2. mos: transfer function, bias, and small signal...
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2. MOS: Transfer Function, Bias, and Small Signal Model
ECE 102, Winter 2011, F. Najmabadi
Reading: Sedra & Smith: Secs. 5.4 & 5.5(Also see Sec. 4.3.7)
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NMOS Characteristic Equations
For PMOS: vGS→ vSG, vDS→ vSD, vtn→ |vtp|, k’n→ k’p, and iD flowing OUT of the drain
PMOS
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NMOS Transfer Function
For vGS < Vtn , NMOS is in cutoff:
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NMOS Transfer Function
o For vGS > Vtn as vGS ↑ → iD ↑ → vDS ↓
o (NMOS in saturation as we started with vDS = VDD > vGS – Vtn )
o iD and vDS can be found from
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NMOS Transfer Function
o As vGS increases vDS becomes smaller until at point B where vDS = vGS – Vtn .
o For larger vGS NMOS is in triode
Exercise: Find VGS|B and VDS|B
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NMOS Transfer Function
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A combination of constant VGS and a signal (vgs)
Bias Bias
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A combination of constant VGS and a signal (vgs)
Bias
Bias and signal
Bias and signal
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Response to the signal appears to be linear!
Bias and signal
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A linear transfer function for the Signal!
vgs
vds
vGS = VGS + vgs
vDS = VDS + vds
iD = IDS + id
Signal and response
Bias
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An Analogy
Total Height, Hb = Bias (HB) + response to signal (hb) Complicated correlation between total height, Hb , and weight
of the boat. Simple correlation between hb and added weight
Hb = HB
Boat
Pool
Bias
Added Weight(signal)
HbHB
Bias + signal
hb
Response
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Bias: HB
Bias + Signal: Hb
Signal & response to signal: hb
Bias: VGS , VDS , ID , VRD
Bias + Signal: vGS , vDS , iD , vRD Signal &
response to signal: vgs , vds , id , vrd
Added Weight(signal)
Hb
HB
Bias + signal
hb
Non-linear correlations among Bias + Signal: vGS , vDS , iD , vRD Simple (and linear) correlation between signal and response to the
signal: vgs , vds , id , vrd
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Important Points!
Signal: We want the response of the circuit to this input.
Bias: State of the system when there is no signal (current and voltages in all elements).o Bias is constant in time (may vary extremely slowly compared to
signal)o Purpose of the bias is to ensure that MOS is in saturation at all times.
Response of the circuit and elements within to the signal is different that the response of the circuit and its elements to Bias (or to Bias + signal):o Different transfer function for the circuito Different iv characteristics for the elements, i.e. relationships among
vgs , vds , id is different than relationships among vGS , vDS , iD .
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Limitations and Constraints
Floating Boat analogy Boat should float at all
times!o Sufficient water in the pool
o Cannot put too much weight (depends on the depth of the water!)
Transistor MOS should be in saturation
at all times!o Bias point in Saturation*
o Signal amplitude cannot become too large (depends on Bias point!)*
* Equations are for NMOS!
VGS > VtnVDS > VGS - Vtn
vGS = VGS + vgs > VtnvDS = VDS + vds > VGS + vgs- Vtn
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Procedure:
1. How to establish a Bias point (bias is the state of the system when there is no signal).o Stable and robust bias point should be resilient to variations in k’,
Vt , … due to temperature and/or manufacturing variability.
2. Find the iv characteristics of the elements for the signal (which can be different than their characteristics equation for bias). o This will lead to different circuit configurations for bias versus
signal
3. Compute circuit response to the signalo Focus on fundamental MOS amplifier configurations
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BIAS(Ensure that MOS is in saturation at all times,
Important parameters are ID and VDS )
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Bias with Gate Voltage
ID = 0.5 k’n (W/L) (VGS – Vtn)2
VDS = VDD – RD ID
This method is NOT desirable as k’, Vt , … are not “well-defined” as bias point (i.e., ID and VDS) can change due to temperature and/or manufacturing variability.o See Exercise 5.33
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Bias with Source Degeneration
Resistor Rs provides negative feedback
Basic Arrangement
VGS = VG – RS ID
Bias with one power supply
VGS = VG – RS ID
Bias with two power supplies
VGS = VSS – RS ID
(KVL: 0+ VGS + RS ID – VSS = 0)
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Resistor Rs provides negative feedback
VGS = VG – RS ID
ID = IS = 0.5 k’n (W/L) (VGS – Vtn)2
o If ID ↑ (because k’n ↑ or Vtn ↓ ) VGS ↓ ID ↓
o If ID ↓ (because k’n ↓ or Vtn ↑ ) VGS ↑ ID ↑
ID Eq.VGS Eq.
VGS Eq. ID Eq.
Negative Feedback:
Feedback is most effective if RS ID >> VGS as
0 = – VGS + VG – RS ID ≈ VG – RS ID or ID ≈ VG /RS
Basic Arrangement
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Example: Find Bias point for Vt =1 V, k’ W/L = 1 mA/V2
GS-KVL: VG = VGS + RS ID
ID = 0.5 k’n (W/L) (VGS – Vtn)2
7 = VGS + 5 (VGS – 1)2
VGS = 2 V , VS = VG – VGS = 5 V
VG = (7)/(7+8) X 15 = 7 V
DS-KVL: 15 = VDS + (RS + RS )ID
VDS = 5 V , VD = VS + VDS = 10 V
Impact of RS: if Vt = 1.5 V (50% change), ID = 0.455mA (9% change)
VD = 10 V
VS = 5 V
VG = 7 V
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Bias in ICs Resistors take too much space on the chip
A “robust” bias has ID and VDS that do not change. One can force ID to be constant using a current source.
