# Frequency response I

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```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
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
5
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
6
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
7
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
8
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
9
Single stage Amplifiers
Figure 1. A Common-Source, Common Drain (Source follower)
and Common Gate Single Stage Amplifier in CMOS transistors
10
```