Frequency Response Review
• Frequency response defined by poles and zeros
– Equal number of poles and zeros, but some may be at
origin or at infinity.
• Zero: increase magnitude slope by 20 dB/decade
– 90 deg phase lag/lead for RHP/LHP zero
1 + !sz
1 + !j!z
H(s) =
Hp (s)
=
Hp (s)
• Pole: decrease
– 90 deg phase lead/lag for RHP/LHP pole
• 20dB/dec = 10x/10x
Hz (s)
Hz (s)
=
H(s) =
1 + !sp
1 + !j!p
Frequency Response Review
• Single RC pole
– pole at freq where Z(C) = R
– 3 dB (√2) attenuation
– 20 dB/decade after pole
– 90° total phase lag, 45° at pole
R
vin
+
vout
-
C
ZC
H(s) =
ZC + R
=
1
sC
1
sC
+R
1
=
1 + sCR
Frequency Response Review
• RC time constant: single pole
in left half plane
– real part is negative.
• Exists over entire s plane, but
evaluated over imaginary axis
1
H(s) =
1+
1
⇥p =
RC
s = + j⇥
s
p
Frequency Response Review
• Single RC pole
– 90° total phase lag, 45° at pole
– Phase lag = atan(ω/ωp)
• Why does phase matter?
– Feedback stability
– Waveform fidelity
R
vin
+
vout
-
C
ZC
H(s) =
ZC + R
=
1
sC
1
sC
+R
1
=
1 + sCR
Single-Pole High Frequency Response
Can split into the low-frequency term and the pole.
Pole associated with a node at ω = 1/(RTCT)
1
sC1
R2
vx
1+ sR2C1
=
=
R2
vi
1
R
+
R1 + R2
1
1+ sR2C1
sC1
R2
R2
1
%
R1 + R2 "
R1R2
C1 '
$1+ s
R1 + R2 &
#
R2
1
=
R1 + R2 (1+ s[R1 || R2 ]C1)
=
€
Single-Pole High Frequency Response
(cont.)
Substituting s=j2πf and using fp=1/(2π[R1||R2]C1)
vx
R2
1
=
vi R1 + R2 "
f %
$1+ j '
fp &
#
This expression has two parts,
the midband gain, R2/(R2+R1), and the high frequency
characteristic, 1/(1+jf/fp).
Frequency Response Review
• High-Pass RC
– Zero at DC, Pole at 1/RC
C
vin
+
R
vout
-
R
H(s) =
ZC + R
R
= 1
sC + R
s
= RC
1 + sRC
Frequency Response Review
C
vin
+
R
vout
-
Transfer Function Analysis
L
s + !Z1
FL =
s + !PL 1
FH
Amid is midband gain between upper
and lower cutoff frequencies.
H
1 + s/!Z1
=
1 + s/!PH1
L
L
... s + !Zk
s + !Z2
s + !PL 2 ... s + !PL k
H
H
... 1 + s/!Zk
1 + s/!Z2
1 + s/!PH2 ... 1 + s/!PHk
FH ⇡ 1 for
,i =1…l
FL ⇡ 1 for
,j =1…k
High-Frequency Response
1
FH ⇡
(1 + s/!P 3 )
!H ⇡ !P 3
ωP3 < all other poles
(1+(ωH 2 /ωZ12 ))(1+(ωH 2 /ωZ22 ))
(1+(ωH 2 /ωP12 ))(1+(ωH 2 /ωP22 ))
1
=
2
€
dominant high-frequency pole ().
With no dominant pole, poles and
zeros interact to determine ωH.
