# The Common-Base Amplifier Basic Circuit DC Solution by lanyuehua

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```									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 + R2 + V − R1
VBB =                          RBB = R1 kR2
R1 + R2

VEE = V −       REE = RE            VCC = V +   RCC = RC
(b) Make an “educated guess” for VBE . Write the loop equation between the VBB and the VEE
nodes. To solve for IC , this equation is
IC             IC
VBB − VEE = IB RBB + VBE + IE REE =              RBB + VBE +    REE
β              α

(c) Solve the loop equation for the currents.

VBB − VEE − VBE
IC = αIE = βIB =
RBB /β + REE /α

(d) Verify that VCB > 0 for the active mode.

VCB = VC − VB = (VCC − IC RCC ) − (VBB − IB RBB ) = VCC − VBB − IC (RCC − RBB /β)

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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 gm , rπ , re , and r0 from the DC solution..
IC             VT               VT              VA + VCE
gm =            rπ =             re =            r0 =
VT             IB               IE                 IC
(c) Replace the circuits looking out of the base and emitter with Thévenin equivalent circuits
as shown in Fig. 4.
RE
vtb = 0        Rtb = 0        vte = vs                Rte = Rs kRE
Rs + RE

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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 ic(sc) .
RE
ic(sc) = −Gme vte = −Gme vs
Rs + RE
1       αr0 + re0              rx
0
Gme =          0           0
re =      + re
Rte + re kr0 r0 + re              1+β
(c) Solve for vo .
RE
vo = −ic(sc) ric kRC kRL = Gme vs           ric kRC kRL
Rs + RE
0
r0 + re kRte
ric =                0
1 − αRte / (re + Rte )

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(d) Solve for the voltage gain.

vo     RE
Av =      =         Gme ric kRC kRL
vs   Rs + RE
(e) Solve for rin .

0          r0 + Rtc
rin = R1 kR2 krie      rie = re    0
re + r0 + Rtc / (1 + β)

(f) Solve for rout .
rout = ric kRC

Example 1 For the CB ampliﬁer in Fig. 1, it is given that Rs = 100 Ω, R1 = 120 kΩ, R2 = 100 kΩ,
RC = 4.3 kΩ, RE = 5.6 kΩ, R3 = 100 Ω, RL = 20 kΩ, V + = 15 V, V − = −15 V, VBE = 0.65 V,
β = 99, α = 0.99, rx = 20 Ω, VA = 100 V and VT = 0.025 V. Solve for Av , rin , and rout .

Solution. Because the dc bias circuit is the same as for the common-emitter ampliﬁer example,
the dc bias values, re , gm , rπ , and r0 are the same.
In the signal circuit, the Thévenin voltage and resistance seen looking out of the emitter are
given by
RE
vte =             vs = 0.9825vs   Rte = Rs kRE = 98.25 Ω
Rs + RE
The Thévenin resistances seen looking out of the base and the collector are

Rtb = 0       Rtc = RC kRL = 3.539 kΩ
0
Next, we calculate re , Gme , ric , and rie .

Rtb + rx                                      1      αr0 + re 0     1
0
re =            + re = 12.03 Ω       Gme =             0           0
=       S
1+β                                     Rte + re kr0 r0 + re     111.4
0
r0 + re kRte                                                r0 + Rtc
0
ric =                0
= 442.3 kΩ           rie = re    0
= 12.83 Ω
1 − αRte / (re + Rte )                                 re   + r0 + Rtc / (1 + β)
The output voltage is given by
RE
vo = Gme (ric kRtc ) vte = Gme (ric kRtc )           vs = 30.97vs
Rs + RE
Thus the voltage gain is
Av = 30.97
The input and output resistances are

rin = R1 kR2 krib = 12.81 Ω        rout = ric kRC = 4.259 kΩ

Approximate Solutions
These solutions assume that r0 = ∞ except in calculating ric . In this case, ic(sc) = i0 = αi0 = βib .
c     e

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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 i0 and ric .
c

