Date post: | 20-Jan-2016 |
Category: |
Documents |
Upload: | anthony-manning |
View: | 223 times |
Download: | 0 times |
EE105 Fall 2007 Lecture 10, Slide 1 Prof. Liu, UC Berkeley
Lecture 10
OUTLINE• BJT Amplifiers (cont’d)
– CB stage with biasing– Emitter follower (Common-collector amplifier)– Analysis of emitter follower core– Impact of source resistance– Impact of Early effect– Emitter follower with biasing
Reading: Chapter 5.3.3-5.4
ANNOUNCEMENTS• Alan Wu will hold an extra lab session tomorrow (9/28), 2-4PM• The post-lab assignment for Experiment #4 has been shortened!• 2 pgs of notes (double-sided, 8.5”×11”) allowed for Midterm #1
EE105 Fall 2007 Lecture 10, Slide 2 Prof. Liu, UC Berkeley
Biasing of CB Stage
SEmE
ECm
in
out
in
X
X
out
in
outv
RRgR
RRg
v
v
v
v
v
v
v
vA
1
inSEmE
Ein
Sin
inX v
RRgR
Rv
RR
Rv
1
• RE is necessary to provide a path for the bias current IE to flow, but it lowers the input impedance.
Em
E
Em
Em
Em
in Rg
R
Rg
Rg
Rg
R
11
1
||1
EE105 Fall 2007 Lecture 10, Slide 3 Prof. Liu, UC Berkeley
Reduction of Input Impedance Due to RE
• The reduction of input impedance due to i1 is undesirable because it shunts part of the input current to ground instead of to Q1 (and RC). Choose RE >> 1/gm , i.e. ICRE >> VT
EE105 Fall 2007 Lecture 10, Slide 4 Prof. Liu, UC Berkeley
Creation of Vb
• A resistive voltage divider lowers the gain.• To remedy this problem, a capacitor is inserted
between the base and ground to short out the resistive voltage divider at the frequency of interest.
EE105 Fall 2007 Lecture 10, Slide 5 Prof. Liu, UC Berkeley
Example of CB Stage with Bias Design a CB stage for Av = 10 and Rin = 50. • Rin = 50≈ 1/gm if RE >> 1/gm Choose RE = 500
• Av = gmRC = 10 RC = 500• IC = gm·VT = 0.52mA• VBE=VTln(IC/IS)=0.899V
• Vb = IERE + VBE = 1.16V• Choose R1 and R2 to provide Vb
and I1 >> IB, e.g. I1 = 52A• CB is chosen so that (1/(+1))(1/CB) is small compared to 1/gm at
the frequency of interest.
VCC = 2.5VIS = 5x10-16 A= 100VA = ∞
EE105 Fall 2007 Lecture 10, Slide 6 Prof. Liu, UC Berkeley
Emitter Follower (Common Collector Amplifier)
EE105 Fall 2007 Lecture 10, Slide 7 Prof. Liu, UC Berkeley
Emitter Follower Core• When the input voltage (Vin) is increased by Vin, the
collector current (and hence the emitter current) increases, so that the output voltage (Vout) is increased.
• Note that Vin and Vout differ by VBE.
EE105 Fall 2007 Lecture 10, Slide 8 Prof. Liu, UC Berkeley
Unity-Gain Emitter Follower• In integrated circuits, the follower is typically realized
as shown below.– The voltage gain is 1 because a constant collector current
(= I1) results in a constant VBE; hence Vout = Vin .
1vA
AV
EE105 Fall 2007 Lecture 10, Slide 9 Prof. Liu, UC Berkeley
Small-Signal Model of Emitter Follower
• The voltage gain is less than 1 and positive.
mE
E
E
in
out
E
outoutinm
outin
E
outm
outin
gR
R
Rrv
v
R
vvvg
r
vv
R
vvg
r
v
vvv
111
1
1
:emitterat KCL
AV
EE105 Fall 2007 Lecture 10, Slide 10 Prof. Liu, UC Berkeley
Emitter Follower as a Voltage Divider
AV
EE105 Fall 2007 Lecture 10, Slide 11 Prof. Liu, UC Berkeley
Emitter Follower with Source Resistance
11
S
mE
E
in
out
Rg
R
R
v
v
AV
EE105 Fall 2007 Lecture 10, Slide 12 Prof. Liu, UC Berkeley
Input Impedance of Emitter Follower• The input impedance of an emitter follower is the
same as that of a CE stage with emitter degeneration (whose input impedance does not depend on the resistance between the collector and VCC).
