cs amplifier with diode connected load 020303[1]
TRANSCRIPT
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CS Amplifier with Diode-Connected Load
• “Diode-Connected” MOSFET is only a name. In BJT if Base and Collector are short-circuited, then the BJT acts exactly as a diode.
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CS Amplifier with Diode-Connected Load
• We want to replace RD with a MOSFET that will operate like a small-signal resistor.
mo
md
Xmo
XXX
gr
gR
Vgr
VIVV
1||
1
1
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Common Source with Diode-Connected NMOS Load (Body Effect included)
(gm gmb)Vx Vx
ro Ix
Vx
Ix
1
gm gmb|| ro
1
gm gmb
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CS with Diode-Connected NMOS Load Voltage Gain Calculation
Av gm1 1
gm2 gmb2
gm1
gm2
1
1
1
1
)/(2
)/(2
22
11
DOXn
DOXnv
ILWC
ILWCA
Av (W / L)1
(W / L)2
1
1
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CS with Diode-Connected NMOS Load Voltage Gain Calculation
Av (W / L)1
(W / L)2
1
1
• If variations of η with the output voltage are neglected, the gain is independent of the bias currents and voltages (as long as M1 stays in Saturation)
• The point: Gain does not depend on Vin (as we so in the case of RD load)
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Amplifier is relatively linear (even for large signal analysis!)
22
2
2
1
)(5.0)(5.0 THoutDDCnTH1inCn VVVL
WVV
L
WOXOX
)()( 221
THoutDDTH1in VVVL
WVV
L
W
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Assumptions made along the way:
)()( 221
THoutDDTH1in VVVL
WVV
L
W
• We neglected channel-length modulation effect
• We neglected VTH voltage dependent variations
• We assumed that two transistors are matching in terms of COX.
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Comment about “cutting off”
• If I1 is made smaller and smaller, what happens to Vout?
• It should be equal to VDD at the end of the current reduction process, or is it?
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Comment about “cutting off” (cont’d)
• If I1 is made smaller and smaller, VGS gets closer and closer to VTH.
• Very near I1=0, if we neglect sub-threshold conduction, we should have VGS≈VTH2, and therefore Vout≈VDD-VTH2 !
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Cutoff conflict resolved:
• In reality, sub-threshold conduction, at a very low current, eventually brings Vout to the value VDD.
• Output node capacitance slows down this transition.
• In high-speed switching, sometimes indeed Vout doesn’t make it to VDD
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Back to large-signal behavior of the CS amplifier with diode-connected NMOS load:
)()( 221
THoutDDTH1in VVVL
WVV
L
W
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Large signal behavior of CS amplifier with diode-connected load
• When Vin<VTH1 (but near), we have the above “imperfect cutoff” effect.
• Then we have a, more or less, linear region.• When Vin exceeds Vout+VTH1 amplifier enters the
Triode mode, and becomes nonlinear.
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CS with Diode-Connected PMOS Load Voltage Gain
2
1
)/(
)/(
LW
LWA
p
nv
No body effect!
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Numerical Example
• Say that we want the voltage gain to be 10. Then:
100)/(
)/(
10)/(
)/(
2
1
2
1
LW
LW
LW
LWA
p
n
p
nv
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Example continued
100)/(
)/(
2
1 LW
LW
p
n
• Typically pn 2
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Example continued
• Need a “strong” input device, and a “weak” load device.
• Large dimension ratios lead to either a larger input capacitance (if we make input device very wide (W/L)1>>1) or to a larger output capacitance (if we make the load device very narrow (W/L)2<<1).
• Latter option (narrow load) is preferred from bandwidth considerations.
50)/(
)/(
2
1 LW
LW
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CS with Diode-Connected Load Swing Issues
)(
||
)/(
)/(
2
1
TH1GS1
TH2GS2
p
nv
VV
VV
LW
LWA
ID1 ID2,
This implies substantial voltage swing constraint. Why?
2
2
2
1
)()( TH2GS2pTH1GS1n VVL
WVV
L
W
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Numerical Example to illustrate the swing problems
• Assume for instance VDD=3V• Let’s assume that VGS1-VTH1=200mV (arbitrary
selection, consistent with current selection)• Assume also that |VTH2|=0.7V (for PMOS load
VTH2=-0.7V)• For a gain of 10, we now need |VGS2|=2.7V (for
PMOS load VGS2<-2.7V)• Therefore because VGD2=0: VDS2=VGS2=-2.7V).• Now VDS1=VDD-VSD2<3-2.7=0.3V, and recall that
VDS1>VGS1-VTH1=0.2V. Not much room left for the amplified signal vds1.
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DC Q-Point of the Amplifier
• VG1 determines the current and the voltage VGS2
• If we neglect the effect of the transistors’ λ, the error in predicting the Q-point solution may be large!
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CS with Diode-Connected Load –How do we take into account λ?
)||||1
21( oomb2m2
m1v rrgg
gA
)||||1
21(1 oo
m2v rr
ggAm
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Example as Introduction to Current Source Load
• M1 is biased to be in Saturation and have a current of I1.
• A current source of IS=0.75I1 is hooked up in parallel to the load – does this addition ease up the amplifier’s swing problems?
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Example as Introduction to Current Source Load
• Now ID2=I1/4. Therefore (from the ratio of the two transconductances with different currents):
2
1
)/(
)/(4
LW
LWA
p
nv
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Example: Swing Issues
)(
||4
TH1GS1
TH2GS2A
VV
VVv
,4 21 DD II
It sure helps in terms of swing.
2
2
2
1
)(4)( TH2GS2pTH1GS1n VVL
WVV
L
W