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