Lecture 5b
Common Emitter Amplifier Frequency Response (2)
Objectives
„ To analyze the CE amplifiers and to calculate different poles and zeros
which determine its frequency response
„ To calculate the dominant poles and the CE amplifier bandwidth
„ To analyze the effect of the emitter resistor on the CE amplifier gain
and bandwidth
Introduction
In the last lectures the frequency response of the amplifier circuits were
examined.
Also, the frequency response of the common source amplifier was calculated
and the dominant poles were determined.
In this lecture the frequency response of the common emitter amplifier will be
considered using the SCTC and OCTC techniques introduced in the last
lecture.
The design trade-off using the emitter resistor will also be explored.
Direct Determination of High-Frequency Poles and Zeros:
Similar to what we did in the last lectures.
The high frequency poles and zeros may be determined by direct analysis.
In this case all the coupling and bypass capacitors may be considered short
circuit.
However, the high frequency model of the transistor should be used since the
transistor internal capacitances can't considered open circuit at high
frequencies.
The following example will illustrate this method for the common emitter
amplifier.
Example 1:
Drive an expression for the high frequency response AvH(s) of the common
emitter amplifier shown in Figure 1. Hence, determine the midband gain Amid,
high frequency poles and zeros, and the high cutoff frequency.
Figure 1: Common Emitter Amplifier Circuit
Solution:
The CE amplifier is redrawn for small signal analysis as shown in Figure 2 by
grounding the DC voltage source VCC and replacing the transistor by its high
small signal model.
The biasing resistors R1 and R2 are combined into RB.
∴
RB
= R1 || R 2
All the coupling and bypass capacitors are considered short circuit at high
frequency.
Figure 2: The CE amplifier in Figure 2 redrawn with the transistor
replaced by its high frequency small signal model
The small-signal model can be simplified by using Thevinin's theorem as
shown in Figure 3.
The simplified circuit is shown in Figure 4.
Figure 3: Using Thevinin's Theorem to simplify the CE amplifier circuit in
Figure 2
v th = v sig
v th 2 =
RB
R sig + R B
and Rth =
v th .rπ
Rth + rx + rπ
R sig R B
R sig + R B
and rth 2 = rπ ( R th + rx )
Also, R L′ = R L || RC || ro
Figure 4: Simplified circuit to calculate the frequency response of the CE
amplifier
Writing the two nodal equations at nodes A and B in the frequency domain,
v be′ ⋅ sC π +
v be′ −v th 2
+ (v be′ −v o )sC μ = 0
Rth 2
vo
+ g mv gs + (v o −v be′ )sC μ = 0
R L′
Solving the last two equations by eliminating v`be yields,
vo
v th2 (s) (sC μ - g m )
R th 2
Δ
(s) =
⎛C
⎛
1
1
Δ = s 2 (C πC μ ) + s ⎜ π + C μ ⎜ g m +
+
⎜ R′
R th 2 R L′
⎝
⎝ L
∴
AvH (s ) =
⎞⎞
1
⎟ ⎟⎟ +
R L′ ⋅ R th 2
⎠⎠
(sC μ - g m )
v o (s )
RB
=
v sig (s )
(R th + rx ) ⋅ ( R sig + R B )
Δ
Let us define
CT
⎛
R′ ⎞
= C π + C μ ⎜1 + g m R L′ + L ⎟
R th 2 ⎠
⎝
∴
Δ = s 2C π C μ +
∴
AvH (s ) =
sC T
R L′
+
1
R L′ ⋅ R th 2
RB
⋅
(R th + rx ) ⋅ (R sig + R B ) ⋅C π ⋅C μ s 2 +
(sC μ - g m )
sC T
1
+
R L′ ⋅C π ⋅C μ R L′ ⋅ R th 2 ⋅C π ⋅C μ
Amid may be found from the last expression by assuming s → 0 .
∴
A mid =
∴
A mid =
A mid
=
- g m R L′ ⋅ R B ⋅ R th 2
( R th + rx ) ⋅ ( R sig + R B )
- g m R L′ ⋅ R B ⋅ rπ
( rπ + R th + rx ) ⋅ ( R sig + R B )
-βo R L′ ⋅ R B
( rπ + R th + rx ) ⋅ ( R sig + R B )
High-frequency response is given by 2 poles, one finite zero and one zero at
infinity.
Finite right-half plane zero, ωZ = + gm/Cμ > ωT can easily be neglected.
For a polynomial s2 + sA1 + A0 with roots a and b, if one root is much larger
than the other one, we can assume that a =A1 and b=A0/A1.
