# Analog and Mixed-Signal Center Edgar Sánchez

```Analog and Mixed-Signal Center
Edgar S&aacute;nchez-Sinencio
How to measure and/or compare wide-bands?
iin
gm
fT =
2p (C gs + C gd )
+
iout
Vin
∼
or
−
Cg =
port 1
gm
w
T
port 2
An equivalent circuit of an MOS transistor as a two-port device.
Cgd
G
D
g mVin
Cgs
Cdb
S
(a) Hybrid- π Model
D
G
gm
wT
g mVin
S
fT Simplified Model valid if Cdb &lt;&lt;
Cgs + Cgd
Small-signal MOS transistor models.
Reference
J.M. Steinger, “Understanding Wide-Band MOS Transistors”, IEEE
Circuits and Devices, pp. 26-31, May 1990.
What Bandwidths are possible?
Vdd
Vdd
Av
M2
Av
+
+
Vin
M1
= −
= −
g
g
m1
(W / L)1
(W / L)
Vin
−
−
max Vout swing ≤ Vdd - VTn
Vin
−
g m1
gm 2
wT
wT
g m1Vin
M2
Vout
−
w
+
gm 2
wT
g m 2Vout
M1
Simple practical CMOS amplifier.
Enhancement-mode inverter.
+
+
+
2
Vout
−
bias
M3
m2
Vout
−
p
=
w
m2
T
=
gm1 gm2 Ao + 1
+
g
w
T
w
T
where
| Ao |= gm1 / gm 2
f p / fT = 1 /(1+ Ao )
A small-signal equivalent circuit of the simple amplifier shown above based
on the fT transistor model.
To obtain an amplifier with the highest possible bandwidth a cascade
of low-gain stages is desirable. How to optimize the cascade gain
Scaled MOS and fT
fT
fT
↑
↑
and/or Cg
↓
If Cg is only the parallel plate
capacitance between the gate and
channel, then
ID
fT
Cg
≈
Cox WL
Furthermore
gm =
fT vs. V on and ID.
if g m
Von
dI d
dVgs
g m = m C ox
W
(Vgs − VT )
L
and
Von = Vgs − VT
fT
g m = m C ox
W
Von
L
then
log I CC
fT
fT of a bipolar transistor vs. log ICC.
=
1 gm
2p C g
≅
1
2p
m Von
L2
To compare fT of MOS transistors from different technologies is by holding
Von constant or by keeping the same current inversion level if.
fT For Various Size Technologies
20
16
fT
Von = Vgs − Vth
12
[Ghz]
=
0.75
2.25
mm
simple theory
measured data
8
1.3m m
4
3 .0 m m
0
1
0
0.4
0.8
1.2
1.6
fT vs. Leff for various CMOS technologies
Remarks
2
Leff
[
2
1
m
2
]
•
Short-channel transistors operated at practical bias votlages do not
behave as square-law devices.
•
tox &lt; 500 &Aring; produce high electric fields in the channel, both in the
longitudinal and transverse directions.
g m ↓ Vdrift ↓
• Simple theory neglects gate-to-drain overlap and assumes an optimistic
Cg.
Wide-Band CMOS Amplifiers
Vdd
M6
M5
M4
M3
M11
M12
1: a
I1
aI1
out −
in +
M1
M2
out +
in −
abI 1
b :1
M9
M7
bias
Mbias
M8
M 10
Gnd
Simple differential wideband amplifier.
Av
=
Gm Ro
≅
Av
=
g m1 (W / L) 6 (W / L )9
g m11(W / L ) 3(W / L ) 7
( g m1ab) / g m11
•
Limited output swing
•
DC bias current and small signal are both amplified by the current
mirrors. This might be solved by using interstage DC substractors.
An alternative structure is next shown
bias 2
MB4
bias 4
−
MB 2
+
I o−
vo
vin
M7
M 11
bias
M9
M1
MB
M2
bias 3
MB1
bias1
MB3
•
−
vin
Vo+
+
Io
•
M3
M8
M4
M6
M5
M 10
•
All current-mirrors are N-type (faster)
•
The output bias current is set by the current source (MB3 and MB4)
•
Optimal current-gain and matching are easier to obtain.
