Lesson #10 Chapter 4 - New Jersey Institute of Technologyjoelsd/electronics/classwork/electronics1/BME373... · BME 372 Electronics I – J.Schesser 274 Small Signal Equivalent Circuits

BME 372 Electronics I –J.Schesser

270

Transistors

Lesson #10Chapter 4


271

Small Signal Equivalent Circuits and Parameters for the BJT

• If the variable portion of the input signal is small (in amplitude), it is possible to approximate the (non-linear) transistor as a linear device by representing it by an equivalent circuit

• Here’s how we do it:– Recall

– Since the base-emitter junction is a forward-biased diode and the current that crosses this junction is IE, then we can use the Shockley equation

1 where1 and

)1( then ; :Define

BBC

EBE

C

BCE

iii

iiii

iii

)1( T

BEVv

ESE eIi


272


• Then:• Before we continue let’s define the following notation:

– A lower case signal (voltage or current) with upper case subscripts represents the total of the DC portion of the signal and the AC portion of the signal

– An upper case signal with upper case subscripts represents the DC portion (the Q-point)

– A lower case signal with lower case subscripts represents the AC portion

iB(t) = IBQ + ib(t)vBE(t) = VBEQ + vbe(t)

)1()1( T

BEVv

ESB eIi


273


T

beBQb

T

beBQBQ

T

beBQbBQ

x

Vtv

BQbBQ

VV

ESBQ

Vtv

VV

ESbBQ

VtvV

ESbBQ

Vv

ESB

VtvIti

VtvIIV

tvItiI

xe

eItiI

eII

eeItiI

eItiI

eIi

T

be

T

BE

T

be

T

BE

T

beBE

T

BE

)()(

)())(1()(

1 :function lexponentia theofexpansion Taylor theof termsfirst two theUsing

)(

)1(

equation,Shockley esatisfy th alsomust espoint valu-Q that theNoting

)()1()(

termlexponentia the tocompared neglible isit since term1- theIgnoring

)1()1()(

)1()1(

)(

)(

)(


274


ip.relationsh thisrepresent tocollector in the sourcecurrent dependent a have weAnd)()(

)()()()(current collector thesince Secondly,

circuit. equivalentour of resistanceinput therepresents andpoint - theoffunction a is resistance This

)()()()()(

:ltageemitter vo-base thecurrent to base theofportion AC theof iprelationsh a is This

tititiItiI

titi

Q

IVr

rtiIVtitvV

tvIti

bc

bBcC

BC

BQ

T

bBQ

Tbbe

T

beBQb


275


• Here is our equivalent circuit:

• Here is a second version equivalent circuit:

B

E

Crπ β ib

o

o

o

rπ β ib = β / rπ vbe= gm vbe

B

E

Co

o

o


276

Small Signal Equivalent CircuitsExample

• We see that we have capacitors and extra resistors.

• The extra resistors represent – the load to the next stage (input

impedance of the next stage)– the source resistance (output

impedance of the previous stage)• The extra capacitors are to:

– protect the DC design of our amplifier

– allow the passing the AC portion of the input signal from the source to our amplifier and on to the load (coupling capacitors)

– eliminate the loading of the emitter resistor for the AC signal (bypass capacitors)

VCC = 15V

RC 1kR1 10k

β = 300

RE 1kR2

500 RL

2k

RS

vin = 0.001sinωt

+

-5k

+vo

--

+vi

--


277

Small Signal Equivalent Circuits - ExampleDC Analysis

VCC = 15V

RC 1kR1 10k

β = 300

RE 1kR2

500 RL 2k

RS

vin+

-5k

• For DC the capacitors are open-circuits so for DC analysis the circuit becomes the same circuit as we have just analyzed

• The components not affected by the DC voltage are “grayed out”.

• The next step is to perform the AC analysis.


278

Small Signal Equivalent Circuits-ExampleAC Analysis

• The next step is to perform the AC analysis by

– Shorting any DC voltage sources– Opening any DC current sources – Shorting the capacitors– Eliminating any resistors which

are shorted by the capacitors (e.g., the emitter resistor

– Connecting the source, source resistance, and load resistance

– Replace the transistor with its small signal equivalent circuit.

• Short circuits are drawn in thick lines, the removed capacitor is “greyed-out”, and the equivalent circuit is in red.

