

C

To

CT

Cm

15061

0664.040

11

21

Transconductance

Input Resistance

KIV

IVV

yr

gIy

C

A

C

CEAo

mC

2.451

22

11

Output Resistance

RTHVTH=0.88 VS

rRL= RC| R3| ro

VbegmVbe

Vout880

vvandR

rvvandRvgv SThThbeLbemout 88.0

rR

rRg

vv

vv

vv

vv

GainVoltageA

rRg

ThLm

S

th

th

be

be

out

S

outv

SThTh

ThbeLbemout

88.0


VVA

A

v

v

ThSthbeS

/139

88.01506880

1506000,100||4300||200,450664.0

For Extra Examples see: Jaeger section 13.6, and

pages 627-630 (top of 630)

Supplement Added to describe more details of theImportant!

Supplement Added to describe more details of the Solution of this Problem

Bipolar Junction Transistors (BJT): Part 5

D t il f A lifi A l iDetails of Amplifier Analysis

Reading:

Jaeger 13.5-13.6, Notes


Detailed Example: Single Transistor Amplifier Analysis

Important!


Notes on slides 14-25 were prepared by a previous student.

Step 1: Determine DC Operating PointRemove the Capacitors

Important!

Remove the Capacitors

Because the impedance of a capacitor is Z = 1/(jωC), capacitors have infinitehave infinite impedance or are open circuits in DC (ω = 0).

Inductors (not present in this circuit) have an impedance Z = jωLimpedance Z = jωL, and are shorts in DC.


Step 1: Determine DC Operating PointDetermine the DC Thevenin Equivalent

Important!Determine the DC Thevenin Equivalent

Replace all connections to the transistor with their Thevenin equivalents.


Step 1: Determine DC Operating PointCalculate Small Signal Parameters

Important!

Calculate Small Signal Parameters

Identify the type of transistor (npn in this example) and draw the base, collector, and emitter currents in their

di i d h i di l+

ICproper direction and their corresponding voltage polarities.Applying KVL to the controlling loop (loop 1):

-

IB

( p )VTHB – IBRTHB – VBE – IERE = 0

Applying KCL to the transistor:

+ -+

VBEB

IE

pp y gIE = IB + IC

Because IC = βIB,

+

-BE

1IE = IB + IC = IB + βIB = IB(1+β)

Substituting for IE in the loop equation:

-


VTHB – IBRTHB – VBE – IB(1+β)RE = 0

Step 1: Determine DC Operating PointPlug in the Numbers

Important!

Plug in the Numbers

VTHB – IBRTHB – VBE – IB(1+β)RE = 0VTHB – VBE – IB(RTHB + (1+ β)RE) = 0

+

ICVTHB VBE IB(RTHB + (1+ β)RE) 0VTHB = 12R1/(R1+R2) = 3 VRTHB = R1 || R2 = 7.5 kΩAssume VBE = 0.7 V

-

IB

BEAssume β for this particular transistor is given to be 100. + -

+VBE3 0 7 I (7500 + (1+100)*1300) 0 B

IE

+

-BE3 – 0.7 – IB(7500 + (1+100)*1300) = 0IB = 16.6 μAIC = βIB = 1.66 mAI = I + I = 1 676 mA

1 -IE = IB + IC = 1.676 mA


Step 1: Determine DC Operating PointCheck Assumptions: Forward Active?

Important!

Check Assumptions: Forward Active?

VC = 12 – ICRC = 12 - (1.66 mA)(4300) = 4.86 VV = I R = (1 67 mA)(1300) = 2 18 V

+

ICVE = IERE = (1.67 mA)(1300) = 2.18 VVB = VTHB – IBRTHB = 3 – (16.6μA)(7500) = 2.88 V

-

Check: V

IB

+ -+

VBE

Check:For an npn transistor in forward active:

VC > VB

VC

VE

VB

B

IE

+

-BE

4.86 V > 2.88 V

VB – VE = VBE = 0.7 V

E

1 -2.88 V – 2.18 V = 0.7 V


Step 2: Convert to AC-Only ModelShort the Capacitors and DC Current Sources

Important!

