c Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of
°
Electrical and Computer Engineering.
The Common-Emitter Amplifier
Basic Circuit
Fig. 1 shows the circuit diagram of a single stage common-emitter amplifier. The object is to solve
for the small-signal voltage gain, input resistance, and output resistance.
Figure 1: Single-stage common-emitter 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.
VBB =
V + R2 + V − R1
R1 + R2
RBB = R1 kR2
VEE = V −
REE = RE
(b) Make an “educated guess” for VBE . Write the loop equation between the VBB and the VEE
nodes.
IC
IC
VBB − VEE = IB RBB + VBE + IE REE =
RBB + VBE +
REE
β
α
(c) Solve the loop equation for the currents.
IC = αIE = βIB =
VBB − VEE − VBE
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: Bias circuit.
Figure 3: Signal circuit.
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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.
(b) Calculate gm , rπ , re , and r0 from the DC solution.
gm =
IC
VT
rπ =
VT
IB
re =
VT
IE
r0 =
VA + VCE
IC
(c) Replace the circuits looking out of the base and emitter with Thévenin equivalent circuits
as shown in Fig. 4.
vtb = vs
R1 kR2
Rs + R1 kR2
Rtb = R1 kR2
vte = 0
Rte = RE kR3
Figure 4: Signal circuit with Thévenin base circuit.
Exact Solution
This solution is based on the exact equivalent circuits developed in the more advanced notes on
the BJT. It treats r0 as a resistor from collector to emitter without the r0 approximations.
(a) Replace the BJT in Fig. 4 with the Thévenin base circuit and the Norton collector circuit
as shown in Fig. 5.
Figure 5: Base and collector equivalent circuits.
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(b) Solve for ic(sc) .
ic(sc) = Gmb vtb = Gmb vs
Gmb =
re0
(c) Solve for vo .
r0 − Rte /β
α
+ Rte kr0 r0 + Rte
vo = −ic(sc) ric kRC kRL = −Gmb vs
ric =
R1 kR2
Rs + R1 kR2
re0 =
Rtb + rx
+ re
1+β
R1 kR2
ric kRC kRL
Rs + R1 kR2
r0 + re0 kRte
1 − αRte / (re0 + Rte )
(d) Solve for the voltage gain.
Av =
R1 kR2
vo
= −Gmb
ric kRC kRL
vs
Rs + R1 kR2
(e) Solve for rin .
rin = R1 kR2 krib
rib = rx + rπ + Rte
(1 + β) r0 + Rtc
r0 + Rte + Rtc
(f) Solve for rout .
rout = ric kRC
(g) Special Case for Rte = 0.
Gmb =
α
re0
ric = r0
rib = rx + rπ
(h) Special Case for r0 = ∞.
Gmb =
re0
α
+ Rte
ric = ∞
rib = rx + rπ + (1 + β) Rte
Example 1 For the CE amplifier of Fig. 1, it is given that Rs = 5 kΩ, 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 the gain Av = vo /vs , the
input resistance rin , and the output resistance rout . The capacitors can be assumed to be ac short
circuits at the operating frequency.
Solution. For the dc bias solution, replace all capacitors with open circuits. The Thévenin
voltage and resistance seen looking out of the base are
VBB =
V + R2 + V − R1
= −1.364 V
R1 + R2
RBB = R1 kR2 = 54.55 kΩ
The Thévenin voltage and resistance seen looking out of the emitter are VEE = V − and REE = RE .
The bias equation for IE is
IE =
VBB − VEE − VBE
= 2.113 mA
RBB / (1 + β) + REE
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To test for the active mode, we calculate the collector-base voltage
¶
µ
¡ +
¢
IE
RBB = 8.521 V
VCB = VC − VB = V − αIE RC − VBB −
1+β
Because this is positive, the BJT is biased in its active mode.
For the small-signal ac analysis, we need r0 and re . To calculate r0 , we first calculate the
collector-emitter voltage
VCE = VCB + VBE = 9.171 V
It follows that r0 and re have the values
r0 =
VA + VCE
= 52.18 kΩ
αIE
re =
VT
= 11.83 Ω
IE
For the small-signal analysis, V + and V − are zeroed and the three capacitors are replaced with
ac short circuits. The Thévenin voltage and resistance seen looking out of the base are given by
vtb = vs
R1 kR2
= 0.916vs
Rs + R1 kR2
Rtb = Rs kR1 kR2 = 4.58 kΩ
The Thévenin resistances seen looking out of the emitter and the collector are
Rte = RE kR3 = 98.25 Ω
Rtc = RC kRL = 3.539 kΩ
Next, we calculate re0 , Gmb , ric , and rib .
re0 =
Gmb =
ric =
re0
Rtb + rx
+ re = 57.83 Ω
1+β
r0 − Rte /β
α
1
S
=
+ Rte kr0 r0 + Rte
157.8
r0 + re0 kRte
= 138.6 kΩ
1 − αRte / (re0 + Rte )
rib = rx + (1 + β) re + Rte
(1 + β) r0 + Rtc
= 10.39 kΩ
r0 + Rte + Rtc
The output voltage is given by
vo = −Gmb × (ric kRtc ) vtb = −Gmb × (ric kRtc ) × 0.916vs = −20.04vs
Thus the voltage gain is
Av = −20.04
The input and output resistances are given by
rin = R1 kR2 krib = 8.73 kΩ
rout = ric kRC = 3.539 kΩ
Approximate Solutions
These solutions use the r0 approximations. That is, it is assumed that r0 = ∞ except in calculating
ric . In this case, ic(sc) = i0c = αi0e = βib .
