Transcript Frequency response I

Frequency response I
• As the frequency of the processed signals
increases, the effects of parasitic capacitance in
(BJT/MOS) transistors start to manifest
• The gain of the amplifier circuits is frequency
dependent, usually decrease with the frequency
increase of the input signals
• Computing by hand the exact frequency response of
an amplifier circuits is a difficult and time-consuming
task, therefore approximate techniques for obtaining
the values of critical frequencies is desirable
• The exact frequency response can be obtained from
computer simulations (e.g SPICE). However, to
optimize your circuit for maximum bandwidth and to
keep it stable one needs analytical expressions of
the circuit parameters or at least know which
parameter affects the required specification and how
1
Frequency response II
• The transfer function gives us the information about the
behavior of a linear-time-invariant (LTI) circuit/system
for a sinusoidal excitation with angular frequency

T ( ) | T ( ) | exp( jT ( )) | T ( ) | exp( j ( ))
• This transfer function is nothing but the ratio between the
Fourier transforms of the output and input signals and it
is also called the frequency response of the LTI circuit.
• | T ( ) | represents the gain magnitude of the frequency
response of the circuit, whereas  ( ) is the phase of
frequency response. Often, it is more convenient to
express gain magnitude in decibels.
• To amplify a signal without distortion, the amplifier gain
magnitude must be the same for all of the frequency
components
2
Frequency response of amplifers
•
•
•
A bode plot shows the the gain magnitude and phase in decibles versus
frequency on logarithmic scale
A few prerequisites for bode plot:
 Laplace transform and network transfer function
 Poles and zeros of transfer function
 Break frequencies
Some useful rules for drawing high-order bode plots:
 Decompose transfer function into first order terms.
 Mark the break frequencies and represent them on the frequency axis
the critical values for changes
 Make bode plot for each of the first order term
 For each first order term, keep the DC to the break frequency constant
equal to the gain at DC
 After the break frequency, the gain magnitude starts to increase or
decrease with a slope of 20db/decade if the term is in the numerator or
denominator
 For phase plot, each first order term induces a 45 degrees of increase
or decrease at the break frequency if the term is in the numerator or
denominator
 Consider frequencies like one-tenth and ten times the break frequency
and approximate the phase by 0 and 90 degrees if the frequency is with
the numerator (or 0 and -90 degrees if in the denominator)
3
 Add all the first order terms for magnitude and phase response
Frequency response of RLC circuits
E.g. 1
1
1
Av ( f ) 
, fb 
1 j 2RCf
2RC
| Av ( f )|db  20 log 1 ( f / f b ) 2
 ( f )   arctan( f / f b )
E.g. 2
Av ( f ) 
1  j 2fR2C
1
1
, f b1 
, fb2 
1  j 2f ( R1  R2 )C
2R2C
2 ( R1  R2 )C
| Av ( f ) |db  20 log 1  ( f / f b1 ) 2  20 log 1  ( f / f b 2 ) 2
 ( f )  arctan( f / f b1 )  arctan( f / f b 2 )
E.g. 3
Av ( f ) 
j( f / fb )
j 2fR2C
R2
1