M 12
Wide Output and Good Linearity Observations
Gm =
RL
gm
1 + gm Rs
vo
vi
Av
gm
= − Gm R
L
=
=
1
1
+ Rs
gm
−R
L
1
+ Rs
gm
Rs
vi
gm
vo
RL
Rs
N-driver Transconductor
P-driver Transconductor
•
In order to make Av independent of gm, it is required that Rs &gt;&gt; 1/gm
•
Large Rs take up large area and have large parasitic capacitances
which limit the amplifier bandwith.
•
It would be desirable to have Av independent of gm with small Rs
Wide Output Swing and Improved Linearity Amplifier
VDD
Ib
m
Ib
Ib
m
Ib
Vx •
2 Rs
m &gt;&gt; 1
VY
id
M2
RL +
M3
M1
in +
M4
in −
RL −
out +
out −
M7
M5
M6
M8
•
•
This circuit is capable to cancel the effect of gm1, gm4
•
We need to force the DC components of Vx and VY to be equal, such that
the current through 2RS is only a function of the differential input
vin+ − vin− = vd , thus id becomes vd/2RS.
•
Condition to satisfy
v gs1
+
vgs 2
=
v gs4
+
vgs3
⇒
Vx
=
VY
Amplifiers with Wide Output Swing and Good Single-Ended Linearity A Circuit
Implementation.
Vdd
bias
Mb 4
Mb1
2 Rs
Mb 2
Mb3
RL −
RL +
M2
M3
out −
out +
in +
M7
M1
M4
M5
M6
in −
Gnd
Resistively loaded MOS amplifier with gm cancellation.
Vgs1 + Vgs2 = Vgs3 + V gs4
i2 RS = (v in+ − v in− ) / 2 Rs = v d / 2 RS
If I b1,b 4 &lt;&lt; I b 2 ,b 3
then Av ≅
RL (W / L) 7
RS (W / L) 5
M8
A Bipolar Amplifier version is next shown.
This Circuit is called Gilbert Gain Cell.
x (I E + I B )
VCC
(1 − x)( I E + I B )
RL+
−
+
vo •
•vo
I1
RL−
Q1
Q1
I 3 = xIE
Q1
I2
Q4
Q2
Q2
Q3
(1 − x) I B
xI B
Q3
IE
xI B = I B 1
IE
(1 − x) I B = I B 2
vin+
I2 IB 2 I4
=
=
I 3 I B 1 I1
I o1 = ( I1 + I 3 ) − ( I 2 + I 4 )
I
I +I
AI = o = B E
I in
IB
− VEE
Q5
RE Q6
RE
vin−
IB
− VEE
v
Av = o
vin
=
RL
I
(1 + E )
RE
IB
The dominant pole is placed at
Reference
w
p
=
wT
AI
&amp; Sons, New York 1984.
An alternative transconductor to handle the resistors is shown below.
• Enhanced Gm
• A single ended is shown here, but a differential version should be used.
• Some applications allow input current in that case the input can be applied to
the source of M1
Vb
M2
.
M5
vin
M4
vo
M1
M3
Reference.- J. Rijns., “ CMOS Low Distortion High-Frequency Variable-Gain Amplifier,” IEEE
J. Solid-State Circuits, Vol. 31, No. 7, July 1996.
Even without source degeneration, the voltage gain can be set by a ratio
of resistors by using a bias current proportional to RL.
Vdd
biasp
M4
M3
Vdd
RL −
RL +
out +
out −
in +
M1
M2
in −
M5
M7
•
M6
I bias
biasn
M8
M9
A MOSFET amplifier with Av=gm1,2RL .
• Simple output common-mode setting network
• Vo ,CM = Vgs5 + Vgs7
• The bias current Ibias should be set such that gm1,m2 is a function
of 1/Rb.
