ESE319 Introduction to Microelectronics

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

2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Basic Single BJT Amplifier Features CE Amplifier Voltage Gain (AV) moderate (-RC/RE)

CC Amplifier low (about 1)

Current Gain (AI)

moderate ( 1)

moderate (  )

Input Resistance

high

high

Output Resistance

high

low

VCVS

CB Amplifier high low (about 1) low

CCCS

high

CE BJT amplifier => CS MOS amplifier CC BJT amplifier => CD MOS amplifier CB BJT amplifier => CG MOS amplifier 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Common Collector ( Emitter Follower) Amplifier R1

ro

C in

V CC

RS vs

vs

vO R2

RE

Voltage Bias Design

Q1

Qamp

Q2

vO

ro

Current Bias Design

In the emitter follower, the output voltage is taken between emitter and ground. The voltage gain of this amplifier is nearly one – the output “follows” the input - hence the name: emitter “follower.” 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Equivalent Circuits

C in RE

RS

vout

vO



vs

R B =R1∥R2 RB Rb VB

V CC vout vO

RE

VCC/2

2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Multisim Bias Check V Rb =I B R B =

Cin

IE R =0.495V 1 B

Cin

iB -

vout



VRb +

Rb

vout

Identical results – as expected! 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Follower Small Signal Analysis - Voltage Gain Circuit analysis: v s =R S r  ib R E i e = R S r   1 R E i b

RS ib

vs

ie

vout v o=v e

RE

Solving for ib vs i b= R S r  1 R E v o=R E i e =R E 1i b R E 1v s v o= R S r  1 R E

for Current Bias Design replace RE with ro||ro = ro/2 >> RE 2009 Kenneth R. Laker, updated 05Oct12 KRL

vo R E r o∥r o AV = = ≈1 v s R S r  R E r o∥r o 1 6

ESE319 Introduction to Microelectronics

Small Signal Analysis – Voltage Gain - cont. vo RE = v s R S r  R E 1 Since, typically: R S r  1

≪ R E (or ro||ro = ro/2)

vo R E AV = ≈ =1 vs RE Note: AV is non-inverting 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Quick Review CE Amplifier

CC Amplifier

CB Amplifier

Voltage Gain (AV) Current Gain (AI) Input Resistance Output Resistance

ANSWERS: Low, Moderate or High

2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Quick Review cont. CE Amplifier Voltage Gain (AV) moderate (-RC/RE) Current Gain (AI)

moderate (β)

Input Resistance

high (RB||βRE)

Output Resistance

high (RC||ro)

2009 Kenneth R. Laker, updated 05Oct12 KRL

CC Amplifier

CB Amplifier

low (about 1)

high (RC/RS)

moderate (β + 1)

low (about 1)

high (RB||βRE)

low (re)

low (re) VCVS

high (RC||ro) CCCS

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ESE319 Introduction to Microelectronics

Of What value is a Unity Gain Amplifier? RS ib

vs

ie

vo

To answer this question, we must examine the small-signal output impedance of the amplifier and its power gain.

RE

2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Emitter Follower Output Resistance −i x i x =−i e =−1i b ⇒ i b= 1

RS

ib vs

0

ie

ix

vx Rout

Assume:

R S r  v x =−i b  R S r  = i x where r  ≫ R S 1 r v x R S r  1 VT R out = = ≈ =r e= = ix gm IC 1 1

VT VT I C =1 mA ⇒ r = = =2500  IB IC

=100 R S =50 

2009 Kenneth R. Laker, updated 05Oct12 KRL

R out ≈r e =

2550 =25.5  100

Recall Rin =r bg =r  1 R E

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ESE319 Introduction to Microelectronics