A (s)= A
F (s)
H
mid H
#
&#
&
)
)
%1+(s/ω
(%1+(s/ω
(
$
'$
'
Z1
Z2
=A
mid #%1+(s/ω )&(#%1+(s/ω )&(
$
P1 '$
P2 '
€
Pole ωH < all other pole and zero frequencies
ω
H
1
≅
1
ω
€
+
€
−
2
−
2
2 ω 2 ω 2 ω 2
P1
P2
Z1
Z2
In general, ω ≅
H
For s=jω, at ωH,
1
1
1
1
− 2∑
∑
2
2
n ω Pn
n ω Zn
High-Frequency Response
Low-Frequency Response
Low-Frequency Response
Estimating low cut-off frequency
without a dominant pole:
A (s)= A
F (s)=
L
mid L
#
&#
&
% s+ ω
(% s+ ω
(
$
'$
'
Z1
Z2
A
mid #% s+ ω &(#% s+ ω &(
$
P1'$
P2 '
For s=jω, at ωL,
€
1
=
2
(ωL2 + ωZ12)(ωL2 + ωZ22)
(ωL2 + ωP12)(ωL2 + ωP22)
ω 2 + ω 2 ) (ω 2ω 2 )
(
Z2 + Z1 Z2
1+ Z1
1
⇒ =
2
ω 2
ω 4
L
L
ω 2 +ω 2
ω 2ω 2
P2 + P1 P2
1+ P1
ω 2
ω 4
L
L
(
) (
Pole ωL > all other pole and zero
frequencies
In general, for n poles and n zeros,
)
Transfer Function Analysis and
Dominant Pole Approximation Example
• Problem: Find midband gain, FL(s) and fL for
" s
%
s$$
+1''
#100
&
A (s)= 2000
L
(0.1s+1)(s+1000)
• Analysis: Rearranging the given transfer function to get it in standard
s(s+100)
form,
A (s)= 200
L
s+10)(s+1000€)
(
Now,
€
and
Zeros are at s=0 and s =-100. Poles are at s= -10, s=-1000
All pole and zero frequencies are low and separated by at least a decade.
Dominant pole is at ω=1000 and fL =1000/2π=
159 Hz. For
s
A (s)= 200
frequencies > a few rad/s:
L
(s+1000)
€
Transfer Function Analysis
Amid is midband gain between upper
and lower cutoff frequencies.
for
,i =1…l
for
,j =1…k
Direct Determination of Low-Frequency
Poles and Zeros: C-S Amplifier
R
D
Vo (s) = Io (s)R = −gm Vgs (s)
R
3
R +(1/sC )+ R 3
RD3
3
3
s
= −gm (R R )
Vgs (s)
3 D s+
1
C (R + R )
D 3
s+C R
1 G
Vg (s) =
V (s)
s+C (R + R )+1 i
1 I G
s+(1/C R )
2 S
Vgs (s) = Vg - Vs =
Vg (s)
1
s+ "
%
C $(1/gm ) R '
2#
S&
3
€
C3
€
C2
€
Direct Determination of Low-Frequency
Poles and Zeros: C-S Amplifier (contd.)
s2 $ s+(1/C R )'
#
2 S &
"
F (s) =
L
"
"
%$
$
'$
1
1
$ s+
'$ s+
$
C (R + R ) ''$$ C (*(1/g )
$
#
1 I G &#
2) m
€
%
%
%
'"
'
1
'$
$
'
s+
+ '$
R - ''$# C3(RD + R3) ''&
S ,&
•The three zero locations are: s = 0, 0, -1/(RS C2).
•The three pole locations are: s = − C (R 1+ R ) ,− C #(1/g1
1 I
G
2%$
,−
&
m ) RS ('
1
C (R + R )
2 D 3
one pole and one zero.
• Each capacitor contributes
€
•Series capacitors C1 and C3 contribute the two zeros at s=0 (DC)
•Parallel combination of C2 and RS creates 3rd zero
•Parallel paths generally create zeros: same polarity: LHP; opposite
polarity: RHP
Direct Determination of Low-Frequency
Poles and Zeros: C-S Amplifier (contd.)
RD
R3
RD + (1/sC3 ) + R3
= gm vgs (s) (RD ||R3 ) X
R D R3
=
RD + (1/sC3 ) + R3
R D + R3
=
RD + (1/sC3 ) + R3
Want poles, zeros in form s + z/p
⇥
⇥
s/(RD + R3 )
RD + R 3
=
s/(RD + R3 ) RD + (1/sC3 ) + R3
s
=
s + 1/C3 (RD + R3 )
vo (s) = gm vgs (s)
RD R3
X
RD + R 3
X
X
X
Short-Circuit Time Constant Method to
Determine ωL
• Lower cutoff frequency for a
network with n coupling and
bypass capacitors is given by:
• RiS is resistance at terminals of
ith capacitor Ci
Direct analysis is precise, but tedious
Midband gain, fL, fH are of more interest
than complete transfer function.
• All other capacitors replaced by
short circuits.
• Product RiS Ci is short-circuit
time constant associated with Ci.
