5 grudnia 2013 2
Amplifier
5 grudnia 2013 3
MOS as two port network
D
G
S
In
Out
VSS
πΌπ· =1
2ππΆππ₯
π
πΏππΊπ β πππ»
2 1 + ππ·π β
β1
2ππΆππ₯
π
πΏππΊπ β πππ»
2
ro
vin + -
vout gmvGS
vGS
Small signal model πππππππππ’ππ‘ππππ ππ =ππΌπ·πππΊπ
π΄
π=1
π·π¦πππππ πππ ππ π‘ππππ π0 =ππΌπ·
πππ·π
β1
π
π΄=
πΉππ’ππ‘ππ‘πππ ππ πΌπ· ππΌπ· = πππππΊπ ππ· = πππ£πΊπ
5 grudnia 2013 4
MOS as two port network
D
G
S
In
Out
VSS
πΌπ· =1
2ππΆππ₯
π
πΏππΊπ β πππ»
2 1 + ππ·π β
β1
2ππΆππ₯
π
πΏππΊπ β πππ»
2
ro
vin + -
vout
vGS
gmvGS
Small signal model
If VGS = constant MOS delivers quasi constant current MOS is working as source current
π0 =ππΌπ·πππ·π
β1
=1
12ππΆππ₯
ππΏππΊπ β πππ»
2β1
πΌπ·
ππΊπ = ππππ π‘πππ‘ β πππΊπ = π£πΊπ = 0 β ππ· = πππ£πΊπ = 0
5 grudnia 2013 5
MOS as two port network
D
G
S
In
Out
VSS
πΌπ· =1
2ππΆππ₯
π
πΏππΊπ β πππ»
2 1 + ππ·π β
β1
2ππΆππ₯
π
πΏππΊπ β πππ»
2
ro
vin + -
vout gmvGS
vGS
Small signal model
If VGS = VDS MOS (connected as diode) is working as active load (resistor)
ππ³ =πππ·πππΌπ·=πππΊπππΌπ·=1
ππ
1ππ
RD = 1M
RD = 5k
5 grudnia 2013 6
One Stage Amplifier
RD = 10k
RD = 100k
V in
VSS
RD
D
S
VDD
ID
V out
7
One Stage Amplifier
πΌπ· =1
2ππΆππ₯
π
πΏππΊπ β πππ»
2 πΌπ· = ππΆππ₯
π
πΏππΊπ β πππ» ππ·π β
ππ·π2
2
π0 = 0 π0 =1
ππΆππ₯ππΏππΊπ β πππ» β ππ·π
8
One Stage Amplifier
πΌπ· =1
2ππΆππ₯
π
πΏππΊπ β πππ»
2 1 + ππ·π
} ~1,01
π0 β1
πΌπ· π0 =
1
ππΆππ₯ππΏππΊπ β πππ» β ππ·π
5 grudnia 2013 9
Offset voltage (DC) shifts the signal (AC) on desidered level
One Stage Amplifier
Out
VSS
VDD
R2
R1 RD ID
In
ππΊπ =π 2
π 1 + π 2ππ·π·
π = πΆπππ π‘ = π πΌ 0 =β π = π‘ππΌ = 0
D
G
S
In
Out
VSS
VDD
RD
The VDD supply voltage acts as an AC ground because its value remains constant with time.
5 grudnia 2013 11
One Stage Amplifier
The pMOS transistor can replace RDβ is working as so called active resistance
D
G
S
In
Out
VSS
RD
VDD
D
G
S
In
Out
VSS
VDD
5 grudnia 2013 12
Gain A is proportional to the load resistance RLoad and transconductance gm
One Stage Amplifier
Gain calculation:
π΄π =π£ππ’π‘
π£ππ
= βdππ·βπ π·
dπ£πΊπ
= βππ β π π·
Out
VSS
VDD
R2
R1 RD ID
In
13
Current Sources
A MOSFET behaves as a current source , when it is in saturation mode.
NMOS draws current from a point to ground (sinks current), whereas PMOS draws current from VDD to a point (sources current).
