The Common-Collector Amplifier

Basic Circuit
Fig. 1 shows the circuit diagram of a single stage common-collector amplifier. The object is
to solve for the small-signal voltage gain, input resistance, and output resistance.

Figure 1: Common-collector 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.
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: 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 with a Thévenin equivalent circuit as
shown in Fig. 4.
R1 kR2
vtb = vs Rtb = R1 kR2
Rs + R1 kR2

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 Figure 4: Signal circuit with Thévenin base circuit.

Exact Solution
(a) Replace the BJT in Fig. 4 with the Thévenin base emitter circuits as shown in Fig. 5.
Solve for ve(oc) .
r0 + Rtc / (1 + β)
ve(oc) = vtb 0 Rtc = 0
re + r0 + Rtc / (1 + β)
Note that the Thévenin resistance Rtc looking out of the collector is zero in the original
circuit. The exact solution gives the correct answer even if Rtc 6= 0.

Figure 5: Base and emitter equivalent circuits.

(b) Solve for vo .

RE kRL R1 kR2 r0 + Rtc / (1 + β) RE kRL

vo = ve(oc) = vs 0
rie + RE kRL Rs + R1 kR2 re + r0 + Rtc / (1 + β) rie + RE kRL
r0 + Rtc Rtb + rx + rπ Rtb + rx
rie = re0 0 re0 = = + re
re + r0 + Rtc / (1 + β) 1+β 1+β
(c) Solve for the voltage gain.
vo R1 kR2 r0 + Rtc / (1 + β) RE kRL
Av = = 0
vs Rs + R1 kR2 re + r0 + Rtc / (1 + β) rie + RE kRL
(d) Solve for rin .
(1 + β) r0 + Rtc
rin = R1 kR2 krib rib = rx + rπ + Rte Rte = RE kRL
r0 + Rte + Rtc

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 (e) Solve for rout .
rout = rie kRE
(f) Special case for r0 = ∞.

ve(oc) = vtb rib = rx + rπ + (1 + β) Rte rie = re0

Example 1 For the CC amplifier in Fig. 1, it is given that RS = 5 kΩ, R1 = 120 kΩ, R2 =
100 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 circuits are the same as for the common-emitter amplifier
example, the bias values, re , gm , and rπ are the same. Because VCE is different, a new value
of r0 must be calculated. The collector-to-emitter voltage is given by
µ ¶
+ IE
VCE = VC − VE = V − VBB − RBB − VBE = 17.01 V
1+β
Thus r0 has the value
VA + VCE
r0 = = 55.93 kΩ
αIE
In the signal circuit, the Thévenin voltage and resistance seen looking out of the base are
given by
R1 kR2
vtb = vs = 0.916vs Rtb = RS kR1 kR2 = 4.58 kΩ
RS + R1 kR2
The Thévenin resistances seen looking out of the emitter and the collector are
Rte = RE kR3 = 4.375 kΩ Rtc = 0
Next, we calculate re0 , ve(oc) , rie , and rib , where Rtc = 0.
Rtb + rx
re0 = + re = 57.83 Ω
1+β
r0 + Rtc / (1 + β)
ve(oc) = vtb = 0.999vtb
re0
+ r0 + Rtc / (1 + β)
r0 + Rtc
rie = re0 0 = 57.77 Ω
re + r0 + Rtc / (1 + β)
(1 + β) r0 + Rtc
rib = rx + (1 + β) re + Rte = 407 kΩ
r0 + Rte + Rtc
The output voltage is given by
Rte
vo = ve(oc) = 0.987 × 0.999 × 0.916vs = 0.903vs
rie + Rte
Thus the voltage gain is
Av = 0.903
The input and output resistances are given by
rin = R1 kR2 krib = 48.1 kΩ rout = rie kRE kRL = 57.02 Ω

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Alternate Solutions
These solutions are exact because the Thévenin resistance Rtc looking out of the collector is
zero. If Rtc 6= 0, replace r0 with an open circuit in all formulas, i.e. let r0 = ∞ In this case,
the solutions are no longer exact, they are approximate.

Simplified T Model Solution

(a) After making the Thévenin equivalent circuits looking out of the base, replace the BJT
with the simplified T model as shown in Fig. 6.

Figure 6: Simplified T model.

(b) Solve for re0 .

Rtb + rx
re0 = + re Rtb = RS kR1 kR2
1+β
(c) Solve for vo .
r0 kRE kRL R1 kR2 r0 kRE kRL
vo = vtb = vs
re0 + r0 kRE kRL 0
Rs + R1 kR2 re + r0 kRE kRL
(d) Solve for the voltage gain.
vo R1 kR2 r0 kRE kRL
Av = = 0
vs Rs + R1 kR2 re + r0 kRE kRL
(e) Solve for rin . Because the base node is absorbed, use the formula for rib .

rin = R1 kR2 krib rib = rx + (1 + β) (re + r0 kRte ) Rte = RE kRL

(f) Solve for rout .

rout = r0 kre0 kRE

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

Av = 0.916 × 0.986 = 0.903

rin = 48.51 kΩ rout = 57.18 Ω

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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.

