C. Hutchens Chap 4 ECEN 3313 Handouts 1 Chapter 4 Bipolar Junction Transistors.

C. Hutchens Chap 4 ECEN 3313 Handouts 1

Chapter 4Bipolar Junction Transistors


BJT I-V Relationships

• Ebers-Moll Model -Table 4.2

• Input and Output Characteristics

• SPICE parameters ----to I-V curvers

• Regions of Opertion

• Biasing the BJT

• Bias Stability

• Small Signal Models

• Single Transistor Amplifiers


Output characteristicsE b e r s M o l l E q u a t i o n s

I = - I ( (E E S

Vn V

Vn V

B E

t

B C

t 1 1) )R C SI

I = - I ( (C C S

Vn V

Vn V

B C

t

B E

t 1 1) )F E SI

N o w L e t u s e x a m t h e e q u a t i o n s i n t h em o s t u s e f u l r e g i o n V B E > 4 n V t a n d V B C

< < 4 n V t , V t = 2 6 m V a t r o o m t e m p .

F O R W A R D B I A S

I B = - ( I C + I E )

I = - I ( ( - I (E E S

Vn V

E S

Vn V

B E

t

B E

t ) ) ) R C SI 1

I = - I ( ( (C C S

Vn V

Vn Vt

B E

t0 1 ) ) ) F E S

E

F E SI I

I = -C F EI

B-E and C-B Junctions


Forward BiasI = -C F EI

I C = F / ( 1 -

F ) I B = F I B w h e r e

F =

F / ( 1 - F )

N o w r e p e a t i n g t h e e x e r c i s e w i t h V B C >4 n V t a n d V B E < < 4 n V t , V t

= 2 6 m V a t r o o mt e m p . W e c a n d e f i n e

R =

R / ( 1 - R )

B J T a r e t y p i c a l l y d e s i g n e d ( P r o c e s sE n g i n e e r e d ) t o k e e p

F > >

R a n d F I E S =

R I C S = I S

CUTOFF-both pn junctions are reversed biased

IE IES - RICS

IC ICS - FIES

VBE and VBC < 0 for npn BJT


Ebers-Moll for computer/hand use 4.13

E b e r s M o l l r e a r r a n g e d

I E - I F - R I R w h e r e I = I (E E S

Vn V

B E

t 1 )

I C I R - F I F w h e r e I = I (C C S

Vn V

B Ct 1 )

V a l u eE q u a t i o n s f o r o b t a i n i n g a n u m e r i c a l s o l u t i o n

E m p e r i c a l I - V c u r v e s

C o m p u t e r m o d e l f o r s i m u l a t i o n

B a s e s f o r a s m a l l s i g n a l m o d e l f o r h a n dc a l c u l a t i o n s


Ebers-Moll for computer/hand useE b e r s M o l l r e a r r a n g e d

I E - I F - R I R w h e r e I = I (F E S

Vn V

B E

t 1 )

I C - I R - F I F w h e r e I = I (R C S

Vn V

B C

t 1 )

1 s t q u a d r a n t o f i n t r e s t V B C > 0 , S i n = 1

V B E > 4 V t , V B C < < 0 A c t i v e L i n “ a m p ”

I I (C F E S

Vn V

B E

t ) ( )I IE B 1

V B E < 4 V t , V B C < < 0 O f f s w i t c h “ a m p ”

I E = I E S + R I C S , I C = I C S +

F I E S

I E = I E S + I S , I C = I C S + I S

“ D i o d e l e a k a g e c u r r e n t s ”

V B E > 4 V t , V B C > 4 n V t S a t . s w i t c hs h o u l d b e a v o i d e d s l o w r e c o v e r y ”

I I ( I (C F E S

Vn V

C S

Vn V

B E

t

B C

t ) )

I I ( I (E E S

Vn V

C S

Vn V

B E

t

B C

t ) )R

V B E > 4 V t , V B C 0 n o n s a t s w i t c h

i s p r e f e r e d .