NMOS or PMOS devices configured as shown above operate as current sources
X
VSS
ID Vb
Y
VDD
ID Vb
πΌπ· =π½
2ππ β ππ
2 πΌπ· =π½
2ππ·π· β ππ β ππ
2
14
Current Sources
NMOS or PMOS devices configured as shown above do not operate as current sources because variation of VX or VY directly changes the gate-source voltage of each transistor, thus changing the drain current considerably.
X
VSS
ID Vb
Y
VDD
ID Vb
15
Common-Source Stage
Common-source stage is when the input is applied to the gate and the output at the drain. Source is the common pin for input and output. For small signals, M1 converts the input voltage variations to proportional drain current changes, and RD transforms the drain currents to the output voltage. The VDD supply voltage acts as an AC ground because its value
remains constant with time. Thus, we simply βgroundβ the supply voltage in small-signal analysis.
vout
VSS
VDD
RD
ID
vin
RD vin
+ -
vout
gmv1
v1
small signals model
r0
16
Common-Source Stage
If channel-length modulation is neglected =0 , the small-signal model yields vin = v1 and vout = βgmv1RD, where The voltage gain of the common-source stage is
DDoxnv
Dm
in
outv
RIL
WCA
Rgv
vA
2
Doxnm IL
WCg 2
Vout
VSS
VDD
RD
ID
Vin
RD vin
+ -
vout
gmV1 v1
small signals model
17
CS Stage with =0
The output impedance of the CS amplifier is simply equal to load
resistance RL. The voltage gain Av is equal
Lmv RgA
RL vin
+ -
vx gmV1 v1
+ -
ix
18
CS Stage with 0
However, channel length modulation affects CS stages as drain current is also a weak function of VDS, for small-signal can be calculated resistance r0
OLout
in
OLmv
rRR
R
rRgA
||
||
DTHGSoxn
DS
Do
IVVL
WCV
Ir
1
2
1
1
2
1
RL vin
+ -
vx gmV1 v1
+ -
ix
ro
19
CS Stage with Current-Source Load
Frequently an active current-source load is used. Because v1= 0 the gm2v1=0.
This is advantageous because a current-source has a high output resistance and can tolerate a small voltage drop across it.
21
211
||
||
OOout
OOmv
rrR
rrgA
Vout
VSS
VDD
ID
Vin
Vb + -
v1
r02
Vout
Vin r01
gm2v1
20
PMOS CS Stage with NMOS as Load
Similarly, with PMOS as input stage and NMOS as the load, the voltage gain is the same as before. Transistor M2 generates a small-signal current equal to gm2vin
)||(212 OOmv
rrgA Vout
VSS
VDD
ID
Vin
Vb
21
CS Stage with Diode-Connected Load
Doxnm IL
WCg 2
We may use a diode-connected MOSFET as the drain load.
With Ξ» = 0, M2 acts as a small-signal resistance equal to 1/gm2
More accurate expression take channel-length modulation into account:
Vout
VSS
VDD
ID
Vin
1/gm2 r02
Vout
Vin r01
ππΏ =ππ£π·ππππ·=ππ£πΊππππ·=1
ππ
π΄π£ = βππ1 β ππΏ = βππ1 β1
ππ2 π΄π£ = β
π1πΏ1π2πΏ2
π΄π£ = βππ1 β ππΏ = βππ11
ππ2| π02 |π01
22
CS Stage with Degeneration
When a CS stage is degenerated, its gain, I/O impedances, and linearity change. The degeneration resistor sustains a fraction of the input voltage
change.
What happend if we put RS resistance between source and ground?
π£ππ = π£1 + πππ£1π π π£1 =π£ππ
1+πππ π
Vout
VSS
VDD
RD
ID
Vin
RS
RD vin
+ -
vout
gmV1 v1
RS
23
CS Stage with Degeneration
Since gmv1 flows through RD, vout= βgmv1RD and vin = v1(1 + gmRs)
What happend if we put RS resistance between source and ground?
