17.6
High-Frequency Limitations of Common-Emitter and Common-Source Amplifiers
1309
17.6 HIGH-FREQUENCY LIMITATIONS OF COMMONEMITTER AND COMMON-SOURCE AMPLIFIERS
Now that the complete hybrid-pi model has been described, we can explore the high-frequency
limitations of the three basic single-stage amplifiers. This section begins with an example of direct
analysis of the common-emitter amplifier; the approximation technique for estimating ω H , the
open-circuit time-constant method, is then presented.
17.6.1 Example of Direct High-Frequency Analysis —
The Common-Emitter Amplifier
The common-emitter amplifier circuit discussed earlier in this chapter is redrawn in Fig. 17.31. At
high frequencies, the impedance of the coupling and bypass capacitors are negligibly small, and
the three capacitors can once again be considered short circuits, resulting in the high-frequency
ac model in Fig. 17.31(b). In this figure, R L represents the parallel combination of R3 and RC
(100 k4.3 k), the emitter is connected directly to ground through bypass capacitor C3 ,
and base bias resistor R B equals the parallel combination of R1 and R2 (30 k10 k). Load
capacitance C L has been added to the circuit for completeness. In most multistage amplifiers,
the common-emitter or common-source amplifiers will be driving an equivalent load capacitance
presented by the input impedance of the next amplifier stage.
Analysis of the circuit in Fig. 17.31 begins by replacing the transistor with its small-signal
model, as in Fig. 17.32. In this figure, the signal source and associated resistors have been replaced
VCC = 12 V
R2
RI
1 kΩ
vI
RC
4.3 kΩ
C2
30 kΩ
C1
0.1 µF
R3
100 kΩ
2 µF
+
vO
–
CL
R1
RE
C3
1.3 kΩ
10 µF
10 kΩ
(a)
Thévenin
transformation
RI
1 kΩ
vi
7.5 kΩ
RB
RL
4.12 kΩ
CL
(b)
Figure 17.31 (a) Common-emitter amplifier. (b) High-frequency ac model for amplifier in (a).
1310
Chapter 17
Frequency Response
Rth
Cµ
rx
v1
rπ
vth
Cπ
gmv1
v2
CL
RL
Figure 17.32 ac Model for a common-emitter amplifier at high frequencies.
Cµ
Rth + rx
vth
rπ
is
v1
is
rπ o
gmv1
v1
rπ o
v2
CL
RL
Cπ
Figure 17.33 Norton source transformation for
Figure 17.34 Simplified small-signal model for the high-frequency
C-E amplifier.
common-emitter amplifier.
by a Thévenin equivalent circuit in which
vth = vi
RB
RI + RB
and
Rth =
RI RB
RI + RB
(17.72)
Direct analysis proceeds using nodal equations. The network in Fig. 17.32 can be reduced to
a two-node problem by using a final Norton transformation of the resistors and voltage source
attached to node v1 , as indicated in Fig. 17.33. The short-circuit current and Thévenin equivalent
resistance for this network are expressed by
is =
vth
Rth + r x
rπ o = rπ (Rth + r x )
and
(17.73)
Figure 17.34 shows the common-emitter amplifier in final simplified form.
Writing and simplifying the nodal equations in the frequency domain for the circuit in
Fig. 17.34 yields Eq. (17.74):
Is (s)
0
=
s(Cπ + Cµ ) + gπ o
−sCµ
−(sCµ − gm )
s(Cµ + C L ) + g L
V1 (s)
V2 (s)
(17.74)
An expression for the output voltage, node voltage V2 (s), can be found using Cramer’s rule:
V2 (s) = Is (s)
(sCµ − gm )
(17.75)
in which represents the determinant of the system of equations given by
= s 2 [Cπ (Cµ + C L ) + Cµ C L ] + s[Cπ g L + Cµ (gm + gπ o + g L ) + C L gπ o ] + g L gπ o (17.76)
From Eqs. (17.75) and (17.76), we see that the high-frequency response is characterized by
two poles, one finite zero, and one zero at infinity. The finite zero appears in the right-half of the
s-plane at a frequency
ωZ = +
gm
> ωT
Cµ
(17.77)
17.6
High-Frequency Limitations of Common-Emitter and Common-Source Amplifiers
1311
The zero given by Eq. (17.77) can usually be neglected because it appears at a frequency above ωT
(for which the model itself is of questionable validity). Unfortunately, the denominator appears
in unfactored polynomial form, and the positions of the poles are more difficult to find. However,
good estimates for both pole positions can be found using the approximate factorization technique
in Sec. 17.6.2. Note that even though there are three capacitors, the circuit only has two poles.