S e e T a b l e 4 . 2 p p 2 3 5


Example Using Table 4.2 Eqs

Example F = 0.995, R = 0.95, Si n= 1, IS = 0.1fA

Determine the transistor currents for VS = VBE = 0.4, 0.6, 0.7 andVCE = 1V.

VBC = VBE-1V = -0.6, -0.4 and -0.3 V respective.

ICS = IS/R = 0.1f (1.05) = 0.105fA

IB = -(IC + IB)

Tabulating

VS IC IE IB

0.4 480pA -483pA 2.4pA0.6 1.05uA -1.06uA 5.3nA0.7 49uA - 49.5uA 248nA0.9 108mA -108.5mA 543uA Note 200


SPICE Example to VERIFY Table 4.2Simulation with SPICE “MODEL PARAMETERS”

IS IS 0.1 x10-15A

F BF fwd current gain 50-250

R BR rev current gain 10-.1VA VA Early voltage 10-200V

GENERAL FORMAT

Title card*** comment linesVBB 11 0 DC 0RB 11 1 330RC 22 2 220VCC 22 0 DC 1Q1 2 1 0 NPNSPEC* QXXXX c b e model_name optional area.MODEL NPNSPEC NPN ( SPICE parameters).DC VBB LIST 0.4 0.6 0.7 0.8 0.9.PRINT DC IC(Q1) IB(Q1) IE(Q1) V(1) V(1,2)*.PLOT*.PROBE.OPTIONS TNOM = 28.6.END


Generating OTC Curves Ex 4.14 pp239


Circuit Modeling of Regions


Physical Base Currents

All of thesefactorscontribute to being < 1


Example Region chk 4.12Given VBB = 2V, VCC = 10V, RB = 22K, RE =100, RC = 2K and F = 0.99 Determine theregion of operation and estimate the Base andcollector currents.

VBB = IB RB + VBE + RE (IB+ IC )

VCC = ICRC + VCE + RE (IB+ IC)

Observe 99 from F/(1-F)

VBB - VBE = IB RB + RE (IB+ IB )

VBB - VBE = 2-.7 =1.3 = IB(RB + RE(1+c)

IB = 1.3/(22k + 100 100) = 40.6uA

VCC - VCE = IB RC + RE (IB+ IB ) “TST for Sat”

IB = (VCC - VCE)/{ RC + RE (1+ )}

IB = (10-0.2)/{2k (99) + 100 x 100} 46mA

This is not consistent with the input equationthe “Active Lin Reqion”

IC = IB 99 x 40.6 = 4.0mA

VCE = VCEQ = VCC - { IB RC + RE (IB+ IB )}

VCEQ 10 - 4.0mA (2k + 100) = 1.6V

Since VCE is > 100-200mV. Thistransistor is biased to be anamplifier.


Simple BJT Switch 4.12VCC = ICRC + VCE, VIN = IB RB + VBE

A switch has two stable Qpts

Logic 1: IB = 0, and VCE = VCC i.e5V

Logic 0 : IB must be such that VCE

= VCEsat i.e. 200mV

To ensure that VCE is small we assume = BF minimum

min IB = IC > (VCC -VCEsat)/RC

IB > (VCC -VCEsat)/( min RC )

RB < (VINmin - VBE)/IB = (VINmin - VBE)/{(VCC -VCEsat)/( min RC )}

EX Given VCC = 5V, RC = 500, VINmin = 4.5V,VCEsat = 200mV, min = 50. Now Assuming VBEsat =0.8V