Short the Capacitors and DC Current Sources

• DC voltage sources are h t ( ltshorts (no voltage

drop/gain through a short circuit).

• DC current sources areDC current sources are open (no current flow through an open circuit).

• Large capacitors are shorts (if C is large, 1/jωC is small).

• Large inductors are• Large inductors are open (if L is large, jωL is large).


Step 2: Convert to AC-Only Model(Optional) Simplify Before Thevenizing

Important!

(Optional) Simplify Before Thevenizing

rrc

4 3 kΩ

rs

r2

rL

30 kΩ

100 kΩ2 kΩ

4.3 kΩ

2 kΩ 4.12 kΩ

rs

r ||r

rc||rL

vsr1

kΩ2 kΩ10 kΩ 7.5 kΩ

r1||r2vs


Step 3: Thevenize the AC-Only Model Important!

|| ||

rthC = rc||rL

4 12 kΩ

2 kΩ 4.12 kΩ

rs

r ||r

rc||rL

rthB = rs||r1||r2

1.58 kΩ

4.12 kΩ

vthC = 0 V

7.5 kΩr1||r2

vsrthE = 0 ΩvthE = 0 V

vthB = vs * (r1||r2)/(rs+[r1||r2])


Step 4: Replace Transistor With Small Signal ModelImportant!

r hB = r ||r1||r2

rthC = rc||rL

4.12 kΩ v hC = 0 V

Important!

rthB rs||r1||r2

1.58 kΩ

r = 0 Ω*

vthC 0 V

After replacing the transistor, apply Ohm’s Law: V = IR to find vout.

rthE = 0 ΩvthE = 0 V

vthB = vs * [(r1||r2)/(rs+r1||r2)]

ro and rthC are in parallel, so that Ohm’s Law becomes: vout = -IR = -(gmvBE)(ro||rthC)

Because rthC = rc||rL

rthB B C

+ +

vout = -(gmvBE)(ro||rc||rL)

vout/vBE = -gm(ro||rc||rL) is the gain from transitor input (vBE)

vthB rthC

E E

-

vBE rπgmvBE

ro

-

voutto transistor/circuit output vout)


TRANSISTOREXTERNAL

Step 5: Calculate Gain and Small Signal ParametersImportant!

rthB B C

+ +

As previously determined: vthB/vs = (r1||r2)/([r1||r2] + rs)

vthB rthC

E E

-

vBE rπgmvBE

ro

-

vout

G i / ( / )( / )( / )

Applying a voltage divider: vBE/vthB = rπ/(rπ+rthB)

G i f tTRANSISTOREXTERNAL

Gain = vout/vs = (vthB/vs)(vBE/vthB)(vout/vBE) Gain factor:vout/vBE = -gm(ro||rc||rL)

Because calculating the DC operating point was done first, we have equations for gm,Because calculating the DC operating point was done first, we have equations for gm, rπ, and ro in terms of previously calculated DC currents and voltages.

Plugging in the numbers:

Gain = vout/vs = -139 V/V


Interpretation/Analysis of Results Important!

Gain = vout/vs = (vthB/vs)(vBE/vthB)(vout/vBE) = -139 V/V vthB/vs = (r1||r2)/([r1||r2] + rs)

v /v = r /(r +r )vBE/vthB = rπ/(rπ+rthB)

vout/vBE = -gm(ro||rc||rL)Both terms are loss factors, i.e. they can never be greater than 1

This term is the gain factorand is responsible for amplifying the signalnever be greater than 1

in magnitude and thus cause the gain to decrease.

amplifying the signal.The AC input signal has been amplified 139 times in magnitude. The negative i i di h h bsign indicates there has been

a phase shift of 180°.


Completing the Small Signal Model of the BJTBase Charging Capacitance (Diffusion Capacitance)

In active mode when the emitter-base is forward biased, the capacitance of the emitter-base junction is dominated by the diffusion capacitance (not

Recall for a diode we started out by saying:

y p (depletion capacitance).

