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Unformatted text preview: 6.012  Microelectronic Devices and Circuits Lecture 20  DiffAmp Anal. I: Metrics, Max. Gain  Outline • Announcements
Announcements  D.P.: No Early effect in large signal analysis; just LECs.
Lec. 21 foils useful; Sp 06 DP foils, too (on Stellar)
Do PS #10: free points while working on D.P. • Review  Differential Amplifier Basics
Difference and commonmode signals: vID = vIN1 vIN 2, vIC = (vIN1+ vIN2)/2
Halfcircuits: half of original with wires shorted or cut (familiar, easy analyses) • Performance metrics  specific to diff. amps.
Difference and commonmode gains Commonmode rejection ratio Input and output voltage swings • Nonlinear loads
The limitation of resistive loads: Gain limited by voltage supply Nonlinear loads: High incremental resistance/small voltage drop • Active loads
Lee load Current mirror load Clif Fonstad, 11/19/09 Lecture 20  Slide 1 Differential Amplifiers  overview of features and properties
Intrinsic advantages and features:
 large difference mode gain
 small common mode gain
 easy to cascade stages; no coupling capacitors  no emitter/source capacitors in CS/CE stages
Performance metrics:
 difference mode voltage gain, Avd
 common mode voltage gain, Avc
 input resistance, Rin
 output resistance, Rout
 common mode input voltage range
 output voltage swing
 DC offset on output
 Power dissipation
Clif Fonstad, 11/19/09 Today
Today ∞
Lec 21 Today
Today
Lec 21
Lec 18
Lecture 20  Slide 2 Differential Amplifiers  commonmode input range
(VC,min ≤ vC ≤ VC,max)
vGS stays constant We have said the output changes very little for commonmode inputs. This is true as long as the vC doesn't push the transistors out of saturation. There are a minimum and maxiumum vC: VC, max: As vC increases, vDS8 and vDS9 decrease until Q8 and Q9 are no longer in saturation. Q4 + Q5 vC
up
vC
down v DS gets
smaller + Q8 vDS +
vGS + vDS  +   + vGS stays constant B Q8, Q9 forced
out of
saturation
if v C too high
vC
Q9 up   VC, min: As vC decreases, vDS10 decreases until Q10 is no longer in saturation. Clif Fonstad, 11/19/09 + 1.5 V
+Q
7
Q6  v DS gets
smaller vDS Q10 +
  vGS vc
down Q10 forced out
of saturation
if v c too low  1.5 V
The node between Q8/Q9 and Q10 moves up
and down with vC. Lecture 20  Slide 3 Differential Amplifiers  output voltage range
+ 1.5 V Fixed + ++
v v SG16 SG12 Q12 v SD12  + Q13
! 0.6V
+
v DS15 +
v GS15  Q15
 AQ16 (VOUT,min ≤ vOUT ≤ VOUT,max) + v SD12 and v SD16 decrease as
vOUT goes up, so Q 12 and/or Q 16
may be forced out of
v SD16
saturation if v OUT is too high  + Q20 vOUT
+
! 0.6V
up
! 0.6V
Q
+ Q18 ! 0.6V +
vOUT
17
! 0.6V
 Q down
21
Q19 +
v DS19
B+
v DS15 and v DS19 decrease
Fixed v GS19  1.5 V as v OUT goes down, so
Q15 and/or Q 19 may be
forced out of saturation if
vOUT is too low As vOUT goes down, Q15 and/or Q19 may go out of saturation;
as vOUT goes up, the same may happen to Q12 and/or Q16.
Clif Fonstad, 11/19/09 Lecture 20  Slide 4 Differential Amplifier Analysis incremental analysis exploiting symmetry and superposition +
vid
 a LEHC:
one half
of sym.
LEC +
vod
 +
vic
 No voltage on
common links, so
incrementally they
are grounded. a LEHC:
one half
of sym.
LEC +
voc
Clif Fonstad, 11/19/09 a LEHC:
one half
of sym.
LEC +
vid
 +
vod
 a LEHC:
one half
of sym.
LEC No current in
common links, so
incrementally they
are open. +
vid
 +
voc
 a LEHC:
one half
of sym.
LEC +
vod = A vdvid
 +
vic
 +
vic
 a LEHC:
one half
of sym.