The three capacitors are connected in a “pi” configuration, and only two of the capacitor voltages
are independent. Once we know two of the voltages, the third is also defined.
17.6.2 Approximate Polynomial Factorization
We estimate the pole locations based on a technique for approximate factorization of polynomials.
Let us assume that the polynomial has two real roots a and b:
(s + a)(s + b) = s 2 + (a + b)s + ab = s 2 + A1 s + A0
(17.78)
If we assume that a dominant root exists — that is, that a b — then the two roots can be
estimated directly from coefficients A1 and A0 using two approximations:
A1 = a + b ∼
=a
a∼
= A1
so,
and
A0
ab ∼ ab
=
=b
=
A1
a+b
a
b∼
=
and
(17.79)
A0
A1
Note in Eq. (17.78) that the s 2 term is normalized to unity.
The approximate factorization technique exemplified by Eqs. (17.78) and (17.79) can be
extended to polynomials having any number of widely spaced real roots. For a third-order polynomial, for example,
(s + a)(s + b)(s + c) = s 3 + A2 s 2 + A1 s + A0
Assuming c b a,
(17.80)
c∼
= A0 /A1
b∼
= A1 /A2 ,
and a ∼
= A2
17.6.3 Poles of the Common-Emitter Amplifier —
The C T Approximation
For the case of the common-emitter amplifier, the smallest root is the most important because it is
the one that limits the frequency response of the amplifier and determines ω H . From Eq. (17.79)
we see that the smaller root is given by the ratio of coefficients A0 and A1 :
g L gπ o
A0
ω P1 ∼
=
=
A1
Cπ g L + Cµ (gm + gπ o + g L ) + C L gπ o
or
(17.81)
ω P1
1
∼
=
rπ o [Cπ + Cµ (1 + gm R L )] + (Cµ + C L )R L
1312
Chapter 17
Frequency Response
The dominant pole is set by two time constants: rπ o times the input capacitance consisting of Cπ
and the Miller amplification of Cµ , plus the load resistance times the total capacitance (Cµ + C L )
connected to the output node. We can also view the lower-frequency pole as determined by resistor
rπ o and the total effective input capacitance C T expressed by Eq. (17.82):
RL
RL
C T = Cπ + Cµ 1 + gm R L +
+ CL
(17.82)
rπ o
rπ o
C T consists of Cπ plus Cµ multiplied by (1 + gm R L + R L /rπ o ) plus an additional term due to
load capacitance C L . The factor multiplying Cµ can be quite large because gm R L represents the
intrinsic voltage gain of the common-emitter amplifier and can be expected to be of the order of
10–20VCC .
As we will see in Chapter 18, the second pole often plays an important role in frequency
compensation of operational amplifiers because the pole can degrade the phase margin of feedback
amplifiers. The location of the second pole of the amplifier is estimated from coefficient A1 alone
(in properly normalized form):
Cπ g L + Cµ (gm + gπ o + g L ) + C L gπ o
ω P2 ∼
=
Cπ (Cµ + C L ) + Cµ C L
or
(17.83)
ω P2
gm
gm
∼
∼
=
=
CL
Cπ + C L
+ CL
Cπ 1 +
Cµ
in which the Cµ gm term has been assumed to be the largest term in the numerator, as is most
often the case for C-E stages with reasonably high gain. We can interpret the last approximation
in Eq. (17.83) in this manner, particularly when Cµ is large. At high frequencies, capacitor Cµ
shorts the collector and base of the transistor together so that C L and Cπ appear in parallel, and
the transistor behaves as a diode with resistance of 1/gm .