Estimate of ωL for C-E Amplifier
Using SCTC method, for C1,
For C2,
rπ + R
th
R = R RiE = R
2S 4
4 β +1
o
rπ +(R R )
I B
=R
4
βo +1
€
€
€
For C3,
R = R +(R RiC)= R +(R ro )
3S 3
C
3
C
≅ R +R
3 C
ω ≅
L
3
1
i =1 RiSCi
∑
f = 735Hz
L
High-Frequency Response
• HF response may be dominated
by ROUT and CLoad
• Often more complex
• Sometimes need to consider
parasitic capacitances
– Wiring
– Devices
RC
vy
vx
CLoad
RE
Frequency-dependent Hybrid-Pi Model
for BJT
Base-emitter capacitance, parallel
with rπ.:
Collector-base capacitance is
voltage dependent:
Cµo is total collector-base
junction capacitance at zero
bias, Φjc is its built-in potential.
•τF is forward transit-time of the
BJT.
•At higher frequency, Cπ reduces vbe
for given input signal current;
reduces current gain
Unity-gain Frequency of BJT
ωβ = 1/ rπ(Cµ + Cπ ) is the beta-cutoff frequency
where
fT = ωT /2π is the unity gain current bandwidth
High-frequency right-half plane zero
ωZ = + gm/Cµ can be neglected.
High-frequency Model of MOSFET
iD (s) = (gm
sCGD )vGS )
(gm sCGD )vGS )
s(CGS + CGD )
iD (s)
) (s) =
iG (s)
✓
◆
s
!T (s)
€
1
=
s
!T (1 + (CGS /CGD ))
= iG (s)
gm
ω =
T C +C
GS GD
Limitations of High-frequency Models
• Above 0.3 fT, behavior of simple pi-models begins to deviate
significantly from the actual device.
• ωT depends on operating current as shown and is not constant as
assumed.
• For given BJT, a collector current ICM exists that yield fTmax.
• For FET in saturation, CGS and CGD are independent of Q-point
current, so
– Efficiency/speed tradeoff
Effect of Base Resistance on Midband
Amplifiers
Base material is resistive.
rx models voltage drop between base
contact and active area of the BJT.
To account for base resistance rx is
absorbed into equivalent pi model and
can be used to transform expressions for
C-E, C-C and C-B amplifiers.
Miller Effect
We desire to replace Cxy with Ceq to
ground. Starting with the definition of
small-signal capacitance:
C=
ΔQ
ΔV
Now write an expression for the change
in charge for Cxy:
€
ΔQ = C phy (ΔVx − ΔVy ) = Cxy (ΔVx − Axy ΔVx )
= CxyΔVx (1− Axy )
€
We can now find and equivalent
capacitance, Ceq:
Cxy ΔVx (1− Axy )
Ceq =
= Cxy (1− Axy )
ΔVx
Miller Effect - Alternate View
• Break cap into two series caps
– size determined by
Miller multiplier
• vx =vi C1/(C1+C2)
vi
+vo C2/(C1+C2)
• vx = 0 if
C1 = CGD(1-AV)
C2 = CGD(1-1/AV)
CGD
RD
C1
vo
vi
vx C2
RD
vo
C-E Amplifier High Frequency Response
using Miller Effect
First, find the simplifed smallsignal model of the C-E amp.
C1, C2, C3 shorted. Focusing on
mid-band and high frequency
C-E Amplifier High Frequency Response
using Miller Effect (cont.)
vb
Rin
r
=
⋅ π
v i Ri + Rin rx + rπ
R1 || R2 || (rx + rπ )
r
=
⋅ π
Ri + R1 || R2 || (rx + rπ ) rx + rπ
Input gain is found as
Ai =
Terminal gain is
v
Abc = c = − gm ro || RC || R3 ≅ − gm RC || R3 ≅ − gm RL
€ vb
CeqB = Cµ (1− Abc ) + Cπ (1− Abe )
Using the Miller effect, we
find the equivalent
capacitance€at the base as:
€
= Cµ (1− [−gm RL ]) + Cπ (1− 0)
= Cµ (1+ gm RL ) + Cπ
C-E Amplifier High Frequency Response
using Miller Effect (cont.)
