AD8223 Analog Devices, AD8223 Datasheet - Page 10
![no-image](/images/manufacturer_photos/0/0/56/analog_devices_sml.jpg)
AD8223
Manufacturer Part Number
AD8223
Description
Single Supply, Rail-to-Rail, Low Cost Instrumentation Amplifier
Manufacturer
Analog Devices
Datasheet
1.AD8223.pdf
(20 pages)
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Part Number:
AD8223AR
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Part Number:
AD8223ARMZ
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
AD8223ARMZ-R7
Manufacturer:
AD
Quantity:
4 866
Part Number:
AD8223ARMZ-R7
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Part Number:
AD8223ARMZ-REEL7
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
AD8223ARZ
Manufacturer:
ADI
Quantity:
5 486
Part Number:
AD8223ARZ
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
AD8223ARZ-R7
Manufacturer:
ADI
Quantity:
5 486
Company:
Part Number:
AD8223ARZ-RL
Manufacturer:
ADI
Quantity:
5 786
Company:
Part Number:
AD8223BRZ
Manufacturer:
MINI
Quantity:
5 000
AD8223
Figure 15. Common-Mode Input vs. Maximum Output Voltage,
Figure 16. Common-Mode Input vs. Maximum Output Voltage,
–10
–12
–14
–10
–20
–30
–1
–2
–3
–4
–5
–6
16
14
12
10
–2
–4
–6
–8
70
60
50
40
30
20
10
6
5
4
3
2
1
0
8
6
4
2
0
0
–14
100
–6
V
–12 –10 –8 –6 –4 –2
S
–5
= ±12V
V
S
–4
G = 1000
G = 100
G = 10
G = 5
= ±5V
Figure 14. Gain vs. Frequency
V
–3
1k
S
= ±2.5V
G = 5, Small Supplies
G = 5, Large Supplies
–2
FREQUENCY (Hz)
–1
OUTPUT (V)
OUTPUT (V)
0
10k
0
2
1
4
+V
2
6
S
= +15V, –V
100k
8
3
V
S
10 12 14
= +5V
4
S
= –5V
5
1M
16
6
Rev. PrA | Page 10 of 20
Figure 17. Common-Mode Input vs. Maximum Output Voltage,
Figure 18. Common-Mode Input vs. Maximum Output Voltage,
–10
–12
–14
140
120
100
–1
–2
–3
–4
–5
–6
16
14
12
10
–2
–4
–6
–8
80
60
40
20
6
5
4
3
2
1
0
8
6
4
2
0
–14
0
–6
1
Figure 19. Positive PSRR vs. Frequency, V
–12 –10 –8 –6 –4 –2
V
–5
S
V
= ±12V
S
–4
= ±5V
10
Preliminary Technical Data
–3
G = 100, Small Supplies
G = 100, Large Supplies
V
S
–2
= ±2.5V
G = 10
FREQUENCY (Hz)
100
–1
OUTPUT (V)
OUTPUT (V)
0
0
2
4
G = 5
1
1k
6
G = 1000
+V
2
S
8 10 12 14 16
= +15V, –V
S
V
3
= ±12 V
S
10k
G = 100
= +5V
4
S
= –5V
5
100k
18
6