Download The Common-Collector Amplifier

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The Common-Collector
Amplifier
The common-collector amplifier: the BJT collector is at smallsignal ground! Examples of this type of amplifier include:
VCC
VCC
COUS
COUS
+
-
vi (t )
VBB
vO (t )
+
-
(a)
(b)
VCC
VEE
VCC
VCC
(c)
vO (t )
vi (t )
(d)
COUS
COUS
vi (t )
+
-
vO (t )
+
-
vO (t )
vi (t )
VEE
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Do you see why each of
these four circuits is a guldurn common-collector
amplifier?
Make dang sure that you do!
Let’s consider first circuit (a). It turns out that for commoncollector amplifiers, the T-model (as opposed to the hybrid-)
typically provides the easiest small-signal analysis.
Using the T-model, we find that the small-signal circuit for
amplifier (a) is:
C
gmvbe  ie
ib
+
-
RB
vi (t )
B
re
+
vbe
-
E
RE
vo (t )
ie
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Let’s determine the open-circuit voltage gain of this small-signal
amplifier:
vooc
Avo =
vi
We therefore must determine the output voltage vo in terms of
input voltage vi
From KVL, we find that:
ic
0  vi  RB ib  vbe  vo
C
gmvbe  ie
ib
+
-
RB
vi (t )
B
re
+
vbe
-
E
RE
And from KCL, we find:
ib = ie - ic
= ie - gm vbe
Where from Ohm’s Law:
ie =
So:
vo - 0 vo
=
RE
RE
vo (t )
ie
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ib = ie - gm vbe =
vo
- gm vbe
RE
Inserting this into the KVL equation above:
vo  vi  RB ib  vbe
v

 vi  RB  o  gmvbe   vbe
 RE

R
 vi  B vo  gm RB vbe
RE
Likewise using KCL and Ohm’s Law:
ie 
vbe vo

re
RE
Or rearranging:
vbe 
re
vo
RE
Inserting this result in the solution above:
R 
vo  vi   B vo  gm RB vbe
 RE 
R 
r
 vi   B vo  gm RB  e
 RE
 RE 

v o

R 
R 
 vi   B vo  gm re  B vo
 RE 
 RE 
R 
 vi   gm re  1   B vo
 RE 
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From this result we can determine the small-signal output
voltage:
1
R 

vo   1  1  gm re  B  vi
RE 

And so the open-circuit voltage gain is:
R 

v
Avo  o   1  1  gm re  B 
RE 
vi

We now note that:
gm re 
1
I
VT IC
 C 
IE VT
IE
Therefore:
1 - gm re = 1 - a = 1 -
β
1
=
β+1 β+1
And so the gain becomes:
æ
v
1 RB
Avo = o = çç1 +
çè
β + 1 RE
vi
We note here that:
1
= 1
β+1
ö÷
÷
÷
÷
ø
1
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We find therefore, that the small-signal gain of this commoncollector amplifier is approximately:
æ
1 RB
Avo = çç1 +
çè
β + 1 RE
@ (1 + 0 )= 1.0
ö÷÷
÷
÷
ø
1
1
The gain is approximately one!
Q: What!? The gain is
equal to one? That’s
just dog-gone silly!
What good is an
amplifier with a gain
of one?
A: Remember, the open-circuit voltage gain is just one of three
fundamental amplifier parameters. The other two are input
resistance Rin and output resistance Rout .
First, let’s examine the input resistance.
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Using the small-signal circuit, we find that:
C
v v
Rin  i  i
ii ib
gmvbe  ie
ib
Using KVL,
0  vi  RB ib  re  RE  ie  0
+
-
B
RB
re
vi (t )
+
vbe
-
E
RE
and adding the fact that ie = (β + 1 )ib , we
vo (t )
ie
find that the small-signal base current is:
ib =
vi
RB + (β + 1 )(re + RE
)
Combining these equations, we find that the input resistance for
this common-collector amplifier is:
Rin =
vi
= RB + (β + 1 )(re + RE
ib
)
Since beta is large, the input resistance is typically large—this
is good!
Now, let’s consider the output resistance Rout of this particular
common-collector amplifier. Recall that the output resistance is
defined as:
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Rout 
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vooc
iosc
where vooc is the open-circuit output voltage and iosc is the
short-circuit output current.
C
Using KVL,
gmvbe  ie
ib
0  vi  RB ib  re ie  0
+
-
B
RB
re
vi (t )
E
- 1
and adding the fact that ib = (β + 1 ) ie , we
RE
+
vbe
-
ie
iosc
0
find that the small-signal emmitter current is:
ie =
vi
- 1
RB (β + 1 )
+ re
And from KCL, this emitter current is likewise the short circuit
output current:
iosc = ie =
vi
- 1
RB (β + 1 )
+ re
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Of course, we already have determined that the open-circuit
output voltage is approximately equal to the input voltage:
vooc  vi
(i.e., Avo @ 1 )
Therefore, we find that the output resistance will be:
Rout =
vooc
iosc
- 1
= RB (β + 1 )
+ re
Since the emitter resistance re is typically small (e.g., re  2.5
if IE  10.0mA ), and  is typically large, we find that the output
resistance of this common-collector amplifier will typically be
small!
Let’s summarize what we have learned about this commoncollector amplifier:
1. The small-signal voltage gain is approximately equal to
one.
2. The input resistance is typically very large.
3. The output resistance is typically very small.
This is just like the op-amp voltage follower !
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The common-collector amplifier is alternatively referred to as
an emitter follower (i.e., the output voltage follows the input
voltage).
The common-collector amplifier is typically used as an output
stage, where it isolates a high gain amplifier with large output
resistance (e.g. a common emitter) from an output load of small
resistance (e.g. an audio speaker).
VCC
VCC
VCC
VCC
COUS
COUS
COUS
vi (t )
COUS
+
-

vO (t )
COUS
common-emitter
gain stage

emitter-follower
output stage
(common-collector)