¡ 0     ¢ i0 ¡ 0     ¢               α
0 − vte = i0 re + Rte = c re + Rte =⇒ i0 = −vte 0
e                           c
α                      re + Rte
0
r0 + re kRte
ric =                0
1 − αRte / (re + Rte )
(c) Solve for vo .
α                        RE       α
vo = −i0 ric kRC kRL = vte
c                       0
ric kRC kRL = vs          0
ric kRC kRL
re + Rte                  Rs + RE re + Rte

(d) Solve for the voltage gain.
vo      Rs      α
Av =      =          0 +R
ric kRC kRL
vs   Rs + RE re   te

(e) Solve for rie and rin .
ve
0
0 − ve = i0 re =⇒ i0 = −
e        e         0
re
ve     0
rie =       = re
−i0
e
0
rin = re kRE
(f) Solve for rout .
rout = ric kRC

Example 2 For Example 1, use the simpliﬁed T-model solutions to calculate the values of Av , rin ,
and rout .

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¡            ¢ ¡           ¢
Av = 0.9825 × 8.978 × 10−3 × 3.511 × 103 = 30.97
rin = 12 Ω      rout = 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 i0 and ric .
c

i0
c      i0 i0                  −vte
0 − vte = ib rx + vπ + i0 Rte =
e            rx + c + c Rte =⇒ i0 =
c
β      gm   α             rx    1   Rte
+    +
β    gm    α
0
r0 + re kRte
ric =                0
1 − αRte / (re + Rte )
(c) Solve for vo .

vte                     RE           1
vo = −i0 ric kRC kRL =                      r kR kR = vs                       r kR kR
c                     rx    1     Rte ic C L      Rs + RE rx    1    Rte ic C L
+      +                            +     +
β    gm      α                      β    gm     α

(d) Solve for the voltage gain.

vo     RE           1
Av =      =                       r kR kR
vs   Rs + RE rx    1    Rte ic C L
+     +
β    gm     α

(e) Solve for rout .
rout = ric kRC

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(f) Solve for rie and rin .
i0
e                         1+β
0 − ve = ib (rx + rπ ) =       (rx + rπ ) =⇒ i0 = −ve
e
1+β                        rx + rπ
ve    rx + rπ
rie =       0
=
−ie    1+β
rin = rie kRE

Example 3 For Example 1, use the π-model solutions to calculate the values of Av , rin , and rout .

¡            ¢ ¡           ¢
Av = 0.9825 × 8.978 × 10−3 × 3.539 × 103 = 30.97
rin = 12 Ω         rout = 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 i0 and ric .
c

i0
c     i0                        −vte
0 − vte = ib rx + i0 (re + Rte ) =
e                       rx + c (re + Rte ) =⇒ i0 =
c
β       α                     rx re + Rte
+
β      α
0
r0 + re kRte
ric =                  0
1 − αRte / (re + Rte )
(c) Solve for vo .
vte                   RE        1
vo = −i0 ric kRC kRL =                 r kR kR = vs                    r kR kR
c                    rx re + Rte ic C L      Rs + RE rx re + Rte ic C L
+                               +
β       α                       β     α

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(d) Solve for the voltage gain.

vo     RE        1
Av =        =                    r kR kR
vs   Rs + RE rx re + Rte ic C L
+
β     α

(e) Solve for rie and rin .
µ              ¶
i0
e                        rx                        −ve
0 − ve =   ib rx + i0 re
e      =     rx + i0 re = i0
e       e           + re       =⇒ i0 =
e      rx
1+β                       1+β                          + re
1+β
ve     rx
rie =       =     + re
−i0
e   1+β
rin = RE krie
(f) Solve for rout .
rout = ric kRC

Example 4 For Example 1, use the T-model solutions to calculate the values of Av , rin , and rout .

¡            ¢ ¡           ¢
Av = 0.9825 × 8.978 × 10−3 × 3.539 × 103 = 30.97
rin = 12 Ω       rout = 4.259 kΩ

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