RB < (4.5- 0.8V)/{ (5-0.2V)/( 50 x 500)}

RB < 185K


Operating and Biasing Constraints

PDISS < IC VCE , VCEmax (Breakdown Voltage), Icmax

Qpt for BJT must be inside the PDISS Curve

IE + IC + IB = 0 Kirchoff Current law for BJT

IC = IB BF in SPICE The magic in transistors

Early Voltage VA or VA in SPICE terminology


Fixed Biased Circuit pp236-54

Constaint Equations

VCE = VCE - IC RC

VB = VCC =IBRB + VBE

VCE & IC define the Qpt

Solving for RC and RB

RC = (VCC - VCEQ)/IC

RB = (VCC-- VBE)/IB

Now given that VCE & IC at the Qpt

at 4V and 6mA respectively.Fined RC and RB. VCC = 10V

RC = (VCC - VCEQ)/IC

RC = (10- 4)/6mA =1k ohm

RB = (VCC-- VBE)/ {IC/ }

RB = (10-- 0.7)/{6mA/200 }

RB = 310k


Temperature Effects and RE pp239Collector dependency

IB and therefore IC are very dependent onTemp VBE vs. Temp –2mV/Co

Sol Add RE; IE RE drop > 3-5X VBE

Early effect – IC dependency on VCE- Reduces Voltage Gain

(1+VCE/VA)

iC = ISevBE/VT(1+VCE/VA)



Analysis of Transistor DC Bias

RC = 2K, RB = 100k, =100Find the Operating Point

Step 1 IB

Step 2 IC

Step 3 VCE

RC = 5K, RB1 = 100K, RB2 = 50K, =100, RE = 3K

Find the Operating Point.Step 1 Thevenin Eqivalent VBB and RB

Step 2 IB = IE/(+ 1)

Step 4 IC IE = (VBB-VRE)/(RE+RB/(+ 1)

Step 5 VCE = VCC – IC - REIE

VBB = IBRB +VBE + IERE


Self Biased with RE

Fixed Bias Constaint Equations

VCE = VCC- IC RC - IERE VCC- IC RC - ICRE

VBias = VCC =IBRB + VBE + IERE IBRB + VBE + ICRE

IC = IB = BF IB

VCE & IC define the Qpt

IC = VCC/( RC + RE)


Self Biased with RE

Self biased Equations

IC = IB = BF IB

VCC = ICRC + VCE + IERE = ICRC + VCEQ + (IC + IB )RE

VCC ICRC + VCE + ICRE “VRE ICRE > 3-4X VBE i.e. 2.0V”

VBB = IBRB + VBE + (IC + IB )RE

VBB IBRB + VBE + ( IB )RE “RB < 0.1 IB RE”

Where RB = RB1||RB2 and VBB = VCC RB2/(RB1 + RB2)


Self Biased Example

VCC ICRC + VCE + ICRE

VBB IBRB + VBE + ( IB )RE

Where RB = RB1||RB2 and VBB = VCC RB2/(RB1 + RB2)

VCE VCC -( ICRC + ICRE) = 12 - (4mA 1K + 2V) = 6V

RE 2V/4mA = 500 select 510.

VBB IBRB + VBE + ( IB )RE

= 20A 10K + 0.7V + 2V = 2.9V

RB = RB1 RB2/( RB1 + RB2) = 10 K andVBB = VCC RB2/( RB1 + RB2) = 2.9V

Example: Given BF = 200, VBE

= 0.7, IC = 4mA, VRE 2V, RC =1k, and VCC = 12V. Select RB

> 9 k and find VCEQ, RE, RB1

and RB2.


Example Con’t

GB = GB1 + GB2 =1/10 K = 10-4 S

RB1 = RB2 ( 1 - VBB/VCC) = 2.9V

GB1 = GB2/(1 - VBB/VCC) = GB2 VCC/(VCC- VBB)

GB = 10-4 S = GB2 VCC/(VCC- VBB)+ GB2

GB2 = GB/( VCC/(VCC- VBB)+ 1) = 10-4/(2.32)

= 0.431x 10-4 S or RB2 = 23.2k Select 24K

GB1 = GB - GB2 = 10-4 S - 0.431x 10-4 = 0.569x 10-4 S

or RB1 = 17.57K Select 18KNow Chk RB = RB1||RB2 = 10.28K


Bias Stability

Ensure Transistor CKT will work over the rangeof specified voltage, tempertature, and transistorvariation.

