Cb Amplifier

The Common-Base Amplifier
Basic Circuit
Fig. 1 shows the circuit diagram of a single stage common-base amplifier. The object is to solve
for the small-signal voltage gain, input resistance, and output resistance.
Figure 1: Common-base amplifier.
DC Solution
(a) Replace the capacitors with open circuits. Look out of the 3 BJT terminals and make Thévenin
equivalent circuits as shown in Fig. 2.
V
BB
=
V
+
R
2
+V
−
R
1
R
1
+R
2
R
BB
= R
1
kR
2
V
EE
= V
−
R
EE
= R
E
V
CC
= V
+
R
CC
= R
C
(b) Make an “educated guess” for V
BE
. Write the loop equation between the V
BB
and the V
EE
nodes. To solve for I
C
, this equation is
V
BB
−V
EE
= I
B
R
BB
+V
BE
+I
E
R
EE
=
I
C
β
R
BB
+V
BE
+
I
C
α
R
EE
(c) Solve the loop equation for the currents.
I
C
= αI
E
= βI
B
=
V
BB
−V
EE
−V
BE
R
BB
/β +R
EE
/α
(d) Verify that V
CB
> 0 for the active mode.
V
CB
= V
C
−V
B
= (V
CC
−I
C
R
CC
) −(V
BB
−I
B
R
BB
) = V
CC
−V
BB
−I
C
(R
CC
−R
BB
/β)
1
Figure 2: DC bias circuit.
Small-Signal or AC Solutions
(a) Redraw the circuit with V
+
= V
−
= 0 and all capacitors replaced with short circuits as shown
in Fig. 3.
Figure 3: Signal circuit.
(b) Calculate g
m
, r
π
, r
e
, and r
0
from the DC solution..
g
m
=
I
C
V
T
r
π
=
V
T
I
B
r
e
=
V
T
I
E
r
0
=
V
A
+V
CE
I
C
(c) Replace the circuits looking out of the base and emitter with Thévenin equivalent circuits
as shown in Fig. 4.
v
tb
= 0 R
tb
= 0 v
te
= v
s
R
E
R
s
+R
E
R
te
= R
s
kR
E
2
Figure 4: Signal circuit with Thévenin emitter circuit.
Exact Solution
(a) Replace the BJT in Fig. 4 with the Thévenin emitter circuit and the Norton collector circuit
as shown in Fig. 5.
Figure 5: Emitter and collector equivalent circuits.
(b) Solve for i
c(sc)
.
i
c(sc)
= −G
me
v
te
= −G
me
v
s
R
E
R
s
+R
E
G
me
=
1
R
te
+r
0
e
kr
0
αr
0
+r
0
e
r
0
+r
0
e
r
0
e
=
r
x
1 +β
+r
e
(c) Solve for v
o
.
v
o
= −i
c(sc)
r
ic
kR
C
kR
L
= G
me
v
s
R
E
R
s
+R
E
r
ic
kR
C
kR
L
r
ic
=
r
0
+r
0
e
kR
te
1 −αR
te
/ (r
0
e
+R
te
)
3
(d) Solve for the voltage gain.
A
v
=
v
o
v
s
=
R
E
R
s
+R
E
G
me
r
ic
kR
C
kR
L
(e) Solve for r
in
.
r
in
= R
1
kR
2
kr
ie
r
ie
= r
0
e
r
0
+R
tc
r
0
e
+r
0
+R
tc
/ (1 +β)
(f) Solve for r
out
.
r
out
= r
ic
kR
C
Example 1 For the CB amplifier in Fig. 1, it is given that R
s
= 100 Ω, R
1
= 120 kΩ, R
2
= 100 kΩ,
R
C
= 4.3 kΩ, R
E
= 5.6 kΩ, R
3
= 100 Ω, R
L
= 20 kΩ, V
+
= 15 V, V
−
= −15 V, V
BE
= 0.65 V,
β = 99, α = 0.99, r
x
= 20 Ω, V
A
= 100 V and V
T
= 0.025 V. Solve for A
v
, r
in
, and r
out
.
Solution. Because the dc bias circuit is the same as for the common-emitter amplifier example,
the dc bias values, r
e
, g
m
, r
π
, and r
0
are the same.
In the signal circuit, the Thévenin voltage and resistance seen looking out of the emitter are
given by
v
te
=
R
E
R
s
+R
E
v
s
= 0.9825v
s
R
te
= R
s
kR
E
= 98.25 Ω
The Thévenin resistances seen looking out of the base and the collector are
R
tb
= 0 R
tc
= R
C
kR
L
= 3.539 kΩ
Next, we calculate r
0
e
, G
me
, r
ic
, and r
ie
.
r
0
e
=
R
tb
+r
x
1 +β
+r
e
= 12.03 Ω G
me
=
1
R
te
+r
0
e
kr
0
αr
0
+r
0
e
r
0
+r
0
e
=
1
111.4
S
r
ic
=
r
0
+r
0
e
kR
te
1 −αR
te
/ (r
0
e
+R
te
)
= 442.3 kΩ r
ie
= r
0
e
r
0
+R
tc
r
0
e
+r
0
+R
tc
/ (1 +β)
= 12.83 Ω
The output voltage is given by
v
o
= G
me
(r
ic
kR
tc
) v
te
= G
me
(r
ic
kR
tc
)
R
E
R
s
+R
E
v
s
= 30.97v
s
Thus the voltage gain is
A
v
= 30.97
The input and output resistances are
r
in
= R
1
kR
2
kr
ib
= 12.81 Ω r
out
= r
ic
kR
C
= 4.259 kΩ
Approximate Solutions
These solutions assume that r
0
= ∞ except in calculating r
ic
. In this case, i
c(sc)
= i
0
c
= αi
0
e
= βi
b
.
