-
Chapter 11 Frequency Response
11.1 Fundamental Concepts 11.2 High-Frequency Models of Transistors 11 3 Analysis Procedure 11.3 Analysis Procedure 11.4 Frequency Response of CE and CS Stages 11.5 Frequency Response of CB and CG Stagesq y p g 11.6 Frequency Response of Followers 11.7 Frequency Response of Cascode Stage 11.8 Frequency Response of Differential Pairs 11.9 Additional Examples
CH 10 Differential Amplifiers 11
-
Chapter Outline
CH 10 Differential Amplifiers 2CH 11 Frequency Response 2
-
High Frequency Roll-off of Amplifier
As frequency of operation increases, the gain of amplifier
CH 10 Differential Amplifiers 3CH 11 Frequency Response 3
As frequency of operation increases, the gain of amplifier decreases. This chapter analyzes this problem.
-
Example: Human Voice I
Natural Voice Telephone System
Natural human voice spans a frequency range from 20Hz to 20KHz, however conventional telephone system passes
CH 10 Differential Amplifiers 4
p y pfrequencies from 400Hz to 3.5KHz. Therefore phone conversation differs from face-to-face conversation. CH 11 Frequency Response 4
-
Example: Human Voice II
Path traveled by the human voice to the voice recorder
Mouth RecorderAir
Path traveled by the human voice to the human ear
Mouth EarAir
SkullSkull
Since the paths are different, the results will also be
CH 10 Differential Amplifiers 5CH 11 Frequency Response 5
different.
-
Example: Video Signal
High Bandwidth Low BandwidthHigh Bandwidth o a d dt
Video signals without sufficient bandwidth become fuzzy as they fail to abruptly change the contrast of pictures from complete white into complete black.
CH 10 Differential Amplifiers 6
complete white into complete black.
CH 11 Frequency Response 6
-
Gain Roll-off: Simple Low-pass Filter
In this simple example, as frequency increases the impedance of C decreases and the voltage divider consists
CH 10 Differential Amplifiers 7
impedance of C1 decreases and the voltage divider consists of C1 and R1 attenuates Vin to a greater extent at the output.
CH 11 Frequency Response 7
-
Gain Roll-off: Common Source
1||
1||out m in DL
V g V RC s
The capacitive load, CL, is the culprit for gain roll-off since
CH 10 Differential Amplifiers 8CH 11 Frequency Response 8
at high frequency, it will “steal” away some signal current and shunt it to ground.
-
Frequency Response of the CS Stage
RV1222
LD
Dm
in
out
CRRg
VV
At low frequency, the capacitor is effectively open and the gain is flat. As frequency increases, the capacitor tends to
CH 10 Differential Amplifiers 9CH 11 Frequency Response 9
g q y , pa short and the gain starts to decrease. A special frequency is ω=1/(RDCL), where the gain drops by 3dB.
-
Example: Figure of Merit
CVVMOF 1...
LCCT CVV
This metric quantifies a circuit’s gain bandwidth and This metric quantifies a circuit s gain, bandwidth, and power dissipation. In the bipolar case, low temperature, supply, and load capacitance mark a superior figure of
itCH 10 Differential Amplifiers 10CH 11 Frequency Response 10
merit.
-
Example: Relationship between Frequency Response and Step Responsep p p
2 2 2
1H s j 0 1 expouttV t V u t
R C
2 2 21 1 1R C
01 1
pout R C
The relationship is such that as R1C1 increases, the bandwidth drops and the step response becomes slower.
CH 10 Differential Amplifiers 11CH 11 Frequency Response 11
bandwidth drops and the step response becomes slower.
-
Bode Plot
11 ss
21011
)( zzss
AsH
21
11pp
When we hit a zero, ωzj, the Bode magnitude rises with a slope of +20dB/dec.
CH 10 Differential Amplifiers 12CH 11 Frequency Response 12
When we hit a pole, ωpj, the Bode magnitude falls with a slope of -20dB/dec
-
Example: Bode Plot
1LD
p CR1
1
The circuit only has one pole (no zero) at 1/(RDCL), so the l d f 0 t 20dB/d
CH 10 Differential Amplifiers 13CH 11 Frequency Response 13
slope drops from 0 to -20dB/dec as we pass ωp1.
-
Pole Identification Example I
11 p CR
12
inSp CR1
LD
p CR
222212 11Dm
in
out RgVV
CH 10 Differential Amplifiers 14CH 11 Frequency Response 14
21 11 ppin
-
Pole Identification Example II
1 1inS
p
Cg
R
1||1
LDp CR
12
CH 10 Differential Amplifiers 15CH 11 Frequency Response 15
mg
-
Circuit with Floating Capacitor
The pole of a circuit is computed by finding the effective resistance and capacitance from a node to GROUND.
CH 10 Differential Amplifiers 16CH 11 Frequency Response 16
The circuit above creates a problem since neither terminal of CF is grounded.
-
Miller’s Theorem
F
AZZ
11
F
AZZ
/112
If Av is the gain from node 1 to 2, then a floating impedance Z can be converted to two grounded impedances Z and Z
vA1 vA/11
CH 10 Differential Amplifiers 17CH 11 Frequency Response 17
ZF can be converted to two grounded impedances Z1 and Z2.