, fb 
1  j 2f ( R1  R2 )C R1  R2 1  j ( f / f b )
2 ( R1  R2 )C
| Av ( f ) |db  12  20 log( f / f b )  20 log 1  ( f / f b ) 2
 ( f )  90  arctan( f / f b )
4
The MOS Transistor
Polysilicon
Aluminum
5
The Gate Capacitance
Polysilicon gate
Source
Drain
xd
n+
xd
Ld
W
n+
Gate-bulk
overlap
Top view
Gate oxide
tox
n+
L
n+
Cross section view
6
Gate Capacitance
G
G
CGC
CGC
D
S
Cut-off
G
CGC
D
S
Resistive
D
S
Saturation
Most important regions in digital design: saturation and cut-off
7
Diffusion Capacitance
Channel-stop implant
N A1
Side wall
Source
ND
W
Bottom
xj
Side wall
LS
Channel
Substrate N A
8
Frequency response of common source MOS amplifer
High-frequency MOS Small-signal
equivalent circuit
MOSFET common Source amplifier
Small-signal equivalent circuit for the MOS common source amplifier
Frequency response analysis shows that there are three break frequencies, and mainly
9
the lowest one determines the upper half-power frequency, thus the -3db bandwidth
Exact frequency response of amplifiers
• Exact frequency analysis of amplifier
circuits is possible following the steps:
Draw small-signal equivalent circuit (replace
each component in the amplifier with its smallsignal circuit)
Write equations using voltage and current
laws
Find the voltage gain as a ratio of polynomial
of laplace variable s
Factor numerator and denominator of the
polynomial to determine break frequencies
Draw bode plot to approximate the frequency
response
10
The Miller Effect
• Consider the situation that an impedance is connected
between input and output of an amplifier
 The same current flows from (out) the top input terminal if an
impedance Z in,Miller is connected across the input terminals
 The same current flows to (in) the top output terminal if an
impedance Z out, Miller is connected across the output terminal
 This is know as Miller Effect
 Two important notes to apply Miller Effect:
 There should be a common terminal for input and output
 The gain in the Miller Effect is the gain after connecting feedback
impedance Z f
11
Graphs from Prentice Hall
Application of Miller Effect
• If the voltage gain magnitude is large (say
larger than 10) compared to unity,
Z out, Miller 
Z f Av
Av  1
 Zf
then
we can perform an approximate analysis by
assuming Z out, Miller is equal to Z f
find the gain including loading effects of Z out,Miller ( z f )
use the gain to find out Z in,Miller
Thus, using Miller Effect, gain calculation and
frequency response characterization would be
much simpler
12
Application of Miller Effect
• If the feedback impedance is a capacitor C f ,then
the Miller capacitance reflected across the input
1

terminal is Z
.
jC (1  A )
• Therefore, connecting a capacitance C f from the
input to output is equivalent to connecting a
capacitance C f (1  Av )
• Due to Miller effect, a small feedback
capacitance appears across the input terminals
as a much larger equivalent capacitance with a
large gain (e.g. | Av | 80 ). At high frequencies,
this large capacitance has a low impedance that
tends to short out the input signal
in, Miller
f
v
13
BJT small-signal models (for BJT amplifiers)
The
r  
model for low-frequency analysis
The
Hybrid  
model for high-frequency analysis
rx
The base-spreading resistance for the base region (very small)
r   VT / I CQ
The dynamic resistance of the base emitter region
r
The feedback resistance from collector to base (very large)
ro  V A / I CQ
Account for the upward slope of the output characteristic
C
The depletion capacitance of the collector-to-base region
C
The diffusion capacitance of the base-to-emitter junction
Note: in the following analysis of the CE, EF and CB amplifier in the next three slides, we will
assume rx  0, r   for simplicity (though they still appear in the small signal models).14
Miller Effect: common emitter amplifier I
15
Miller Effect: common emitter amplifier II
Assume the current flowing through C  is very small compared to g m v , then
'
the gain will be Avb '   g m RL considering the input terminal of the amplifier at
b’ to ground.
Applying the miller effect for the amplifier, the following simplified circuit can be
obtained:
C
g m v
Thus, the total capacitance from terminal b’ to ground is given as follows
(neglect the miller capacitance from output terminal c to ground):
CT  C  C (1  g m RL' )
Rs'  Rs || R1 || R2 || r
1
fb 
2Rs' CT
The break frequency, thus the -3db frequency is set by the RC lowpass filter
(other voltage controlled current source, resistance does not contribute to the
16
break frequency). C  is a main limiting factor for -3db bandwidth.
Emitter-follower amplifier
• Using Miller Effect, we
obtain
the
above
equivalent circuit. If
neglecting rx , r , It
shows that the break
frequency is
fb 
1
2 RT CT
CT  C  
C
1  g m RL'
RT  Rs' || [r (1  g m RL' )]
Rs'  Rs || RB
RL'  ro || RE || RL
17
Common base amplifier
• What about amplifier that
do not have capacitance
connected directly from
output to the input?
For approximate analysis, we can neglect rx ( short circuit ), r , ro (open circuit ) the simplified
equivalent circuit can be shown in (c). Derive the transfer function for this circuit, it shows two
break frequencies (with typical values, f b1 is approximately -3db bandwidth)
1
1
'
'
f b1 
,
f

,
R

R
||
R
||
r
||
(
1
/
g
),
R
 Rc || RL
18
b
2
s
s
E

m
L
2 C Rs'
2 C  RL'