• This circuit looks too messy, let’s redraw

RC 1kR1 10k

β = 300

RE 1kR2

500 RL 2k

RS

+

-5k

vin = 0.001sinωt

rπβ ib


279


RC

1k

R110k

R2500

RL

2k

RS

+

-5k

vin = 0.001sinωt

rπ β ib+vi

--

+vo

--

RL’RB

500

RS

+

-

vin = 0.001sinωt

rπ β ib+vi

--

+vo

--

1 21 2

1 2

1 2' || 6671 2

5 10|| 3.335 10

26 184414.1

'

' 109

|| 1.19

' '

c LL c L

c L

B

T

BQ

o b L

i b

o Lv

i

Bin B

B

o b L i iLo v

L L L L

ii

in

i

R R kR R RR RR R kR R R k

R RV mrI

v i Rv i r

v RAv r

R rZ R r kR r

v i R v vRi AR R R r RviZ

iA

64.4

7000

i

o inLv v

ii L

in

i v

vZRA Avi R

ZG A A

ii ib

ioii

ib

Zin


280

Small Signal Equivalent Circuits-ExampleAC Analysis Alternative Method for Ai

1 21 2

1 2

1 2' || 6671 2

5 10|| 3.335 10

26 184414.1

By current division

( )

( )

1 3.33001 2

c LL c L

c L

B

T

BQ

Co b

C L

Bb i

B

C Bo i

C L B

o C Bi

i C L B

R R kR R RR RR R kR R R k

R RV mrI

Ri iR R

Ri iR r

R Ri iR R R ri R RAi R R R r

kk k

3 64.363.33 1.844

kk k

RLRB

500

RS

+

-

vin = 0.001sinωt

rπ β ib+vi

--

+vo

--

ii ib

Zin

io

RC

Note they are the same:'in BL

i vL L B

C L

C L B

L B

C B

C L B

Z R rRA AR r R R rR R

R R R rr R R r

R RR R R r


281


RL’RB

500

RS

+

-

vin = 0.001sinωt

rπ β ib+vi

--

+vo

--

mVtvvRZ

ZAvvA

vv

vv

vvA

RZZ

vv

vRZ

Zv

vv

vv

vvA

krR

rRrRZ

rR

vvA

io

sin

inv

in

iv

in

i

i

o

in

ovs

sin

in

in

i

in

sin

ini

in

i

i

o

in

ovs

B

BBin

L

i

ov

sin4.764.76

4.76

7.

19.1||

109'

ii ib

Zin


282


kRivZ

Rvii

iRvi

ivZ

c

x

xo

c

xxb

b

c

xx

x

xo

1

;0

RC

RB

500

RS

rπ β ib+vi

--

Ib=0

Z0+vx

--

ix


283

Small Signal Equivalent CircuitsEmitter Follower Example

• We see that the load is across the emitter resistor

• The extra resistors and capacitors are represent – the load to the next stage (input

impedance of the next stage)– the source resistance (output

impedance of the previous stage)

– Coupling capacitors• Our analysis plan

– First, DC analysis to determine Q-point and equivalent circuit parameters

– AC analysis to calculate the gains

vin

VCC = 20V

R1 100k

β = 200

RE 2kR2

10k

RL

1k

RS

+

-100k +

vo

--

+vi

--

VBEQ = 0.7 V


284

Small Signal Equivalent CircuitsEmitter Follower Example DC Analysis

• Remove the circuit elements which are not affects by the DC voltages

• Opening the coupling capacitors

VVRR

RV

kRRR

CCTH

TH

10

50||

12

2

21

VCC = 20V

R1 100k

β = 200

RE 2kR2

10k

RL

1k

RS

vin

+

-100k +

vo

--

+vi

--

VBEQ = 0.7 V

100k

VCC = 20V

RE 2k

RTH

VTH

β IB

VCC = 20V

RE 2k

RTHVTHVBE

12606.20

26 ;7.11

14.4)1( ;12.4

6.202)2001(50

7.010)1(

)1(

m

IVrVRIVV

mAIImAII

AkkRR

VVI

RIVRIRIVRIV

BEQ

TEEQCCCEQ

BQEQBQCQ

ET

BEQTHBQ

EBQBEQTHBQEEQBEQTHBQTH

• Next, use the DC equivalent circuit for the active region and then write KVL for the base circuit