4
Figure 6: Simplified T model circuit.
Simplified T Model Solution
(a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the
BJT with the simplified T model as shown in Fig. 6.
(b) Solve for i
0
c
and r
ic
.
0 −v
te
= i
0
e
¡
r
0
e
+R
te
¢
=
i
0
c
α
¡
r
0
e
+R
te
¢
=⇒i
0
c
= −v
te
α
r
0
e
+R
te
r
ic
=
r
0
+r
0
e
kR
te
1 −αR
te
/ (r
0
e
+R
te
)
(c) Solve for v
o
.
v
o
= −i
0
c
r
ic
kR
C
kR
L
= v
te
α
r
0
e
+R
te
r
ic
kR
C
kR
L
= v
s
R
E
R
s
+R
E
α
r
0
e
+R
te
r
ic
kR
C
kR
L
(d) Solve for the voltage gain.
A
v
=
v
o
v
s
=
R
s
R
s
+R
E
α
r
0
e
+R
te
r
ic
kR
C
kR
L
(e) Solve for r
ie
and r
in
.
0 −v
e
= i
0
e
r
0
e
=⇒i
0
e
= −
v
e
r
0
e
r
ie
=
v
e
−i
0
e
= r
0
e
r
in
= r
0
e
kR
E
(f) Solve for r
out
.
r
out
= r
ic
kR
C
Example 2 For Example 1, use the simplified T-model solutions to calculate the values of A
v
, r
in
,
and r
out
.
5
A
v
= 0.9825 ×
¡
8.978 × 10
−3
¢
×
¡
3.511 × 10
3
¢
= 30.97
r
in
= 12 Ω r
out
= 4.259 kΩ
π Model Solution
(a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the
BJT with the π model as shown in Fig. 7.
Figure 7: Hybrid-π model circuit.
(b) Solve for i
0
c
and r
ic
.
0 −v
te
= i
b
r
x
+v
π
+i
0
e
R
te
=
i
0
c
β
r
x
+
i
0
c
g
m
+
i
0
c
α
R
te
=⇒i
0
c
=
−v
te
r
x
β
+
1
g
m
+
R
te
α
r
ic
=
r
0
+r
0
e
kR
te
1 −αR
te
/ (r
0
e
+R
te
)
(c) Solve for v
o
.
v
o
= −i
0
c
r
ic
kR
C
kR
L
=
v
te
r
x
β
+
1
g
m
+
R
te
α
r
ic
kR
C
kR
L
= v
s
R
E
R
s
+R
E
1
r
x
β
+
1
g
m
+
R
te
α
r
ic
kR
C
kR
L
(d) Solve for the voltage gain.
A
v
=
v
o
v
s
=
R
E
R
s
+R
E
1
r
x
β
+
1
g
m
+
R
te
α
r
ic
kR
C
kR
L
(e) Solve for r
out
.
r
out
= r
ic
kR
C
6
(f) Solve for r
ie
and r
in
.
0 −v
e
= i
b
(r
x
+r
π
) =
i
0
e
1 +β
(r
x
+r
π
) =⇒i
0
e
= −v
e
1 +β
r
x
+r
π
r
ie
=
v
e
−i
0
e
=
r
x
+r
π
1 +β
r
in
= r
ie
kR
E
Example 3 For Example 1, use the π-model solutions to calculate the values of A
v
, r
in
, and r
out
.
A
v
= 0.9825 ×
¡
8.978 × 10
−3
¢
×
¡
3.539 × 10
3
¢
= 30.97
r
in
= 12 Ω r
out
= 4.259 kΩ
T Model Solution
(a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the
BJT with the T model as shown in Fig.??.
Figure 8: T model circuit.
(b) Solve for i
0
c
and r
ic
.
0 −v
te
= i
b
r
x
+i
0
e
(r
e
+R
te
) =
i
0
c
β
r
x
+
i
0
c
α
(r
e
+R
te
) =⇒i
0
c
=
−v
te
r
x
β
+
r
e
+R
te
α
r
ic
=
r
0
+r
0
e
kR
te
1 −αR
te
/ (r
0
e
+R
te
)
(c) Solve for v
o
.
v
o
= −i
0
c
r
ic
kR
C
kR
L
=
v
te
r
x
β
+
r
e
+R
te
α
r
ic
kR
C
kR
L
= v
s
R
E
R
s
+R
E
1
r
x
β
+
r
e
+R
te
α
r
ic
kR
C
kR
L
7
(d) Solve for the voltage gain.
A
v
=
v
o
v
s
=
R
E
R
s
+R
E
1
r
x
β
+
r
e
+R
te
α
r
ic
kR
C
kR
L
(e) Solve for r
ie
and r
in
.
0 −v
e
= i
b
r
x
+i
0
e
r
e
=
i
0
e
1 +β
r
x
+i
0
e
r
e
= i
0
e
µ
r
x
1 +β
+r
e
¶
=⇒i
0
e
=
−v
e
r
x
1 +β
+r
e
r
ie
=
v
e
−i
0
e
=
r
x
1 +β
+r
e
r
in
= R
E
kr
ie
(f) Solve for r
out
.
r
out
= r
ic
kR
C
Example 4 For Example 1, use the T-model solutions to calculate the values of A
v
, r
in
, and r
out
.
A
v
= 0.9825 ×
¡
8.978 × 10
−3
¢
×
¡
3.539 × 10
3
¢
= 30.97
r
in
= 12 Ω r
out
= 4.259 kΩ
8