Overall Transfer Function for the Common-Emitter Amplifier
An overall expression for the gain of the common-emitter amplifier can be obtained by combining
Eqs. (17.72) to (17.76), (17.81), and (17.83):
Vo (s) =
Vth (s)
Rth + r x
g L gπ o
(sCµ − gm )
s
s
1+
1+
ω P1
ω P2
and
s
1−
Vth (s)
ω
Z
Vo (s) =
(−gm R L rπ o ) s
s
Rth + r x
1+
1+
ω P1
ω P2
(17.84)
For C L = 0, ω Z and ω P2 are both greater than ωT , and Eq. (17.84) can be approximated by
Vth (s) (gm R L rπ o )
Vo (s) ∼
=−
s
Rth + r x
1+
ω P1
(17.85)
17.6
1313
High-Frequency Limitations of Common-Emitter and Common-Source Amplifiers
Recognizing that rπ o = rπ (Rth + r x )/(Rth + r x + rπ ) yields
βo R L
Vo (s) ∼
Rth + r x + rπ
Avth (s) =
=− s
Vth (s)
1+
ω P1
(17.86)
or
Avth (s) ∼
=
Amid
s
1+
ω P1
(17.87)
in which
Amid = −
βo R L
Rth + r x + rπ
and
ω P1 =
1
rπ o C T
(17.88)
Equation (17.87) indicates that the high-frequency behavior of the common-emitter amplifier can
be modeled by a single dominant pole, as in the circuit in Fig. 17.35.
Rth + rx
vth
rπ
v1
CT
gmv1
vo
RL
Figure 17.35 Dominant-pole model for the common-emitter amplifier at high frequencies.
Before we leave this section, it should be noted again that we have neglected the right-halfplane zero ω Z that appears in Eq. (17.84). Since ω Z is high in frequency (beyond ωT ), it has
minimal impact on the magnitude of the transfer function. However, the additional phase shift
that this zero introduces at high frequency can degrade the stability of feedback amplifiers that
we study in Chapter 18.
EXAMPLE 17.6
HIGH-FREQUENCY ANALYSIS OF THE COMMON-EMITTER AMPLIFIER
Find the midband gain and upper-cutoff frequency of a common-emitter amplifier.
PROBLEM Find the midband gain and upper-cutoff frequency of the common-emitter amplifier in Fig. 17.31
using the C T approximation, assuming βo = 100, f T = 500 MHz, Cµ = 0.5 pF, r x = 250 ,
and a Q-point of (1.60 mA, 3.00 V). Find the additional poles and zeros of the common-emitter
amplifier. Assume C L = 0.
SOLUTION Known Information and Given Data: Common-emitter amplifier circuit in Fig. 17.31;
Q-point = (1.60 mA, 3.00 V); βo = 100, f T = 500 MHz, Cµ = 0.5 pF, and r x = 250 ;
expressions for the gain, poles, and zeros are given in Eqs. (17.81) through (17.88).
Unknowns: Values for Amid , f H , ω Z 1 , ω P1 , and ω P2
Approach: Find the small-signal parameters for the transistor. Find the unknowns by substituting
the given and computed values into the expressions developed in the text.
1314
Chapter 17
Frequency Response
Assumptions: Small-signal operation in the active region; VT = 25.0 mV; C L = 0
Analysis: The common-emitter stage is characterized by Eqs. (17.82), (17.83), and (17.88).