LEC +
voc = A vc vic
Lecture 20  Slide 5 Looking at the design problem circuit:
Lesson  Draw the difference and common mode half circuits.
roLL dm Difference mode half circuit: roCM dm
Q17 +
vid
 Q8 +
roQ 16 roLL cm Common mode half circuit: Q12 vic
 Q8
2roQ 10 vod
 RLOAD roCM cm
Q17 + Q20 Q20
+ Q12
roQ 16 voc RLOAD  We have reduced the transistor count from 22 to 4, and we see that our
complex amplifier is a just cascade of 4 singletransistor stages.
Clif Fonstad, 11/19/09 Lecture 20  Slide 6 V+ What's with these active, nonlinear loads?
Why doesn't the design problem use resistors? RSL Linear Resistor Loads:
the limit on maximum stage gain
 with linear resistor loads we must
make a compromise between the
voltage gain and the size of the
output voltage swing.
Maximum voltage gains
g CO gmv gs  go +
v out
 vout
 +
vin
IBIAS d +
v in = v gs + Q CS
G SL
(=1/RSL ) V MOSFET : Av ,max = gm RSL = 2 [ ID RSL ] max
2 ID RSL
$
[VGS # VT ] [VGS # VT ] min Bipolar : Av,max = gm RSL = [I R ]
qIC RSL
$ C SL max
kT
VThermal s,b s,b
" What are [ICRSL]max, [IDRSL]max, and [VGS  VT]min ? !
Clif Fonstad, 11/19/09 * For a MOSFET, gm = (2KID/α)1/2 = K(VGS  VT)/α = 2ID/(VGS  VT) Lecture 20  Slide 7 Resistor Loads: cont.
 What are [IDRSL]max, [ICRSL]max, and [VGS  VT]min? [IDRSL]max, [ICRSL]max:
 [IDRSL]max and [ICRSL]max are
determined by the desired
voltage swing at the output
and/or by the commonmode input voltage range.
 The ultimate limit is the power supply. [VGS  VT]min:
 [VGS  VT]min is set by how close
to threshold the gate can safely
be biased before the strong
inversion, drift model fails. We
will say more about this shortly
(Slide 23).
Clif Fonstad, 11/19/09 V+
RSL
CO
Q
+
vin
 +
vout
 IBIAS
CS
V Lecture 20  Slide 8 Current Source Loads: Incrementally large resistance
Relatively small quiescent voltage drop   transistors with a DC input voltage, i.e. set up as sources/sinks  MOSFET: g
+
v gs = 0
sv bs = 0
+b +
V REF
+
V REF
 d
gmv gs
=0 go = " I D = Bipolar:
+
V REF
 b
+
V REF gmv !
=0 e
 Clif Fonstad, 11/19/09 +
v !=
g! ! 0
 s go
s, b, g c gmb v bs
=0 d c e go
e, b go ID
VA go IC
go = " IC =
VA
Lecture 20  Slide 9 V+ Current Source Loads:
the limit on the maximum
stage gain IStage Load
CO  current source loads eliminate the
need to compromise between the
voltage gain and the output
voltage swing
d +
v in = v gs gmv gs +
v out
 go  IBIAS
V[VGS # VT ] Av ,max MOSFET :
Bipolar : !lif Fonstad, 11/19/09
C 2 VA ,eff
2 ID
gm
=
=
$
go + gsl
ID VA ,Q + ID VA ,SL
[VGS # VT ] min Av,max " with VA ,eff " CS gsl s,b s,b vout
 +
vin
 Maximum voltage gains
g + Q VA ,eff
gm
qIC kT
=
=
=
go + gsl
IC VA ,Q + IC VA ,SL
Vt VA ,QVA ,SL [V A ,Q + VA , SL ] , Vt " kT
q Typically VA,eff >> [ID RSL]max * For a MOSFET, gm = (2KID/α)1/2 = K(VGS  VT)/α = 2ID/(VGS  VT) Lecture 20  Slide 10 Current Source Loads: the maximum stage gain, cont.  the similarity in the results for BJT's
and MOSFETs operating in strong
inversion extends to MOSFETs
operating subthreshold and in
velocity saturation, also: g d +
v in = v gs gmv gs go  +
v out
 gsl s,b s,b The MOSFET LEC: the same for all. Maximum voltage gains
MOSFET sub  threshold : iD = IS ,s" t e( vGS "VT ) nVt , gm = VA ,eff
gm
ID nVt
=
=
=
go + gsl
ID VA ,Q + ID VA ,SL
nVt Av,max MOSFET w. velocity saturation : ID
(vGS " VT ) VA ,QVA ,SL [V A ,Q Clif Fonstad, 11/19/09 *
*
iD = W ssat Cox (vGS " VT ), gm = W ssat Cox = VA ,eff
ID (vGS " VT )
gm
=
=
#
go + gsl
ID VA ,Q + ID VA ,SL
(vGS " VT ) min Av,max with VA ,eff " ID
nVt + VA ,SL ]
Lecture 20  Slide 11 Current Source Loads: Example  biasing a sourcecoupled
pair differential amplifer stage
Want: Build: V+ V+
Q1 ILOAD Q4
+
vIN1
 +
vO1
 Q3 Q2 ILOAD +
vO2
 Q5
+
vIN2
 R1 Q4
+
vIN1
 +
vO1
 +
vO2
 Q5
+
 IBIAS
Q7 VNote: I LOAD = I BIAS /2 Q6
V Note: W1 = W2 = W3
W7 = 2W6 This is nice…can we do even better?