ReqB = (Rth + rx ) || RinB
The total equivalent
resistance at the base is
The total capacitance and
resistance at the collector is
€
Because of interaction
through Cµ, the two RC€
time constants interact, €
giving rise to a dominant
pole
€
ω p1 =
€
Jaeger/Blalock
8/10/10
= (Ri || R1 || R2 + rx ) || rπ = rπ 0
CeqC = Cµ + CL
ReqC = ro || RC || R3 ≅ RC || R3 ≅ RL
1
ω p1 =
rπ 0 [Cµ (1+ gm RL ) + Cπ ] + RL [Cµ + CL ]
RL
CT = [Cµ (1+ gm RL )+ Cπ ]+
[Cµ + CL ]
rπ 0
1
1
=
rπ 0CT r ([C (1+ g R ) + C ] + RL [C + C ])
π0
µ
m L
π
µ
L
rπ 0
Microelectronic Circuit Design, 4E
McGraw-Hill
Chap 17-31
Direct High-Frequency Analysis: C-E
Amplifier
R = R R =100kΩ 4.3kΩ
L 3 C
€
+
-
R = R R = 30kΩ10kΩ
B 1 2
The small-signal model can be simplified
by using Norton€source transformation.
Direct High-Frequency Analysis: C-E
Amplifier (Pole Determination)
From nodal equations and Cramer’s
rule
(sCµ - gm )
V (s) = Is (s)
C
Δ
$
'
$
'
Δ = s2&&Cπ &Cµ +C ) +Cπ C ))
%
L(
L(
%
$
$
'
'
+s&&Cπ g +Cµ & g + gµ + gπ ) +C gπo )) + g gπo
L
L
% L
(
L
%
(
€
fH(s) given by 2 poles, one finite zero
and one zero at infinity.
For a polynomial s2+sA1+A0 with
roots a and b, a =A1 and b=A0/A1.
First pole (smallest root) limits
frequency response and determines ωH.
Second pole is important for phase
response in feedback amplifiers.
Direct High-Frequency Analysis: C-E
Amplifier (Pole Determination)
From nodal equations and Cramer’s
rule
(sCµ - gm )
V (s) = Is (s)
C
Δ
$
'
$
'
Δ = s2&&Cπ &Cµ +C ) +Cπ C ))
%
L(
L(
%
$
$
'
'
+s&&Cπ g +Cµ & g + gµ + gπ ) +C gπo )) + g gπo
L
L
% L
(
L
%
(
€
fH(s) given by 2 poles, one finite zero
and one zero at infinity.
For a polynomial s2+sA1+A0 with
roots a and b, a =A1 and b=A0/A1.
First pole (smallest root) limits
frequency response and determines ωH.
Second pole is important for phase
response in feedback amplifiers.
Direct High-Frequency Analysis: C-E
Amplifier (Overall Transfer Function)
Dominant pole model at high
frequencies for C-E amplifier is as
shown.
Direct High-Frequency Analysis: C-E
Amplifier (Example)
• Problem: Find midband gain, poles, zeros and fL.
• Given data: Q-point= ( 1.60 mA, 3.00V), fT =500 MHz, βo =100, Cµ
=0.5 pF, rx =250Ω, CL =0
• Analysis: gm =40IC =40(0.0016) =64 mS, rπ = βo/gm =1.56 kΩ.
R = R R =100kΩ 4.3kΩ= 4.12kΩ
L 3 C
f
R = R R = 7.5kΩ1kΩ= 882Ω
B I
th
gm
= 512 MHz
P2 2π (C +C )
π
L
=
€
€
C = Cπ
T
€
R
L (C +C )=156pF
µ L
πo
A
#
&
+Cµ %1+ gm R ( +
$
L' r
vth
€
= A A = 0.512(−264)= −135
i bc
High Frequency Poles for the C-B Amplifier
1
1+ gm Ri
v
Aec = c = gm RiC || RL ≅ gm RL
ve
Ai ≅
€
RiC = ro (1+ gm rπ || RI )
Since Cµ does not couple input and
€ output, input and output poles can be
found
directly.
€
€
CeqC = Cµ + CL
CeqE = Cπ
1
ReqE =
|| RE || Ri
gm
1
g
€ ω p1 =
≅ m
1
( || RE || Ri )Cπ Cπ
gm
ReqC = RiC || RL ≅ RL
€p 2 =
ω
€
€
1
1
≅
(RiC || RL )(Cµ + CL ) RL (Cµ + CL )
Spice Simulation of Example C-E
Amplifier
Estimation of ωH using the Open-Circuit
Time Constant Method
At high frequencies, impedances of
coupling and bypass capacitors are small
enough to be considered short circuits.