βo R L
gm
Amid = −
ωZ = +
Rth + r x + rπ
Cµ
1
1
1
gm
1
1+
ω P1 =
ω P2 =
+
+
rπ o C T
R L Cµ
Cπ
gm rπ o
gm R L
RL
rπ o = rπ (Rth + r x )
C T = Cπ + Cµ 1 + gm R L +
rπ o
The values of the various small-signal parameters must be found:
βo
100
gm = 40IC = 40(0.0016) = 64.0 mS
rπ =
=
= 1.56 k
gm
0.064
gm
0.064
Cπ =
− Cµ =
− 0.5 × 10−12 = 19.9 pF
2π f T
2π(5 × 108 )
R L = RC R3 = 4.3 k100 k = 4.12 k
Rth = R B R I = 7.5 k1 k = 882 rπ o = rπ (Rth + r x ) = 1.56 k(882 + 250 ) = 656 Substituting these values into the expression for C T yields
RL
C T = Cπ + Cµ 1 + gm R L +
rπ o
4120
= 19.9 pF + 0.5 pF 1 + 0.064(4120) +
656
= 19.9 pF + 0.5 pF(1 + 264 + 6.28) = 156 pF
and
1
1
=
= 1.56 MHz
2πrπ o C T
2π(656 )(156 pF)
1
1
gm
1
1+
=
+
+
R L Cµ
Cπ
gm rπ o
gm R L
1
0.064
1
1
=
+
1+
+
(4.12 k)(0.5 pF) 19.9 pF
0.064(656) 0.064(4120)
ω P2
f P2 =
= 603 MHz
2π
gm
0.064
fz =
=
= 20.4 GHz
2πCµ
2π(0.5 pF)
f P1 =
ω P2
βo R L
100(4120)
= −153
=−
Rth + r x + rπ
882 + 250 + 1560
Remember that the overall gain from vs to the output will be reduced to Amid = 0.882(−153) =
−135 since vth = 0.882vs .
Avth = −
Check of Results: We have found the desired information. By double checking, the calculations
appear correct. Let us use the gain-bandwidth product as an additional check: |Amid f H | =
239 MHz, which does not exceed f T .
17.6
High-Frequency Limitations of Common-Emitter and Common-Source Amplifiers
1315
Discussion: The dominant pole is located at a frequency f P1 = 1.56 MHz, whereas f P2 and
f Z are estimated to be at frequencies above f T (500 MHz). Thus, the upper-cutoff frequency
f H for this amplifier is determined solely by f P1 : f H ∼
= 1.56 MHz. Note that this value of f H is
less than 1 percent of the transistor f T and is consistent with the concept of GBW product. We
should expect f H to be no more than f T /Amid = 3.3 MHz for this amplifier. Note also that f P1
and f P2 are separated by a factor of almost 1000, clearly satisfying the requirement for widely
spaced roots that was used in the approximate factorization.
It is important to keep in mind that the most important factor in determining the value of C T
is the term in which Cµ is multiplied by gm R L . To increase the upper-cutoff frequency f H of
this amplifier, the gain (gm R L ) must be reduced; a direct trade-off must occur between amplifier
gain and bandwidth.
Computer-Aided Analysis: SPICE can be used to check our hand analysis, but we must
define the device parameters that match our analysis. We set BF = 100 and IS = 5 fA, but
let VAF default to infinity. The base resistance r x must be added by setting SPICE parameter
RB = 250 . Cµ is determined in SPICE from the value of the zero-bias collector-junction
capacitance CJC and the built-in potential φ jc . The Q-point from SPICE gives VC E = 2.70 V,
which corresponds to VBC = 2.0 V if VB E = 0.7 V. In SPICE, VJC faults to 0.75 V and MJC
defaults to 0.33 (see Sec. 17.4). Therefore, to achieve Cµ = 0.5 pF, CJC is specified as
20 V 0.33
CJC = 0.5 pF 1 +
= 0.768 pF
0.75 V
Cπ is determined by the SPICE forward transit-time parameter TF as defined by Eqs. (5.39) and
(17.50):
TF =
Cπ
19.9 pF
= 0.311 ns
=
gm
64 mS
After adding these values to the transistor model, we perform an ac analysis using FSTART =
100 Hz and FSTOP = 10 MHz with 20 frequency points per decade. The SPICE simulation
results in the graph below yield Amid = −135 (42.6 dB) and f H ∼
= 1.56 MHz, which agrees
closely with our hand calculations. Checking the device parameters in SPICE, we also find
r x = 250 , Cπ = 19.9 pF, and Cµ = 0.499 pF, as desired.
Av (dB)
+45
+40
fH
+35
+30
+25
+20
+100
+1000
+10K
+100K
Frequency (Hz)
+1Meg
+10Meg
1316
Chapter 17
Frequency Response
Exercise: Use SPICE to recalculate f H for V A = 75 V. For V A = 75 V and r x = 0?