The desired input and output impedances, gainand bandwidth are maintained.

The maximum power hyperbola is note violated.


Collector Current VariationW e m u s t r e m e m b e r t h a t I C O , o r B F , a n d V B E

a r e f u n c t i o n s o f T e m p e r a t u r e a n d p r o c e s s .

SI

I

I

IIC

C O

C

C O

SI

V

I

VVC

B E

C

B E

SI I

IC

F

C

F

N o w t h e t o t a l i n c r e m e n t a l c h a n g e i n c u r r e n t c a nb e w r i t t e n a s

I S I S V SC T I C O V B E F


Collector Current VariationN o w w r i t i n g t h e i n p u t b i a s l o o p e q u a t i o n , s u b s i t u t i n ga n d s o l v i n g f o r I C .

V B B = I B R B + V B E + I E E R E ( 1 )

T h r o u g h m a n i p u l a t i o n o f t h e E b e r s - M o l l e q u a t i o n s

I C = F I B + (

F + 1 ) I C O w h e r e I C O ( 1 - R

F ) I C S

“ c o l l e c t o r c u r r e n t w i t h b a s e o p e n ”

I B = ( I C - ( F + 1 ) I C O ) /

F ( 2 )

F r o m I C + I B + I E = 0 w h e r e I E E = - I E

I E E = { ( F + 1 ) /

F } ( I C - I C O ) ( 3 )

S u b s t i t u t i n g ( 2 ) a n d ( 3 ) i n t o ( 1 ) a n d s o l v i n g f o r I C

IV V I R R

R RCF B B B E C O F B E

B F E

( ) ( ) ( )

( ) ( )

1

1


Collector Current Variation

N o wS

I

I

I

I

R R

R R

R RR R

IC

C O

C

C O

F B E

B F E

B E

B

FE

( ) ( )

( ) ( )

( )

( )

1

1

k e e p i n g R B s m a l l a n d R E s m a l l

SI

V

I

V R RVC

B E

C

B E

F

B F E

( ) ( )1 “ U s i n g B - E l o o p ”

k e e p R E l a r g e a s p o s s i b l e

S

I I V V R R

R RI

C

F

C

F

F B B B E B E

B F E

( ) ( )

( ) ( )12

“ U s i n g B - E l o o p ”

“ N o t e d u e t o d e n o m i n a t o r S I i s f r e q u e n t l yn e g l e c t e d ”


ExampleEX Given the selfbiased circuit and VBE = 0.7, IB =15uA “designed” at 27Co and shifts to , BF=255,BR=6, VA=75, IS=14fA, find the change in IC for Tempchange 27 to 50Co. From SPICE output curves ICO

=55.5fA, VBE = -0.03 V, and F = 30.

T h e T h e v e n i n e q u i v a l e n t v o l t a g e V B B = 2 . 7 9 a n d R B =R B 1 | | R B 2 = 7 . 6 7 K .

SI

I

I

I

R RR R

KKI

C

C O

C

C O

B E

B

FE

( )

( )

( . )

( . ).

7 6 7 5 1 07 6 7

2 5 5 5 1 01 5 2

SI

V

I

V R R Km SV

C

B E

C

B E

F

B F E

( ) ( ) . ( ) ( ).

1

2 5 5

7 6 7 2 5 6 5 1 01 8 5

S

I I V V R R

R RI

C

F

C

F

F B B B E B E

B F E

( ) ( )

( ) ( )102

I S I S V S f A m S u AC T I C O V B E F ( . ) . ( ) ( . )1 5 2 5 5 5 1 8 0 0 3 7 9

N o w V C E Q = I C ( R C + R E ) = 7 9 u A ( 1 . 5 1 K ) = 1 2 0 m V


PSPICE.TEMP 27 50

Worst case analysis .WCASE

.model device_name NPN (BF=250 DEV 20 ...

