Lecture 24
ANNOUNCEMENTS
• The last HW assignment (HW#12) will be due 12/6
• Prof. Liu will be away on Tuesday 12/4 (no office hour that day)
OUTLINE
• MOSFET Differential Amplifiers
Reading: Chapter 10.3-10.6
EE105 Fall 2007
Lecture 24, Slide 1
Prof. Liu, UC Berkeley
Common-Mode (CM) Response
• Similarly to its BJT counterpart, a MOSFET differential pair
produces zero differential output as VCM changes.
V X  VY  VDD
EE105 Fall 2007
Lecture 24, Slide 2
I SS
 RD
2
Prof. Liu, UC Berkeley
Equilibrium Overdrive Voltage
• The equilibrium overdrive voltage is defined as VGS-VTH
when M1 and M2 each carry a current of ISS/2.
VGS  VTH equil 
EE105 Fall 2007
Lecture 24, Slide 3
I SS
W
 n Cox
L
Prof. Liu, UC Berkeley
Minimum CM Output Voltage
• In order to maintain M1 and M2 in saturation, the commonmode output voltage cannot fall below VCM-VTH.
• This value usually limits voltage gain.
VDD
EE105 Fall 2007
Lecture 24, Slide 4
I SS
 RD
 VCM  VTH
2
Prof. Liu, UC Berkeley
Differential Response
EE105 Fall 2007
Lecture 24, Slide 5
Prof. Liu, UC Berkeley
Small-Signal Response
• For small input voltages (+DV and -DV), the gm values are
~equal, so the increase in ID1 and decrease in ID2 are ~equal in
magnitude. Thus, the voltage at node P is constant and can
be considered as AC ground.
I EE
I D1 
 DI
2
I D2
I EE

 DI
2
DVP  0
 DI D1  g m DV ; DI D 2   g m DV
EE105 Fall 2007
Lecture 24, Slide 6
Prof. Liu, UC Berkeley
Small-Signal Differential Gain
• Since the output signal changes by -2gmDVRD when the input
signal changes by 2DV, the small-signal voltage gain is –gmRD.
• Note that the voltage gain is the same as for a CS stage, but
that the power dissipation is doubled.
EE105 Fall 2007
Lecture 24, Slide 7
Prof. Liu, UC Berkeley
Large-Signal Analysis
I D1  I D 2
EE105 Fall 2007
4 I SS
1
W
2
  n Cox Vin1  V in 2 
 Vin1  Vin 2 
W
2
L
 n Cox
L
Lecture 24, Slide 8
Prof. Liu, UC Berkeley
Maximum Differential Input Voltage
• There exists a finite differential input voltage that completely
steers the tail current from one transistor to the other. This
value is known as the maximum differential input voltage.
Vin1  Vin 2
EE105 Fall 2007
max

2 VGS  VTH equil
Lecture 24, Slide 9
Prof. Liu, UC Berkeley
MOSFET vs. BJT Differential Pairs
• In a MOSFET differential pair, there exists a finite differential
input voltage to completely switch the current from one
transistor to the other, whereas in a BJT differential pair that
voltage is infinite.
MOSFET Differential Pair
EE105 Fall 2007
BJT Differential Pair
Lecture 24, Slide 10
Prof. Liu, UC Berkeley
Effect of Doubling the Tail Current
• If ISS is doubled, the equilibrium overdrive voltage for each
transistor increases by 2 , thus DVin,max increases by 2 as
well. Moreover, the differential output swing will double.
EE105 Fall 2007
Lecture 24, Slide 11
Prof. Liu, UC Berkeley
Effect of Doubling W/L
• If W/L is doubled, the equilibrium overdrive voltage is lowered
by 2 , thus DVin,max will be lowered by 2 as well. The
differential output swing will be unchanged.
EE105 Fall 2007
Lecture 24, Slide 12
Prof. Liu, UC Berkeley
Small-Signal Analysis
• When the input differential signal is small compared to
4ISS/nCox(W/L), the output differential current is ~linearly
proportional to it:
4 I SS
1
W
W
I D1  I D 2   n Cox Vin1  Vin 2 
  n Cox I SS Vin1  Vin 2 
W
2
L
L
 n Cox
L
• We can use the small-signal model to prove that the
change in tail node voltage (vP) is zero:
g m1v1  g m 2 v2  0
vin1  vin2
 v1  vP  v2  vP 
EE105 Fall 2007
 v1  v2
Lecture 24, Slide 13
Prof. Liu, UC Berkeley
Virtual Ground and Half Circuit
• Since the voltage at node P does not change for small input
signals, the half circuit can be used to calculate the voltage gain.
vP  0
Av   g m RD
EE105 Fall 2007
Lecture 24, Slide 14
Prof. Liu, UC Berkeley
MOSFET Diff. Pair Frequency Response
• Since the MOSFET differential pair can be analyzed using its
half-circuit, its transfer function, I/O impedances, locations of
poles/zeros are the same as that of the half circuit’s.
EE105 Fall 2007
Lecture 24, Slide 15
Prof. Liu, UC Berkeley
Example
 p, X
1

