741 op-amp where we are going:. typical cmos amplifier
DESCRIPTION
Drain CharacteristicsTRANSCRIPT
741 Op-AmpWhere we are going:
Typical CMOS Amplifier
Drain Characteristics
Current Sources
Ever wonder howwe make one of these?
GND
Vb M5
Vout
Iout
CurrentSink
V1
Vdd
M6
Iout
CurrentSource
How “good” a current source?
Current versus Drain Voltage
Not flat due to Early effect (channel length modulation)
Id = Id(sat) (1 + (Vd/VA) )
Id = Id(sat) eVd/VA
or
Ic = Ic(sat) (1 + (Vc/VA) )
Ic = Ic(sat) eVc/VARout10A
GND
Iout
Current Mirrors
GNDGND
Iin
Vb M5Mb
Vout
Iout
Iout = ( (W/L)5 / (W/L)b ) Iin
nFET Current Mirror
A good way to generate a bias current
pFET Current Mirror
Iout = ( (W/L)7 / (W/L)4 ) Iin
Vdd VddVb
Iin
Iout
M7M4
Current Mirror
GNDGND
Iin
Vb
M5
Mb
Vout1
Iout1
GND
M6
Vout2
Iout2
GND
M7
Vout3
Iout3
Iout = ( (W/L)5 / (W/L)b ) Iin Iout / Iin =
( (W/L)6 / (W/L)b )
Iout / Iin =
( (W/L)7 / (W/L)b )
Basic One-Transistor Circuits
Common GateCommon Source Source Follower
The fundamental two-transisor circuit: Differential Pair
Common BaseCommon Emitter Emitter Follower
Signal Flow in Transistors
Rules of Thumb• The collector or drain can never be an input terminal.• The base or gate can never be an output terminal.In addition it is important to note polarity reversals on these signal paths.• The base-collector or gate-drain path inverts. • All other paths are noninverting.(This of course assumes that there are no reactive elements causing phase shifts)
(Never is too strong a word)
Spectrum of Amplifier “Loads”Vdd
GND
R1
Vout
Vin
10A
Vdd
GND
Vout
Vin
Vb
Vdd
GND
Vout
Vin
Ideal CurrentSource Load
Transistor CurrentSource Load
ResistiveLoad
Remember: On-chip resistors are expensive
Multiple Transistor Configurations
GND
10A
Vdd
GND
Vout
Vin
500A
Vdd
100pA
Vdd
Vout
Vin
GND
Vout
Vin
JFETs as well….
SubthresholdMOS
Above thresholdMOS
BJT
Source or Emitter FollowerIdeal current source
Iref = Ieo e(Vin -Vout )/UT Vout = -UT ln(Ibias/Ieo) + Vin
Vout = VinIref = Iref eVin -Vout )/UT
100A
Vdd
GND
Vout
Vin
10nA
Vdd
GND
Vout
Vin
BJT or Subthreshold MOSFET:
Vout = Vin – sqrt(2 Iref / K) )
(SubVT MOS: Vout = Vin )
or
MOS (Above VT MOS ):
Iref = (K/2) ( Vin - Vout - VT )2
Small-Signal Analysis (CD or CC)
gmV
r
GND GND
Vout
ro
+ V -Vin
(Vin - Vout ) / r+ (Vin - Vout ) gm = Vout / ro
Vout/Vin = 1/(1 + [(r / ro)/(1 + r gm)])
BJT (ro >> r) MOS (r = 0)
Vout/Vin ~ 1/(1 + [(r / ro)/(r gm)]) = 1/(1 + 1 / (r gm) ) = 1/(1 + UT / VA ) ~ 1
Vout (1 + ro gm) ~ ro gmVin
Vout/Vin = 1/(1 + 1 / (r gm) ) = 1/(1 + UT / VA ) ~ 1
Common Drain or Emitter
Iref 100A
Vdd
GND
Vout
Vin
Ibias = Ico eVin/UT eVout /VA
Vout = -VA ln(Ibias/Ico) + ( VA / UT) Vin
100pA
Vdd
GND
Vout
Vin
Iref
Ideal current sourceBJT or Subthreshold MOSFET:
MOS (Above VT MOS ):
Vout = ( VA / UT) Vin
Ibias = Ibias eVin/UT eVout/VA
Ibias = (K/2) ( Vin - VT )2 (1 + (Vout/VA) )
Operating region decreases (Vout > Vin - VT)
Derive using quadratic functions:
Common Drain
Amplifies the input signal at the output
Vout = ( VA / UT) Vin
Ibias = Ibias eVin/UT eVout/VA
100pA
Vdd
GND
Vout
Vin
Ibias
Input conductance = 0
Common Drain
We must account for the other current source:
Vout = ( (VAn // VAp) UT) Vin
Id = Ibias e-Vout/VAp
= Ibias eVin/UT eVout/VAn
Vb
Vdd
GND
Vout
M6
M7Vin
Ibias
Common Drain
What about above-threshold operation:
Ibias = (K/2) ( Vin - VT )2 (1 + (Vout/VA) )
100A
Vdd
GND
Vout
Vin
Ibias
Operating region decreases (Vout > Vin - VT)
Derive using quadratic functions:
Common Base
Amplifies the input signal at the output (non-inverting gain)
Ibias 100A
Vdd
Vout
Vin
Common Base / Common Gate
Assuming an ideal current source:
Ibias = Ico e (Vb -Vin )/UT eVout /VA
Vout = -VA ln(Ibias/Ico) + (VA / UT) Vin (VA / UT) Vb
Vb
Gain = VA / UT = Av
Common Gate
Using a subthreshold MOSFET :
100pA
Vdd
Vout
Ibias
Vin
Vb
Ibias = Io e (Vb -Vin )/UT eVout /VA
Vout = -VA ln(Ibias/Io) + (VA / UT) Vin ( VA / UT) Vb
Gain = VA / UT = Av
Problem: Large input current
Cascode CircuitsUse a common-gate/base transistor to: 1. Improve the output resistance of another transistor.2. Reduce the Gate-to-Drain capacitance effect of another transistor.
Input resistance of common-gate is low Source is nearly fixed if connected to the drain of a transistor
Vdrain
Vb
GND
V1
Vgate
Cascode CircuitsVdrain
Vbias
GND
V1
Vgate
Fixes the voltage at V1 or isolates V1 from the output
GND
Vgate
Vdrain
Idrain = Io e (Vbias -V1 )/UT eVdrain /VA
= Io e Vgate/UT eV1 /VA V1 ~ Vbias - Vgate + (UT/VA) Vdrain
Drain is fixed
Idrain = Io e Vgate/UT e Vbias /VA eVdrain / (Av VA )
Cascode Common-Drain Amp
GND
Ibias
VbMb
GND
V1
Vdd
Vout
biasp
biasn
One Pole
HighOutput
Resistance / DC Gain