ESE319 Introduction to Microelectronics

Common Base BJT BJT Amplifier

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

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 1 ESE319 Introduction to Microelectronics Basic Single BJT Amplifier Features

CE Amplifier CC Amplifier CB Amplifier Voltage (A ) moderate (-R /R ) low (about 1) high V C E

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

Input Resistance high high low

Output Resistance high low high CE BJT amplifier => CS MOS amplifier CC BJT amplifier => CD MOS amplifier CB BJT amplifier => CG MOS amplifier

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 2 ESE319 Introduction to Microelectronics

Common Collector ( Emitter Follower) Amplifier

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

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 3 ESE319 Introduction to Microelectronics

Emitter Follower

Then, choose/specified I , and i E C the rest of the design follows: Vb

i vout B V E VCC /2−0.7 i E RE= = I E I E

For an assumed  = 100: Split bias voltage drops about As with CE bias design, stable op. equally across the pt. => RB≪1 RE , i.e. V (or V ) and V (or V ). CE CB Re B R R For simplicity,choose: R =R ∥R = 1 =1 E ≈10 R B 1 2 2 10 E V CC R R V B= ⇒ 1= 2 R =R =20 R 2 1 2 E 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 4 ESE319 Introduction to Microelectronics

Typical Design

Choose: I E=1mA i VCC=12V C And the rest of the design Vb i follows immediately: B vout i V 12 2 0.7 E R E / − 5.3k E= = −3 =  I E 10 Use standard sizes!

RE=5.1k 

R1=R2=100 k 

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 5 ESE319 Introduction to Microelectronics

Equivalent Circuits

RB=R1∥R2

Rb vout vout <=>

V /2 CC

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 6 ESE319 Introduction to Microelectronics

Multisim Bias Check

I V =I R = E R =0.495V Rb B B 1 B

i B - vout Rb vout VRb <=> +

Identical results – as expected!

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 7 ESE319 Introduction to Microelectronics

Emitter Follower Small Signal Circuit

Mid-band equivalent circuit:

RB 50 vsig ' = vsig = vsig≈vsig vout RBRS 50.05 Rb 50 R =R ∥R = R ≈R TH S B 50.05 S S

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.

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 8 ESE319 Introduction to Microelectronics Follower Small Signal Analysis - Voltage Gain Circuit analysis:

vsig = RSr 1REib

i Solving for i b b vsig i = i vout b e RSr 1 RE

RE 1vsig vout= RSr 1 RE

vout RE vsig AV= = ≈1 vsig RSr  R 1 E

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 9 ESE319 Introduction to Microelectronics

Small Signal Analysis – Voltage Gain - cont.

v R out = E vsig RSr  R 1 E

vout Since, typically:

RSr  ≪R 1 E

vout RE AV= ≈ =1 vsig RE

Note: AV is non-inverting 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 10 ESE319 Introduction to Microelectronics Blocking Capacitor - C - Selection B i + b Use the base current expression:

vbg=r  ibRE iE =r 1ib vbgRb vbg ib= - r 1 RE R 50 k R R B= ≫ S vbg in r r 1 R 1 R 101 5.1 k 515k RS=50 bg= =   E≈  E= ⋅ =  ib 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: 515 R =R R ∥r ≈R ∥r = 50 k =45.6 k i n S B bg B bg 51550

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 11 ESE319 Introduction to Microelectronics C – Selection cont. B

Choose CB such that its reactance is ≤ 1/10 of Rin at  min: 1 R = i n Assume the lowest frequency C B 10 is 20 Hz: 10 2 min=220≈125=1.25⋅10 C B ≥ min Ri n =100 10⋅10−7 C B ≥ ≈1.73 F R ≈46 k 1.25⋅0.46 i n

Pick CB = 2  F (two 1  F caps in parallel), the nearest standard value in the RCA Lab. We could be (unnecessarily) more precise

and include Rs as part of the total resistance in the loop. It is very

small compared to Rin. 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 12 ESE319 Introduction to Microelectronics

Final Design

2.0 uF

vout

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 13 ESE319 Introduction to Microelectronics

Multisim Simulation Results

20 Hz. Data

1 Khz. Data

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 14 ESE319 Introduction to Microelectronics

Of What value is a Unity Gain Amplifier?

To answer this question, i b we must examine the of the

i vout amplifier and its power e gain.