∴
ωP 1 =
A0
A1
∴
ωP 1 ≅
1
Rth 2CT
Assume that the midband gain is very large
⎛
R′ ⎞
( i.e. g m R L′ ⎜ 1 + L ⎟ , and
⎝ R th 2 ⎠
g m R L′C μ C π )
∴
ωP 2 ≅
gm ⎛
1
1 ⎞
+
⎜1 +
⎟
C π ⎝ g m R th 2 g m R L′ ⎠
∴
ωP 2 ≅
gm
Cπ
Smallest root that gives first pole limits frequency response and determines
ωH.
Second pole is important in frequency compensation as it can degrade phase
margin of feedback amplifiers.
Assuming ωp1 ωp2 (ωp1 is the dominanat pole)
∴
ωH ≅ ωP 1 =
1
Rth 2CT
Dominant pole model at high frequencies for CE amplifier is shown in Figure
5.
Figure 5: Equivalent circuit assuming dominant pole model
Example 2:
Find the midband gain, high frequency poles and zeros, ωH, and the
bandwidth for the CE amplifier in Example 1. Given the following transistor
parameters fT = 500 MHz, βo =100, Cμ = 0.5 pF, rx =250 Ω, and VA = 10V.
Solution:
Referring to example 2 solution in Lecture 5a, the Q-point is (1.66 mA, 2.7 V)
IC
VT
∴
gm =
∴
g m = 40 ⋅ I C
rπ
= 66.4mS
= 1.5k Ω
ro =
Cπ =
VA
IC
=
gm
2π f T
100
= 60.2k Ω
1.66
− C μ = 20.6pF
R L′ = R L || RC || ro
R L′ = 100 || 4.3 || 60.2 = 3.86kΩ
Rth = R B R sig = 7.5 1 = 0.88kΩ
R th 2 = rπ ( R th + rx ) = 0.64 k Ω
CT
⎛
R′ ⎞
= C π + C μ ⎜ 1 + g m R L′ + L ⎟
R th 2 ⎠
⎝
CT
= 152.3pF
ωP 1 =
1
Rth 2CT
= 10.26M rad/sec
ωP 2 ≅
gm ⎛
1
1 ⎞
+
⎜1 +
⎟
C π ⎝ g m R th 2 g m R L′ ⎠
ωP 2 ≅ 3.31 G rad/sec
ωz =
gm
Cμ
= 132.8 G rad/sec
It is clear that ωp1 is much lower than ωp2 and ωz which means that ωp1 is the
dominant pole.
∴
ωH ≅ ωP 1 = 10.26 M rad/sec
fH
=
ωH
= 1.63 MHz
2π
BW
.
= f H −f L
BW
.
= 1.63MHz − 170 Hz
A
A
≅ 1.63MHz
- βo R L′ ⋅ R B
( rπ + R th + rx ) ⋅ ( R sig + R B )
mid
=
mid
= − 129.5 V/V
Gain-Bandwidth Product Limitations of the C-E Amplifier
As we can see from the expressions of the midband gain and the CE amplifier
bandwidth, Rth is appearing in both expressions.
If Rth is reduced to zero in order to increase the bandwidth, then Rth2 would not
be zero but would be limited to approximately rx.
GBW = Av ωH
⎛
⎞⎛ 1 ⎞
βo R L′
≤ ⎜
⎟⎜
⎟
⎝ R th + rx + rπ ⎠ ⎝ R th 2C T ⎠
If Rth = 0, rx << rπ so that rx = Rth2 and CT
= C μ ( g m R L′ )
∴
GBW ≤
1
rx C μ
If we used the same values in Example 2
⇒
GBW ≤ 8G rad/sec
The Actual GBW in Example 2 is 1.334G rad/sec.
Open-Circuit Time Constant Method to Determine ωH
As mentioned in the last lectures the open-circuit time constants
associated with the transistor capacitances may be used to simplify the
determination of the high cutoff frequency of the amplifier.
In the next example the high cutoff frequency of the CE amplifier will be
determined using the OCTC method.
Example 3:
Derive an expression for the high cutoff frequency for the circuit in Example 2
using the OCTC method.
Solution:
Using OCTC method for the circuit in Figure 4
For Cπ the resistance seen across the capacitor terminals may be calculated
from Figure 6.
R πO = R th 2
So the time constant associated with Cπ equals Cπ· Rth2
Figure 6: Circuit used to calculate the open-circuit time constant
associated with Cπ
For Cμ the resistance seen across the capacitor terminals may be calculated
from Figure 7.
