lecture 13 biasing and loading single stage fet amplifiers
TRANSCRIPT
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Lecture 13
Biasing and Loading Single Stage FET Amplifiers
The Building Blocks of Analog Circuits - III
In this lecture you will learn:
• Current biasing of circuits • Current sources and sinks for CS, CG, and CD circuits
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
A Poor Man’s Current Source
What if one wanted to bias a device with a current source?
BIASI
………but one only had a voltage source?
Solution: Use a large voltage source with a large resistor in series!
BIASVR
I
BIAS BIAS
D
V VI
R R R
DR
DRPoor man’s current source
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
DDV
R
SR
+
-
+
-
BIASV
sv
The Common Source Amplifier
Rrgv
Ri
v
vA om
in
d
in
outv ||
Open circuit voltage gain:
+
-
outOUT vV
DI
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Source Amplifier: Problems
Rrgv
vA om
in
outv ||
Open circuit voltage gain:
To achieve large gain one needs:1) A large DC current bias ID in order to get a large gm
2) A large value of the resistor R
In saturation:
DSnDnGS
Dm VIk
VI
g
12
Both the above requirements cannot be met easily simultaneously: Not easy to realize large resistors in micro-chips A large resistor R will limit the maximum value of the DC current bias ID (because
the potential drop IDR can become large enough to put the FET in the linear region where gm and the gain will be small)
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Digital Logic Inverter: Problems
To achieve fast switching one needs:1) A small resistor (because then the RCL charging time will be smaller when the
output switches from “0” to “1”)
00
OUTV
INVTNV
DDV TNIN VV
Cut-off
Saturation
Linear
But A small resistor will require a very large current (to achieve a large potential drop
across it) when the input is “1”, and the NFET is turned on, and the output need to be low or “0” – and this means more power dissipation
When the input is “1”, and the NFET is turned on, there is constant static power dissipation
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Source Amplifier with a Current SourceDDV
BIASI
SR
+
-sv
+
-
outOUT vV
DI
What is needed is an ideal current source in the drain that can supply a large DC current and at the same time has a large small signal resistance
The incremental (or differential or small signal) resistance looking into an ideal current source is infinite
+
-BIASV
omin
outv rg
v
vA
An ideal current source, of course, does not exist
But one can certainly do much better than using a resistor in the drain
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Source Amplifier with a Current Source: DC Biasing
BIASI
DSnTNGSoxnBIASD VVVCL
WII 1
22
00
DSV
DI
1OUTV 2OUTV
gmVBIAS
Suppose for VBIAS1 the output was VOUT1
Suppose VBIAS1 is changed to a smaller value VBIAS2
VBIAS = VBIAS2 – VBIAS1 < 0
The output VOUT2 can be obtained graphically, as shown
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Source Amplifier with a Current Source: DC Biasing
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The CS Amplifier with a Current Source: Small Signal Analysis
Gate Drain
gsmvg
+gsv-
gi
Source
oroutv+
-inv+
-
di
bsmbvg
+bsv-
Base
0
omin
outv rg
v
vA
Open circuit voltage gain:
oout rR
Output resistance:
Large
Large
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Large Signal Model of a Current SourceThe output resistance of the current sourceBIASI
Large signal circuit model
ocr
A large signal model of a current source roc is large for a good current source and infinite for an ideal current source
A current source
BIASI
Ideal current source
IOUT
VOUT
IBIAS
ocOUT
OUT
rdV
dI 1
Slope
ocr BIASIOUTV
OUTI+
-
One can have any voltage at the at the output terminals of a good current source and the current delivered will remain mostly constant
ocr
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Small Signal Model of a Current Source
BIASISmall signal circuit model
ocr
A current source
IOUT
VOUT
IBIAS
ocOUT
OUT
rdVdI 1
Slope
ocr BIASIOUTV
OUTI+
-
The output resistance of the current source
ocr BIASI
Ideal current source
Large signal circuit model
ocr
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Large Signal Model of a Current SinkThe output resistance of the current sinkBIASI
Large signal circuit model
ocr
A large signal model of a current source roc is large for a good current source and infinite for an ideal current source
A current sink
BIASI
Ideal current source
IOUT
VOUT
IBIAS
ocOUT
OUT
rdV
dI 1
Slope
ocr BIASIOUTV
OUTI+
-
One can have any voltage at the at the output terminals of a good current sink and the current dinked will remain mostly constant
ocr
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Small Signal Model of a Current Sink
BIASISmall signal circuit model
ocr
A current sink
IOUT
VOUT
IBIAS
ocOUT
OUT
rdV
dI 1
Slope
ocr BIASIOUTV
OUTI+
-
The output resistance of the current source
ocr BIASI
Ideal current source
Large signal circuit model
ocr
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
PFET Loaded NFET Common Source Amplifier
DDV
BIASI
SR
+
-sv
+
-
outOUT vV
DI
+
-BIASV
An ideal current source, of course, does not exist
But one can certainly do much better than using a resistor in the drain
Use a PFET!
