AD810AR Analog Devices Inc, AD810AR Datasheet - Page 9
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AD810AR
Manufacturer Part Number
AD810AR
Description
IC CURR-FDBK AMP VIDEO LP 8-SOIC
Manufacturer
Analog Devices Inc
Datasheet
1.AD810ARZ.pdf
(16 pages)
Specifications of AD810AR
Slew Rate
1000 V/µs
Mounting Type
Surface Mount
Rohs Status
RoHS non-compliant
Applications
Current Feedback
Number Of Circuits
1
-3db Bandwidth
80MHz
Current - Supply
6.8mA
Current - Output / Channel
60mA
Voltage - Supply, Single/dual (±)
5 V ~ 36 V, ±2.5 V ~ 18 V
Package / Case
8-SOIC (0.154", 3.90mm Width)
No. Of Amplifiers
1
Bandwidth
80MHz
No. Of Pins
8
Settling Time
125ns
Operating Temperature Max
85°C
Peak Reflow Compatible (260 C)
No
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Part Number:
AD810AR
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
AD810ARM-REEL7
Manufacturer:
AD
Quantity:
5 510
Company:
Part Number:
AD810ARM-REEL7
Manufacturer:
XILINX
Quantity:
5 510
Part Number:
AD810ARZ
Manufacturer:
ADI/亚德诺
Quantity:
20 000
REV. A
Figure 33. Closed-Loop Gain and Phase vs. Frequency
G = –1, R
and 2.5 V
Figure 35. –3 dB Bandwidth vs. Supply Voltage,
Gain = –1, R
V
100
IN
Figure 31. Inverting Amplifier Connection
–1
–2
–3
–4
–5
90
80
70
60
50
40
30
20
1
0
HP8130
PULSE
GENERATOR
1
L
= 150 , R
G = –1
R
V
2
O
L
R
= 150
= 250mV p-p
L
G
PHASE
= 150
GAIN
4
F
SUPPLY VOLTAGE – ±Volts
R
R
6
R
V
= 681
10
F
F
F
2
3
3
S
FREQUENCY – MHz
= 681
= 500
= 1k
= ±15V
±2.5V
8
AD810
±5V
–V
+V
7
4
R
S
S
F
10
for 15 V, 620
0.1µF
0.1µF
12
6
100
PEAKING
PEAKING
GAIN = –1
R
14
L
Typical Characteristics, Inverting Connection–
= 150
V
TEKTRONIX
P6201 FET
PROBE
±2.5V
V
±5V
16
O
S
R
TO
1.0dB
= ±15V
L
0.1dB
18
for 5 V
V
1000
O
180
135
–45
–90
90
45
0
–9–
Figure 34. Closed-Loop Gain and Phase vs. Frequency,
G = –1, R
5 V and 2.5 V
Figure 32. Small Signal Pulse Response, Gain = –1,
R
Figure 36. –3 dB Bandwidth vs. Supply Voltage,
Gain = –1, R
F
V
= 681 , R
V
IN
O
180
160
140
120
100
–1
–2
–3
–4
–5
80
60
40
20
1
0
90
0%
100
10
1
L
= 1 k , R
1V
1V
G = –1
R
V
2
L
O
PHASE
L
L
GAIN
= 1k
= 250mV p-p
= 1 k
= 150 , V
4
F
= 681
SUPPLY VOLTAGE – ±Volts
6
R
R
R
10
F
F
V
F
FREQUENCY – MHz
S
= 649
= 500
= 1k
= ±15V
±2.5V
8
S
±5V
= 5 V
for V
10
S
12
20nS
= 15 V, 620
100
PEAKING
PEAKING
14
GAIN = –1
R
L
= 1k
±2.5V
V
±5V
16
S
AD810
= ±15V
1.0dB
0.1dB
18
1000
for
180
135
90
45
0
–45
–90