Yes, with active loads. Consider…
Clif Fonstad, 11/19/09 Lecture 20  Slide 12 Active Loads: Loads that don't just sit there and look pretty. First example: the current mirror load V+
Q1 Q2 Signal actively fed from
left side to right side,
and applied inputs to
"stage load" MOSFETs. +
Q3
+
vI1
VClif Fonstad, 11/19/09 I BIAS ,
rob Q4
+
vI2
 vOUT
 Now "single ended," i.e.
only one output, but it
is twice as large:
vout = 2vout1
Load selfadjusting; circuit
forces ILOAD = IBIAS/2.
Lecture 20  Slide 13 Active Loads: The current mirror load, cont.
V+
Large differentialmode
gain, small commonmode
gain.
Also provides high gain
conversion from doubleended to singleended
output.
The circuit is no
longer symmetrical, so
halfcircuit techniques
can not be applied.
The full analysis is
found in the course
text. We find: Q1 Q2 id id 2id
id id
+
vid/2
 Q4 Q3 V Differencemode inputs Clif Fonstad, 11/19/09 v out ,d I BIAS ,
rob + +
vid/2
 vOUT
 2 gm 3
=
v id 2
( go2 + go 4 + gel )
Lecture 20  Slide 14 ! Active Loads: The current mirror load
V+
Q1 Commonmode inputs v out ,c = gob
v ic
2 gm 2 Q2 ic ic 0
ic ic
+
vic
 ! With both inputs: v out Q4 Q3 V I BIAS ,
rob +
vic
 +
vOUT
 2 gm 3
(v in1 " v in 2 ) " gob (v in1 + v in 2 )
=
2
2 gm 2
2
(go 2 + go4 + gel ) Note: In D.P. the output goes to the base of two BJTs; gel ≠ 0 and can
be important.
Clif Fonstad, 11/19/09
Lecture 20  Slide 15 ! What if we want an active load and yet stay differential? Active Loads  The Lee load
A load for a fullydifferential stage
that looks like a
large resistance in
differencemode
and small resistance in commonmode)
The conventional
schematic is drawn
here, but the coupling of the load and
what is happening
is made clearer by
redrawing the
circuit (next slide.)
Clif Fonstad, 11/19/09 Q1 Q5
+
vI1
 V+
Q2 Q3 +
+
vO1 vO2
 V I BIAS ,
rob Q4 Q6
+
vI2
 Normal format
Lecture 20  Slide 16 Active Loads  The Lee load. cont.
Drawn as on the
right we see that
the load MOSFETs
on each side are
driven by both outputs. The result is
different if the two
outputs are equal
and opposite (diffmode operation) or
if they are equal
(commonmode).
The next few slides
give the results for
each mode. Q1 Q3
+ vO1
 Q5
+
vI1
 V+
Q2 + vO2
+
 vO1  +
vO1
 V Q4
+ vO2
 +
vO2
 Q6
+
vI2
 I BIAS ,
rob Drawn to highlight crosscoupling
and demonstrate symmetry
Clif Fonstad, 11/19/09 Lecture 20  Slide 17 V+
Q1 The Lee load:
analysis for
differencemode
inputs Q3
+ vod +
  vod
vod +
vod
 Q5
+
vid
 +
 Q4
+ vod
 +
vod
 V LEHC: differencemode
+
v id /2 = v gs5
 Q2 Q6
+
vid
 I BIAS ,
rob
+ gm5v id /2 go5 go1
gm1v od /2 gm3v od /2 go3 v od /2
 gel goLLd
Clif Fonstad, 11/19/09 Lecture 20  Slide 18 The Lee load: analysis for differencemode inputs, cont
LEHC: differencemode
+
v id /2 = v gs5
 +
gm5v id /2 +
v id /2 = v gs5
 go5 gm3v od /2 gm1v od /2 go3 v od /2
 gel +
gm5v id /2 g go5 gm1 go1 gm1 go1 v od /2
 gel d +
v id /2 = v gs5
s,b go1 +
gm5v gs5 go5 goLLd v od /2
 gel goLLd = 2 go1 s,b v od
" gm 5
Avd =
=
v id ( go 5 + 2 go1 + ge!