RS [CGS1  (1  g m1 / g m 3 )CGD1 ]
 p ,Y 
 p ,out
EE105 Fall 2007
1



g m3 
  C DB1  CSB3 
CGS 3  CGD1 1 
g m1 



1

RD CGD3  C DB3 
1
g m3
Lecture 24, Slide 16
Prof. Liu, UC Berkeley
Half Circuit Example 1
0
Half circuit for small-signal analysis
 1

Av   g m1 
|| rO3 || rO1 
 g m3

EE105 Fall 2007
Lecture 24, Slide 17
Prof. Liu, UC Berkeley
Half Circuit Example 2
 0
Half circuit for small-signal analysis
RDD 2
Av  
1 g m   RSS 2
EE105 Fall 2007
Lecture 24, Slide 18
Prof. Liu, UC Berkeley
MOSFET Cascode Differential Pair
Half circuit for small-signal analysis
Av   g m1rO3 g m3 rO1
EE105 Fall 2007
Lecture 24, Slide 19
Prof. Liu, UC Berkeley
MOSFET Telescopic Cascode Amplifier
Half circuit for small-signal analysis
Av   g m1  g m3 rO3 rO1  || ( g m5 rO5 rO 7 )
EE105 Fall 2007
Lecture 24, Slide 20
Prof. Liu, UC Berkeley
CM to DM Conversion Gain, ACM-DM
• If finite tail impedance and asymmetry are both present, then
the differential output signal will contain a portion of the
input common-mode signal.
DVCM  DVGS  2DI D RSS 
 DI D 
DI D
 2DI D RSS
gm
DVCM
1
 2 RSS
gm
DVout1  DI D RD
DVout 2  DI D RD  DRD 
DVout  DVout1  DVout 2  DI D DRD
DVout
DRD

DVCM 1 / g m   2 RSS
EE105 Fall 2007
Lecture 24, Slide 21
Prof. Liu, UC Berkeley
MOS Diff. Pair with Active Load
• Similarly to its BJT counterpart, a MOSFET differential pair
can use an active load to enhance its single-ended output.
EE105 Fall 2007
Lecture 24, Slide 22
Prof. Liu, UC Berkeley
Asymmetric Differential Pair
• Because of the vast difference in magnitude of the resistances
seen at the drains of M1 and M2, the voltage swings at these
two nodes are different and therefore node P cannot be viewed
as a virtual ground…
EE105 Fall 2007
Lecture 24, Slide 23
Prof. Liu, UC Berkeley
Thevenin Equivalent of the Input Pair
vThev   g mN roN (vin1  vin 2 )
RThev  2roN
EE105 Fall 2007
Lecture 24, Slide 24
Prof. Liu, UC Berkeley
Simplified Diff. Pair w/ Active Load
vout
 g mN (rON || rOP )
vin1  vin 2
EE105 Fall 2007
Lecture 24, Slide 25
Prof. Liu, UC Berkeley