1100 - Vos and Ib - slides

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Input Offset Voltage (VOS) &
Input Bias Current (IB)
TIPL 1100
TI Precision Labs – Op Amps
Presented by Ian Williams
Prepared by Art Kay and Ian Williams
Input Offset Voltage - VOS
VOUT
VOS = 50!V
2
Input Offset Voltage - VOS
Input offset from
mismatch of input
transistors
VOUT
3
Offset Voltage Specs and Distribution
OPA827
4
Simulate Input Offset Voltage
V
U1 OPA350
-
VM1 -150uV
Vcm 2.5
+
+
VEE 5
R1 1k
+
+
V
VM2 -150uV
Vload 2.5
5
Drift Slope – Positive and Negative
OPA835
OPA835
For this example VOS drift is defined as:
( )
!V os
Vos T1 " Vos ( 25C)
!T
T1 " 25C
6
Drift Slope – Common Definition
Vosi at Temp
160
140
(-40C, 100uV)
Vosi (uV)
120
(85C, 150uV)
100
80
60
40
(25C, 30uV)
20
0
-40
-20
0
20
40
60
80
100
Temp (degC)
! Vos
!T
! Vos
!T
=
=
( )
( )
Vos T1 ! Vos( 25C) + Vos T
T12 ! Vos( 25C)
T 1 ! T2
100"V ! 30"V + 150"V ! 30"V
85C ! ( !40C)
= 1.52
"V
°C
7
Application Example
R2 99k
+
Vosi
0.1mV
Vout = (1mV+ 0.1mV)*(100)
= 110mV
+
R3 1k
R1 1k
VOS introduces
10% error!
Vin 1m
8
Input Offset Drift Calculations
R2 99k
+
Vosi drift
1.5uV/C x (T – 25C)
+
Vosi
0.1mV
Vout
+
R3 1k
R1 1k
Temp (°C)
VOS Initial +
Drift +1.5uV/C
VOS Initial +
Drift -1.5uV/C
-25 °C
25uV
175uV
0 °C
62.5uV
137.5uV
25 °C
100uV
100uV
50 °C
137.5uV
62.5uV
85 °C
190uV
10uV
125 °C
250uV
-50uV
Vin 1m
Vosi = Vosi_room + Vosi_drift" ( T ! 25C)
Example calculations:
!V
Vosi = 100!V + 1.5
" [ ( 25C) ! 25C] = 100!V
C
At 25C
!V
Vosi = 100!V + 1.5
" [ ( 125C) ! 25C] = 250!V
C
At 125C
9
Range of Offset - !V to mV
Op Amp
VOS (max)
(high grade)
VOS Drift (max)
(high grade)
Technology
OPA333
10 !V
0.05 !V/°C
Zero Drift CMOS
OPA277
20 !V
0.15 !V/°C
Precision Bipolar
OPA188
25 !V
0.085 !V/°C
Auto-Zero CMOS
OPA192
25 !V
0.5 !V/°C
CMOS
OPA211
50 !V
1.5 !V/°C
Precision Bipolar
OPA827
150 !V
1.5 !V/°C (typ.)
JFET input, Bipolar, Precision
OPA350
500 !V
4 !V/°C (typ.)
CMOS
OPA835
1.85 mV
13.5 !V/°C
High Speed Bipolar
LM741
3.00 mV
15 !V/°C
Bipolar commodity (lower cost)
10
Input Bias Current - IB
150 nA
VOUT
210 nA
Input bias offset current:
IB_OS = IB1 – IB2 = 210 nA – 150 nA = 60 nA
11
Simple Bipolar, No IB Cancellation
Vcc
R1
R2
Ib1
Vin1
Vin2
Ib2
Q1
Q2
Bias current in bipolar amplifiers is
from input transistor base current.
It is typically larger than in FET-input
amplifiers and it flows into the input
terminals.
IS1
12
Bipolar with IB Cancellation
Vcc
Ib
Cancel
Circuit
R1
R2
Ib1
Vin1
Vin2
!
Q1
!
IS1
Ib2
Q2
The input bias currents are mirrored and
summed back in to cancel the bias current.
This has the effect of significantly reducing
input IB. Note that when this is done, IB can
flow in both directions. Also, IB_OS is no longer
smaller then IB.
IB_OS = IB1 – IB2
Ib
Cancel
Circuit
13
Bias Current for CMOS
Vcc
Vcc
Bias current in FET-input
amplifiers is mainly from leakage
into ESD protection diodes.
R1
R2
Q1
Q2
Ib1
Vin1
Vcc
IS1
Vin2
Ib2
14
IB over Temperature
OPA350
OPA277
CMOS amplifier:
In this case you see a dramatic
increase in bias current at 25 °C.
Note the logarithmic graph, which
doubles every 10 °C.
Bipolar amplifier:
In this case you see a dramatic
increase in bias current at 75 °C.
15
IB Calculation – OPA211 at High Temp.
Using nodal analysis
Point for nodal analysis
Vin
R1
R1 1k
R2 99k
Rf
+ Ib = 0
Using superposition set Vin=0V
-
Ib 200nA
R3 1k
+
Vin 1m
Vin ! Vout
Vin
Vin %
"
Vout = Rf ( # Ib +
+
&
R1
Rf
$
'
Ib 200nA
Ib 150nA
+
0
0 %
Vout = Rf ( "# Ib +
+
= Ib ( Rf
Rf &
R1
$
'
In this example
Vout_Ib = ( 200nA) ( ( 99k! ) = 20mV
99
Vout_vin = 1mV( "#
+ 1 %& = 100mV
$1
'
Vout_total = 20mV_100mV = 120mV
16
Range of Bias Current – fA to nA
Op Amp
IB (max)
(high grade)
IB at max
temp.
Technology
OPA129
100 fA
20 pA (typ.)
Difet – Ultra Low Bias Current
OPA627
5 pA
1nA
Difet – Precision High Speed
OPA350
10 pA
500 pA (typ.)
CMOS
OPA827
50 pA
50 nA max
JFET input, Bipolar, Precision
OPA333
70 pA
150 pA (typ)
Zero Drift CMOS
OPA277
1 nA
2 nA (max)
Precision Bipolar
OPA211
125 nA
200 nA
Precision Bipolar
OPA835
400 nA
530 nA
High Speed Bipolar
LM741
80 nA
0.2 !A (max)
Bipolar commodity (lower cost)
17
Thanks for your time!
Please try the quiz.
18
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