Open-circuit time constants associated
with impedances of device capacitances
are considered instead.
where Rio is resistance at terminals of
ith capacitor Ci with all other
capacitors open-circuited.
For a C-E amplifier, assuming CL =0
x
High-Frequency Analysis: C-S Amplifier
Analysis similar to the C-E case yields
the following equations:
C
T
€
R
"
%
= C +C $1+ gm R ' + L
GS GD#
L& R
th
gm
P2 C +C
GS L
+g
ω = m
Z C
GD
ω
€
€
=
(C +C )
GD L
C-S Amplifier High Frequency Response
with Source Degeneration Resistance
First, find the simplifed smallsignal model of the C-A amp.
Recall that we can define an effective
gm to account for the unbypassed
source resistance.
gm
gm '=
1+ gm RS
C-S Amplifier High Frequency Response with
Source Degeneration Resistance (cont.)
vg
RG
Ai = =
v i Ri + RG
Input gain is found as
=
Terminal gain is
Agd =
R1 || R2
Ri + R1 || R2
vd
−g R || R3
= −gm '(RiD || RD || R3 ) ≅ m D
vg
1+ gm RS
€C = C (1− A ) + C (1− A )
Again, we use the Miller
effect to find the equivalent
capacitance €
at the gate as:
eqG
GD
gd
GS
gs
[−gm RL ]
g R
) + CGS (1− m S )
1+ gm RS
1+ gm RS
g R || RL
CGS
= CGD (1+ m D
)+
1+ gm RS
1+ gm R3
= CGD (1−
C-S Amplifier High Frequency Response with
Source Degeneration Resistance (cont.)
The total equivalent resistance at the
gate is
ReqG = RG || RI = Rth
The total capacitance and resistance
at the collector is
CeqD = CGD + CL
€
Because of interaction through C
GD,
ReqD = RiD || RD || R3 ≅ RD || R3 = RL
the two RC time constants interact,
giving rise to the dominant pole:
ω p1 =
€
1
g R
CGS
R
€
Rth [CGD (1+ m L ) +
+ L (CGD + CL )]
1+ gm RS 1+ gm RS Rth
And from previous analysis:
gm '
gm
ω p2 =
=
(CGS + CL ) (1+ gm RS )(CGS + CL )
€
ωz =
+gm '
+gm
=
CGD (1+ gm RS )(CGD )
C-E Amplifier with Emitter
Degeneration Resistance
Analysis similar to the C-S case yields
the following equations:
rπ 0 = ReqB = (Rth + rx ) ||[rπ + (β + 1)RE ]
R =R R
L C 3
€
ω p1 =
1
rπ 0CT
€
1
=
rπ 0 ([Cµ (1+
gm RL
Cπ
R
)+
] + L [Cµ + CL ])
1+ gm RE 1+ gm RE
rπ 0
ω p2 ≅
gm
2π (1+ gm RE )(Cπ + CL )
€
ωz =
€
€
+gm
2π [1+ gm RE ][Cµ ]
Gain-Bandwidth Trade-offs Using
Source/Emitter Degeneration Resistor
Adding source resistance to the CS
amp caused gain to decrease and
dominant pole frequency to increase.
Agd =
v d −gm RD || R3
=
vg
1+ gm RS
1
ω p1 =
Rth [CGD (1+
gm RL
CGS
R
)+
+ L (CGD + CL )]
1+ g€m RS 1+ gm RS Rth
However, decreasing the gain also
decreased the frequency of the
€second pole.
Increasing the gain of the C-E/C-S
stage causes pole-splitting, or€
increase of the difference in
frequency between the first and
second poles.
gm
ω p2 =
(1+ gm RS )(CGS + CL )
High Frequency Poles for the C-G Amplifier
Similar to the C-B, since CGD does not couple the input and
output, input and output poles can be found directly.