Vout
VSS
VDD
RD
ID
Vin
RS
RD vin
+ -
vout
gmV1 v1
RS
π΄π£ =π£ππ’π‘π£ππ= β
πππ π·1 + πππ π
= βπ π·
π π +1ππ
24
Example of CS Stage with Degeneration
Transistor M2 serves as a diode-connected device, presenting an impedance of 1/gm2. The gain is calculated if RS is replaced with 1/gm2:
21
11
mm
Dv
gg
RA
Compute the voltage gain of the circuit if Ξ» = 0.
A diode-connected device degenerates a CS stage.
Vout
VSS
VDD
RD
ID
Vin
Vout
VSS
VDD
RD
ID
Vin
1/gm2
25
CS Core with Biasing
Degeneration is used to stabilize bias point, and a bypass capacitor can be used to obtain a larger small-signal voltage gain at the frequency of interest.
S
m
D
G
v
Rg
R
RRR
RRA
1||
||
21
2121 ||RRR in
Such a circuit no longer exhibits an infinite input impedance
Dm
G
v RgRRR
RRA
21
21
||
||
Vout
VSS
VDD
R2
R1 RD
RS
ID
Vin
Vout
VSS
VDD
R2
R1 RD
RS
ID
Vin
Vout
VSS
VDD
R2
R1 RD
RS
ID
Vin
RG
C2
S
m
Dv
Rg
RA
1
Source follower
27
Source Follower Stage
1v
A
If the gate voltage of M1 is raised by a small amount, Vin, the gate-source voltage tends to increase, thereby raising the source current and hence the output voltage. Thus, Vout βfollowsβ Vin. Since the dc level of Vout is lower than that of Vin by VGS, we say the follower can serve as a βlevel shiftβ circuit. We expect this topology to exhibit a subunity gain
Vout
VSS
VDD
ID
vin
RL
28
Source Follower Core
The voltage gain is therefore positive and less than unity. It is desirable to maximize RL (and rO).
The source follower can be analyzed as a resistor divider.
LO
m
LO
in
outV
Rrg
Rr
v
vA
||1
||
LOmout Rrvgv ||1
outin vvv 1
resistor divider
r0 vin
+ -
gmV1 v1
RL vout vin
+ -
1/gm
RL||r0
vout
29
Source Follower Example
In this example, M2 acts as a current source.
If rO1||rO2 Β» 1/gm1, then Av β 1.
21
1
21
||1
||
OO
m
OO
v
rrg
rrA
Vout
VSS
VDD
ID
vin
Vb
Vout
vin
r02
r01
30
Source Follower with Biasing
RG sets the gate voltage to VDD, whereas RS sets the drain current.
The quadratic equation above can be solved for ID.
22
1THSDDDoxnD VRIV
L
WCI
GSSDDD VRIV
Vout
VSS
VDD
vin
RL
RG C1
C2
31
Supply-Independent Biasing
If Rs is replaced by a current source, drain current ID becomes independent of supply voltage.
Vout
VSS
VDD
vin
RG C1
C2 Vout
VSS
VDD
Vin
RG C1
C2
Vb
5 grudnia 2013 32
Differential
Amplifier
5 grudnia 2013 33
The differential pair is composed of two equal transistors. The goal of the differential amplifier is to compare two analog signals, and to
amplify their difference.
Current mirror 1:1
nMOS differential pair
Differential Amplifier
Vout+
VSS
VDD
V- V+
Vout-
ID ID
5 grudnia 2013 34
Vout = Vout+ - Vout- = K(Vin+ - Vin -)
Usually, the gain K is high, ranging from 10 to 1000. The consequence is that the differential amplifier output saturates very rapidly, because of the supply
voltage limits.
Current mirror 1:1
nMOS differential pair
In-
In+ Out
Differential Amplifier
Vout
VSS
VDD
V- V+
ID ID
5 grudnia 2013 35
If one input is set to reference voltage, the out amplified the difference of input with respect to reference voltage
Vout
VSS
VDD
Vin
Vref
Reference voltage stage
nMOS differential pair
ID ID
In-
In+ Out
Differential Amplifier
5 grudnia 2013 36
Differential Amplifier
5 grudnia 2013 37
Differential Amplifier