Lec 9 CCandCBDesigns 08

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL

Common Base BJT AmplifierCommon Collector BJT Amplifier

Common Collector (Emitter Follower) Configuration Common Base Configuration Small Signal Analysis Design Example Amplifier Input and Output Impedances



Basic Single BJT Amplifier Features

CE Amplifier CC Amplifier CB AmplifierVoltage Gain (AV) moderate (-RC/RE) low (about 1) high

Current Gain (AI) moderate ( ) moderate ( ) low (about 1)

Input Resistance high high low

Output Resistance high low high

1

CE BJT amplifier => CS MOS amplifierCC BJT amplifier => CD MOS amplifierCB BJT amplifier => CG MOS amplifier



Common Collector ( Emitter Follower) Amplifier

In the emitter follower, the output voltage is taken be-tween emitter and ground. The voltage gain of this ampli-fier is nearly one the output follows the input - hence the name: emitter follower.

vout



Split bias voltage drops aboutequally across the transistorV

CE (or V

CB) and VRe (or VB).

For simplicity,choose:

R1=R2

For an assumed = 100:

RB=R1R2=R12=1

RE1010 RE

R1=R2=20 REVB=VCC

2

RE=VEI E

=VCC /20.7

I E

Then, choose/specified IE, and the rest of the design follows:

Vb

iB

iC

iE

vout

As with CE bias design, stable op. pt. => RB1RE , i.e.

Emitter Follower Biasing



Typical DesignChoose: I E=1mA

VCC=12VAnd the rest of the designfollows immediately:

RE=VEI E

=12 /20.7103

=5.3k

Use standard sizes!

R1=R2=100 k

RE=5.1k

vout

Vb

iB

iC

iE



Equivalent Circuits

vout

vout

VCC

/2

Rb

RB=R1R2



Multisim Bias Check

Identical results as expected!

Rb

+

-VRb

V Rb= I B RB=I E

1RB=0.495V

iB

Rbvout vout



Small signal mid-band circuit - where CB has negligible reactance(above min). Thevenin circuit consisting of RS and RB shows effect of RB negligible, since it is much larger than RS.

Emitter Follower Small Signal Circuit

Mid-band equivalent circuit:

vsig '=RB

RBRSvsig=

5050.05

vsigvsig

RTH=RSRB=50

50.05RSRS

Rb

vout



Follower Small Signal Analysis - Voltage GainCircuit analysis:

ib=vsig

RSr1RE

vout=RE 1vsig

RSr1RE

AV=voutvsig

=RE vsig

RSr 1

RE

1

vsig=RSr1RE ib

vout

ib

ie

Solving for ib



Small Signal Analysis Voltage Gain - cont.voutvsig

=RE

RSr 1

RE

Since, typically:

RSr1

RE

AV=voutvsig

RERE

=1

Note: AV is non-inverting

vout



ib=vbg

r1RE

Use the base current expression:

To obtain the base to ground resistance of the transistor:This transistor input resistance is in parallel with the 50 k RB, forming the total amplifier input resistance:

Ri n=RSRBr bgRBr bg=515

5155050 k=45.6 k

vbg=r ibRE iE=r1 ibvbg

+

-

Rin r bg=

vbgib

=r1RE1RE=1015.1 k=515k

Rb

RB=50 kRSRS=50

Blocking Capacitor - CB - Selection

ib



CB Selection cont.

Ri n46 k=100

min=220125=1.25102

Assume the lowest frequencyis 20 Hz:

C B10107

1.250.461.73 F

Choose CB such that its reactance is 1/10 of Rin at min:

C B10

min Ri n

1C B

=Ri n10

Pick CB = 2 F (two 1 F caps in parallel), the nearest standard value in the RCA Lab. We could be (unnecessarily) more preciseand include Rs as part of the total resistance in the loop. It is verysmall compared to Rin.



Final Design

2.0 uF

vout



Multisim Simulation Results

20 Hz. Data

1 Khz. Data



Of What value is a Unity Gain Amplifier?

To answer this question,we must examine theoutput impedance of theamplifier and its powergain.

vout

ib

ie



Emitter Follower Output Resistance

vx

ix

Rout

0ib

i x=ib ib=1 ib ib=ix1

vx=ibRsr=RSr1

i x

Rout=vxi x=

RSr1

r

1

Assume:I C=1 mA r=

VTI B

=VTI C

=2500

=100 RS=50

Rout2550100

=25.5

vout

RB=50 kRSR

out is the Thevenin resistance looking

into the open-circuit output.



