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
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 Gain (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 Biasing
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 transistor 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 RBRS 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 = RSr 1REib
i Solving for i b b vsig i = i vout b e RSr 1 RE
RE 1vsig vout= RSr 1 RE
vout RE vsig AV= = ≈1 vsig RSr 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 RSr R 1 E
vout Since, typically:
RSr ≪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 ibRE iE =r 1ib 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 51550
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=220≈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 output impedance 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=−1ib ⇒ib= b 1 0 vout RSr vx=−ib Rsr = 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=−1i x i sc=i x Av*vsig vsig=RS ibr ib A = 1 Rin V <=> AV vsig RSr + Rout= = i x 1 Rin=RSr 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 transistors, R would be out sig out v A v 1 R oc V sig rms 25.25 lower - close to 12 . out= = = = i sc iscrms 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 scrms 0.0396 Rin=RSr 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: RLRout 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 = = RoutRL 75 1 2 psig =vsig isig ≈ vsig 2 2 5000 pload =vload iload = vsig 225 pload 25000 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 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!
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 eRS 470 Ohm j C IN
Z RE2∥r e in ideally vRe2= vsig for ≥ min RE2∥r eRS r e 1 1 RSr e 10 ≪ RSRE2∥r e ⇒ = ⇒C IN = minC IN minC IN 10 min RSr 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 RSr e≫1⇒C IN≥ = F 2minRSr e 220⋅75
10 C = ≈1062 F ! Not too practical! IN 125.6⋅75
Must choose smaller value of CIN.
1. Choose: minC IN RSr 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 RSr 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 RSr 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 RSr 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 ⇒ ≪1RSr 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 220 1001502500
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 RSr e R S 1 R R >> R ' ' C E2 S vout =RC ic= RC ie= vsig 1 RSr e v R C C out C 100 5100 Assume: B= IN =∞ AV= = = ≈67 vsig 1 RSr e 101 5025
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