Answer: 1.67 MHz; 1.96 MHz
Exercise: Repeat the calculations in Ex. 17.6 if a load capacitance CL = 3 pF is added to
the circuit.
Answers: 1.39 MHz, 71.6 MHz (A significant decrease in f P2 !)
Exercise: Find the midband gain and the frequencies of the poles and zeros of the
common-emitter amplifier in Ex. 17.6 if the transistor has fT = 500 MHz, but Cµ = 1 pF.
Answers: −153, 835 kHz, 578 MHz, 10.2 GHz
17.6.4 Gain-Bandwidth Product Limitations
of the Common-Emitter Amplifier
The important role of r x in ultimately limiting the frequency response of the C-E amplifier should
also not be overlooked. If the Thévenin equivalent source resistance Rth were reduced to zero
in an attempt to increase the bandwidth, then rπ o would not become zero but would be limited
approximately to the value of r x . Let us look at the asymptotic behavior of the gain-bandwidth
product of the C-E amplifier,
GBW = |Av ω H | ≤
βo R L
Rth + r x + rπ
1
rπ o C T
(17.89)
using these approximations:
Rth = 0, r x rπ
so that
rπ o ∼
= r x and C T ∼
= Cµ (gm R L )
(17.90)
1
r x Cµ
(17.91)
Substituting these values in Eq. (17.89) yields
GBW ≤
The product of the base resistance r x and the collector-base capacitance Cµ places an upper
bound on the gain-bandwidth product of the C-E amplifier. From Eq. (17.91), we can see that
one important consideration in the design of transistors for very high frequency operation is
minimization of the r x Cµ product.
Exercise: Compare the gain-bandwidth product of the amplifier in Ex. 17.6 to the limit in
Eq. (17.91) and to the fT of the transistor.
Answers: 239 MHz < 1.27 GHz, 239 MHz ∼
= 500 MHz/2
17.7
1317
Miller Multiplication
C2
RI
C1
M
1 kΩ
vi
0.1 µF
RG
RD
0.1 µF
R3
vo
100 kΩ
4.3 kΩ
996 Ω
243 kΩ
RS
Rth
C3
1.3 kΩ
vth
v1
10 µF
CGD
CGS
RL
v2
4.12 kΩ
gmv1
(b)
(a)
Figure 17.36 (a) Common-source amplifier. (b) The high-frequency small-signal model.
17.6.5 Dominant Pole for the Common-Source Amplifier
Analysis of the C-S amplifier in Fig. 17.36 mirrors that of the common-emitter amplifier. The
small-signal model is similar to that for the C-E stage, except that both r x and rπ are absent from
the model. For Fig. 17.36(b),
Rth = R I RG
R L = R D R3
vth = vi
RG
R I + RG
(17.92)
The expressions for the finite zero and poles of the C-S amplifier can be found by comparing
Fig. 17.36(b) to Fig. 17.34:
ω P1 =
ω P2
1
Rth C T
and
1
gm
=
+
RL CG D
CG S
RL
C T = C G S + C G D 1 + gm R L +
Rth
1
1
gm
1+
ωZ =
+
gm Rth
gm R L
CG D
(17.93)
Exercise: What is the upper-cutoff frequency for the amplifier in Fig. 17.36 if CGS = 10 pF,
CG D = 2 pF, and gm = 1.23 mS? What are the positions of the second pole and the zero?
What is the fT of this transistor?
Answers: 5.26 MHz; 58.7 MHz, 97.9 MHz; 16.3 MHz
17.7 MILLER MULTIPLICATION
The amplification of Cµ by the voltage gain of the C-E amplifier is an example of Miller multiplication. Miller originally derived an expression for the input admittance of the general amplifier
shown in Fig. 17.37, in which capacitor C is connected between the input and output terminals
of an inverting amplifier.
Let us calculate the input admittance of this amplifier. The output voltage vo is equal to the
negative gain times the input voltage vi , and the input current to the amplifier is the admittance
of the capacitor (sC) multiplied by the total voltage across the capacitor:
Vo (s) = −AVI (s)
and
Is (s) = sC [VI (s) − Vo (s)]
(17.94)