• Redraw the circuit and replace the base circuit with its Thevinen’s equivalent


285

Small Signal Equivalent CircuitsEmitter Follower Example AC Analysis

• The next step is to perform the AC analysis by– Shorting any DC voltage

sources– Opening any DC current

sources – Shorting the capacitors– Connecting the source, source

resistance, and load resistance– Replace the transistor with its

small signal equivalent circuit.• And Redraw to simplify

R1 100k

β = 200

RE 2kR2

10k

RL

1k

RS

vin

+

-100k

+vo

--

+vi

--

VBEQ = 0.7 V

rπβ ib


286


R1

100kRE2k

R210k

RL

1k

RS

vin

+

-100k

+vo

--

+vi

--

rπ

β ib

ie= (1+β)ibib

RB10k

RL’

667

RS

vin

+

-

50k

+vo

--

+vi

--

rπ

β ib

ie= (1+β)ibib

1 2|| 50' || 667

(1 ) '

(1 ) '(1 ) '

(1 ) '(1 200) 667 .991

(1 200) 667 1260

(1 ) ' 134

|| 36.5

;

B

L E L

o b L

oi b o o

L

o Lv

i L

v

iit L

b

in B it

o v i io i

L L in

oi

R R R kR R Rv i R

vv i r v r vR

v RAv R r

A

vZ r R ki

Z R Z kv A v vi iR R ZiAi

36.2

35.8

inv

i L

v i

ZAR

G A A

Zin

ii iO

Zit

ii


287


(1 200) 667 .991(1 200) 667 1260

0.991

36.5 0.7836.5 10

0.991 0.78 0.78

ov

i

o o i ivs

s i s s

i in

s in s

o o ivs

s i s

vAvv v v vAv v v v

v Zv Z R

v v vAv v v

RB10k

RL’

667

RS

vin

+

-

50k

+vo

--

+vi

--

rπ

β ib

ie= (1+β)ibib

Zin Zit

10k

RS

vin

+

-+vi

--

Zin


288


6.461

'||)'()1(1

1

rRRrRR

Z SE

SE

o

RB10k

RE

2k

RS

50k

rπ

β ib

ie= (1+β)ibib

ZO +vx

--

ix

RS’

))'()1(1

1(

')1(

)'(33.8||'

)1(

rRRiv

rRv

Rvi

rRivkRRR

iRvi

Rvi

ivZ

SE

xx

S

x

E

xx

Sbx

BSS

b

E

xe

E

xx

x

xo


289

The BJT as a Digital Switch

VCC

RC

RB

LoadIo

Vo

+

-

0.E+00

1.E-03

2.E-03

3.E-03

4.E-03

5.E-03

6.E-03

7.E-03

0 1 2 3 4 5 6 7 8 9 10

v CE volts

i C amps

Vin

+

-

in

In Saturation, if

.2

Then, .2

BEQo CE CC C C CC C B CC C

B

o CE

V VV V V R i V R i V R V

RV V

CCo VVVV

Then,7.

if Cutoff,In

in

• Operating between cutoff and saturation (i.e., bypassing the active region), the BJT acts like an inverter.

• From this behavior, logic circuits such as NOR gates can be developed

0 μa

iB=50 μa

40 μa

30 μa

10 μa

20 μa


290

The BJT as a Digital Switch

CCo VVVV

Then,7.

if Cutoff,In

in

VCC= 3

RC

RB

LoadIo

Vo

+

-Vin

+

-

7.77.2

2.35

2.5

7.23

2.

if ,SaturationIn

in

in

in

V

VV

VR

VVRVV

o

B

BEQCCCo

2k

5k

β =10

Vo

Vin1.4

VCC=3

VCC=30.7

0.2

=10

=100


291

Switching and Timing

VOL

VOL

VOH

VOH

(VOH + VOL)/2

(VOH + VOL)/2

50%

10%

90%

100%

0%

trtf

td tPLH

vi

vo

tr = rise time

tf = fall time

td = delay time (propagation delay from High to Low)

ts = storage time (propagation delay from Low to High)


292

Homework

• Probs. 4.40, 4.42, 4.43, 4.45, 4.46, 4.51, 4.53, 4.54, 4.56,

Lesson #10 Chapter 4 - New Jersey Institute of Technologyjoelsd/electronics/classwork/electronics1/BME373... · BME 372 Electronics I – J.Schesser 274 Small Signal Equivalent Circuits

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