VG = 0
ID = I
VGS is set by
I = ID = 0.5 k’n (W/L) (VGS – Vtn)2
VDS = VD – VS
VD = VDD – RD ID
VS = VG – VGS = – VGS
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Current Mirrors (or Current Steering Circuits)
Identical MOS:Same k’n and Vt
Circuit works as long as Q2 is in saturationVDS2 > VGS - Vt
Q1 is always in saturationVDS1 = VGS > VGS - Vt
Since VGS1 = VGS2 = VGS :
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An implementation of a Current Mirror
Identical MOS:Same k’n and Vt
Circuit works as long as Q2 is in saturationVDS2 > VGS - Vt
Bias point of Q1 is uniquely set by:
Since VGS1 = VGS2 = VGS :
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Examples of Current Steering circuits
Current steering circuit can bias several transistors A PMOS current mirror
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An implementation of current steering circuit to bias several transistors in an IC
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SMALL SIGNAL MODEL
2. Find the iv characteristics of the elements for the signal (which can be different than their characteristics equation for bias).
This will lead to different circuit configurations for bias versus signal
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Bias and signal
RD: VRDIRD = ID
MOS: VGS, ID, VDS
VDD: VDD
RD: vRD = VRD + vrdiRD = iD = ID + id
MOS: vGS = VGS + vgsvDS = VDS + vds
iD = ID + id
VDD: VDD
“Signal-only” circuit is different!Bias
Signal only
RD: vrdird = id
MOS: vgs, id, vds
No signal here!
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Signal Model for linear circuit elements Independent voltage source (e.g., VDD)
o No signal: effectively grounded
Independent current sourceo No signal: effectively open circuit (Careful about current mirrors as they
are NOT “ideal” current sources, channel width modulation was ignored!)
Resistors, capacitors, inductoro Remain the same:
Dependent sourceso Remain the same with the control parameter related to the signal!
Non-linear Elements: o Different!
iR = IR + irvR = VR + vr = RIR + vrvR = R iR = R (IR + ir ) = RIR + R irvR = RIR + vr = RIR + R irvr = R ir
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Diodes: signal response is non linear but can be linearized when signal is small
vD
iD
VD
ID
vd
id?
vd
id R = nVT/ID
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Formal derivation of small signal model
f(∙)aAA xXx += aAA yYy +=
)( AA xfy =
( ) ( ) ...!2
)()()( 2)2(
)1( +−⋅+−⋅+= AAA
AAAA XxXfXxXfXf
...!2
)()()( 2)2(
)1( +⋅+⋅+= aA
aAA xXfxXfXf
aAA xXfXf ⋅+≈ )()( )1(
)( AA XfY =
f(∙)AX AY
g(∙)Ax Ay
2)2(
)1(
!2)()( a
AaA xXfxXf ⋅>>⋅
)()(2 )2(
)1(
A
Aa Xf
Xfx ⋅<<
Small signal means:
aAaa xXfxgy ⋅== )()( )1(
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Derivation of diode small signal model
−⋅= 1T
DnVv
SD eIi
−⋅= 1)( TnV
x
S eIxf
−⋅== 1)( T
DnVV
SDD eIVfI
dT
SDd
T
nVV
Sd
VxT
nVx
SdDd v
nVIIv
nVeIv
nVeIvVfi
T
D
D
T
⋅
+=⋅
⋅
=⋅
⋅
=⋅=
=
)()1(
dT
Dd
T
SDd v
nVI
vnV
IIi ⋅
≈⋅
+=
vd
id R = nVT/ID
D
dd r
vi =D
TD I
nVr ≈
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Derivation of MOS small signal model
f(∙, ∙)Ax
AzAy
f(∙, ∙)AX
AZAY
aAA xXx +=
aAA yYy +=
),( AAA yxfz =
),( AAA yxfz =
...)(),()(),(),(,,
+−⋅∂
∂+−⋅
∂∂
+= AAYX
AAYX
AA Yyy
yxfXxx
yxfYXfAAAA
aYX
aYX
A yy
yxfxx
yxfZAAAA
⋅∂
∂+⋅
∂∂
+≈,,
),(),(
aYX
aYX
a yyfx
xfz
AAAA
⋅∂∂
+⋅∂∂
=,,
),( AAA YXfZ =
AAa Zzz −=
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Derivation of MOS small signal model
iG = 0iD = 0.5 k’n (W/L) (vGS – Vtn)2 (1 + λ vDS) = f (vGS , vDS)iD = f (x, y) with x ↔vGS and y ↔ vDS
dsVV
gsVV
d vyfv
xfi
DSGSDSGS
⋅∂∂
+⋅∂∂
=,,
Do I
r⋅
=λ
1tnGS
Dm VV
Ig−⋅
=2
For λ vDS << 10=
+⋅=
g
o
dsgsmd
irvvgi
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MOS “circuit” small signal model0 and =+⋅= g
o
dsgsmd i
rvvgi
PMOS circuit model for small signals is identical to NMOS in gm formula replace VGS – Vtn with VSG - |Vtp|
Do I
r⋅
=λ
1
tnGS
Dm VV
Ig−⋅
=2
1)(
2>>
−=
tnGSom VV
rgλ
id
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PMOS “circuit” small signal model is identical to NMOS
0 and =+⋅= go
sdsgmd i
rvvgi
For PMOS small signal model, id flows into the drain
Do I
r⋅
=λ
1
tpSG
Dm VV
Ig−⋅
=2
id
vsg gmvsg
id
=