Common Base BJT Amplifier Common Collector BJT Amplifierese319/Lecture_Notes/Lec_9_CCandCBDesign… · Common Base BJT Amplifier Common Collector BJT Amplifier ... Design Example

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 (-R

C/R

E) 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=R1∥R2=R1

2=1

RE

10≈10 RE

R1=R2=20 REVB=VCC

2

RE=VE

I E=

VCC /2−0.7I 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. => RB≪1RE , i.e.

Emitter Follower Biasing



Typical DesignChoose: I E=1mA

VCC=12VAnd the rest of the designfollows immediately:

RE=VE

I E=12 /2−0.7

10−3 =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=R1∥R2



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

vsig≈vsig

RTH=RS∥RB=50

50.05RS≈RS

Rb

vout



Follower Small Signal Analysis - Voltage GainCircuit analysis:

ib=vsig

RSr1RE

vout=RE 1vsig

RSr1RE

AV=vout

vsig=

RE vsig

RSr

1RE

≈1

vsig=RSr1RE ib

vout

ib

ie

Solving for ib



Small Signal Analysis – Voltage Gain - cont.vout

vsig=

RE

RSr

1RE

Since, typically:

RSr

1≪RE

AV=vout

vsig≈

RE

RE=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=RSRB∥r bg≈RB∥r bg=515

5155050 k=45.6 k

vbg=r ibRE iE=r1 ibvbg

+

-

Rin r bg=

vbg

ib=r1RE≈1RE=101⋅5.1 k=515k

Rb

RB=50 k≫RS

RS=50

Blocking Capacitor - CB - Selection

ib



CB – Selection cont.

Ri n≈46 k=100

min=220≈125=1.25⋅102

Assume the lowest frequencyis 20 Hz:

C B≥10⋅10−7

1.25⋅0.46≈1.73 F

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

C B≥10

min Ri n

1C B

=Ri n

10

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=−ix

1

vx=−ibRsr=RSr

1i x

Rout=vx

i x=

RSr

1≈

r

1

Assume:I C=1 mA⇒ r=

VT

I B=

VT

I C=2500

=100 RS=50

Rout≈2550100

=25.5

vout

RB=50 k≫RSR

out is the Thevenin resistance looking

into the open-circuit output.



Multisim Verification of Rout

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

Rout=voc

i 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 x

vsig=RS ibr ib

Rout=AV vsig

i x=

RSr

1Rin=RSr1RE∥RL≈1RL

+

-voc=AV vsig



Equivalent Circuits with Load RL

vout

ib

ie

RL

+-

RL||Revload+

-

Rout=vsig rms

i sc rms= 1

0.0396=25.25

<=> RinZin=

vsig

ie'

Rin=RSr1RE∥RL≈1RL

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 vsig

RLRout=

5075

vsig=23

vsig

i load=AV vsig

RoutRL=

vsig

75

pload=vload i load=2

225vsig

2

i sig=ib=vsig

Rin≈

vsig

1RE∥RL≈

vsig

101⋅50

psig=vsig isig≈1

5000vsig

2

+-

vload

-vth=G vsig

Rout≈25

RL≈50+

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

C=∞

AV≈1iloadRinvsig

Av*vsig

G pwr=pload

psig= 25000

225=44.4≫1

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 IN

Determine CIN

:

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

≪RSRE2∥r e⇒1

minC IN=

RSr e

10⇒C IN=

10minRSr e

vRe2=RE2∥r e

RE2∥r eRS1

jC IN

vsig

vRe2=RE2∥r e

RE2∥r eRSvsig

(let ) C B=∞∞

ib

ic

ie

r e=r

1

re



Determine C IN cont.

minC IN RSr e≫1⇒C IN≥10

2minRSr e=

10220⋅75

F

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

C IN=10

125.6⋅75≈1062F ! 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=vsig

ie'

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

RS1

1 r1

jC B Z in=

vsig

ie' =RS

11 r

1jC B

ideally Z in≈RSr

1⇒ 1C B

≪1RSr for ≥min

Zin

RE2

>> RS

ib'



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

i'e Choose:

C B≥10

min 1RSr F

vout

Z in≈RSr

1⇒ 1C B

≪1RSr

For ≥min

C B≥10

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

' =

1RC

RSr evsig

AV=vout

vsig=

1

RC

RSr e=100

1015100

5025≈67

∞

Assume: C B=C IN=∞

RE2

>> RS

ib'



Multisim Simulation

1062 uF



Multisim Frequency Response

20 Hz. response

1 KHZ. Response

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Common Base BJT Amplifier Common Collector BJT Amplifierese319/Lecture_Notes/Lec_9_CCandCBDesign… · Common Base BJT Amplifier Common Collector BJT Amplifier ... Design Example