DDV
SR
+
-sv
+
-
outOUT vV
DI
+
-BIASV
A PFETload
BV
A PFET approximation to a current source
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Current Sink:
I I roc
IOUT
VOUT
One can have any voltage at the at the output terminals of a good current sink and the current sinked will remain mostly constant
Current Sinks and NFET Transistors
OUTV
OUTI
DI
OUTI
OUTV
oo
OUT
OUT
rg
dV
dI 1
Slope
Characteristics in saturation resemble that of a non-ideal current source!
IOUT
VOUT
IBIAS
ocOUT
OUT
rdV
dI 1
Slope
+
-
+
-
roc
VB
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Current Sources:
I I roc
IOUT
VOUT
One can have any voltage at the at the output terminals of a good current source and the current delivered will remain mostly constant
Current Sources and PFET Transistors
OUTV
OUTI
DI
Characteristics in saturation resemble that of a non-ideal current source!
IOUT
VOUT
IBIAS
ocOUT
OUT
rdV
dI 1
Slope
OUTI
OUTV
oo
OUT
OUT
rg
dV
dI 1
Slope
+
-
+
-roc
VB
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
PFET Loaded NFET: Large Signal Analysis
00
DDV
DI
OUTV
M2
M1
DDOUTDS
OUTDS
VVV
VV
2
1
21 DDD III
BIASVBV
DC value of the output voltage (Both FETs in saturation)
Instead of the resistive load line, we now have the full ID vs VDS curves of the PFET
The biasing voltages need to be selected carefully, otherwise one of the transistors can go into the linear region!!
PFET NFET
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
00
DDV
DI
OUTV
BIASV
BV
Instead of the resistive load line, we now have the full ID vs VDS curves of the PMOS
The biasing voltages need to be selected carefully, otherwise one of the transistors can go into the linear region!!
PFET NFET
DC value of the output voltage (PFET in the linear region)
PFET Loaded NFET: Large Signal Analysis
DDOUTDS
OUTDS
VVV
VV
2
1
21 DDD III
M2
M1
1
ID2
ID =
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
OUTnTNBIASn
DSnTNGSn
D VVVk
VVVk
I 12
12
21
211
DDOUTpTPDDBp
DSpTPGSp
D VVVVVk
VVVk
I 12
12
22
222
NFET
PFET
Assuming the biasing is correct, both the FETs are in saturation
Equate the drain current magnitudes of the two FETs
Then the only unknown will be VOUT ; solve for it
Verify that the obtained value of VOUT does indeed result in both the transistors being in saturation
Equating:
OUTDDpTPDDBp
OUTnTNBIASn
DDD
VVVVVk
VVVk
III
12
12
22
21
PFET Loaded NFET: Large Signal Analysis
M2
M1
1
ID2
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
PFET Loaded NFET: Transfer Curve
M2
M1
00
OUTV
INVTNV
DDV TNIN VV
I
II
IV
I: VIN < VTNM1 cut-off
II: VIN > VTN & VOUT > VIN - VTN & VOUT > VB - VTPM1 saturation, M2 Linear
III: VIN > VTN & VOUT > VIN - VTN & VOUT < VB - VTPM1 saturation, M2 saturation
IV: VIN > VTN & VOUT < VIN - VTN & VOUT < VB - VTPM1 linear, M2 saturation
TPB VV III Increasing VIN
I
II
IV
III
1
VB
11
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Gate1 Drain1
11 gsm vg
+1gsv
-
1gi
Source1
1oroutv+
-inv+
-
1di
Gate2 Drain2
22 gsm vg
+2gsv
-
2gi
Source2
2or
2di
PFET
NFET
Looking into the drain end of the PFET, what does one see?
