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
ESE319 Introduction to Microelectronics
22008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
32008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
42008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
52008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
62008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
Equivalent Circuits
<=>vout
vout
VCC
/2
Rb
RB=R1∥R2
ESE319 Introduction to Microelectronics
72008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
Multisim Bias Check
Identical results – as expected!
<=>Rb
+
-VRb
V Rb= I B RB=I E
1RB=0.495V
iB
Rbvout vout
ESE319 Introduction to Microelectronics
82008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
102008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
112008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
122008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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.
ESE319 Introduction to Microelectronics
132008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
Final Design
2.0 uF
vout
ESE319 Introduction to Microelectronics
142008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
Multisim Simulation Results
20 Hz. Data
1 Khz. Data
ESE319 Introduction to Microelectronics
152008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
162008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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.
ESE319 Introduction to Microelectronics
172008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
182008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
192008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
202008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
The Common Base Amplifier
Voltage Bias Design Current Bias Design
vout vout
ESE319 Introduction to Microelectronics
212008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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.
ESE319 Introduction to Microelectronics
222008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
232008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
242008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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
ESE319 Introduction to Microelectronics
252008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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'
ESE319 Introduction to Microelectronics
262008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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'
ESE319 Introduction to Microelectronics
272008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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'
ESE319 Introduction to Microelectronics
282008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
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'
ESE319 Introduction to Microelectronics
292008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
Multisim Simulation
1062 uF
ESE319 Introduction to Microelectronics
302008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 01Oct08 KRL
Multisim Frequency Response
20 Hz. response
1 KHZ. Response