Download - Bode Plots & Frequency Response
Bode Plots & Frequency Response
Read Chapter 11 of Razavi
1
2
1 2
1 2
( )
1 1 1
( )
1 1 1
We must factor polynomials in variable
where
z z znn
m
p p pm
Transfer Function H s
s s s
bH s
a s s s
s
s j
General Form of H(s)
3
( )( ) ( ); Let ( ) ( )
,
( )
( ) 1( )
( )
, ( )
st st
st st st
st st
di tRi t L v t i t Ie and v t Ve
dtTherefore
Re sLe I Ve
i t IH s
v t V R sL
So Ie H s Ve
i(t) v(t)
R
L
Example: R-L Circuit
4
1( ) 1
( )( ) 1
where 2 p p
i t I RH ssv t V R sLRL
Rf
L
Example: R-L Circuit
|H(f)|
f fp
0 dB
Slope = + 6 dB/octave & 20 dB/decade
+ -3 dB
f fp
Slope = - 45/decade +
90
45
0
0.1 fp 10 fp
These are log-log plots – frequency is logarithmic
5
R-L and R-C Circuits
Low-pass
High-pass
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Bode Plots for R-L and R-C Circuits Summary
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1 / 2 1 / 2
1 12 2
1 14 4
1 15 5
1 110 10
( ) ( )
1 20 log(1) 0
2 20 log( 2 ) 3
2 20 log(2) 6
4 20 log(4) 12
5 20 log(5) 14
10 20 log(10) 20
20 log( ) 3
20 log( ) 6
20 log( ) 12
20 log( ) 14
20 log( ) 20
dBH s H s
dB
dB
dB
dB
dB
dB
dB
dB
dB
dB
dB
Some Logarithm values convenient for Circuit Analysis
8
Low-pass Filter R-C Network’s Bode Plot
9
1
1
1 1
( )Suppose we have ( ) , then
( )
( ) 20 log 20 log 20 log
10( 100): ( )
( 1)
dB
A S zH s
S p
H s A S z S p
sExample H s
s
A = 10 Pole at s = -1 Zero at s = -100
100 1
20 dB 40 dB 0 dB
( ) 10H s 1( )
( 1)H s
s
( ) ( 100)H s s
frequency
Now add all three together
Example
10
10( 100)( )
( 1)
sH s
s
frequency
0.01 0.1 1 10 100 1000 104 105
60 dB
40 dB
20 dB
0 dB
80 dB
lim ( ) 20s
H s dB
0lim ( ) 60s
H s dB
|H(s
)|
All together now
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Phase in Bode Plots
The transfer function H(s) is a phasor.
For a rational function H(s) we add the phases from the
numerator and subtract the phases from the denominator:
( )( ) ( ) ( ) ( ),
( )
Example:
( )
N jH j H j N j D j
D j
H j
1 1( )( ) tan tan
( )
j zH j
j p z p
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• The capacitive load, CL, is the culprit for gain roll-off since at high frequency, it will “steal” away some signal current and shunt it to ground.
1||out m in D
L
V g V RC s
Poles are Associated with Nodes in Circuits
13
• The circuit only has one pole (no zero) at 1/(RDCL), so the slope drops from 0 to -20dB/dec as we pass ωp1.
LD
pCR
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Corresponding Bode Plot for Previous Circuit
14
inS
pCR
11
LD
pCR
12
2
2
22
1
2 11 pp
Dm
in
out Rg
V
V
Example of a Circuit with Two Poles
15
• For a MOS, there exist oxide capacitance from gate to channel, junction capacitances from source/drain to substrate, and overlap capacitance from gate to source/drain.
Origin of Capacitances in MOSFET – I
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• 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 CGD.
Origin of Capacitances in MOSFET – II
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• At high frequency, capacitive effects come into play. Cb represents the base charge, whereas C and Cje are the junction capacitances.
b jeC C C
Origin of Capacitances in BJT – I
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Origin of Capacitances in BJT – II
• Since an integrated bipolar circuit is fabricated on top of a substrate, another junction capacitance exists between the collector and substrate, namely CCS.
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Example of Capacitors in BJT Circuit
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• 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 the signal path.
• Miller’s theorem helps to decompose floating capacitors into grounded elements.
• Bipolar and MOS devices exhibit various capacitances that limit the speed of circuits.
Summary
But . . . We will soon deal with Miller’s Theorem to simplify circuits analysis.