€
CeqS = CGS
1
ReqS =
|| R4 || RI
gm
1
g
≅ m
€ ω p1 =
1
( || R4 || RI )CGD CGD
gm
€
CeqD = CGD + CL
ReqD = RiD || RL ≅ RL
ω€p 2 =
€
1
1
≅
(RiD || RL )(CGD + CL ) RL (CGD + CL )
High Frequency Poles for the C-C Amplifier
Ai =
vb
Rin
=
v i Ri + Rin
Abe =
ve
g R
= m L
vb 1+ gm RL
€
g R
CeqB = Cµ (1− Abc ) + Cπ (1− Abe ) = Cµ (1− 0) + Cπ (1− m L )
1+ gm RL
€
Cπ
= Cµ +
1+ gm RL
ReqB = Ri || Rin = [(Ri || RB )+ rx ] || [rπ + ( β +1)RL ] = (Rth + rx ) || [rπ + ( β +1)RL ]
€
€
€
CeqE = Cπ + CL
ReqE = RiE || RL ≅ [1/gm +
(Rth + rx )
] || RL
β +1
High Frequency Poles for the C-C Amplifier (cont.)
The low impedance at the output makes the input and output time
constants relatively well decoupled, leading to two poles.
ω p1 =
1
([Rth + rx ] ||[rπ + (β + 1)RL ])(Cµ +
Cπ
)
1+ gm RL
ω p2 =
1
1
≅
[RiE || RL ][Cπ + CL ] [(1/g + Rth + rx ) || R ][C + C ]
m
L
π
L
β +1
The feed-forward high-frequency path through Cp leads
to a zero in the C-C response. Both
the zero and the
€
second pole are quite high frequency and are often
neglected, although their effect can be significant with
large load capacitances.
€
ωz ≅
gm
Cπ
High Frequency Poles for the C-D Amplifier
Similar the the C-C
amplifier, the high
frequency response is
dominated by the first
pole due to the low
€
impedance at the output
of the C-C amplifier.
€
€
1
ω p1 =
Rth (CGD +
ω p2 =
ωz ≅
CGS
)
1+ gm RL
1
1
≅
[RiS || RL ][CGS + CL ] [1/gm || RL ][CGS + CL ]
gm
CGS
Summary of the Upper-Cutoff Frequencies of
the Single-Stage Amplifiers (pg.1037)
Frequency Response: Differential
Amplifier
CEE is total capacitance at emitter node
of the differential pair.
Differential mode half-circuit is similar
to a C-E stage. Bandwidth is determined
by the
product. As emitter is a
virtual ground, CEE has no effect on
differential-mode signals.
For common-mode signals, at very low
frequencies,
Transmission zero due to CEE is
Frequency Response: Differential
Amplifier (contd.)
As REE is usually designed to be large,
Common-mode half-circuit is similar to a
C-E stage with emitter resistor 2REE.
OCTC for Cπ and Cµ is similar to the C-E
stage. OCTC for CEE/2 is:
Frequency Response: CommonCollector/ Common-Base Cascade
REE is assumed to be large and neglected.
The intermediate node pole is neglected since
the impedance is quite low. We are left with
the input pole for a C-D and the output pole of
a C-B stage.
1
ω pB1 =
([Rth + rx1 ] ||[rπ 1 + (β + 1)RL1 ])(Cµ1 +
1
=
([Rth + rx1 ] ||[2rπ 1 ])(Cµ1 +
ω pC 2 ≅
€
1
RC (Cµ + CL )
Cπ 2
)
2
Cπ 1
)
1+ gm RL
Frequency Response: Cascode Amplifier
There are two important poles, the input pole for
the C-E and the output pole for the C-B stage.
The intermediate node pole can usually be
neglected because of the low impedance at the
input of the C-B stage. RL1 is small, so the second
term in the first pole can be neglected. Also note
the RL1 is equal to 1/gm2.
1
1
=
rπ 0CT r ([C (1+ gm1RL1 ) + C ] + RL1 [C + C ])
π 01
µ1
π1
µ1
L1
1+ gm1RE1
rπ 0
1
1
=
≅
g
1/gm 2
rπ 01 ([Cµ1 (1+ m1 ) + Cπ 1 ] +
[Cµ1 + Cπ 2 ]) rπ 01 (2Cµ1 + Cπ 1 )
gm 2
rπ 01
ω pB1 =
ω pC 2 ≅
€
1
RL (Cµ 2 + CL )
Frequency Response: MOS Current
Mirror
This is very similar to the C-S stage
simplified model, so we will apply the
C-S equations with relevant changes.
ω
=
1
P1 (1/g )C
m1 T
=
r
1/g (C
+C
+C
1+ g r + o2 C
)
m1 GS1 GS2 GD2
m2 o2 1/g
GD2
m1
=
1
2C
+2C
r
GS1 g
GD2 o2
m1
(
€
)
Assumes matched transistors.
