741 op-amp

741 Op-AmpWhere we are going:

Typical CMOS Amplifier

Subthreshold MOSFETs

0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.910

-11

10-10

10-9

10-8

10-7

10-6

Gate voltage (V)

Dra

in c

urr

en

t (A

)

= 0.58680 Io = 1.2104fA

In linear scale, we have a quadraticdependence

In log-scale, wehave an exponentialdependence

G

S

D

nFET

G

S

D

B

pFET

MOSFET Current-Voltage Curves

1

/)(0

//)(0

///0

TSG

TdsTSG

TDTSTG

uVV

uVuVV

uVuVuVDS

eI

eeI

eeeII

Saturation

4 Tds UV

eeIIuVVuVV TdgTSg //

0

1 //0

TSdTSg uVVuVVeeI

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/VA

Rout10A

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 Vdd

Vb

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 )

Diode-Capacitor Dynamics

C (dVi/dt) = I

in - Ico exp(V

i/U

T)

Iout

= Ico

exp(Vi/U

T)

(C / Iout

) (d Iout

/dt) = Iin - I

out

C (d Iout

/dt) = Iout

( Iin - I

out )

GND

Iin

GNDGND

Iout

Vi

C

Basic One-Transistor Circuits

Common GateCommon Source Source Follower

The fundamental two-transisor circuit: Differential Pair

Common BaseCommon Emitter Emitter Follower

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

Above Threshold MOSFET Equations

I = (K/2) ( ((Vg - VT) - Vs)2 - ((Vg - VT ) - Vd) 2 )

Saturation: Qd = 0

I = (K/2) ( ((Vg - VT) - Vs)2

If = 1 (ignoring back-gate effects):

I = (K/2) ( 2(Vgs - VT) Vds - Vds2 )

Gummel Plots

0.1 0.2 0.3 0.4 0.5 0.6 0.710

-12

10-11

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

Base-Emitter Voltage (V)

Cur

rent

s

Ic: n=1, Is = 5.52fA

Ib: n=1.019, Is = 0.048fA

Small-Signal Modeling

gmV ro

V3

V2V2

r

V1 +

V

-

V3

V2

V1

V3

V2

V1

gm ror

BJT

Above VTMOSFET

Sub VTMOSFET

Av

(UT ) / I

I I

I / UT

I / UT

2I /(V1-V2 -VT)

VA / I

VA / I

VA / I

VA / UT

VA / UT

2VA/(V1-V2 -VT)

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

Basic One-Transistor Circuits

Source Follower or Emitter Follower

Buffers (Isolates) the input to (from) the output

Assuming an ideal current source:

Ibias = Ieo e(Vin -Vout )/UT

Vout = -UT ln(Ibias/Ieo) + Vin

Vout = Vin

Ibias = Ibias eVin -Vout )/UT

100A

Vdd

GND

Vout

Vin

Basic One-Transistor CircuitsAssuming an ideal current source:

Ibias = Io eVin/UT e

-Vout/UT

Vout = UT ln(Ibias/Io) + Vin

Vout = Vin

Ibias = Ibias eVin/UT e

-Vout/UT

10nA

Vdd

GND

Vout

Vin

If we use a transistor as a current source:

Id = Ibias eVout/VA = Io e

Vin/UT e-Vout/UT

Vout = UT ln(Ibias/Io) + ( // (VA/UT))Vin

MOS Follower Circuits

Source Degeneration

GND

Vout

Vin

GND

Vout

Vin

CircuitElement

Why do this?

• Higher Linearity• Possible Stability

Why not do this? gm

• Lower Bandwidth• Higher Noise / f

Source Degeneration

GND

Vout

Vin

GND

Vout

Vin

V1

Neglect VA of Q1 and assume matched devices:

Q1

II = Ieo e

V1 /UT = Ieo e(Vin - V1 + Vout/Av )/UT

2 V1 = Vin + Vout / Av

I = Ieo e(Vin + Vout/Av )/(2 UT)

A similar result for MOSFETs