Multisim Verification of Rout RS

Rout

vs Rin

Av*vsig AV = 1

i x=−1i b

i sc =i x



Rin =r bg =R S r 1 R E ≈1 R E Rin =R S r 1 R E ≈ 1 R E

Thevenin equivalent for the short-circuited emitter follower. If β = 200, as for most good NPN transistors, Rout would be lower - close to 12 Ω. 2009 Kenneth R. Laker, updated 05Oct12 KRL

v oc R out = i sc vvsigs = =R RSSiibbr ri bi b

vs +

v ocv=ocA=v V v ss -

Rout i sc =i x

vAs v RS r R S r  V=sig RRoutout= =i = 1 sc i x 1

=100

Multisim short circuit check (  = 100, vo = vs): v oc Av s rms 1 V1   v v oc V s rms = = = == =22.25  RRoutout= =25.25 39.61 mA iiscsc i iscscrms 0.0396 rms  12

ESE319 Introduction to Microelectronics

Emitter Follower Power Gain R S C in ib

+

ib

Rb ie v bg vbg RB

vs

-

Rin

vo RE

Consider the case where a RL=50 Ω load is connected through an infinite capacitor to the emitter of the follower. Using its Thevenin equivalent:

RL

r bg

Rin

@ midband where Av = 1

R L AV v s 50 2 vo= = v= v R L∥R E R out 75 s 3 s AV v s vs i o= = R out R L∥R E 75 2 2 p o =v o i o = vs 225 2009 Kenneth R. Laker, updated 05Oct12 KRL

is

vs

R out≈ 25

i

Rin

o + - A vth=G v v s vsig

C=∞ +

vo

R E∥R L=50 

-

50  load is in parallel with 5.1k  vs vs vs vs REb = and ≈ dominates: ≈ i s =i ≈ R in

1 R L∥R E 101⋅50 R E∥R L =5.1 k ∥50 ≈50  1 2 p s =v s i s ≈ vs 5000 p o 25000 A pwr = = =44.4≫1 ps 225

5000

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ESE319 Introduction to Microelectronics

Emitter Follower Biasing – Typical Design R1 C in RS vs

VB Vb iB

R2

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

iC

iE

vO =v E vout

RE

V CC

V E V CC /2−0.7 RE= = IE IE

For an assumed  = 100: Split bias voltage drops about equally across the transistor VCE (or VCB) and VRe (or VB). For simplicity,choose: V CC ⇒ R 1=R 2 V B= 2 2009 Kenneth R. Laker, updated 05Oct12 KRL

As with CE bias design, stable op. pt. => R B ≪1 R E , i.e. R 1 1 R B =R 1∥R 2= = R E ≈10 R E 2 10 R 1=R 2 =20 R E 14

U

ESE319 Introduction to Microelectronics

Typical Design - Cont. Given: R out =r e =25  V CC =12 V And the rest of the design follows: VT I E ≈ I C = =1 mA re

V E 12/ 2−0.7 RE= = =5.3 k  −3 IE 10 Use standard sizes

100 kΩ

C in v O =v E

12V

100 kΩ 5.1 kΩ

R E =5.1 k  R1 =R 2=100 k  2009 Kenneth R. Laker, updated 05Oct12 KRL

15

ESE319 Introduction to Microelectronics R S =50  R S C in ib

Blocking Capacitor - Cin - Selection Rb ie vbg

vs

v bg R B -

Rin

ib

+

Rin

Use the base current expression: v bg =r  i b R E i E =r  1 R E  ib

vo RE

r bg

v bg i b= r  1 R E

v bg r bg = =r  1 R E ≈1 R E =101⋅5.1 k =515 k  ib

To obtain the base to ground resistance of the transistor: This transistor input resistance is in parallel with RB = 50 k , forming the total amplifier input resistance: 515 Rin= R S  R B∥r bg ≈ R B∥r bg = 50 k =45.6 k ≈ R B =50 k  51550 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Cin – Selection cont. Choose Cin such that its reactance is ≤ 1/10 of RB at fmin: RB 1 Assume fmin = 20 Hz = 2 f C in 10 with R B =50 k  10 C in ≥ 2  f min R B 10 C in ≥ =1.59 F 3 2 ⋅20⋅50⋅10

Pick Cin = 3.3 µ F ), the nearest standard value in the Detkin Lab. We could be (unnecessarily) more precise and include rbg and Rs as part of the total resistance in the loop. 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Final Design

C in

R1

3.3 uF

RS vs

2009 Kenneth R. Laker, updated 05Oct12 KRL

V CC R2

vo RE

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ESE319 Introduction to Microelectronics