R μo =
vx
R′
= Rth 2 ⋅ (1 + g m R L′ + L )
ix
Rth 2
So the time constant associated with Cμ equals Cμ· Rth2 (1 + gmR`L + R`L/ Rth)
Figure 7: Circuit used to calculate the open-circuit time constant
associated with Cμ
The high cutoff frequency is calculated using the OCTC as:
ωH
≅
1
2
∑ R ioC i
=
1
Rπo C π + R μ o C μ
=
1
rth 2 C T
i =1
Which is the same expression found before using direct calculations in
Example 3 of lecture 5a.
Gain-Bandwidth Trade-off Using Emitter Resistor
The emitter resistance of the common emitter amplifier may be used to trade
off between the gain and the amplifier bandwidth.
In this case no bypass capacitor is used.
The small signal circuit is shown in Figure 8.
Where Rth and vth are the same as calculated in Example 3 of lecture 5a.
v th = v sig
RB
R sig + R B
and Rth =
R sig R B
R sig + R B
Figure 8: Small signal CE Amplifier circuit at high frequency with no
bypass capacitor
The midband gain may be calculated from Figure 8 by assuming the transistor
capacitors to be open circuit.
A mid
for
= −
βo R L′ .R B
( R th + rx + rπ + ( βo + 1) R E ) ⋅ ( R sig + R B )
rπ >> R th + rx , rB >> R sig and g m ⋅ R E
A mid ≅ −
>> 1
R L′
RE
The midband voltage gain decreases as the emitter resistance increases and
the bandwidth of CE amplifier will correspondingly increase.
To find the effect of RE on the amplifier bandwidth the high cutoff frequency
may be calculated as follows:
Using OCTC method for the circuit in Figure 8
For Cπ the resistance seen across the capacitor terminals may be calculated
as rπ || Req.
Where Req is calculated from Figure 9.
Req =
Rth + rx + R E
vx
=
ix
1+ g m RE
R π O = rπ Req ≅
Rth + rx + R E
1+ g m RE
So the time constant associated with Cπ equals Cπ · Rπo
Figure 9: Circuit used to calculate the open-circuit time constant
associated with Cπ
For Cμ the resistance seen across the capacitor terminals may be calculated
from Figure 10.
Figure 10: Circuit used to calculate the open-circuit time constant
associated with Cμ
To simplify the calculation, the test source ix is first split into two equivalent
sources as shown in Figure 11 and then superposition is used to find vx = (vb
- vc).
Figure 11: Modified circuit used to calculate the open-circuit time
constant associated with Cμ
Assuming that βo >>1 and ( R th + rx
R μo =
)
<<
( rπ + ( βo + 1)R E )
vx
ix
⎛
g m R L′
R L′ ⎞
R μ o = (R th + rx ) ⎜ 1 +
+
⎟
⎝ 1 + g m R E R th + rx ⎠
So the time constant associated with Cμ equals Cμ · Rμo
The high cutoff frequency is calculated using the OCTC as:
ωH
≅
1
2
∑ R ioC i
i =1
ωH ≅
1
⎛ Cπ
⎛
⎛
R L′
RE ⎞
g m R L′
(R th + rx ) ⎜⎜
+
⎜1 +
⎟ + C μ ⎜1 +
⎝ 1 + g m R E R th + rx
⎝ 1 + g m R E ⎝ R th + rx ⎠
⎞⎞
⎟ ⎟⎟
⎠⎠
If we used the same amplifier discussed in Example 2 with CE = 0, we will
obtain:
A mid ≅ −
ωH
ωH
≅
R L′
RE
=
−3.86
= − 2.97 V/V
1.3
1
⎛ Cπ
(R th + rx ) ⎜⎜
⎝ 1+ g m RE
⎞
⎛
g m R L′
R L′ ⎞ ⎞
+
⎟ + C μ ⎜1 +
⎟ ⎟⎟
⎠
⎝ 1 + g m R E R th + rx ⎠ ⎠
1
=
⎛ 20.6 pF ⎛
1.3
66.4 × 3.86
3.86 ⎞ ⎞
⎞
⎛
(0.88 + 0.25)k Ω ⎜
+
⎜1 +
⎟ + 0.5 pF ⎜ 1 +
⎟⎟
⎝ 1 + 66.4 * 1.3 0.88 + 0.25 ⎠ ⎠
⎝ 1 + 66.4 ×1.3 ⎝ 0.88 + 0.25 ⎠
⎛
RE
⎜1 +
⎝ R th + rx
ωH = 211 M rad/sec
Note that this is a higher bandwidth on the expense of a lower midband gain.