.WCASE DC IC(Q1) YMAX HI VARY DEV DEVICESQ

BF = BF + DEV = 250 + 20 = 270 &

BF = BF - DEV = 250 - 20 = 230

* worst case analysis with PSPICEVCE 1 0 15VIB 0 2 60uAQ1 1 2 0 NPNBJT.model NPNBJT NPN(BF=250 DEV 20 VA= 75).DC DEC 0 0.001 15 IB 0 50u 10u.PROBE.END


Small Signal Model pp 256


Model parameters can befound either1. analytically or2. graphically from exp. data

ac = hfe


Small Signal Model pp257i C = I S e v B E / n V t ( 1 + V C E / V A ) e q - 4 . 2 7 a u g m e n t e d t o i n c l u d e t h e e a r l ye f f e c t .

I S a n d V A I N S P I C E

M o d e l p a r a m e t e r s c a n b e f o u n d a n a l y t i c a l l y

gI

V

I

n V

I

V

I

VmC

B E

C

t

C Q

t

B

B E

g rI

V

I

V

I

V

I

Vc e c eC

C E

C

C E

S

V

A

C

A

B En V t

1 /

N o w f r o m I b = i C / a n d e q - 4 . 2 7

g rI

V

I

V

I

n V

I

V

gb e b e

B E

B

B E

C

t

C Q

t

m 1 /

B

r = r b e = / g m T a b l e 4 . 3 p p 2 7 1 K n o w


Small Signal Model Gain pp259-60

gm = ICQ /Vt , gce =ICQ/VA = 1/rce

vin is imposed across vbe and r

vout (gce + GC) + gm vbe where vbe = vin

Solving for vo/vin = - gm /(Gc + gce)

A - gm Rc Voltage Gain for Gc >> gce

See Ex 4.9 Application of ss models


Application of Small Signal Model pp262Step 1 replace all Cs and DC power supplies if presentwith a short circuit.

Step 2 replace the BJT with the hybrid model.

Step 3 Analyze to find Ri , Ro or AV and Ai.

Step 4 Using DC analysis determine ICQ and assume ICQ

= IEQ. You must go to the data sheet for “Beta” or hfe.

Step 5 Solve for hybrid model parameters using ICQ =IEQ and Vt above.

Step 6 Insert the results of step 5 into the equations ofstep 3 above.

Step 7 Verify with SPICE.

Start by looking at example 4.9 and 4.10


Small Signal Model with Re pp285

Ri = vb/ib = ? Assume rce can be neglected. ie = ib + ib

At e ve = gm(ve – vb) + ieRe and noting ib = (vb - ve)/r,

Solving for

Ri= vb/ib = r + ( + 1) Re or r + (gm r + 1) Re Note = gm r

Re YIELDS THE DESIRED AMP INPUT RESISTANCE!


Small Signal Model with Re pp285

Av = vb/vo = ? Assume rce can be neglected. ie = ib + ib

At e; ve = gm(ve – vb) + ieRe and noting ib = (vb - ve) /r, voGC + gm vbe =0

Solving for Avc = vo/vb = vo/vin AND Ave = vo/ve = vo/vin

Avc = gm RC/[1 + gm Re] Ave = 1/[1 + gm Re]

THE PRICE OF RE IS REDUCED GAIN.THIS IS YOUR ENGINEERING TRADE OFF,

HIGHER RI BUT WITH LOWER GAIN.


