AD629A AD [Analog Devices], AD629A Datasheet - Page 11

• Manufacturer Part Number: AD629A
• Description: High Common-Mode Voltage Difference Amplifier
• Manufacturer: AD [Analog Devices]

Error Budget Analysis Example 2
This application is similar to the previous example except that
the sensed load current is from an amplifier with an ac common-
mode component of ± 100 V (frequency = 500 Hz) present on
the shunt (Figure 38). All other conditions are the same as
Error Source
ACCURACY, T
TEMPERATURE DRIFT (85°C)
RESOLUTION
REV. A
Figure 37. Error Budget Analysis Example 1. V
Full-Scale, V
Power-Line Interference
SHUNT
Initial Gain Error
Offset Voltage
Gain
Offset Voltage
Noise, Typ, 0.01–10 Hz, µV p-p
CMR @ 60 Hz
Nonlinearity
AC CMR @ 500 Hz
CURRENT
OUTPUT
1
10 AMPS
200V
TO GROUND
POWER LINE
CM
DC
60Hz
CM
A
= 25°C
= 200 V DC. R
Table IV. AD629 vs. INA117 AC Error Budget Example 2 (V
–V
S
REF(–)
–IN
+IN
0.1 F
SHUNT
1
2
3
4
NC = NO CONNECT
380k
(0.0005 × 10) ÷ 10 V × 10
(0.001 V ÷ 10 V) × 10
10 ppm/°C × 60°C
(20 µV/°C × 60°C) × 10
15 µV ÷ 10 V × 10
(141 × 10
(10
(141 × 10
AD629
21.1k
380k
AD629
= 1 Ω , 1 V p-p 60 Hz
–5
× 10 V) ÷ 10 V × 10
380k
20k
–6
–6
× 200 V) ÷ 10 V × 10
× 1 V) ÷ 10 V × 10
8
7
6
5
NC
REF(+)
IN
6
0.1 F
= 10 V
+V
V
6
OUT
S
6
/10 V
6
6
–11–
6
before. Note that the same kind of power line interference can
happen as detailed in Example 1. However, the ac common-
mode component of 200 V p-p coming from the shunt is much
larger than the interference of 1 V p-p, so that this interference
component can be neglected.
6
SHUNT
Figure 38. Error Budget Analysis Example 2. V
Full-Scale, V
INA117
(0.0005 × 10) ÷ 10 V × 10
(0.002 V ÷ 10 V) × 10
10 ppm/°C × 60°C
(40 µV/°C × 60°C) × 10
25 µV ÷ 10 V × 10
(500 × 10
(10
(500 × 10
CURRENT
OUTPUT
1
–5
× 10 V) ÷ 10 V × 10
10 AMPS
TO GROUND
100V AC CM
POWER LINE
–6
–6
Total Resolution Error:
× 200 V) ÷ 10 V × 10
× 1 V) ÷ 10 V × 10
Total Accuracy Error:
60Hz
CM
CM
= ± 100 V at 500 Hz, R
Total Drift Error:
= 100 V @ 500 Hz)
6
–V
Total Error:
6
S
6
/10 V
REF(–)
6
6
–IN
+IN
0.1 F
6
1
2
3
4
6
NC = NO CONNECT
380k
380k
21.1k
AD629
AD629
2,820
4,166
2,846
Error, ppm of FS
500
100
600
600
120
720
SHUNT
14
10
380k
20k
2
= 1 Ω
AD629
8
7
6
5
NC
REF(+)
IN
0.1 F
INA117
10,000
10,063
11,603
= 10 V
500
200
600
240
V
700
840
+V
OUT
50
10
S
3