'dvdQC

D

DDiffusion

Neglect charge injected from the base into the

Sum up all the minority carrier h ith

'

''

dvdt

dtdQ

DD xvxv

D

D

D base to t eemitter due to p+ emitter in pnp

charges on either side of the junction

0011 dxeenqAdxeepqAQ nT

DpT

DL

xV

v

poL

xV

v

noD

h d i d l i l h


Excess charge stored is due almost entirely to the charge injected from the emitter.


Th BJT t lik ffi i t “ i h ” A j it i•The BJT acts like a very efficient “siphon”: As majority carriers from the emitter are injected into the base and become “excess minority carriers”, the Collector “siphons them” out of the base.

•We can view the collector current as the amount of excess charge in the base collected by the collector per unit time.

•Thus, we can express the charge due to the excess hole us, we c e p ess e c ge due o e e cess o econcentration in the base as:

FCB iQ or the excess charge in the base depends on the magnitude ofor the excess charge in the base depends on the magnitude of current flowing and the “forward” base transport time, F, the average time the carriers spend in the base.

•It can be shown (see Pierret section 12 2 2) that:•It can be shown (see Pierret section 12.2.2) that:

whereD

W

BF ,

2

2

ECE 3040 - Dr. Alan DoolittleGeorgia TechtcoefficiendiffusioncarrierMinorityD

widthregionneutralQuasiBaseW

B


Thus, the diffusion capacitance is,

CB iWQC

2

poQBE

C

BpoQ

BE

BB

IvDv

QC

intint 2

mFT

CFB g

VI

C

The upper operational frequency of the transistor is limited by the forward base transport time: f

21

Note the similarity to the Diode Diffusion capacitance we found previously:

F2


timetransittheis

IqALnLp

wheregCS

npopnottdDiffusion

Completing the Small Signal Model of the BJTBase Charging Capacitance (Total Capacitance)

In active mode for small forward biases the depletion capacitance of the base-emitter junction can contribute to the total capacitance

V

CC

EB

jEojE

1

total capacitance

where

V baseemitterforbi

,

1

junctionBEtheforvoltageinbuiltV

cecapacidepletionbiaszeroC

baseemitterforbi

jEo

tan

CCC

Thus, the total emitter-base capacitance is:


jEB CCC

Completing the Small Signal Model of the BJTBase Charging Capacitance (Depletion Capacitance)

In active mode when the collector-base is reverse biased, the capacitance of the collector-base junction is dominated by the depletion capacitance (not diffusion capacitance)depletion capacitance (not diffusion capacitance).

CC o

h

VV

basecollectorforbi

CB

1

junctionCBtheforvoltageinbuiltV

cecapacidepletionbiaszeroCwhere

o

tan,

junctionCBtheforvoltageinbuiltV basecollectorforbi


Completing the Small Signal Model of the BJTCollector to Substrate Capacitance (Depletion Capacitance)

In some integrated circuit BJTs (lateral BJTs in particular) the device has a capacitance to the substrate wafer it is fabricated in This results from a “buried” reverse biased junction Thusin. This results from a buried reverse biased junction. Thus, the collector-substrate junction is reverse biased and the capacitance of the collector-substrate junction is dominated by the depletion capacitance (not diffusion capacitance). Emitterp p ( p )

VC

CCS

CSCS

1

p

p-collectorn-base

where

V substratecollectorforbi

,

1 p-collector

n-substrate

junctionsubstrateCtheforvoltageinbuiltVcecapacidepletionbiaszeroC

substratecollectorforbi

CS

tan


Completing the Small Signal Model of the BJTParasitic Resistances

•rb = base resistance between metal interconnect and B- E junction•r = parasitic collector resistancerc parasitic collector resistance •rex = emitter resistance due to polysilicon contact•These resistance's can be included in SPICE sim lations b t are s all ignored in handsimulations, but are usually ignored in hand calculations.


Completing the Small Signal Model of the BJTComplete Small Signal Model


lecture 20 bipolar junction transistors (bjt): part 4 small signal

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