Multisim Simulation Results

20 Hz Data

Av = 0.995 1 kHz Data 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

The Common Base Amplifier

Voltage Bias Design

2009 Kenneth R. Laker, updated 05Oct12 KRL

Current Bias Design

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ESE319 Introduction to Microelectronics

Common Base Configuration Both voltage and current biasing follow the same rules as those applied to the common emitter amplifier. As before, insert a blocking capacitor in the input signal path to avoid disturbing the dc bias. The common base amplifier uses a bypass capacitor – or a direct connection from base to ground to hold the base at ground for the signal only! RECALL: The common emitter amplifier (except for intentional RE feedback) holds the emitter at signal ground, while the common collector circuit does the same for the collector. 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Voltage Bias Common Base Design CE Amp

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. 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Mid-band Small Signal Analysis Input Impedance

RC ro

is

Zin

C in =∞

Note: is = - ie RE >>re

2009 Kenneth R. Laker, updated 05Oct12 KRL

v Re =r e∥R E i s v Re VT Z in = =r e∥R E ≈r e = is IC vo Z out = =RC∥r o ic

Current Gain ic Ai = =≈1 ie Voltage Gain v s =−i e R S −r e∥R E i e

RC 1 v o =−RC i c =− RC i e = vs  R S r e∥R E v o 1 RC Av = ≈ v s  R S r e

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ESE319 Introduction to Microelectronics

Common Base Small Signal Analysis - Cin ib

vo

RC

Determine Cin: (let C b =∞ )

4.7 k Ohm

ic

∞ Cb R B re

25 Ohm

NOTE: RB is shorted

v Re

ie RS C in RE 470 Ohm

v Re =

R E∥r e R S  vs

by Cb = ∞ 1 2 f min C in

R E∥r e

re ≪ R S R E∥r e ⇒

2009 Kenneth R. Laker, updated 05Oct12 KRL

1 j2  f C in

R E∥r e ideally v Re = R ∥r R v s E e S

vs

r re= 1

for f ≥ f min

R S r e 10 = ⇒C in = 10 2  f min C in 2  f min R S r e  1

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ESE319 Introduction to Microelectronics

Determine CIN cont. A suitable value for Cin for a 20 Hz fmin with re = 25 Ω and RS = 50 Ω:

10 10 2 f min C in  R S r e ≫1⇒ C in ≥ = F 2 f min  R S r e  2  20⋅75 10 C in = ≈1062  F ! 125.6⋅75

Not Practical!

Must choose smaller value of Cin. Choose: 2 f min C in R S r e =1 1 C in = ≈106.2  F 125.6⋅75 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

ignore RB

Small-signal Analysis - Cb Determine Cb: (let C in =∞ ) Note the ac reference current reversals (due to vs polarity)! 1  j 2 f C b 1 v Re = vs 1 R E∥r e  R S j 2 f C b 1 R E∥r e 

R E∥r e ideally v Re = R ∥r R v s for f ≥ f min E e S re 1 1 10 ≪ re ⇒ = ⇒ C b= 2 f min C b 1 2  f min C b 1 10 2  f min r e 1 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Determine - CB cont. i'b

i'c vout

vo

R E∥r e≈r e

is

R

4.7 Ck Ohm ib'

i'e

vs

ignore RB

2009 Kenneth R. Laker, updated 05Oct12 KRL

Choose (conservatively): 10 C b= F 2  f min  1r e  for fmin = 20 Hz i.e.

10 C b= =31.8  F 2  20  10025 

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ESE319 Introduction to Microelectronics

Multisim Simulation

4.7 k Ohm

vO

31.8 uF 1060 uF

470 Ohm

vs

v o 1 RC RC 4700 AV = = ≈ = =62.7 v s  R S r e R S r e 5025 2009 Kenneth R. Laker, updated 05Oct12 KRL

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ESE319 Introduction to Microelectronics

Multisim Frequency Response

20 Hz response

vertical axis is a linear scale

1 %

1 kHz Response 2009 Kenneth R. Laker, updated 05Oct12 KRL

Av sim=63.3A v theory =62.7 29