SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 1/14
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 2/14
EE 311
Electronic Devices & Circuits 2
Lecture 25
10.3
High-Frequency Response of
the CS Amplifier
Kwang W. Oh, Ph.D., Associate Professor
SMALL (Sensors & MicroActuators Learning Lab)
Department of Electrical Engineering
University at Buffalo, The State University of New York
113C Davis Hall, Buffalo, NY 14260-1920
Tel: (716) 645-1025, Fax: (716) 645-3656
kwangoh@buffalo.edu, http://www.SMALL.Buffalo.edu
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 3/14
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 4/14
High-Frequency Performance
The CS Amplifier
● Objective is to identify the mechanism that limits high-frequency
performance.
 As well as find AM (the midband gain).
● Figure 10.18(a) shows
high-frequency equivalentcircuit model of a CS
amplifier.
● MOSFET is replaced with
model of Figure 10.18(b).
● It may be simplified using
Thevenin’s theorem.
● Also, bridging capacitor
(Cgd) may be redefined.
● Cgd gives rise to much
larger capacitance Ceq.
 The multiplication effect
that it undergoes is
known as the Miller
Effect.
Figure 10.18(d): The
frequency-response plot, which
is that of a low-pass, singletime-constant circuit.
Observe that the gain does not
fall off at low frequencies, and
the midband gain AM extends
down to zero frequency.
Figure 10.18: Determining the high-frequency response of the CS amplifier: (a) equivalent circuit; (b)
the circuit of (a) simplified at the input and the output; (c) the equivalent circuit with Cgd replaced at the
input side with the equivalent capacitance Ceq; (d) the frequency response plot, which is that of a lowpass, single-time-constant circuit.
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EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 5/14
Low-pass filter:
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 6/14
●
, where,
Example 10.3
∥
∥
150 ∥ 15 ∥ 15
1
7.14
7.14V/V
7.14kΩ
●
● Midbandgain 10.50 ⇒ 10.50
⇒ 10.43 A
,
Miller Effect
1
4.7
●
where
(c)
1
⇒ 10.53
2
Figure 10.18: Determining the high-frequency response of the CS amplifier: (a) equivalent circuit; (b)
the circuit of (a) simplified at the input and the output; (c) the equivalent circuit with Cgd replaced at
the input side with the equivalent capacitance Ceq; (d) the frequency response plot, which is that of
a low-pass, single-time-constant circuit.
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 7/14
∥
1
●
●
∥
7.14
150 ∥ 15 ∥ 15
7.14V/V
⇒A
0.4
1
●
∥
●
3.26
1
4.26pF
∥ 4700
● Upper3dBfrequency 10.53 .
.
. 7.14
.
∥
100 ∥ 4700
Upper3dBfrequency 10.53 ∥
1
●
●
∥
7.14
1
1
2
2
4.26
10
97.9
10
382kHz
Exercise 10.9
150 ∥ 15 ∥ 15
7.14V/V
7.14kΩ
● Midbandgain 10.50 7.14
3.26pF
.
/
4.7
4.7 100
⇒A
1
●
1
1
●
97.9kΩ
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 8/14
7.14kΩ
4.7
4.7
1
●
●
Exercise 10.8
● Midbandgain 10.50 ●
7.14
●
4.7
7.14
7V/V
100
1
0.4
1
3.26pF
1 3.26 4.26pF
∥
●
7.14
100 ∥ 4700
.
7.12V/V
8.14
8.14
97.9kΩ
● Upper3dBfrequency 10.53
.
7.14
⇒ 1.63
⇒ 1MHz
1
8.14
⇒
0.077pF
SUNY–Buffalo | Electrical Engineering
EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 9/14
VB
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EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 10/14
EE 311
Electronic Devices & Circuits 2
Lecture 25
10.3.2
High-Frequency Response of
the CE Amplifier
Kwang W. Oh, Ph.D., Associate Professor
SMALL (Sensors & MicroActuators Learning Lab)
Department of Electrical Engineering
University at Buffalo, The State University of New York
113C Davis Hall, Buffalo, NY 14260-1920
Tel: (716) 645-1025, Fax: (716) 645-3656
kwangoh@buffalo.edu, http://www.SMALL.Buffalo.edu
The CE Amplifier
● Figure 10.19(a) shows
high-frequency
equivalent circuit of a CE
amplifier.