(b) Solve for i0e .

i0e
vtb = ib (Rtb + rx + rπ ) + i0e r0 kRE kRL = (Rtb + rx + rπ ) + i0e r0 kRE kRL
1+β
vtb
=⇒ i0e =
Rtb + rx + rπ
+ r0 kRE kRL
1+β
(c) Solve for vo .
vtb
vo = i0e r0 kRE kRL = r0 kRE kRL
Rtb + rx + rπ
+ r0 kRE kRL
1+β
R1 kR2 r0 kRE kRL
= vs
Rs + R1 kR2 Rtb + rx + rπ
+ r0 kRE kRL
1+β
(d) Solve for the voltage gain.
vo R1 kR2 r0 kRE kRL
Av = =
vs Rs + R1 kR2 Rtb + rx + rπ
+ r0 kRE kRL
1+β
(e) Solve for rib and rin .

vb = ib (rx + rπ ) + i0e r0 kRE kRL = ib (rx + rπ ) + (1 + β) ib r0 kRE kRL

= ib [rx + rπ + (1 + β) r0 kRE kRL ]
vb
rib = = rx + rπ + (1 + β) r0 kRE kRL
ib

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 rin = R1 kR2 krib
(f) Solve for rout .First, solve for the open-circuit output voltage. This is the output
voltage with RL = ∞.
r0 kRE
vo(oc) = vtb
Rtb + rx + rπ
+ r0 kRE
1+β
Next, solve for the short-circuit output current. This is the output current with RL = 0.
The output current is given by
vo vtb r0 kRE kRL vtb r0 kRE
io = = =
RL Rtb + rx + rπ RL Rtb + rx + rπ R + r0 kRE
+ r0 kRE kRL + r0 kRE kRL L
1+β 1+β
Now, let RL = 0 to obtain
vtb
io(sc) =
Rtb + rx + rπ
1+β
The output resistance is given by
µ ¶
vo(oc) r0 kRE Rtb + rx + rπ Rtb + rx + rπ
rout = = = kr0 kRE
io(sc) Rtb + rx + rπ 1+β 1+β
+ r0 kRE
1+β
Note this is simply re0 kr0 kRE , an answer that is obvious using the simplified T model.
Example 3 Use the π-model solutions to calculate the values of Av , rin , and rout for Example
1.

Av = 0.916 × 0.986 = 0.903

rin = 48.51 kΩ rout = 57.18 Ω

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.
(b) Solve for i0e .
i0e
vtb = ib (Rtb + rx ) + i0e (re + r0 kRE kRL ) = (Rtb + rx ) + i0e (re + r0 kRE kRL )
1+β
vtb
=⇒ i0e =
Rtb + rx
+ re + r0 kRE kRL
β
(c) Solve for vo .
vtb
vo = i0e r0 kRE kRL = r0 kRE kRL
Rtb + rx
+ re + r0 kRE kRL
β
R1 kR2 r0 kRE kRL
= vs
Rs + R1 kR2 Rtb + rx
+ re + r0 kRE kRL
β

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 Figure 8: T model circuit.

(d) Solve for the voltage gain.

vo R1 kR2 r0 kRE kRL
Av = =
vs Rs + R1 kR2 Rtb + rx
+ re + r0 kRE kRL
β
(e) Solve for rib and rin .
vb = ib rx + i0e (re + r0 kRE kRL ) = ib rx + (1 + β) ib (re + r0 kRE kRL )
= ib [rx + (1 + β) (re + r0 kRE kRL )]
vb
rib = = rx + (1 + β) (re + r0 kRE kRL )
ib

rin = R1 kR2 krib

(f) Solve for rout . First, solve for the open-circuit output voltage. This is the output
voltage with RL = ∞.
r0 kRE
vo(oc) = vtb
Rtb + rx
+ re + r0 kRE
β
Next, solve for the short-circuit output current. This is the output current with RL = 0.
The output current is given by
vo vtb r0 kRE kRL vtb r0 kRE
io = = =
RL Rtb + rx RL Rtb + rx R + r0 kRE
+ re + r0 kRE kRL + re r0 kRE kRL L
1+β 1+β
Now, let RL = 0 to obtain
vtb
io(sc) =
Rtb + rx
+ re
1+β
The output resistance is given by
µ ¶ µ ¶
vo(oc) r0 kRE Rtb + rx Rtb + rx
rout = = + re = + re kr0 kRE
io(sc) Rtb + rx 1+β 1+β
+ re + r0 kRE
1+β

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This is the same answer obtained from the simplified T model.

Example 4 Use the T-model solutions to calculate the values of Av , rin , and rout for Ex-
ample 1.

Av = 0.916 × 0.986 = 0.903

rin = 48.51 kΩ rout = 57.18 Ω