-
Miller Multiplication
With Miller’s theorem, we can separate the floating capacitor However the input capacitor is larger than the
CH 10 Differential Amplifiers 18CH 11 Frequency Response 18
capacitor. However, the input capacitor is larger than the original floating capacitor. We call this Miller multiplication.
-
Example: Miller Theorem
in CRR
11 out
CR
11
1 FDmS CRgR 1 F
DmD CRg
R
1
CH 10 Differential Amplifiers 19CH 11 Frequency Response 19
-
High-Pass Filter Response
1212
12
1
11
CRCR
VV
in
out
The voltage division between a resistor and a capacitor can be configured such that the gain at low frequency is reduced
CH 10 Differential Amplifiers 20
reduced.
CH 11 Frequency Response 20
-
Example: Audio Amplifier
nFCi 6.79 nFCL 8.39
200/1100i
gKR
In order to successfully pass audio band frequencies (20 Hz-20 KHz), large input and output capacitances are needed.
200/1mg
CH 10 Differential Amplifiers 21CH 11 Frequency Response 21
-
Capacitive Coupling vs. Direct Coupling
Capacitive Coupling Di t C liCapacitive Coupling Direct Coupling
Capacitive coupling, also known as AC coupling, passes AC signals from Y to X while blocking DC contents.
CH 10 Differential Amplifiers 22
This technique allows independent bias conditions between stages. Direct coupling does not.CH 11 Frequency Response 22
-
Typical Frequency Response
Lower Corner Upper Corner
CH 10 Differential Amplifiers 23CH 11 Frequency Response 23
-
High-Frequency Bipolar Model
b jeC C C
At high frequency, capacitive effects come into play. Cb represents the base charge, whereas C and Cje are the
CH 10 Differential Amplifiers 24CH 11 Frequency Response 24
je junction capacitances.
-
High-Frequency Model of Integrated Bipolar Transistor
Since an integrated bipolar circuit is fabricated on top of a
CH 10 Differential Amplifiers 25CH 11 Frequency Response 25
substrate, another junction capacitance exists between the collector and substrate, namely CCS.
-
Example: Capacitance Identification
CH 10 Differential Amplifiers 26CH 11 Frequency Response 26
-
MOS Intrinsic Capacitances
For a MOS, there exist oxide capacitance from gate to channel junction capacitances from source/drain to
CH 10 Differential Amplifiers 27CH 11 Frequency Response 27
channel, junction capacitances from source/drain to substrate, and overlap capacitance from gate to source/drain.
-
Gate Oxide Capacitance Partition and Full Model
The gate oxide capacitance is often partitioned between The gate oxide capacitance is often partitioned between source and drain. In saturation, C2 ~ Cgate, and C1 ~ 0. They are in parallel with the overlap capacitance to form CGS and C
CH 10 Differential Amplifiers 28CH 11 Frequency Response 28
CGD.
-
Example: Capacitance Identification
CH 10 Differential Amplifiers 29CH 11 Frequency Response 29
-
Transit Frequency
gf mT 2mTgf 2
C
fT2GS
T Cf2
Transit frequency, fT, is defined as the frequency where the current gain from input to output drops to 1
CH 10 Differential Amplifiers 30CH 11 Frequency Response 30
current gain from input to output drops to 1.
-
Example: Transit Frequency Calculation
THGSnT VVLf 2232
VVVnmL
10065
GHfsVcm
mVVV
n
THGS
226)./(400
1002
CH 10 Differential Amplifiers 31
GHzfT 226
CH 11 Frequency Response 31
-
Analysis Summary
The frequency response refers to the magnitude of the transfer function.
Bode’s approximation simplifies the plotting of the frequency response if poles and zeros are known.
In general, it is possible to associate a pole with each node In general, it is possible to associate a pole with each node in the signal path.
Miller’s theorem helps to decompose floating capacitors i t d d l tinto grounded elements.
Bipolar and MOS devices exhibit various capacitances that limit the speed of circuits.p
CH 10 Differential Amplifiers 32CH 11 Frequency Response 32
-
High Frequency Circuit Analysis Procedure
Determine which capacitor impact the low-frequency region of the response and calculate the low-frequency pole (neglect transistor capacitance).(neglect transistor capacitance).
Calculate the midband gain by replacing the capacitors with short circuits (neglect transistor capacitance).
Include transistor capacitances. Merge capacitors connected to AC grounds and omit those
that play no role in the circuit.t at p ay o o e t e c cu t Determine the high-frequency poles and zeros. Plot the frequency response using Bode’s rules or exact
l ianalysis.
CH 10 Differential Amplifiers 33CH 11 Frequency Response 33
-
Frequency Response of CS Stagewith Bypassed Degeneration
1sCRRgV 11
SmbS
bSDm
X
out
RgsCRsCRRgs
VV
I d i h idb d i i C i In order to increase the midband gain, a capacitor Cb is placed in parallel with Rs.
The pole frequency must be well below the lowest signal
CH 10 Differential Amplifiers 34
p q y gfrequency to avoid the effect of degeneration.
CH 11 Frequency Response 34
-
Unified Model for CE and CS Stages
CH 10 Differential Amplifiers 35CH 11 Frequency Response 35
-
Unified Model Using Miller’s Theorem
CH 10 Differential Amplifiers 36CH 11 Frequency Response 36