741 op-amp
Post on 31-Dec-2015
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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
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