Bias circuit for the previous amplifier that produces
gm =
1
Rb
Vdd
g m1, 2
MB4
MB3
biasp
1 :1
Ib
Ib
biasn
MB1
W
4
L
W
L
Rb
1 / Rb
Av
=
Av
= + (4 K
g m1, 2 RL
I B )1/ 2 RL
RL
Rb
bias M 1 and M 2 such that
1
g m1, 2 =
Rb
Av
MB2
=
=
Gnd
Vgs2
=
Vgs1
+
I b Rb , Vgs
=
Id
K
−
VT
Then
Ib
K
Ib
=
Rb
=
1 Ib
+ I b Rb
2 K
1
1
=
2
4 KRb 2 m C ox (W / L ) Rb2
=
1
gm
=
1
2 m C o (W / L ) I b
W 
W 
=



 =
 L  B2  L  9
W 

L

 1, 2
2
W 
W 
W 
= 


 = 

L
L
L

 B 3, B 4 
3 
4
LV Transconductance Amplifier
iR=vin/R
VDD
M5
M12
M6
M14
ix = - iR
Vbias
IB
IB
IB
IB
B
M1
vin
ib
M3
R
X
ON
A
ix
iR
M13
M2
IB
M8
Fig. 1
M4
M9
io+
M10
M11
VSS
Min Supply voltage =
Max{3VD SAT, VT+2VD SAT}
OP
io-
i
&gt; max{ R} =(max{vin}-VA)/R
v+
OP
X
Figure 1
ON
Single-ended
v-
Amplifier Design Using Linear transconductor with Resistor Input”, IEEE
Trans on Circuits and Systems-II, vol 47, pp776-778, August 2000
io-
OP
X
Figure 1
ON
Fig. 2
Refe.- E.K.F. Lee, “Low-Voltage OpAmp Design and Differential Difference
io+
Pseudo Differential
VDD
Transimpedance
amplifier
Vbias
CC
vo
io+
v+
Figure 2
io-
v-
R1
R1
VSS
Voltage-Current-Voltage Amplifier
io1+
v1+
v1-
Figure 2
io2+
v2+
v2-
io1-
Figure 2
io2-
Transimpedance
amplifier
vo
Self-Biased CMOS Differential
Amplifiers
• Capable of supplying switching currents greater
than the quiescent bias current.
• Double differential mode gain (+6 dB)
• Less sensitive of active-region biasing to
variations in processing, temperature, and power
supply.
• Properties suitable for precision comparators.
• A Transconductance for linear applications using
double source degeneration is possible.
CMOS Differential OP AMP
Vb i a s P
A2 ≅ −
M3
Vin+
g M 1B
g o1 + g o1 B
VinM1A
M1B
Vb i a s P
Vb i a s P
M3
Vout
M3
M1A
Vin+
Vin+
Vin-
Vout
VH
M1B
Vin-
Vb i a s
Vout
M2A
VbiasN
M2B
Vin+
VL
M2A
Vout
VbiasN
M4
M2B
M4
VinM3 &amp; M4 OHMIC
VbiasN
M4
gM 2 B
A1 ≅ −
g o1 + g o 2B
Application of CSDA
(gM 1 + gM 2 )
g o1 + g o 2 B
External
TTL Input
Vo
Vin
-
CSDA
Vin+
Note: connect an
node Vb i a s to improve
speed.
V ref
(midpoint of
= 1.4V standard TTL
input voltage)
Is a source degeneration OTA
developed using this architecture ?
How can we make a transconductance amplifier with the
combined P-N differential pairs?
Ib
Ib
Rp
Mp
Mp
-
vin+
+
Rn Rn
Mn
Ib
Vo
Vcm
Vcm
vin−
Mn
Ib
• Increased effective transconductance
• Rp and Rn can be implemented with
transistors in ohmic region.
• Vcm signal can be used for the CMFB
• The differential output current is
next
iod
vd = vin+ − vin−
where
g mn , p = 2
W
 m n , p Cox I B
 L  n, p

n
=
1 −
N n + 1 
g m Vd
+



vd VDSAT 

2( N n + 1) 
2
g mp vd   vd VDSAT 

1 − 
N p   2(N p + 1)




2



1
2
1
2
A possible transistor level implementation is:
VDD
Vp
2 g mn
Nn =
b 3 (VGS3 − Vtn )
vin−
2 g mp
Np =
b 4 (VGS 4 − Vt p )
M4
Mp
Vgp i
cd
vin+
-
vo +
Vgn
M3
Mn
M3 V
cm
Gm =
Vn
Ib
Ib
VSS
gmp
gmn
Gm =
+
Nn + 1 N p + 1
HD3
d1, 2 =
1 
=
 
32  
2
2
 vd VDSATp  
vd VDSAT 
 
 d1 + 


N n + 1
 N p + 1  
g m n, p (N p,n + 1)
g m n (N p + 1) + g m p ( N n + 1)
∂i
∂ vin
vo = 0
```