PFET Loaded NFET: Small Signal Analysis
SR+
-sv
M2
M1
1
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Gate1 Drain1
11 gsm vg
+1gsv
-
1gi
Source1
1oroutv+
-inv+
-
1di
Drain2
2or
2di
NFET
Looking into the drain end of the PFET, what does one see?
PMOS Loaded NMOS: Small Signal Analysis
SR+
-sv
M2
M1
1
12
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
11 gsm vg
+1gsv
-
1gi
1oroutv+
-inv+
-
1di
2or
211 || oomin
outv rrg
v
vA
Open circuit voltage gain:
21 || ooout rrR
Output resistance:
PFET Loaded NFET: Small Signal Analysis
M2
M1
1
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
DDV
SR
+
-sv
+
-
outOUT vV
1DI
+
-BIASV
PFETload
2BV
3BV
M2
M1
M3
This load has a larger resistance looking into it
PFET Loaded NFET: The Cascode PFET Load
DDV
SR
+
-sv
+
-
outOUT vV
1DI
+
-BIASV
PFETload
2BVM2
M1
DRBetter
This topology is called the “Cascode” current load
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Gate2 Drain2
22 gsm vg
+2gsv
-
2gi
Source2
2or
2di
PFET
Gate3 Drain3
33 gsm vg
+3gsv
-
3gi
Source3
3or
3di
PFET
Cascode PEFT Load: Small Signal Analysis
PFETload
1
ID2
ID3
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Gate2 Drain2
22 gsm vg
+2gsv
-
2gi
Source2
2or
2di
PFET
Drain3
3or
3di
Cascode PFET Load: Small Signal Analysis
PFETload
1
ID2
ID3
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Gate2 Drain2
22 gsm vg
+2gsv
-
2gi
Source2
2or
2di
3or
Resistance looking into the current source is:
322223232232 1 oomomoooomoo rrgrgrrrrgrr
Cascode PFET Load: Small Signal Analysis
PFETload
1
Typically a good approximation
ID2
ID3
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
2di
Resistance looking into the current source is:
322223232232 1 oomomoooomoo rrgrgrrrrgrr
Cascode PFET Load: Small Signal Analysis
PFETload
1
Typically a good approximation
ID2
ID3
322
2232 1
oom
omoo
rrg
rgrr
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
11 gsm vg
+1gsv
-
1gi
1oroutv+
-inv+
-
1di
322 oom rrg
1132211 || omoomomin
outv rgrrgrg
v
vA
Open circuit voltage gain:
13221 || ooomoout rrrgrR
Output resistance:
NFET
Cascode PFET Load: Small Signal Analysis
PFETload
Typically a good approximation
Typically a good approximation
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Source Amplifier: General TopologyV
BIASI
OUTV
INVBIASI
OUTV
INV
V
V
V
NFET PFET
V + is the most positive voltage in the circuitV - is the most negative voltage in the circuit
Large input resistanceLarge output resistanceLarge voltage gainLarge current gain
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Gate Amplifier: General TopologyV
BIASI
OUTV
INV BIASI
OUTV
V
V
V
INV
Can have small input resistanceLarge output resistanceLarge voltage gainSmall (unity) current gain
GV
GV
NFET PFET
ECE 315 – Spring 2007 – Farhan Rana – Cornell University
The Common Drain Amplifier: General TopologyV
BIASI
OUTV
INV
BIASI
OUTV
INV
V
V
V
Large input resistanceSmall output resistanceSmall (less than unity) voltage gain
NFET PFET
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ECE 315 – Spring 2007 – Farhan Rana – Cornell University
Chip-Based Current Sources and Voltage Sources
We have realized a decent load but ……
How does one generate voltage levels on a chip for biasing??
Current and voltage biasing of circuits require transistor based current and voltage sources on a chip!
What are the good figures of merit of chip-based voltage and current sources?