Common Collector pp290

Using previous equations

Ri = r + ( + 1) Re or r + (gm r + 1) Re Ave = 1/[1 + gm Re]

and Noting Re >> 1/gm

Ri= Rs + r + (gm r + 1) Re Re(gm r + 1) = Re ( + 1)

Ave = 1

Re MUST BE >> 1/ gm

AND Re ( + 1) should be > Rs + r

WHEN ONE SOLVES for Ro

Ro = 1/gm || RE|| (Rs+ r)/


Re MUST BE >> 1/ gm

AND Re ( + 1) should be > Rs + r

WHEN ONE SOLVES for Ro

Ro = 1/gm || RE|| (Rs+ r)/

Observations regarding impedances and theTransistor

looking into the baseEmitter to gnd side resistors inc. by

looking into the emitterEmitter to base to gnd side resistors dec. by

Common Collector Con’t


Small Signal Model Con’t Add CsQ n = W 2 / 2 D n i C =

F i C F

T F S P I C E

Cd Q

Vg

i

Vd en

B EF m F

C

t

J u n c t i o n C a p d u e B - E a n d C - B d i o d e s

C j e 2 C J E 0 f o r B - E

C j c C J C 0 / 2 = C f o r C - E

N o w t o t a l b a s e - e m i t t e r c a p a c i t o r s y m b o l i c a l l y i s

C = C j e + C d e 2 C J 0 +

F i C o r

C = 2 C J E + T F i C i n S P I C E m o d e l p a r a m e t e r s

N o w t o t a l c o l l e c t o r e m i t t e r c a p a c i t o r s y m b o l i c a l l y i s

C C J C / 2


Small Signal Model Cuttoff fIc = gm v + (0- v) sC (1)

(v - vin )gx + v (g +s C + sC ) = 0 (2)

Or Ib = v (g + sC +sC ) (3)

hfe = Ic/Ib = {gm v - vssC }/{v (g + s(C + C )}

hfe = {gm - sC }/ {g + s(C + C )}

hfe(s) = [gm/ g]{1 - sC/ gm }/ {1 + s(C + C )/g}

We now have a pole a zero and a DC term gm/ g = 0 the “DC” small signal current BF pole @ g/(C + C ) = [gm/0] /(C + C ) In SPICE

gm = ICQ /Vt, C = TF (ICQ /Vt)+ 2CJE, C = CJC/2

Ic

THIS IS VERY IMPORTANTthe LINK BETWEENANALYSIS- THE SIMULATORand OUR MODEL. SPICEMODELS ARE 98% REALITYAND TRANSFERABLE.


.model Q2N2222 NPN(Is=14.34f Xti=3 Eg=1.11 Vaf=74.03 Bf=255.9Ne=1.307+ Ise=14.34f Ikf=.2847 Xtb=1.5 Br=6.092 Nc=2 Isc=0 Ikr=0 Rc=1+ Cjc=7.306p Mjc=.3416 Vjc=.75 Fc=.5 Cje=22.01p Mje=.377 Vje=.75+ Tr=46.91n Tf=411.1p Itf=.6 Vtf=1.7 Xtf=3 Rb=10)* National pid=19 case=TO18* 88-09-07 bam creation

Small Signal Model Cuttoff f

0 dB

See Fig 4.73 pp 325 also. This slide presents a very key concept. In practice in broad band amplifier design one CANNOT expect to apply a transistor beyond T….


Small Signal Model EX 4.76This is what its is about

Understand and know for exam!!!

SPICE verification with 1kHz sine and Re = 0 and Re = 100 and selecting 0 = 100Use VBE =0.7 (high for this case), IC

IE 0.84mA and neglecting “Early effect”From model card for 2N2222 and hand analysis

gm = ICQ /Vt = 34mS, r =0/ gm = 3.7K

also C = TF (ICQ /Vt)+ 2CJE, C = CJC/2 can be calculated

RI = RB|| [r + (0+1)Re] 3.57K and 13.1K

Av = gm (RI/( RI + Rs) gm (RC||rl)/(1+ Regm) = -44.7V/V and –21.9V/V

Verify this with PSPICE See example.

C. Hutchens Chap 4 ECEN 3313 Handouts 1 Chapter 4 Bipolar Junction Transistors.

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Transcript of C. Hutchens Chap 4 ECEN 3313 Handouts 1 Chapter 4 Bipolar Junction Transistors.