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 15 ESE319 Introduction to Microelectronics

Emitter Follower Output Resistance

−ix i i x=−ib−ib=−1ib ⇒ib= b 1 0 vout RSr  vx=−ib Rsr = i x i  1 x v x v R r r R x S   out Rout= = ≈ i x 1 1 R is the Thevenin resistance looking out RB=50k ≫RS into the open-circuit output. Assume: V V 2550 T T R 25.5 I C =1 mA⇒r = = =2500 out≈ =  I B I C 100 =100 RS=50

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 16 ESE319 Introduction to Microelectronics Multisim Verification of R out Rout

i x=−1i x i sc=i x Av*vsig vsig=RS ibr  ib A = 1 Rin V <=> AV vsig RSr  + Rout= = i x 1 Rin=RSr 1 RE∥RL≈1 RL v A v oc= V sig i i - sc= x Thevenin equivalent for the short-circuited emitter follower. Multisim short circuit check If  was 200, as for most good (  = 100, v = v ): NPN , R would be out sig out v A v 1 R oc V sig rms 25.25 lower - close to 12  . out= = = =  i sc iscrms 0.0396

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 17 ESE319 Introduction to Microelectronics

Equivalent Circuits with Load R L

i b Rout

vout vsig Zin= ' ie + i Rin + Av*vsig vload RL||Re e <=> - RL -

vsig rms 1 Rout= = =25.25 i scrms 0.0396 Rin=RSr 1 RE∥RL≈1 RL

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 18 ESE319 Introduction to Microelectronics Emitter Follower Power Gain Consider the case where a R = 50 load is connected through an infinite L  capacitor to the emitter of the follower we designed. Using its Thevenin

equivalent: isig Rout≈25 C=∞ + A ≈1 + iload R 50 V vsig Rin vload L≈  - Av*vsigvth=G vsig -

R A v 50 2 50  load is in parallel with 5.1k  v = L V sig = v = v load sig sig RE and dominates: RLRout 75 3 v v v i i sig sig sig A v v sig= b= ≈ ≈ V sig sig Rin 1 RE∥RL 101⋅50 iload = = RoutRL 75 1 2 psig =vsig isig ≈ vsig 2 2 5000 pload =vload iload = vsig 225 pload 25000 G pwr = = =44.4≫1 psig 225

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 19 ESE319 Introduction to Microelectronics

The Common Base Amplifier

vout vout

Voltage Bias Design Current Bias Design

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 20 ESE319 Introduction to Microelectronics Common Base Configuration Both voltage and current biasing follow the same rules as those applied to the 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!

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.

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 21 ESE319 Introduction to Microelectronics

Voltage Bias Common Base Design

4.7 k Ohm 47k Ohm vout

5.1k Ohm 470 Ohm

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.

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 22 ESE319 Introduction to Microelectronics

Common Base Small Signal Analysis - C IN i' i b b 4.7 k Ohm Determine C :(let C = ∞ ) IN B ∞ i c Find a equivalent impedance for I'c i e the input circuit, R , C , and R : i'e S IN E2 RE2∥r e r v v  Re2= sig r = i 1 e 1 sig RE2∥r eRS  470 Ohm j C IN

Z RE2∥r e in ideally vRe2= vsig for  ≥ min RE2∥r eRS r e 1 1 RSr e 10 ≪ RSRE2∥r e ⇒ = ⇒C IN = minC IN minC IN 10 min RSr e

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 23 ESE319 Introduction to Microelectronics

Determine C IN cont. A suitable value for C for a 20 Hz lower frequency: IN 10 10 minC IN RSr e≫1⇒C IN≥ = F 2minRSr e  220⋅75

10 C = ≈1062 F ! Not too practical! IN 125.6⋅75

Must choose smaller value of CIN.

1. Choose: minC IN RSr e=1 or

2. Choose larger min

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 24 ESE319 Introduction to Microelectronics

Small-signal Analysis - C B i' b i'c Determine C : (let C = ∞ ) B IN Note the reference current i b i reversals (due to v polarity)! ib' c sig

' 1 ' vsig=RS ie r  ib j C B i'e   R >> R E2 S ' i i e ' 1 e vsig=RS ie r  j C B 1 Z   in v ' 1 sig ie= vsig Determine Zin= ' 1 ie 1 RSr  j C B

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 25 ESE319 Introduction to Microelectronics

Determine – CB i' b i'c ' 1 ie= vsig ib' 1 1 RSr  j C B

i'e v i' = sig R >> R e E2 S 1 1 RS r  1 j C B Z   in v 1 1 Z = sig =R  r  in ' S 1  j C ie  B  r 1 ideally  Zin≈RS ⇒ ≪1 RSr  for  ≥ min 1 C B

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 26 ESE319 Introduction to Microelectronics

Determine - C B cont. i' b i'c vout For  ≥ min r  1 ib' Zin≈RS ⇒ ≪1RSr  1 C B

Choose: i'e R >> R 10 E2 S C ≥ F B 1 R r min   S 

i.e. 10 C ≥ =10.5 F B 220 1001502500 

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 27 ESE319 Introduction to Microelectronics

Small-signal Analysis – Voltage Gain

' 1 1 ∞ ib' i ≈ v = v e r sig sig   RSr e  R  S 1 R R >> R ' '  C E2 S vout =RC ic= RC ie= vsig 1 RSr e v R C C out  C 100 5100 Assume: B= IN =∞ AV= = = ≈67 vsig 1 RSr e 101 5025

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 28 ESE319 Introduction to Microelectronics

Multisim Simulation

1062 uF

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 29 ESE319 Introduction to Microelectronics

Multisim Frequency Response

20 Hz. response

1 KHZ. Response

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL 30