Multisim Verification of Rout

Multisim short circuit check( = 100, vout = vsig):

Rout=voci sc

=AV vsig rms

isc rms= 1

0.0396=25.25

Thevenin equivalent for the short-circuited emitter follower. If was 200, as for most good NPN transistors, Rout would be lower - close to 12 .

Rout

Av*vsigAV = 1

i sc=i x

i sc=i x

Rin

i x=1i xvsig=RS ibr ib

Rout=AV vsig

i x=

RSr1Rin=RSr1RERL1RL

+

-voc=AV vsig



Equivalent Circuits with Load RL

vout

ib

ie

RL

+-

RL||Revload+

-

Rout=vsig rms i sc rms

= 10.0396

=25.25

RinZin=

vsigie

'

Rin=RSr1RERL1RL

Rout

Av*vsig



Emitter Follower Power GainConsider the case where a R

L = 50 load is connected through an infinite

capacitor to the emitter of the follower we designed. Using its Thevenin equivalent:

vload=RL AV vsigRLRout

=5075

vsig=23

vsig

i load=AV vsig

RoutRL=

vsig75

pload=vload i load=2

225vsig

2

i sig=ib=vsigRin

vsig1RERL

vsig

10150

psig=vsig isig1

5000vsig

2

+-

vload

-vth=G vsig

Rout25

RL50+

50 load is in parallel with 5.1k RE and dominates:

C=

AV1iloadRinvsig Av*vsig

G pwr=ploadpsig

= 25000225

=44.41

isig



The Common Base Amplifier

Voltage Bias Design Current Bias Design

vout vout



Common Base ConfigurationBoth voltage and current biasing follow the same rules asthose applied to the common emitter amplifier.

As before, insert a blocking capacitor in the input signal pathto avoid disturbing the dc bias.

The common base amplifier uses a bypass capacitor or adirect connection from base to ground to hold the base atground for the signal only!

The common emitter amplifier (except for intentional REfeedback) holds the emitter at signal ground, while the commoncollector circuit does the same for the collector.



We keep the same bias that we established for the gain of 10 common emitter amplifier.All that we need to do is pick the capacitor values and calculate the circuit gain.

vout

470 Ohm

4.7 k Ohm47k Ohm

5.1k Ohm

Voltage Bias Common Base Design



Common Base Small Signal Analysis - C INDetermine C

IN:

Find a equivalent impedance for the input circuit, RS, CIN, and RE2:

i'b

I'ci'e

isig

Zin

4.7 k Ohm

470 Ohm

ideally for min

1minC IN

RSRE2r e1

minC IN=

RSr e10

C IN=10

minRSr e

vRe2=RE2r e

RE2r eRS1

jC IN

vsig

vRe2=RE2r e

RE2r eRSvsig

(let ) C B=

ib

ic

ie

r e=r

1

re



Determine C IN cont.

minC IN RSr e1C IN10

2minRSr e=

1022075

F

A suitable value for CIN for a 20 Hz lower frequency:

C IN=10

125.6751062F ! Not too practical!

Must choose smaller value of CIN.1. Choose: minC IN RSr e=1

or2. Choose larger min



Small-signal Analysis - C Bi'b i'c

i'e

Note the reference currentreversals (due to v

sig polarity)!

vsig=RS ie'r 1jC B ib'

vsig=RS ie'r 1jC B ie

'

1

ie' =

1

1RSr 1

jC B

vsigZin=vsigie

'

ic

ie

ib

Determine

Zin

Determine CB: (let )C IN=

RE2

>> RS

ib'



Determine CBi'b i'c

i'e

ie' =

1

1RSr 1

jC B

vsig

ie' =

vsig

RS11 r 1jC B

Z in=vsigie

' =RS11 r 1jC B

ideally Z inRSr 1

1C B

1RSr for min

Zin

RE2

>> RS

ib'



Determine - C B cont.i'b i'c

i'e Choose:

C B10

min 1RSr F

vout

Z inRSr 1

1C B

1RSr

For min

C B10

220 1001502500 =10.5 F

i.e.

RE2

>> RS

ib'



Small-signal Analysis Voltage Gain

ie'

1

RSr 1

vsig=1

RSr e vsig

vout=RC ic' =RC ie

' =1

RCRSr e

vsig

AV=voutvsig

=1

RCRSr e

=100101

51005025

67

Assume: C B=C IN=

RE2

>> RS

ib'



Multisim Simulation

1062 uF



Multisim Frequency Response

20 Hz. response

1 KHZ. Response

Documents

Lec 9 CCandCBDesigns 08