Frequency Response: Multistage
Amplifier
•Problem:Use open-circuit and short-circuit time constant methods to
estimate upper and lower cutoff frequencies and bandwidth.
•Approach: Coupling and bypass capacitors determine low-frequency
response, device capacitances affect high-frequency response.
At high frequencies, ac model for multi-stage amplifier is as shown.
Chap 17-55
Frequency Response: Multistage
Amplifier Parameters
Parameters and operation point information for the
example multistage amplifier.
Frequency Response: Multistage
Amplifier (SCTC Estimate of ωL)
SCTC for each of the six independent coupling and bypass capacitors are
"
% "
%
calculated as follows:
R = $R
R ' +$R R '
R = R +(R R )=10kΩ+1MΩ ∞
1S I
G in1
=1.01MΩ
5S
#
"
= $R
# C2
C2 O2 &
#
B3 in3&
% "
"
r ' + $$ R $r +(β +1)(R
o2 & # B3 # π 3
o3
E3
=18.4kΩ
€
"
% "
%
R = $R R ' +$R
R '
3S # D1 O1& # B2 in2 &
"
% "
%
= $ R r ' + $ R r ' = 2.69kΩ
# D1 o1& # B2 π 2 &
€
n
1
ω ≅ ∑
= 3330rad/s
L
i =1 RiS Ci
ω
f = L = 530Hz
L 2π
€
€
%%
R )'''
L &&
Frequency Response: Multistage
Amplifier (High-Frequency Poles)
High-frequency pole at the gate of M1: Using our equation for the C-S
input pole:
1
fp1 =
RL1
(CGD1 + CL1 )]
Rth1
RL1 = RI12 || rπ 2 || ro1 = 598Ω || (2.39kΩ + 250Ω) ||12.2kΩ = 469 Ω
CL1 = Cπ 2 + Cµ 2 (1+ gm 2 RL 2 )
2πRth1[CGD1(1+ gm1RL1 )+ CGS1 +
€RL 2 = RI 23 || Rin 3 || ro2 = RI 23 ||[rx 3 + rπ 3 + (β 03 + 1)(RE 3 || RL )] || ro2 = 3.33kΩ
CL1 = 39pF +1pF[1+ 67.8mS(3.33kΩ)] = 266pF
fp1 =
1
469Ω
2π 9.9kΩ[1pF(1+ 0.01S(469Ω)]+ 5pF +
(1pF + 266pF)]
9.9kΩ
= 689 KHz
Frequency Response: Multistage
Amplifier (High-Frequency Poles cont.)
High-frequency pole at the base of Q2: From the detailed analysis of the
C-S amp, we find the following expression for the pole at the output of the
M1 C-S stage:
fp2 =
CGS1gL1 + CGD1(gm1 + gth1 + gL1 )+ CL1gth1
2π [CGS1(CGD1 + CL1 )+ CGD1CL1 ]
For this particular case, CL1 (Q2 input capacitance) is much larger than the
other capacitances, so fp2 simplifies to:
€
€
CL1gth1
1
fp2 ≅
≅
2π [CGS1CL1 + CGD1CL1 ] 2πRth1(CGS1 + CGD1 )
1
fp2 =
= 2.68 MHz
2π 9.9kΩ(5pF +1pF)
Frequency Response: Multistage
Amplifier (High-Frequency Poles cont.)
High-frequency pole at the base of Q3: Again, due to the pole-splitting
behavior of the C-E second stage, we expect that the pole at the base of Q3
will be set by equation 16.95:
gm 2
fp3 ≅
C
2π [Cπ 2 (1+ L 2 )+ CL 2 ]
Cµ 2
The load capacitance of Q2 is the input capacitance of the C-C stage.
Cπ 3
50pF
CL 2 = Cµ 3 +
= 1pF +
= 3.55 pF
€ 1+ gm 3 RE 3 || RL
1+ 79.6mS(3.3kΩ || 250Ω)
fp3 ≅
67.8mS[1kΩ (1kΩ+ 250Ω)]
= 47.7 MHz
3.55pF
2π [39pF(1+
)+ 3.55pF]
1pF
Frequency Response: Multistage
Amplifier (fH estimate)
There is an additional pole at the output of Q3, but it is expected to be at a
very high frequency due to the low output impedance of the C-C stage. We
can estimate fH from eq. 16.23 using the calculated pole frequencies.
fH =
1
= 667 kHz
1
1
1
+
+
fp12 fp22 fp32
The SPICE simulation of the circuit on the next slide shows an fH of 667
KHz and an fL of 530 Hz. The phase and gain characteristics of our
€ high frequency response is quite close to that of the SPICE
calculated
simulation. It was quite important to take into account the pole-splitting
behavior of the C-S and C-E stages. Not doing so would have resulted in a
calculated fH of less than 550 KHz.