Ex
xin Rr
i
vR )1(
AV
EE105 Fall 2007 Lecture 10, Slide 13 Prof. Liu, UC Berkeley
Effect of BJT Current Gain• There is a current gain of (+1) from base to emitter.• Effectively, the load resistance seen from the base is
multiplied by (+1).
EE105 Fall 2007 Lecture 10, Slide 14 Prof. Liu, UC Berkeley
Emitter Follower as a Buffer • The emitter follower is suited for use as a buffer
between a CE stage and a small load resistance, to alleviate the problem of gain degradation.
speaker221 )1( RrRin
1inCmv RRgA speakerRRgA Cmv
EE105 Fall 2007 Lecture 10, Slide 15 Prof. Liu, UC Berkeley
Output Impedance of Emitter Follower• An emitter follower effectively lowers the source
impedance by a factor of +1, for improved driving capability.
• The follower is a good “voltage buffer” because it has high input impedance and low output impedance.
Es
mout R
R
gR ||
1
1
EE105 Fall 2007 Lecture 10, Slide 16 Prof. Liu, UC Berkeley
Emitter Follower with Early Effect• Since rO is in parallel with RE, its effect can be easily
incorporated into the equations for the voltage gain and the input and output impedances.
OE
m
sout
OEin
m
SOE
OEv
rRg
RR
rRrR
gR
rR
rRA
||||1
1
||1
11
||
||
EE105 Fall 2007 Lecture 10, Slide 17 Prof. Liu, UC Berkeley
Emitter Follower with Biasing• A biasing technique similar to that used for the CE
stage can be used for the emitter follower. • Note that VB can be biased to be close to VCC because
the collector is biased at VCC.
EE105 Fall 2007 Lecture 10, Slide 18 Prof. Liu, UC Berkeley
Supply-Independent Biasing• By putting an independent current source at the
emitter, the bias point (IC, VBE) is fixed, regardless of the supply voltage value.
EE105 Fall 2007 Lecture 10, Slide 19 Prof. Liu, UC Berkeley
Summary of Amplifier Topologies• The three amplifier topologies studied thus far have
different properties and are used on different occasions.
• CE and CB stages have voltage gain with magnitude greater than one; the emitter follower’s voltage gain is at most one.
EE105 Fall 2007 Lecture 10, Slide 20 Prof. Liu, UC Berkeley
Amplifier Example #1• The keys to solving this problem are recognizing the
AC ground between R1 and R2, and using a Thevenin transformation of the input network.
SE
m
S
C
in
out
RRR
Rg
RRRR
vv
1
1
1
2
11
||||
CE stage Small-signal equivalent circuit
Simplified small-signal equivalent circuit
EE105 Fall 2007 Lecture 10, Slide 21 Prof. Liu, UC Berkeley
Amplifier Example #2• AC grounding/shorting and Thevenin transformation
are needed to transform this complex circuit into a simple CE stage with emitter degeneration.
S
m
S
C
in
out
RRR
Rg
RRR
v
v
1
1
21 1
1
||
EE105 Fall 2007 Lecture 10, Slide 22 Prof. Liu, UC Berkeley
Amplifier Example #3• First, identify Req, which is the impedance seen at the
emitter of Q2 in parallel with the infinite output impedance of an ideal current source.
• Second, use the equations for a degenerated CE stage with RE replaced by Req.
1
1 1
2
R
gR
meq
111 1
21
121
Rgg
RA
RrrR
mm
Cv
in
EE105 Fall 2007 Lecture 10, Slide 23 Prof. Liu, UC Berkeley
Amplifier Example #4
Sm
Cv
Rg
RRA
1|| 1
• Note that CB shorts out R2 and provides a ground for R1, at the frequency of interest.
R1 appears in parallel with RC; the circuit simplifies to a simple CB stage with source resistance.
EE105 Fall 2007 Lecture 10, Slide 24 Prof. Liu, UC Berkeley
• Note that the equivalent base resistance of Q1 is the parallel connection of RE and the impedance seen at the emitter of Q2.
Amplifier Example #5
EB
mmin R
R
ggR ||
1
1
1
11
21