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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 i0e .
R1 kR2
1
vtb
= vs
i0e = 0
0
re + Rte
Rs + R1 kR2 re + Rte
(b) Solve for i0c and ric .
i0c = αi0e = vs
ric =
(c) Solve for vo and Av = vo /vs .
r0 + re0 kRte
1 − αRte / (re0 + Rte )
vo = −ic(sc) ric kRC kRL = vs
Av =
Note that this is of the form
R1 kR2
α
0
Rs + R1 kR2 re + Rte
R1 kR2
α
× −ric kRC kRL
0
Rs + R1 kR2 re + Rte
α
vo
R1 kR2
=
× −ric kRC kRL
vs
Rs + R1 kR2 re0 + Rte
Av =
vo
vtb
i0
i0
vo
=
× e × 0c × 0
vs
vs
vtb ie
ic
(d) Solve for rout .
rout = ric kRC
(d) Solve for rib and rin . Because the base node is absorbed, use the formula for rib .
rib = rx + (1 + β) (re + Rte )
rin = R1 kR2 krib
Example 2 Use the simplified T-model solutions to calculate the values of Av , rin , and rout for
Example 1.
¡
¢ ¡
¢
Av = 0.916 × 6.343 × 10−3 × −3.451 × 103 = −20.05
rib = 1.103 kΩ
rin = 9.173 kΩ
ric = 138.6 kΩ
rout = 4.171 kΩ
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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 i0c and ric .
vtb = ib (Rtb + rx ) + vπ + i0e Rte =
i0c
i0
i0
vtb
(Rtb + rx ) + c + c Rte =⇒ i0c =
1
Rtb + rx
Rte
β
gm
α
+
+
β
gm
α
ric =
(c) Solve for vo .
r0 + re0 kRte
1 − αRte / (re0 + Rte )
vtb
× −ric kRC kRL
1
Rtb + rx
Rte
+
+
β
gm
α
R1 kR2
1
× −ric kRC kRL
= vs
1
Rte
Rs + R1 kR2 Rtb + rx
+
+
β
gm
α
vo = i0c RC kRL =
(d) Solve for the voltage gain.
Av =
1
vo
R1 kR2
=
× −ric kRC kRL
1
Rte
vs
Rs + R1 kR2 Rtb + rx
+
+
β
gm
α
This is of the form
Av =
vo
vtb
i0
vo
=
× c × 0
vs
vs
vtb
ic
(e) Solve for rib and rin .
vb = ib (rx + rπ ) + i0e Rte = ib (rx + rπ ) + (1 + β) ib Rte = ib [rx + rπ + (1 + β) Rte ]
rib =
vb
= rx + rπ + (1 + β) Rte
ib
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rin = R1 kR2 krib
(f) Solve for rout .
rout = ric kRC
Example 3 Use the π-model solutions to calculate the values of Av , rin , and rout for Example 1.
gm = 0.0837
rπ = 1.183 kΩ
¢ ¡
¢
¡
Av = 0.916 × 6.343 × 10−3 × −3.451 × 103 = −20.05 = −20.05
rib = 11.03 kΩ
rin = 9.173 kΩ
ric = 138.6 kΩ
rout = 4.171 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. 8.
Figure 8: T model circuit.
(b) Solve for i0c and ric .
vtb = ib (Rtb + rx ) + i0e (re + Rte ) =
i0c
i0
vtb
(Rtb + rx ) + c (re + Rte ) =⇒ i0c =
re + Rte
R
+
r
β
α
x
tb
+
β
α
ric =
(c) Solve for vo .
r0 + re0 kRte
1 − αRte / (re0 + Rte )
vtb
× −ric kRC kRL
Rtb + rx re + Rte
+
β
α
R1 kR2
1
× −ric kRC kRL
= vs
Rs + R1 kR2 Rtb + rx re + Rte
+
β
α
vo = −i0c RC kRL =
8
(d) Solve for the voltage gain.
Av =
1
vo
R1 kR2
× −ric kRC kRL
=
vs
Rs + R1 kR2 Rtb + rx re + Rte
+
β
α
Note that this is of the form
Av =
vo
vtb
i0
vo
=
× c × 0
vs
vs
vtb
ic
(e) Solve for rib and rin .
vb = ib rx + i0e (re + Rte ) = ib rx + (1 + β) ib (re + Rte ) = ib [rx + (1 + β) (re + Rte )]
rib =
vb
= rx + (1 + β) (re + Rte )
ib
rin = R1 kR2 krib
(f) Solve for rout .
rout = ric kRC
Example 4 Use the T-model solutions to calculate the values of Av , rin , and rout for Example 1.
¡
¢ ¡
¢
Av = 0.916 × 6.343 × 10−3 × −3.451 × 103 = −20.05 = −20.05
rib = 11.03 kΩ
rin = 9.173 kΩ
ric = 138.6 kΩ
rout = 4.171 kΩ
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