Frequency Response: Multistage
Amplifier (SPICE Simulation)
Intro to RF Amplifiers
• Amplifiers with narrow bandwidth are often required in radio
frequency (RF) applications to be able to select one signal from a large
number of signals.
• Frequencies of interest > unity gain frequency of op amps, so active
RC filters can’t be used.
• These amplifiers have high Q (fH and fL close together relative to center
frequency)
• These applications use resonant RLC circuits to form frequency
selective tuned amplifiers.
The Shunt-Peaked Amplifier
• As the frequency goes up, the
gain is enhanced by the
increasing impedance of the
inductor.
Av(s)=
€
(−gmR)(1+ sL/R)
where C = C +C
L GD
2
1+ sRC + s LC
• The gain improvement can be
plotted as a function of
parameter, m, defined below:
Avn(s)=
1+ ms
1+ s+ ms2
where L = mR2C
€
The Shunt-Peaked Amplifier
Single-Tuned Amplifiers
• RLC network selects the
frequency, parallel combination
of RD, R3 and ro set the Q and
bandwidth.
• Neglecting right-half plane
zero,
Single-Tuned Amplifiers (contd.)
• At center frequency, s = jωo,
Av = Amid.
Use of tapped Inductor- Auto
Transformer
CGD and ro can often be small enough to
degrade characteristics of the tuned
amplifier. Inductor can be made to work
as an auto transformer to solve this
problem.
These results can be used to transform the
resonant circuit and higher Q can be
obtained and center frequency doesn’t
shift significantly due to changes in CGD.
Similar solution can be used if tuned
circuit is placed at amplifier input instead
of output
Multiple Tuned Circuits
CS Amp with Inductive Degeneration
• Typically need to match input resistance to
antenna impedance at center frequency,
usually 50 ohms.
• Using our follower analyses, the input
impedance is found as:
Z (s)= Zgs + Zs +(gm Zgs )Zs
in
Z (s)=1/sC + sLs + Req
in
GS
where Req = +gm L /C
S GS
• The following slide shows a complete low
noise CS amp where a series inductor
€
resonates
with the input capacitance to
leave only the r esistance at the center
frequency.
Complete Cascode LNA
Mixer Introduction
• A mixer is a circuit that
multiplies two signals to
produce sum and difference
frequencies:
S0 = S •S = sinω t •sinω t
2 1
2
1
cos(ω − ω )t − cos(ω − ω )t
2 1
2 1
=
2
• A filter is usually used to reject
either the sum or difference
€
frequency to implement upconversion or down-conversion.
Single-Balanced Mixer
• This basic mixer form is
essentially a switched circuit
that 'chops' the sine wave input
with a square wave function
v (t)= Asinω t
1
1
1 2
1
ss(t)= +
∑
sinnω t
2
2 π n odd n
€
Single-Balanced Mixer Output Spectra
cos(nω − ω )t − cos(nω + ω )t
A
A
2 1
2 1
vo(t)= sinω t +
∑
1 π n odd
2
n
€
Differential Pair as Single-Balanced
Mixer
%
cos(nω − ω )t − cos(nω + ω )t (*
4 '
2 1
2 1 *
vo(t)= ∑
R sinnω t + I R
'I
2 1 C
n odd nπ ' EE C
2
*)
&
€
Gilbert Multiplier as a Double-Balanced
Mixer
• The Gilbert Multiplier is an
extension of the differential
single-balanced mixer.
• The input polarity is reversed
on the second diff pair and the
signal v1 selects between the
two diff pairs.
• The currents are summed in the
load resistors and the DC
component is zero.
• Only sum and difference
frequencies are present at the
output.
Gilbert Multiplier Mixer Spectra
R
2 %
(
vo(t)=Vm C ∑
'&cos(nω c − ω m )t − cos(nωc + ω m )t *)
R n odd nπ
1
€