模拟集成电路设计讲义CMOS (1)

1 ω 3dB = ------RC
C
H ( ω ) dB
0dB 20dB/Decade -10dB -20dB 0
log ( ω ) 0.1 ω 3dB ω 3dB 10 ω 3dB
θ
-π/4 -π/2
1 ∝ --ω
ω 3dB
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
Convert H( ω) to polar coordinates, |H(ω)| < θ
H ( ω ) = [ H ( ω ) ( H∗ ( ω ) ) ] Im{H(ω) } θ = atan ------------------------ R { H ( ω ) } e
1/2
1/2 1 = -------------------------2 ω - 1 + ------- ω 3dB
Bode Plot Magnitude
H ( ω ) d B = 20 ⋅ log H ( ω )
-3dB 20dB/Decade 6dB/Octave
θ
θ
-π/4 -π/2 -3π/4 -π
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
R1 ν IN C1 1 C2 R2 ν OUT
FR-10
If we have 180 degree phase shift we have a problem. (+)
Bode Plots (Cont.)
C ν IN ν OUT R R j ω RC j ω RC -------------------= ----------------------- = ------------------1 1 + j ω RC ω 1 + j -------R + ---------jωC 1 ------RC
Robert W. Brodersen
Analog Circuit Design
EECS140
Lectures on FREQUENCY RESPONSE
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
University of California Berkeley
College of Engineering Department of Electrical Engineering and Computer Science
VOUT ( ω ) 1 ⁄ jωC V OUT ( ω ) 1 ⁄ jωC H ( ω ) = -----------------= --------------------------V IN ( ω ) R + 1 ⁄ jωC 1 ⁄ jωC ----------------------H( ω) = 1 + j ω RC
FR-12
Zero at zero frequency, pole at 1/RC
H ( ω ) dB
1
1 ω p = ------RC log ( ω )
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
ω » ω3dB
1/2
ωp = 1/RC = ω3dB
ROBERT W. BRODERSEN
ω
1 H ( ω ) ≈ ---------------2 ω ------- ω 3 dB
ω3dB = -------ω
6dB/Octave - drops by 2 every time frequency doubles
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
( H ( ω ) ⋅ H∗ ( ω ) )
1 /2
FR-2
1 1 ⋅ ---------------------- ----------------------= 1 + j ω RC 1 – j ω RC 1 = ----------------------------2 1 + ( ω RC)
+
ν OUT
The negative feedback will turn into positive feedback. Can’t have positive feedback in a loop with gain > 1
ν IN
(-)
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.) 2 Poles
FR-7
R1 ν IN
Ideal Unity gain Buffer R2 1 C1 C2 νOUT
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
0
FR-11 3 poles on top of each other
ω p1 = ω p2 = ω p3 = ω p 10 ω p 0.1 ω p ωp log ( ω )
θ
-π/4 -π/2 -3π/4 -π -5π/4 -3π/2
0
FR-4
0.1 ω 3 dB θ
-π/4 -π/2
log ( ω ) ω 3dB 10 ω 3dB
Linear Approximation
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
ROBERT W. BRODERSEN
FR-3
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
1 Re { H ( ω ) } = -------------------------2 ω 1 + ------- ω 3 dB ω – -------- ω 3dB Im { H ( ω ) } = -------------------------2 ω 1 + ------- ω 3 dB ω - θ ( ω ) = arcቤተ መጻሕፍቲ ባይዱtan – ------- ω 3d B ω = – arc tan -------- ω3dB
p1 p2 p1 p2
FR-8
20 ⋅ log H ( ω ) = 20 ⋅ log ( H ω ( ω ) ⋅ H ω ( ω ) ) = 20 ⋅ log H ω ( ω ) + 20 ⋅ log H ω ( ω ) ω p1 ω p2
p1 p2
1 --ω 1 -----2 ω
ROBERT W. BRODERSEN
0 Phase Magnitude
FR-5
90 Phase difference Magnitude Change
t = 0 R 1 ----------------------H(ω) = 1 + j ω RC C
t = 0
H ( ω ) = ( H ( ω ) ⋅ H∗ ( ω ) )
1/2
ROBERT W. BRODERSEN
LECTURES ON FREQUENCY RESPONSE
Two Poles:
1 H ( ω ) = ------------------ ω 1 + j ------ ω p1
Bode Plots (Cont.)
1 ⋅ ------------------ = H ω ( ω ) ⋅ H ω ( ω ) ω 1 + j ------ ω p2
ROBERT W. BRODERSEN
1 2
N(ω) IF H ( ω ) = ------------- then D(ω) R e ⋅ { H ( ω ) } = R e ⋅ { N ( ω ) ⋅ D∗ ( ω ) } I M ⋅ { H ( ω ) } = I M ⋅ { N ( ω ) ⋅ D∗ ( ω ) }
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.) 1 Pole Summary R
1 H ( ω ) = -------------------ω 1 + j -------ω 3d B
FR-6
= H ω ⋅ H ω ⋅ exp ( j [ θ ω + θω ] )
p1 p2 p1 p2
= H ( ω ) ⋅ exp ( j θ ( ω ) ) θ ( ω ) = θ ω + θω