1)
Clif Fonstad, 11/19/09 Note: In D.P., the outputs go to MOSFET gates so gel = 0. Lecture 20  Slide 19 V+
Q1 Q3 Q2 Q4 The Lee load:
+ analysis for commonmode inputs  +
voc
 Q5
+
voc
 v ic  Q6
+
vI2
 I BIAS ,
rob +
v gs5
 voc +
voc
 V LEHC: commonmode
+ + + voc
+
 voc  voc +
gm5v gs5 go5
go1
gm1v oc gm3v oc go3 v oc gel gob /2
Clif Fonstad, 11/19/09  goLLc Lecture 20  Slide 20 The Lee load: analysis for commonmode inputs, cont
LEHC: commonmode
+ + + v gs5 gm5v gs5  v ic go5
gm1 go1 gm1 go1 v oc gel gob /2
 g +
v ic d +
v gs5
 +
gm5v gs5 s,b s,b goLLc = 2 ( gm1 + go1 ) go5
goLLc v oc gel " 2 gm1 gob /2
  !
v oc
" gob
gob
Avc =
=
#"
v ic 2[2( gm1 + go1 ) + ge1 ]
4 gm1
Clif Fonstad, 11/19/09 Note: In D.P., the outputs go to MOSFET gates so gel = 0. Lecture 20  Slide 21 Achieving the maximum gain: Comparing linear resistors, current sources, and active loads Maximum Gains MOSFET (SI) Bipolarlike
(BJT and SubVT MOS) Linear resistor loads " Current source loads " Active loads Difference mode $ Common mode $ 2 [ ID RSL ] max [vGS # VT ] min
2 VA ,eff
[vGS # VT ] min
2 VA ,eff [vGS # VT ] min
[vGS # VT ] min
2 VA ,bias " [IC RSL ] max
" $ n Vt
VA ,eff
n Vt
VA ,eff n Vt
n Vt
$
VA ,bias Observations:
 Nonlinear (current source) loads typically yield much higher gain than
linear resistors, i.e. VA,eff >> [IDRSL]max.
 The bias point is not ! important to BJTtype stage gain.
as
 An SI MOSFET should be biased just above threshold for highest gain.
 For active loads what increases Avd, decreases Avc.
Clif Fonstad, 11/19/09 Lecture 20  Slide 22 Achieving the maximum gain: (vGSVT)min = ?
For SIMOSFETs, maximizing the voltage gain (Av or Avd) requires
minimizing (VGSVT). What is the limit? Sub  threshold :
Av
1
=
VA n Vt
Strong inversion :
Av
2
=
VA (VGS " VT )
Av/VA is a smooth
curve, so clearly
(VGSVT)min > 2nVt. ? Note: n = 1.25 was assumed. Clif Fonstad, 11/19/09 Lecture 20  Slide 23 6.012  Microelectronic Devices and Circuits Lecture 20  DiffAmp Analysis I  Summary • Performance metrics  specific to diff. amps.
Difference and commonmode gains: Avd = vod/vid, Avc = voc/vic
Commonmode rejection ratio: CMRR = Avd/Avc
Commonmode input range • Nonlinear loads
Transistors biased in their constant current regions:
MOSFETs in saturation BJTs in their FAR • Active loads
Current mirror load:
Achieves double to singleended conversion without loss of gain
Has high resistance for differencemode signals
Has low resistance for commonmode signals Lee Load:
Maintains differential signals Has high resistance for differencemode signals Has low resistance for commonmode signals Clif Fonstad, 11/19/09 Lecture 20  Slide 24 MIT OpenCourseWare
http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits
Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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This note was uploaded on 11/07/2011 for the course COMPUTERSC 6.012 taught by Professor Charlesg.sodini during the Fall '09 term at MIT.
 Fall '09
 CharlesG.Sodini

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