 BJT is replaced.
 This figure applies to
both discrete and IC
amps.
● This figure may be
simplified using
Thevenin’s theorem.
● Cin is simply sum of C
and Miller capacitance
C(1+gmRL’)
Please prove this by yourself!!!
Figure 10.19 Determining the high-frequency response of the CE amplifier: (a) equivalent circuit; (b)
the circuit of (a) simplified at both the input side and the output side; (c) equivalent circuit with C
replaced at the input side with the equivalent capacitance Ceq; (d) sketch of the frequency-response
plot, which is that of a low-pass STC circuit.
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EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 11/14
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EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 12/14
Example 10.4
10.9(a)
●
∥
● Low-pass filter:




, where,
10.59 ∥
∥
10.58 Millercapacitance:
10.55 ∥
1
1
∥
●
40mA/V
●
2.5kΩ
●
100kΩ
8pF
● (10.40 & 10.41)
⇒
●
●
∥
●
10.56
8pF 1pF 7pF
100 ∥ 8 ∥ 5 3kΩ
∥
● Midband gain:
∥
.
∥


[From 6th Ed] Fig. 9.4
10.54 32dB
∥
⇒ 10.57 Figure 10.19 Determining the high-frequency response of the CE amplifier: (a) equivalent circuit; (b)
the circuit of (a) simplified at both the input side and the output side; (c) equivalent circuit with C
replaced at the input side with the equivalent capacitance Ceq; (d) sketch of the frequency-response
plot, which is that of a low-pass STC circuit.
.
.
40
∥
1
●
3
3
39V/V ⇒ 20 log
7
1 1
40
128pF
∥
●
∥
2.5 ∥ 0.05
100 ∥ 5
1.65kΩ
● Upper 3-dB frequency:
754kHz
.
Coupling capacitors are used in order
not to disturb the dc bias currents and
voltages, through a large capacitor
and
. In analog circuits, a
coupling capacitor is used to connect
two circuits such that only the AC
signal from the first circuit can pass
through to the next while DC is
blocked. This technique helps to
isolate the DC bias settings of the two
coupled circuits.
The purpose of the bypass capacitor
is to create an AC GROUND at the
emitter. The capacitor is a bypass
capacitor bypassing AC noise.
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EE 311 Electronic Devices & Circuits 2
Lecture 25 | Chapter 10 | 4/5 | 13/14
Exercise 10.10
10.4,
19.5V/V
●
∥
●
∥
[From 6th Ed] Fig. 9.4
100 ∥ 8 ∥
●
7.4 ∥
∥
.
.
.
.
.
.
40
∥
1.5kΩ
7.4 ∥
0.013
⇒
1
●
1.5
[From 6thSUNY–Buffalo
Ed] Figure 9.5:
Analysis Engineering
of the
| Electrical
low-frequency
response
of the
CE amplifier
EE 311
Electronic
Devices
& Circuits 2
of Fig. 9.4:
(a) the25effect
of CC110
is |determined
Lecture
| Chapter
4/5 | 14/14
with CE and CC2 assumed to be acting as
perfect short circuits; (b) the effect of CE is
determined with CC1 and CC2 assumed to be
acting as perfect short circuits; (c) the effect
of CC2 is determined with CC1 and CE
assumed to be acting as perfect short
circuits; (d) sketch of the low-frequency gain
under the assumptions that CC1, CE, and CC2
do not interact and that their break (or pole)
frequencies are widely separated.
1p 1
∥
→
•
Reflecting ∥
intoemitterneeds1/
1.9kΩ
7p
∥
∥
Fig. 9.4
⇒
→
•
1/s
1
1/s
40
1
68pF
∥
●
∥
2.5 ∥ 0.05
100 ∥ 5
1.65kΩ
1/s
● Upper 3-dB frequency:
.
•
1.42MHz
(d)
→
1 1